From b3607db55c9ad4f2c77c71ea56131da23762b7d1 Mon Sep 17 00:00:00 2001 From: lucianommartins Date: Tue, 20 May 2025 19:03:33 -0700 Subject: [PATCH] updating Get_started_thinking.ipynb notebook --- quickstarts/Get_started_thinking.ipynb | 1816 +++++------------------- 1 file changed, 390 insertions(+), 1426 deletions(-) diff --git a/quickstarts/Get_started_thinking.ipynb b/quickstarts/Get_started_thinking.ipynb index c8f149fcb..0ee233a78 100644 --- a/quickstarts/Get_started_thinking.ipynb +++ b/quickstarts/Get_started_thinking.ipynb @@ -11,7 +11,7 @@ }, { "cell_type": "code", - "execution_count": null, + "execution_count": 1, "metadata": { "cellView": "form", "id": "906e07f6e562" @@ -112,7 +112,7 @@ }, { "cell_type": "code", - "execution_count": 1, + "execution_count": 2, "metadata": { "id": "IbKkL5ksQYq1" }, @@ -121,7 +121,7 @@ "name": "stdout", "output_type": "stream", "text": [ - "\u001b[?25l \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m0.0/196.3 kB\u001b[0m \u001b[31m?\u001b[0m eta \u001b[36m-:--:--\u001b[0m\r\u001b[2K \u001b[91m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m\u001b[91m╸\u001b[0m \u001b[32m194.6/196.3 kB\u001b[0m \u001b[31m111.8 MB/s\u001b[0m eta \u001b[36m0:00:01\u001b[0m\r\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m196.3/196.3 kB\u001b[0m \u001b[31m3.8 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m\n", + "\u001b[?25l \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m0.0/196.3 kB\u001b[0m \u001b[31m?\u001b[0m eta \u001b[36m-:--:--\u001b[0m\r\u001b[2K \u001b[91m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m\u001b[91m╸\u001b[0m \u001b[32m194.6/196.3 kB\u001b[0m \u001b[31m10.6 MB/s\u001b[0m eta \u001b[36m0:00:01\u001b[0m\r\u001b[2K \u001b[90m━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━\u001b[0m \u001b[32m196.3/196.3 kB\u001b[0m \u001b[31m5.0 MB/s\u001b[0m eta \u001b[36m0:00:00\u001b[0m\n", "\u001b[?25h" ] } @@ -143,7 +143,7 @@ }, { "cell_type": "code", - "execution_count": 2, + "execution_count": 3, "metadata": { "id": "0H_lRdlrQYq3" }, @@ -167,7 +167,7 @@ }, { "cell_type": "code", - "execution_count": 3, + "execution_count": 4, "metadata": { "id": "X3CAp9YrQYq4" }, @@ -181,7 +181,7 @@ }, { "cell_type": "code", - "execution_count": 4, + "execution_count": 5, "metadata": { "id": "zNIfaceKFC82" }, @@ -201,7 +201,7 @@ }, { "cell_type": "code", - "execution_count": 5, + "execution_count": 6, "metadata": { "id": "B0Z9QzC3Q2wX" }, @@ -224,7 +224,9 @@ "\n", "In each of them you can select different models to see how this new model compares to its predecesors.\n", "\n", - "In some cases, you'll still get the good answer from the other models, in that case, re-run it a couple of times and you'll see that Gemini thinking models are more consistent thanks to their thinking step." + "In some cases, you'll still get the good answer from the other models, in that case, re-run it a couple of times and you'll see that Gemini thinking models are more consistent thanks to their thinking step.\n", + "\n", + "**Note:** For now, thinking budget is a feature available for Gemini 2.5 Flash model only." ] }, { @@ -242,19 +244,19 @@ }, { "cell_type": "code", - "execution_count": 10, + "execution_count": 7, "metadata": { "id": "ffa2fd81d26e" }, "outputs": [ { "data": { - "text/markdown": "Given your criteria:\n1. **Aquatic mammal**\n2. **Doesn't live in the sea** (freshwater or brackish)\n3. **Smaller than a cat**\n\nThis is a fun challenge, as many freshwater aquatic mammals are quite large (beavers, otters, capybaras, hippos, freshwater dolphins).\n\n**What could that be?**\n\nThe most likely candidate is a **Muskrat**.\n\n**Why it fits:**\n\n* **Aquatic Mammal:** Yes, they are rodents highly adapted to aquatic life.\n* **Doesn't Live in the Sea:** They primarily inhabit freshwater wetlands, marshes, ponds, lakes, and slow-moving rivers.\n* **Smaller Than a Cat:** Muskrats typically weigh 2-4 pounds and are about 16-25 inches long (including their tail), making them noticeably smaller than most domestic cats.\n\nOther possibilities, though less common or slightly borderline:\n\n* **Water Shrew:** Some species are semi-aquatic and *much* smaller than a cat, but might stretch the definition of \"aquatic mammal\" for some, as they are mostly terrestrial but hunt in water.\n* **Platypus:** It fits all criteria (aquatic, freshwater, smaller than a cat), but is highly specific to Australia/Tasmania and very unique (egg-laying mammal).\n\n**How could you make sure (in 20 questions)?**\n\nAssuming the first few questions established the mammal, aquatic, freshwater, and smaller than a cat criteria, here are some follow-up questions to narrow it down to a Muskrat:\n\n1. **Does it have a long, scaly, almost hairless tail?** (Yes for Muskrat)\n2. **Is it a type of rodent?** (Yes for Muskrat)\n3. **Does it primarily eat plants, like roots, stems, and leaves?** (Yes for Muskrat, though they are omnivores and will eat small animals too)\n4. **Does it build a dome-shaped lodge out of vegetation in the water, similar to a beaver, but smaller?** (Yes for Muskrat)\n5. **Is it commonly found in North America?** (Yes for Muskrat)", + "text/markdown": "Given your clues, this is a fun challenge!\n\nWhat you are looking for could very likely be a **Water Shrew**.\n\nHere's why:\n* **Aquatic Mammal:** Water shrews are semi-aquatic, highly adapted to life in and around water. They dive, swim, and hunt underwater.\n* **Doesn't live in the sea:** They are found in freshwater habitats like streams, rivers, ponds, and marshes.\n* **Smaller than a cat:** They are tiny, much smaller than even the smallest cat, typically only a few inches long (excluding tail).\n\n**Other possibilities that fit the criteria, though less common or slightly different:**\n\n* **Platypus:**\n * Aquatic Mammal: Yes.\n * Doesn't live in the sea: Yes, freshwater rivers and lakes in Australia.\n * Smaller than a cat: Yes, a platypus is typically smaller than a domestic cat.\n * *Why less likely as a first guess for \"general\" 20 questions:* Its unique egg-laying nature and venomous spurs would be very specific questions.\n\n* **Muskrat:**\n * Aquatic Mammal: Yes, semi-aquatic rodent.\n * Doesn't live in the sea: Yes, freshwater and brackish marshes, ponds, lakes, rivers.\n * Smaller than a cat: It's debatable. Some are roughly cat-sized, others slightly larger, but generally they *can* be considered smaller than a large cat.\n\n**How you could make sure (using 20 questions):**\n\nYou'd want to ask questions that narrow down the possibilities and confirm the unique characteristics.\n\n1. **\"Does it live in freshwater?\"** (To confirm \"doesn't live in the sea\" and exclude purely marine mammals).\n2. **\"Is it significantly smaller than a house cat, like mouse-sized or rat-sized?\"** (To distinguish from muskrats which are closer to rat-sized, or very small otters).\n3. **\"Does it lay eggs?\"** (This is a key question to distinguish a Platypus, which is a monotreme, from all other mammals).\n4. **\"Does it have a flat, beaver-like tail?\"** (No, for shrews and platypus, yes for muskrats, although muskrat tails are more rounded than a beaver's, they are still somewhat flattened).\n5. **\"Is it typically venomous?\"** (Another unique trait of the male Platypus).\n6. **\"Does it primarily eat insects and small invertebrates, like worms or insect larvae?\"** (This would strongly suggest a Water Shrew or Platypus, rather than a muskrat which is mostly vegetarian).\n7. **\"Does it have very tiny eyes and a long, pointed snout?\"** (Strongly indicative of a Water Shrew).\n8. **\"Is it native to Australia?\"** (For the Platypus).\n9. **\"Does it build large lodges or dams?\"** (No, this would rule out beavers and other larger aquatic rodents).\n10. **\"Does it have prominent, gnawing front teeth?\"** (No, for shrews and platypus; yes, for muskrats).\n\nBy asking these types of questions, you could quickly home in on the correct answer.", "text/plain": [ "" ] }, - "execution_count": 10, + "execution_count": 7, "metadata": {}, "output_type": "execute_result" } @@ -285,7 +287,7 @@ }, { "cell_type": "code", - "execution_count": 11, + "execution_count": 8, "metadata": { "id": "6d09a98f06e6" }, @@ -295,9 +297,9 @@ "output_type": "stream", "text": [ "Prompt tokens: 59\n", - "Thoughts tokens: 1618\n", - "Output tokens: 487\n", - "Total tokens: 2164\n" + "Thoughts tokens: 1485\n", + "Output tokens: 738\n", + "Total tokens: 2282\n" ] } ], @@ -321,19 +323,19 @@ }, { "cell_type": "code", - "execution_count": 12, + "execution_count": 9, "metadata": { "id": "602bf11f5b78" }, "outputs": [ { "data": { - "text/markdown": "This is a fun challenge!\n\nGiven the criteria:\n* **Aquatic mammal**\n* **Doesn't live in the sea** (implies freshwater)\n* **Smaller than a cat**\n\nThe most likely answer is a **River Otter Pup** (specifically a newborn or very young one).\n\nHere's why and how you could make sure, framed as 20 Questions:\n\n**Why a River Otter Pup?**\n\n* **Aquatic Mammal:** Yes, otters are mammals that spend a lot of time in water.\n* **Doesn't Live in the Sea:** River otters (like the North American river otter, European otter, etc.) live in freshwater habitats: rivers, lakes, wetlands.\n* **Smaller Than a Cat:** A newborn otter pup is tiny, often weighing only a few ounces and fitting in the palm of your hand, much smaller than even a small house cat. Even at a few weeks old, they're still considerably smaller than an adult cat.\n\n**How could you make sure? (20 Questions - focusing on properties related to a river otter pup vs. other possibilities)**\n\nLet's assume I'm asking the questions to you, the \"guesser.\"\n\n1. Is it typically found in North America? (Yes - if North American River Otter)\n2. Does it have fur? (Yes)\n3. Is its primary diet fish? (Yes - though pups are still nursing)\n4. Is it born blind? (Yes - otter pups are altricial, meaning born underdeveloped)\n5. Does it live in a den or burrow? (Yes)\n6. Is it typically born in a litter? (Yes - 1-5 pups)\n7. Does it have webbed feet? (Yes - developing quickly)\n8. Is it able to swim immediately after birth? (No - they need to be taught)\n9. Does it eat plants? (No - otters are carnivores)\n10. Is it typically solitary as an adult? (Often, though family groups persist for a while)\n11. Does it have whiskers? (Yes)\n12. Is it a type of rodent? (No)\n13. Is it domesticated? (No)\n14. Does it lay eggs? (No)\n15. Is it primarily nocturnal? (Often, but can be active during the day)\n16. Does it build dams? (No - that's a beaver)\n17. Is it an endangered species? (Depends on the specific river otter species/region, some are, some aren't globally)\n18. Does it communicate with chirps or whistles? (Yes, young otters vocalize)\n19. Is it usually black or brown in color? (Yes, typically brown)\n20. Does it require its mother's milk to survive for the first few months? (Yes, absolutely)\n\n**Why other options are less likely:**\n\n* **Beaver kit:** Too large, even as a kit, and primarily herbivorous.\n* **Muskrat kit:** Possible for size, but they're rodents, not usually considered \"mammals\" in the same vein as otters, and their lifestyle is more marsh/wetland-focused.\n* **Mink kit:** Also a possibility for size and aquatic nature, but mink are more semi-aquatic and less strictly \"aquatic mammal\" compared to otters. They also aren't as widely known for freshwater dominance.\n* **Platypus:** While aquatic, doesn't live in the sea, and smaller than a cat, it's an egg-laying mammal (monotreme), which usually gets specified if that's the answer. Plus, only found in Australia.\n\nTherefore, a **River Otter Pup** is the strongest candidate!", + "text/markdown": "This is a fun riddle! Based on your criteria, the aquatic mammal that doesn't live in the sea and is smaller than a cat is most likely a:\n\n**Otter (specifically a river otter or Asian small-clawed otter)**\n\nHere's why and how you could make sure:\n\n**Why an Otter?**\n\n* **Aquatic Mammal:** Otters are definitely aquatic mammals, spending a significant portion of their lives in and around water.\n* **Doesn't Live in the Sea:** While there are sea otters, river otters and Asian small-clawed otters live in freshwater environments like rivers, lakes, wetlands, and estuaries.\n* **Smaller than a Cat:** This is the key distinguishing factor. A typical adult domestic cat weighs around 8-10 pounds.\n * **River otters** can be larger, but their body mass is often comparable to a larger domestic cat (though longer).\n * **Asian Small-Clawed Otters** (Aonyx cinereus) are the *smallest* otter species, typically weighing between 2-5 kg (4.4-11 lbs), making them very comparable to or even smaller than many domestic cats in terms of bulk, especially when considering their sleek bodies. They are often kept as pets in some regions, which emphasizes their manageable size.\n\n**How to Make Sure (Questions to ask in the 20 Questions game):**\n\nYou'd want to ask questions that narrow down the possibilities and confirm these specific attributes:\n\n1. **\"Is it an aquatic mammal?\"** (Yes - confirms first criteria)\n2. **\"Does it live in freshwater?\"** (Yes - distinguishes from sea creatures like dolphins, whales, seals that primarily live in the ocean)\n3. **\"Is it smaller than a house cat?\"** (Yes - this is the crucial differentiator that would rule out larger freshwater mammals like beavers, muskrats, or even larger river otters, and points strongly to Asian small-clawed otters or younger river otters).\n\n**Other Possibilities (and why they're less likely):**\n\n* **Beaver:** Too large.\n* **Muskrat:** Could fit the size, but is a rodent, not typically considered a \"mammal\" in the same way otters are in common discourse for this game (though technically it is a mammal). Less \"aquatic\" in its general perception than an otter.\n* **Water Shrew:** Too small, more insectivore than \"mammal\" in the typical sense of this game, and often overlooked in this context.\n* **Capybara (young):** A young capybara might fit the size, but adults are too large. They are also rodents.\n\nTherefore, the **otter (specifically a smaller freshwater species like the Asian small-clawed otter)** is the most fitting answer.", "text/plain": [ "" ] }, - "execution_count": 12, + "execution_count": 9, "metadata": {}, "output_type": "execute_result" } @@ -369,7 +371,7 @@ }, { "cell_type": "code", - "execution_count": 13, + "execution_count": 10, "metadata": { "id": "5dcd275d0b42" }, @@ -380,8 +382,8 @@ "text": [ "Prompt tokens: 59\n", "Thoughts tokens: None\n", - "Output tokens: 832\n", - "Total tokens: 891\n" + "Output tokens: 618\n", + "Total tokens: 677\n" ] } ], @@ -405,19 +407,19 @@ }, { "cell_type": "code", - "execution_count": 14, + "execution_count": 11, "metadata": { "id": "PZw41-lsKKMf" }, "outputs": [ { "data": { - "text/markdown": "Here's a step-by-step calculation of the maximum bending stress for the given cantilever beam:\n\n**1. Define Given Parameters:**\n\n* Length (L) = 3 m\n* Width (b) = 0.1 m\n* Height (h) = 0.2 m\n* Young's Modulus (E) = 200 GPa = $200 \\times 10^9$ Pa\n* Uniformly Distributed Load (w) = 5 kN/m = 5000 N/m\n* Point Load (P) = 10 kN = 10000 N\n\n**2. Calculate the Area Moment of Inertia (I):**\n\nFor a rectangular cross-section, the area moment of inertia about the neutral axis (which is at the centroid for bending) is:\n\n$I = \\frac{b \\cdot h^3}{12}$\n$I = \\frac{0.1 \\cdot (0.2)^3}{12}$\n$I = \\frac{0.1 \\cdot 0.008}{12}$\n$I = \\frac{0.0008}{12}$\n$I = 6.6667 \\times 10^{-5} \\text{ m}^4$\n\n**3. Determine the Maximum Bending Moment (M_max):**\n\nFor a cantilever beam, the maximum bending moment occurs at the fixed support. We need to sum the moments caused by the distributed load and the point load.\n\n* **Moment due to distributed load (M_w):**\n The equivalent point load for the distributed load is $w \\cdot L$, acting at the centroid of the distributed load (L/2 from the fixed end).\n $M_w = (w \\cdot L) \\cdot \\frac{L}{2} = \\frac{w \\cdot L^2}{2}$\n $M_w = \\frac{5000 \\cdot (3)^2}{2} = \\frac{5000 \\cdot 9}{2} = \\frac{45000}{2} = 22500 \\text{ N} \\cdot \\text{m}$\n\n* **Moment due to point load (M_p):**\n The point load acts at the free end, so its moment arm is L.\n $M_p = P \\cdot L$\n $M_p = 10000 \\cdot 3 = 30000 \\text{ N} \\cdot \\text{m}$\n\n* **Total Maximum Bending Moment (M_max):**\n Since both loads cause bending in the same direction (downwards, creating tension on top and compression on bottom for a cantilever), we sum their moments.\n $M_{max} = M_w + M_p$\n $M_{max} = 22500 + 30000 = 52500 \\text{ N} \\cdot \\text{m}$\n\n**4. Calculate the Maximum Distance from the Neutral Axis (y):**\n\nFor a rectangular cross-section, the maximum distance from the neutral axis to the outermost fiber is half of the height.\n\n$y = \\frac{h}{2}$\n$y = \\frac{0.2}{2} = 0.1 \\text{ m}$\n\n**5. Calculate the Maximum Bending Stress (σ_max):**\n\nThe bending stress formula is:\n\n$\\sigma_{max} = \\frac{M_{max} \\cdot y}{I}$\n$\\sigma_{max} = \\frac{52500 \\cdot 0.1}{6.6667 \\times 10^{-5}}$\n$\\sigma_{max} = \\frac{5250}{6.6667 \\times 10^{-5}}$\n$\\sigma_{max} = 78749166.67 \\text{ Pa}$\n\nConvert to MPa (MegaPascals):\n$\\sigma_{max} = \\frac{78749166.67}{10^6} \\text{ MPa}$\n$\\sigma_{max} = 78.75 \\text{ MPa}$ (approximately)\n\n**Final Answer:**\n\nThe maximum bending stress (σ_max) in the beam is approximately **78.75 MPa**.", + "text/markdown": "Here's how to calculate the maximum bending stress for the cantilever beam:\n\n**1. Define Given Parameters and Convert Units:**\n\n* Length of the beam (L) = 3 m\n* Width of the cross-section (b) = 0.1 m\n* Height of the cross-section (h) = 0.2 m\n* Modulus of Elasticity (E) = 200 GPa = 200 * 10^9 N/m²\n* Uniformly distributed load (w) = 5 kN/m = 5 * 10^3 N/m\n* Point load (P) = 10 kN = 10 * 10^3 N\n\n**2. Calculate the Moment of Inertia (I) for a Rectangular Cross-Section:**\n\nThe formula for the moment of inertia for a rectangular cross-section about its neutral axis is:\nI = (b * h^3) / 12\n\nI = (0.1 m * (0.2 m)^3) / 12\nI = (0.1 * 0.008) / 12\nI = 0.0008 / 12\nI = 6.6667 * 10^-5 m^4\n\n**3. Determine the Location of Maximum Bending Moment:**\n\nFor a cantilever beam, the maximum bending moment always occurs at the fixed support.\n\n**4. Calculate the Maximum Bending Moment (M_max):**\n\nThe bending moment at the fixed end due to a uniformly distributed load is (w * L^2) / 2.\nThe bending moment at the fixed end due to a point load at the free end is P * L.\n\nM_max = (w * L^2) / 2 + P * L\n\nM_max = (5 * 10^3 N/m * (3 m)^2) / 2 + (10 * 10^3 N * 3 m)\nM_max = (5 * 10^3 * 9) / 2 + 30 * 10^3\nM_max = 45 * 10^3 / 2 + 30 * 10^3\nM_max = 22.5 * 10^3 + 30 * 10^3\nM_max = 52.5 * 10^3 N.m\n\n**5. Determine the Distance from the Neutral Axis to the Outermost Fiber (y):**\n\nFor a rectangular cross-section, the outermost fiber is at a distance of h/2 from the neutral axis.\n\ny = h / 2\ny = 0.2 m / 2\ny = 0.1 m\n\n**6. Calculate the Maximum Bending Stress (σ_max):**\n\nThe formula for bending stress is:\nσ = (M * y) / I\n\nσ_max = (M_max * y) / I\n\nσ_max = (52.5 * 10^3 N.m * 0.1 m) / (6.6667 * 10^-5 m^4)\nσ_max = 5.25 * 10^3 / (6.6667 * 10^-5)\nσ_max = 78.75 * 10^6 N/m²\nσ_max = 78.75 MPa\n\n**Therefore, the maximum bending stress (σ_max) in the cantilever beam is 78.75 MPa.**", "text/plain": [ "" ] }, - "execution_count": 14, + "execution_count": 11, "metadata": {}, "output_type": "execute_result" } @@ -453,7 +455,7 @@ }, { "cell_type": "code", - "execution_count": 15, + "execution_count": 12, "metadata": { "id": "7b59d32afd90" }, @@ -464,8 +466,8 @@ "text": [ "Prompt tokens: 96\n", "Thoughts tokens: None\n", - "Output tokens: 974\n", - "Total tokens: 1070\n" + "Output tokens: 792\n", + "Total tokens: 888\n" ] } ], @@ -489,19 +491,19 @@ }, { "cell_type": "code", - "execution_count": 16, + "execution_count": 13, "metadata": { "id": "1783ef351094" }, "outputs": [ { "data": { - "text/markdown": "To calculate the maximum bending stress ($\\sigma_{max}$) in a cantilever beam, we need to use the bending stress formula:\n\n$\\sigma = \\frac{M \\cdot y}{I}$\n\nWhere:\n* $\\sigma$ = bending stress\n* M = bending moment\n* y = distance from the neutral axis to the extreme fiber (where stress is maximum)\n* I = moment of inertia of the cross-section\n\nFor maximum stress, we need the maximum bending moment ($M_{max}$) and the distance to the outer fiber.\n\nLet's break down the calculation:\n\n**1. Given Data:**\n* Length (L) = 3 m\n* Width (b) = 0.1 m\n* Height (h) = 0.2 m\n* Young's Modulus (E) = 200 GPa = 200 * 10^9 Pa (Note: E is not needed for stress calculation, only for deflection)\n* Distributed load (w) = 5 kN/m = 5000 N/m\n* Point load (P) = 10 kN = 10000 N\n\n**2. Calculate Geometric Properties:**\n\n* **Distance to the extreme fiber (y):**\n For a rectangular cross-section, the neutral axis is at the center, so y = h/2.\n y = 0.2 m / 2 = 0.1 m\n\n* **Moment of Inertia (I):**\n For a rectangular cross-section, I = (b * h^3) / 12.\n I = (0.1 m * (0.2 m)^3) / 12\n I = (0.1 m * 0.008 m^3) / 12\n I = 0.0008 m^4 / 12\n I = 6.6667 * 10^-5 m^4 (approximately)\n\n**3. Calculate the Maximum Bending Moment ($M_{max}$):**\nFor a cantilever beam, the maximum bending moment occurs at the fixed support (the wall). The moments due to the distributed load and the point load add up at this point.\n\n* **Moment due to point load P at the free end ($M_P$):**\n $M_P = P \\cdot L$\n $M_P = 10000 \\text{ N} \\cdot 3 \\text{ m}$\n $M_P = 30000 \\text{ N.m}$\n\n* **Moment due to uniformly distributed load w ($M_w$):**\n The distributed load acts as a single equivalent force (w * L) at the centroid of the load (L/2 from the fixed support).\n $M_w = (w \\cdot L) \\cdot (L/2) = w \\cdot L^2 / 2$\n $M_w = (5000 \\text{ N/m} \\cdot (3 \\text{ m})^2) / 2$\n $M_w = (5000 \\text{ N/m} \\cdot 9 \\text{ m}^2) / 2$\n $M_w = 45000 \\text{ N.m} / 2$\n $M_w = 22500 \\text{ N.m}$\n\n* **Total Maximum Bending Moment ($M_{max}$):**\n $M_{max} = M_P + M_w$\n $M_{max} = 30000 \\text{ N.m} + 22500 \\text{ N.m}$\n $M_{max} = 52500 \\text{ N.m}$\n\n**4. Calculate the Maximum Bending Stress ($\\sigma_{max}$):**\nNow, substitute the values into the bending stress formula:\n\n$\\sigma_{max} = \\frac{M_{max} \\cdot y}{I}$\n$\\sigma_{max} = \\frac{52500 \\text{ N.m} \\cdot 0.1 \\text{ m}}{6.6667 \\times 10^{-5} \\text{ m}^4}$\n$\\sigma_{max} = \\frac{5250 \\text{ N.m}^2}{6.6667 \\times 10^{-5} \\text{ m}^4}$\n$\\sigma_{max} = 78749999.99 \\text{ N/m}^2$\n\nConvert to MPa (1 MPa = 10^6 N/m^2):\n$\\sigma_{max} = 78.75 \\times 10^6 \\text{ Pa}$\n$\\sigma_{max} = 78.75 \\text{ MPa}$\n\nThe maximum bending stress in the beam is **78.75 MPa**.", + "text/markdown": "To calculate the maximum bending stress ($\\sigma_{max}$) in the cantilever beam, we need to follow these steps:\n\n1. **Calculate the Moment of Inertia (I) for the rectangular cross-section.**\n2. **Determine the maximum bending moment (M_max) in the beam.** For a cantilever, this occurs at the fixed support.\n3. **Use the bending stress formula.**\n\n---\n\n**Given Data:**\n* Length, L = 3 m\n* Width, b = 0.1 m\n* Height, h = 0.2 m\n* Young's Modulus, E = 200 GPa (Not directly needed for stress calculation, but for deflection)\n* Uniformly distributed load, w = 5 kN/m\n* Point load at free end, P = 10 kN\n\n---\n\n**Step 1: Calculate the Moment of Inertia (I)**\nFor a rectangular cross-section, the moment of inertia about the neutral axis (which is at the centroid for symmetric sections) is:\n$I = \\frac{bh^3}{12}$\n$I = \\frac{(0.1 \\text{ m}) (0.2 \\text{ m})^3}{12}$\n$I = \\frac{0.1 \\times 0.008}{12}$\n$I = \\frac{0.0008}{12}$\n$I = 6.6667 \\times 10^{-5} \\text{ m}^4$\n\n**Step 2: Determine the Maximum Bending Moment (M_max)**\nFor a cantilever beam, the maximum bending moment occurs at the fixed support. We will sum the moments due to the distributed load and the point load.\n\n* **Moment due to uniformly distributed load ($M_w$):**\n $M_w = \\frac{wL^2}{2}$\n $M_w = \\frac{(5 \\text{ kN/m}) (3 \\text{ m})^2}{2}$\n $M_w = \\frac{5 \\times 9}{2}$\n $M_w = 22.5 \\text{ kN} \\cdot \\text{m}$\n\n* **Moment due to point load ($M_P$):**\n $M_P = P \\times L$\n $M_P = (10 \\text{ kN}) \\times (3 \\text{ m})$\n $M_P = 30 \\text{ kN} \\cdot \\text{m}$\n\n* **Total Maximum Bending Moment ($M_{max}$):**\n $M_{max} = M_w + M_P$\n $M_{max} = 22.5 \\text{ kN} \\cdot \\text{m} + 30 \\text{ kN} \\cdot \\text{m}$\n $M_{max} = 52.5 \\text{ kN} \\cdot \\text{m}$\n\nTo use this in the stress formula, convert to N·m:\n$M_{max} = 52.5 \\times 10^3 \\text{ N} \\cdot \\text{m}$\n\n**Step 3: Calculate the Maximum Bending Stress ($\\sigma_{max}$)**\nThe bending stress formula is:\n$\\sigma = \\frac{My}{I}$\nThe maximum bending stress ($\\sigma_{max}$) occurs at the furthest distance from the neutral axis, which for a rectangular section is $y = h/2$.\n$y_{max} = h/2 = 0.2 \\text{ m} / 2 = 0.1 \\text{ m}$\n\nNow, substitute the values:\n$\\sigma_{max} = \\frac{M_{max} y_{max}}{I}$\n$\\sigma_{max} = \\frac{(52.5 \\times 10^3 \\text{ N} \\cdot \\text{m}) (0.1 \\text{ m})}{6.6667 \\times 10^{-5} \\text{ m}^4}$\n$\\sigma_{max} = \\frac{5250 \\text{ N} \\cdot \\text{m}^2}{6.6667 \\times 10^{-5} \\text{ m}^4}$\n$\\sigma_{max} = 78,749,375 \\text{ N/m}^2$\n\nTo express this in Megapascals (MPa), recall that 1 MPa = $10^6$ N/m$^2$:\n$\\sigma_{max} = 78.75 \\times 10^6 \\text{ N/m}^2$\n$\\sigma_{max} = 78.75 \\text{ MPa}$\n\n---\n**Final Answer:**\nThe maximum bending stress in the cantilever beam is **78.75 MPa**.", "text/plain": [ "" ] }, - "execution_count": 16, + "execution_count": 13, "metadata": {}, "output_type": "execute_result" } @@ -539,7 +541,7 @@ }, { "cell_type": "code", - "execution_count": 17, + "execution_count": 14, "metadata": { "id": "3dfac516c27b" }, @@ -549,9 +551,9 @@ "output_type": "stream", "text": [ "Prompt tokens: 96\n", - "Thoughts tokens: 1326 / 4096\n", - "Output tokens: 1109\n", - "Total tokens: 2531\n" + "Thoughts tokens: 1455 / 4096\n", + "Output tokens: 1079\n", + "Total tokens: 2630\n" ] } ], @@ -585,7 +587,7 @@ }, { "cell_type": "code", - "execution_count": 18, + "execution_count": 15, "metadata": { "id": "MIcXWXqyzCjQ" }, @@ -597,7 +599,7 @@ "" ] }, - "execution_count": 18, + "execution_count": 15, "metadata": {}, "output_type": "execute_result" } @@ -611,19 +613,19 @@ }, { "cell_type": "code", - "execution_count": 19, + "execution_count": 16, "metadata": { "id": "Lb9o7AeDwVyZ" }, "outputs": [ { "data": { - "text/markdown": "Let's break down the image and find the area of the overlapping region.\n\n1. **Analyze the Circle:**\n * The numbers '3' within the circle indicate its radius. We can see three radii marked '3'.\n * So, the radius of the circle (R) is 3.\n\n2. **Analyze the Triangle:**\n * The triangle is a right-angled triangle because two of its sides meet at a 90-degree angle, and this vertex is at the center of the circle.\n * The lengths of the two sides forming the right angle (the legs) are both marked '3'.\n\n3. **Identify the Overlapping Region:**\n * The center of the circle is at the vertex of the right angle of the triangle.\n * The two legs of the triangle extend outwards from this center, and their lengths (3) are exactly equal to the radius of the circle (3).\n * This means that the part of the circle that overlaps with the triangle is precisely a **sector** of the circle.\n * The central angle of this sector is the right angle of the triangle, which is 90 degrees.\n\n4. **Calculate the Area of the Overlapping Region (the Sector):**\n * The area of a full circle is given by the formula A = πR².\n * A_circle = π * (3)² = 9π.\n * A sector's area is a fraction of the full circle's area, determined by its central angle.\n * Fraction = (Central Angle) / 360°\n * Fraction = 90° / 360° = 1/4.\n * Area of Overlap = (Fraction) * A_circle\n * Area of Overlap = (1/4) * 9π\n * Area of Overlap = **(9/4)π**\n\nThe area of the overlapping region is (9/4)π.", + "text/markdown": "The image shows a circle and a right-angled triangle.\n\n1. **Identify the radius of the circle:** The numbers '3' within the circle indicate that the radius of the circle is `r = 3`.\n\n2. **Identify the shape of the overlapping region:** The overlapping region is a sector of the circle. Its two straight sides are radii of the circle, and its curved side is part of the circle's circumference.\n\n3. **Determine the central angle of the sector:** The two radii that form the straight sides of the sector are also sides of the green triangle. These two sides of the triangle are perpendicular to each other (one is vertical, one is horizontal, clearly forming a right angle). Therefore, the central angle of the sector is `θ = 90` degrees.\n\n4. **Calculate the area of the sector:**\n The formula for the area of a sector is:\n `Area = (θ / 360°) * πr²`\n\n Substitute the values:\n `Area = (90° / 360°) * π * (3)²`\n `Area = (1/4) * π * 9`\n `Area = 9π / 4`\n\nThe area of the overlapping region is **9π/4**.", "text/plain": [ "" ] }, - "execution_count": 19, + "execution_count": 16, "metadata": {}, "output_type": "execute_result" } @@ -661,7 +663,7 @@ }, { "cell_type": "code", - "execution_count": 20, + "execution_count": 17, "metadata": { "id": "F2YeBqzC0J_i" }, @@ -673,7 +675,7 @@ "" ] }, - "execution_count": 20, + "execution_count": 17, "metadata": {}, "output_type": "execute_result" } @@ -696,19 +698,19 @@ }, { "cell_type": "code", - "execution_count": 24, + "execution_count": 18, "metadata": { "id": "Vt2dSjeqA9ZC" }, "outputs": [ { "data": { - "text/markdown": "You've presented an interesting puzzle! However, there are **four** pool balls in the image, not three.\n\nThe balls are numbered:\n* 7 (maroon)\n* 9 (yellow and white striped)\n* 11 (maroon and white striped)\n* 13 (orange and white striped)\n\nIf you meant to use **three** of those four balls, here's how you can sum up to 30:\n\n* **7 + 9 + 13 = 29** (Close, but not 30)\n* **7 + 11 + 13 = 31** (Close, but not 30)\n* **9 + 11 + 13 = 33** (Over 30)\n* **7 + 9 + 11 = 27** (Under 30)\n\n**Therefore, you cannot select any three of these four specific numbered balls to sum up to exactly 30.**\n\nPerhaps there's a trick to the question, or maybe a different interpretation of the image or numbers?", + "text/markdown": "That's a classic riddle! You have **four** pool balls in the image, not three.\n\nHere's how you can make 30:\n\n* **Turn the 9 upside down to make it a 6.**\n\nThen, add the numbers:\n\n* **7 + 11 + 6 + 13 = 37**\n\nWait, that's not 30! Let's re-examine the image. You have balls 7, 9, 11, and 13.\n\nIf the riddle is strictly about using **three** of those specific balls, and the answer needs to be exactly 30, it's impossible with the numbers provided (7, 9, 11, 13). Any combination of three of these numbers will either be too low or too high:\n\n* 7 + 9 + 11 = 27\n* 7 + 9 + 13 = 29\n* 7 + 11 + 13 = 31\n* 9 + 11 + 13 = 33\n\n**However, the most common form of this riddle, when presented with a 6 or 9 ball, involves the trick of flipping the 9.**\n\nIf you are allowed to *only* use three balls and *flip the 9 to a 6*, then:\n\n* **11 + 13 + (flipped 9 becomes 6) = 11 + 13 + 6 = 30**\n\nThis is the standard solution to the \"sum to 30 with pool balls\" riddle. It relies on the visual ambiguity of the number 9.", "text/plain": [ "" ] }, - "execution_count": 24, + "execution_count": 18, "metadata": {}, "output_type": "execute_result" } @@ -743,19 +745,19 @@ }, { "cell_type": "code", - "execution_count": 25, + "execution_count": 19, "metadata": { "id": "6fnVAJuv530a" }, "outputs": [ { "data": { - "text/markdown": "This is a classic riddle!\n\nThere are actually **four** pool balls in the image: 7, 9, 11, and 13.\n\nHere's how you can make 30 using three of them:\n\n1. **Turn the '9' ball upside down.** When a 9-ball is turned upside down, it looks like a **6**.\n2. Now you can use the **6**, **11**, and **13** balls.\n\n**6 + 11 + 13 = 30**", + "text/markdown": "This is a classic riddle!\n\nThe trick lies in rotating one of the balls. If you rotate the **9-ball** upside down, it becomes a **6-ball**.\n\nThen you can use these three balls:\n* The **11-ball**\n* The **13-ball**\n* The **6-ball** (which was originally the 9-ball)\n\n11 + 13 + 6 = **30**", "text/plain": [ "" ] }, - "execution_count": 25, + "execution_count": 19, "metadata": {}, "output_type": "execute_result" } @@ -796,14 +798,14 @@ }, { "cell_type": "code", - "execution_count": 26, + "execution_count": 20, "metadata": { "id": "91072ecec725" }, "outputs": [ { "data": { - "text/markdown": "To obtain 565 with the numbers 10, 8, 3, 7, 1, and 5, using each number only once and common operations, here's a solution:\n\n1. Multiply 8 by 7:\n 8 * 7 = 56\n\n2. Multiply the result (56) by 10:\n 56 * 10 = 560\n\n3. Add the remaining number 5 to 560:\n 560 + 5 = 565\n\nSo, the solution is:\n**(8 * 7) * 10 + 5 = 565**\n\nNumbers used: 8, 7, 10, 5. (Numbers 3 and 1 were not needed, which is allowed as per the rules of only using a number once, not necessarily all numbers.)", + "text/markdown": "Here's how to obtain 565 using the numbers 10, 8, 3, 7, 1, and 5, using each number once:\n\n1. **Add 10 and 7:**\n 10 + 7 = 17\n\n2. **Subtract 3 from the result:**\n 17 - 3 = 14\n\n3. **Multiply by 8:**\n 14 * 8 = 112\n\n4. **Add 1:**\n 112 + 1 = 113\n\n5. **Multiply by 5:**\n 113 * 5 = 565\n\nSo the full expression is:\n**( (10 + 7 - 3) * 8 + 1 ) * 5 = 565**", "text/plain": [ "" ] @@ -848,7 +850,7 @@ }, { "cell_type": "code", - "execution_count": 27, + "execution_count": 21, "metadata": { "id": "qtEhB8xC8QAF" }, @@ -886,7 +888,7 @@ }, { "cell_type": "code", - "execution_count": 28, + "execution_count": 22, "metadata": { "id": "KcgYIrda8kAo" }, @@ -903,7 +905,7 @@ }, { "data": { - "text/markdown": "Here's a detailed thought process to solve this logic puzzle:\n\n1. **Understand the Goal:**\n * Match: Person (Alice, Bob, Carol)\n * Match: House Color (Red, Green, Blue)\n * Match: Pet (Cat, Dog, implicitly a third type for the remaining person, or perhaps not specified but we need to deduce who owns what).\n\n2. **Set up a Structure:** A table is usually the best way to organize this.\n\n | Person | House Color | Pet |\n |--------|-------------|--------|\n | Alice | | |\n | Bob | | |\n | Carol | | |\n\n Or, since we have three fixed categories (Person, House, Pet), we can use a grid. Let's try filling in facts directly first.\n\n3. **Process Each Clue Systematically:**\n\n * **Clue 1: \"Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\"**\n * Confirms 1:1 mapping. No immediate deductions, but good context.\n\n * **Clue 2: \"The person who lives in the red house owns a cat.\"**\n * Relationship: Red House <-> Cat.\n * Table update (mental or written):\n * (Red House) -> Cat\n\n * **Clue 3: \"Bob does not live in the green house.\"**\n * Direct exclusion: Bob ≠ Green.\n * Table:\n * Bob: House ≠ Green\n\n * **Clue 4: \"Carol owns a dog.\"**\n * Direct assignment: Carol -> Dog.\n * Table:\n * Carol: Pet = Dog\n\n * **Clue 5: \"The green house is to the left of the red house.\"**\n * Spatial relationship: Green - Red (Implies Blue is somewhere else, likely right of Red or not involved in this specific left/right). This clue might be a red herring if the *order* isn't needed, but it could also be crucial if we need to deduce a third house's position. For now, let's just note it. It doesn't directly assign a person or pet.\n\n * **Clue 6: \"Alice does not own a cat.\"**\n * Direct exclusion: Alice ≠ Cat.\n * Table:\n * Alice: Pet ≠ Cat\n\n4. **Combine and Deduce:**\n\n * **From Clue 2 (Red House <-> Cat) and Clue 6 (Alice ≠ Cat):**\n * If Alice doesn't own a cat, then Alice cannot live in the red house.\n * So, Alice ≠ Red House.\n\n * **From Clue 4 (Carol -> Dog):**\n * If Carol owns a dog, then Carol cannot own a cat.\n * Therefore, Carol cannot live in the red house (because the red house owner has a cat).\n * So, Carol ≠ Red House.\n\n * **Now we have:**\n * Alice ≠ Red House\n * Carol ≠ Red House\n * There are only three people (Alice, Bob, Carol) and three houses (Red, Green, Blue).\n * If neither Alice nor Carol lives in the red house, then **Bob must live in the red house.**\n\n * **Dedication 1: Bob lives in the red house.**\n * Table:\n * Bob: House = Red\n\n * **From Dedication 1 (Bob -> Red House) and Clue 2 (Red House <-> Cat):**\n * Since Bob lives in the red house, and the red house owner has a cat, then **Bob owns a cat.**\n * Table:\n * Bob: Pet = Cat\n\n * **Summary after Bob's deductions:**\n * Alice: House ≠ Red, Pet ≠ Cat\n * Bob: House = Red, Pet = Cat\n * Carol: House ≠ Red, Pet = Dog\n\n * **Now consider Carol:**\n * Carol's pet is a dog (Clue 4).\n * Carol does not live in the red house (deduced above).\n * There are three houses: Red, Green, Blue.\n * We know Bob is in Red.\n * Carol cannot be in Red.\n * From Clue 3: Bob ≠ Green House (which we already know since Bob is Red).\n * What about Carol? Carol has a dog.\n * The green house is to the left of the red house. This doesn't directly assign Carol.\n\n * **Let's re-evaluate the house assignments for Alice and Carol:**\n * Houses remaining for Alice and Carol: Green, Blue.\n * People remaining: Alice, Carol.\n\n * **Recall Bob's status:** Bob is in the Red House and owns the Cat.\n\n * **Look at Alice:**\n * Alice's pet is not a cat. (Clue 6)\n * Alice's house is not Red. (Deduction)\n * This leaves Alice with either the Green or Blue house, and a pet that is not a cat (which we already know Bob has).\n\n * **Look at Carol:**\n * Carol's pet is a dog. (Clue 4)\n * Carol's house is not Red. (Deduction)\n * This leaves Carol with either the Green or Blue house.\n\n * **Crucial step: Who owns the *remaining* pet?**\n * Pets: Cat, Dog, ??? (There are only three pets mentioned: cat, dog, and then a default third, or maybe the problem implies only two *types* of pets). The problem statement only specifies cat and dog. Let's assume the third person has neither a cat nor a dog, or their pet isn't specified. However, usually these puzzles imply *all* items in a category are unique.\n\n * We have:\n * Bob: Red, Cat\n * Carol: Dog\n\n * Therefore, Alice must have the remaining pet. What is it? The problem *doesn't* specify a third pet for Alice. It only says \"Alice does not own a cat.\" This implies Alice could own a dog, or something else. But Carol owns the dog. So, **Alice owns *neither* a cat nor a dog.** The puzzle doesn't ask for *what* her pet is, just who owns it. This is a point to note. *Correction: The problem asks \"what pet do they own?\". This implies we need to figure out a specific pet for Alice, or confirm there's no third pet type.* Let's re-read the problem very carefully. \"Alice, Bob, and Carol each live in a different house... The person who lives in the red house owns a cat. Carol owns a dog. Alice does not own a cat.\" This strongly implies only two *types* of pets are explicitly named in the clues (cat, dog). If there's a third pet, it's not mentioned. If it's a three-way unique mapping, then Alice must have the third unstated pet. Or perhaps the puzzle implicitly means there are only two pets owned between them, and one person has no pet, or a pet not mentioned. This is a weak point in the puzzle design.\n\n * Let's assume the standard interpretation: three unique people, three unique houses, three unique pets (even if the third isn't named explicitly). This is a common way these puzzles are structured. However, if *only* cat and dog are mentioned, and we've assigned them, it's possible the third person doesn't own a pet, or their pet isn't relevant.\n * *Self-correction:* The common way these puzzles work is that if 3 items are mentioned for one category (e.g., people), then 3 items for another (houses), it implies 3 for the third (pets) *if* the categories are balanced. Here, \"cat\" and \"dog\" are specifically mentioned. We have:\n * Bob: Cat\n * Carol: Dog\n * Alice: Not Cat. Since Carol has the dog, Alice cannot have the dog. This means Alice has *no* pet, or a pet that isn't named. This is unusual.\n\n * Let's re-examine if the \"pet\" category is meant to be exhaustive. \"The person who lives in the red house owns a cat.\" \"Carol owns a dog.\" \"Alice does not own a cat.\" These are the only pet-related clues. It's possible there are only two pets specified and the third person simply doesn't own one of those, or owns something else. Most puzzles of this type specify all items.\n * *Let's assume the puzzle implies a simple 1:1:1 mapping where the third pet is *implicitly* defined by exclusion.* This is how most logic puzzles work.\n * Pets: Cat, Dog, (Implicit Third Pet for Alice)\n * If Bob has a Cat and Carol has a Dog, then Alice has the (Implicit Third Pet). The puzzle doesn't ask *what* that pet is, just \"what pet do they own?\". So we might just say \"Alice owns neither a cat nor a dog\" or \"Alice owns the third pet.\" This seems like a potential ambiguity.\n\n * Let's go back to houses:\n * Bob: Red House\n * Alice: Not Red, Not Cat\n * Carol: Not Red, Dog\n\n * Remaining houses: Green, Blue.\n * Remaining people: Alice, Carol.\n\n * **Clue 3: Bob does not live in the green house.** (Already covered, Bob is Red).\n * **Clue 5: The green house is to the left of the red house.** This means the red house *isn't* the leftmost house. This clue only establishes relative position, not direct assignment to a person. It might be a distractor or for a spatial question not asked.\n\n * Let's check if there's any link between house color and pets for Alice/Carol.\n * Carol owns a dog.\n * Alice does not own a cat.\n\n * If Carol lives in the Green house, then Alice lives in the Blue house.\n * If Carol lives in the Blue house, then Alice lives in the Green house.\n\n * Is there any way to distinguish Green from Blue for Alice/Carol?\n * No direct clue states who lives in Green or Blue.\n * No direct clue states what pet lives in Green or Blue.\n\n * Let's re-evaluate the initial deduction:\n * Bob: Red, Cat (Solid)\n * Carol: Dog (Solid)\n * Alice: Not Cat (Solid). Since Carol has the dog, Alice also doesn't have the dog. So Alice has the \"third pet\" (whatever that is).\n\n * We have:\n * Alice: ??? House, (Third Pet)\n * Bob: Red House, Cat\n * Carol: ??? House, Dog\n\n * Houses remaining: Green, Blue.\n * People remaining: Alice, Carol.\n\n * Is there a clue that links Alice or Carol to Green or Blue?\n * Clue 3: Bob does not live in the green house. (We know Bob is Red)\n * Clue 5: The green house is to the left of the red house. (Spatial, doesn't link people directly to houses)\n\n * This is where I'm stuck on assigning Green/Blue to Alice/Carol. Let me review everything again very carefully.\n\n * **Facts derived so far:**\n 1. Bob lives in the Red house.\n 2. Bob owns a cat.\n 3. Carol owns a dog.\n 4. Alice does not own a cat.\n 5. Alice does not own a dog (because Carol does).\n 6. Alice lives in neither the Red house (Bob does) nor the Green house (if Bob doesn't live in Green and Bob lives in Red, it's consistent. But does Alice live in Green?).\n\n * Let's use a deduction table for houses:\n\n | Person | Red | Green | Blue | Pet |\n |--------|-----|-------|------|-------|\n | Alice | No | ? | ? | Not C, Not D |\n | Bob | Yes | No | No | Cat |\n | Carol | No | ? | ? | Dog |\n\n * From \"Bob does not live in the green house\" (Clue 3) -> Bob is not Green. (Already have Bob as Red). This clue is redundant *after* Bob is assigned Red.\n\n * Okay, if Bob is Red, then Alice and Carol must be Green or Blue.\n * If Alice is Green, then Carol is Blue.\n * If Alice is Blue, then Carol is Green.\n\n * Is there anything that forces Alice to be in Green/Blue or Carol to be in Green/Blue?\n * \"The green house is to the left of the red house.\"\n * Green - Red - Blue (possible arrangement)\n * Blue - Green - Red (possible arrangement)\n\n * This spatial clue *might* be useful if there was a clue like \"The person in the leftmost house owns X\" or \"The person in the middle house owns Y.\" But there isn't.\n\n * This feels like a solvable puzzle, so there must be a missing link.\n\n * Let's assume the standard logic puzzle format where all properties are distributed uniquely.\n * People: Alice, Bob, Carol\n * Houses: Red, Green, Blue\n * Pets: Cat, Dog, ??? (Let's call Alice's pet \"Other Pet\")\n\n * **Recap:**\n * Bob: Red House, Cat\n * Carol: Dog, (Not Red House)\n * Alice: Not Cat, Not Dog (so Other Pet), (Not Red House)\n\n * Remaining Houses for Alice/Carol: Green, Blue\n * Remaining People for Green/Blue: Alice, Carol\n\n * Is there *any* interaction left?\n * \"Bob does not live in the green house.\" We used this to confirm Bob is not Green, but it is *also* a statement about the Green House.\n * We know Bob is Red.\n * This means the Green House must be occupied by either Alice or Carol.\n\n * What if \"Bob does not live in the green house\" also implies something about the *person* Bob? It already led us to Bob being Red.\n\n * Could it be that the information about the *green house* being to the left of the *red house* is critical *with* the \"Bob doesn't live in the green house\" clue?\n * Red house owner = Cat owner = Bob.\n * Green house owner = Not Bob.\n * Carol owns a dog.\n * Alice does not own a cat.\n\n * Consider the houses: _ _ _\n * Position 1, Position 2, Position 3\n * Green is to the left of Red. So we can't have Red as the leftmost.\n * Possibilities:\n * Green - Red - Blue\n * Blue - Green - Red\n\n * This spatial relationship doesn't tell us *who* lives in Green or Blue. It only tells us the relative positions of Green and Red.\n\n * Let's assume there's no third pet type. What if the question is simply \"who lives in each house, and what pet do they own?\" And Alice has no pet, or an unknown pet? That seems unlikely for a standard logic puzzle.\n\n * Let's reconsider the interpretation of the pet rule.\n * \"The person who lives in the red house owns a cat.\"\n * \"Carol owns a dog.\"\n * \"Alice does not own a cat.\"\n\n * From this, we know:\n * Bob (Red House) owns a cat.\n * Carol owns a dog.\n * Alice does not own a cat. Since Carol owns the dog, Alice *cannot* own the dog either.\n * This leaves Alice with no pet of the specified types.\n * This implies either:\n 1. Alice has a pet not mentioned (e.g., a fish).\n 2. Alice has no pet.\n 3. The problem is underspecified for Alice's pet.\n\n * Typically, if it asks \"what pet do they own?\", it expects a specific answer. So, the \"third pet\" implication is strong. However, the puzzle does not provide enough information to *name* that pet. If the answer for Alice's pet is \"None\" or \"Unknown,\" then the question is slightly flawed. Let's assume it means \"the pet *they are assigned* in the puzzle context.\"\n\n * Let's check the constraint propagation carefully.\n * **P1: Red House owner has a cat.**\n * **P2: Bob != Green House.**\n * **P3: Carol owns a dog.**\n * **P4: Green House is left of Red House.**\n * **P5: Alice != Cat.**\n\n * From P1 & P5: Alice != Red House.\n * From P1 & P3: Carol != Red House (because Carol has a dog, not a cat).\n * Therefore, Bob must live in the Red House. (Because Alice and Carol don't).\n * **Deduction 1: Bob lives in the Red House.**\n\n * From Deduction 1 & P1: Bob owns a cat.\n * **Deduction 2: Bob owns a cat.**\n\n * Now we have:\n * Bob: Red House, Cat\n * Carol: Dog, (House: Green or Blue, Not Red)\n * Alice: Not Cat, (House: Green or Blue, Not Red). Since Carol has the dog, Alice also does not have the dog. So, Alice has the *third* unmentioned pet type (or no pet).\n\n * Houses remaining for Alice and Carol: Green, Blue.\n * Is there *any* clue that links Alice or Carol to a specific remaining house?\n * P2: \"Bob does not live in the green house.\" This is consistent with Bob living in the red house. It doesn't help distinguish Alice/Carol for Green/Blue.\n * P4: \"The green house is to the left of the red house.\" This is about relative position, not absolute assignment to a person.\n\n * This implies that the specific assignment of Alice/Carol to Green/Blue might not be uniquely determined by the given clues, or I'm missing a very subtle point.\n\n * Let's assume the question is solvable, and there *is* a unique solution.\n * What if the \"Bob does not live in the green house\" means *more* than just confirming Bob is not in Green?\n * If Bob is in Red, then the Green House must be occupied by either Alice or Carol.\n * Does the \"to the left of\" matter?\n * Consider the houses: [Green] - [Red] - [Blue] (Order 1)\n * Consider the houses: [Blue] - [Green] - [Red] (Order 2)\n\n * If Order 1:\n * Green: Alice or Carol\n * Red: Bob (Cat)\n * Blue: Carol or Alice\n\n * If Order 2:\n * Blue: Alice or Carol\n * Green: Alice or Carol\n * Red: Bob (Cat)\n\n * This spatial info seems irrelevant to assigning people to houses unless a property of position (e.g., leftmost, middle) was given.\n\n * Let me re-read the problem description very slowly for any hidden implications.\n * \"Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\" (Sets up 3x3 mapping)\n * \"The person who lives in the red house owns a cat.\" (Red=Cat)\n * \"Bob does not live in the green house.\" (Bob != Green)\n * \"Carol owns a dog.\" (Carol=Dog)\n * \"The green house is to the left of the red house.\" (Green < Red)\n * \"Alice does not own a cat.\" (Alice != Cat)\n\n * Let's re-verify the \"Bob in Red\" deduction.\n 1. Red house owner has cat.\n 2. Alice doesn't own cat. So Alice != Red.\n 3. Carol owns dog. So Carol != Cat owner. So Carol != Red.\n 4. Only Bob is left. So Bob is Red house owner. (Confirmed, this is solid)\n 5. Bob owns a cat. (Confirmed, this is solid)\n\n * Now, who is in Green and Blue?\n * Houses: Green, Blue\n * People: Alice, Carol\n * Pets: Dog (Carol), Not Cat (Alice)\n\n * **Could the \"Bob does not live in the green house\" be the key to breaking the tie between Alice and Carol for Green/Blue?**\n * We know Bob is Red. This statement is *consistent* with Bob being Red.\n * It also implies that *whoever lives in the green house is not Bob*. We already know that as Bob is in Red.\n\n * This feels like the problem is underspecified if the solution requires Alice to be Green and Carol to be Blue (or vice-versa). There is no clue left that distinguishes the Green house from the Blue house for Alice or Carol based on their pets or any other property.\n\n * *What if \"Bob does not live in the green house\" is meant to be interpreted as \"Bob is NOT the one who lives in the green house\"?*\n * If Bob is NOT the one in Green, then the one in Green is either Alice or Carol.\n * This doesn't help us choose between Alice and Carol.\n\n * *Could the \"left of\" clue have some indirect meaning?*\n * Green is left of Red.\n * Red is Bob.\n * So Green is left of Bob.\n * This still doesn't tell us if Alice or Carol is in Green.\n\n * Let's consider if there's an implicit rule about the pets. We have a cat and a dog. What about Alice?\n * Alice is not a cat owner.\n * Carol is a dog owner. So Alice is not a dog owner.\n * This means Alice must own the *third* pet. The puzzle doesn't name it. This is fine if the answer is \"the third pet\" or \"a pet not specified.\"\n\n * What if the clue \"Bob does not live in the green house\" *only* serves to exclude Bob from Green, leaving Alice and Carol for Green and Blue, and there's no further distinction? This would mean the solution is not unique for Alice/Carol's house assignments. But logic puzzles *always* have a unique solution.\n\n * There must be something I'm overlooking.\n * Let's list all exclusions:\n * Alice: No Cat, No Dog, No Red House\n * Bob: No Green House\n * Carol: No Red House (because Dog, not Cat)\n\n * Let's try a grid:\n\n | | Red | Green | Blue | Cat | Dog | Other Pet |\n |--------|---------|---------|---------|---------|---------|-----------|\n | Alice | No (AC) | | | No (A!=C)| No (C=D)| Yes |\n | Bob | Yes (D1)| No (C3) | | Yes (D2)| No (B=C)| No |\n | Carol | No (CR) | | | No (CR) | Yes (C4)| No |\n | House | | | | | | |\n | Red | | | | Yes (C2)| No | No |\n | Green | | | | No (Red=Cat)| No | No |\n | Blue | | | | No | No | No |\n\n (AC = Alice not Cat owner implies not Red house owner)\n (C3 = Clue 3: Bob != Green)\n (C4 = Clue 4: Carol = Dog)\n (D1 = Deduction 1: Bob = Red)\n (D2 = Deduction 2: Bob = Cat)\n (CR = Carol owns Dog, so not Cat, so not Red house)\n\n * From this, we have:\n * Bob: Red, Cat.\n * Carol: Dog. (Not Red house)\n * Alice: Not Cat, Not Dog (so, Other Pet). (Not Red house)\n\n * Houses left: Green, Blue\n * People left: Alice, Carol\n\n * Is there any connection between Green/Blue and the *type* of pet?\n * Red has Cat.\n * Green has NOT Bob.\n * This means Green has Alice OR Carol.\n\n * Could it be that the puzzle expects us to know the *exact* type of pet for Alice? \"what pet do they own?\"\n * If the pets are only Cat and Dog, and Bob has Cat, Carol has Dog, then Alice has *no* pet. This is unusual phrasing for these puzzles if true. \"Alice does not own a cat\" could mean she owns a dog or something else. But Carol owns the dog. So Alice owns neither. This leads to the \"Alice has no pet\" or \"Alice has unknown pet\" dilemma.\n\n * Let's consider if \"Bob does not live in the green house\" also implies \"The green house does not have Bob's pet.\"\n * Green house does not have Bob's pet (Cat). We already know Red has the cat.\n\n * This feels like a simple puzzle, yet this last step for Alice/Carol's house is elusive.\n\n * What if \"Bob does not live in the green house\" is meant to be a direct assignment for *someone else* to the green house? No, it's an exclusion.\n\n * Let's assume the house colors are also unique in some way that links to Alice or Carol.\n * Red = Bob.\n * Green < Red.\n * Blue is the third house.\n\n * If we look at remaining items:\n * People: Alice, Carol\n * Houses: Green, Blue\n * Pets: Dog (for Carol), \"Other\" (for Alice)\n\n * Is there a connection between a person and a specific *remaining* house color?\n * No.\n * Is there a connection between a specific *remaining* house color and a *remaining* pet?\n * No. (Only Red is linked to Cat).\n\n * The only remaining piece of information is the spatial relationship (Green left of Red). This provides no assignment of a *person* to Green or Blue.\n\n * Could the problem be ambiguous for Alice/Carol's house?\n * Scenario 1: Alice lives in Green, Carol lives in Blue.\n * Alice: Green, (Other Pet)\n * Bob: Red, Cat\n * Carol: Blue, Dog\n * Check: Alice not cat (yes). Bob not green (yes, he's red). Carol dog (yes). Green left of Red (consistent with Green-Red-Blue order, or Blue-Green-Red where Green is left of Red). This all works.\n\n * Scenario 2: Alice lives in Blue, Carol lives in Green.\n * Alice: Blue, (Other Pet)\n * Bob: Red, Cat\n * Carol: Green, Dog\n * Check: Alice not cat (yes). Bob not green (yes, he's red). Carol dog (yes). Green left of Red (consistent). This also works.\n\n * This indicates the puzzle *is* underspecified for Alice/Carol's house assignment.\n * However, usually these puzzles have a unique solution. Let me check if there is an implied convention for pets.\n * \"Alice, Bob, and Carol... The person who lives in the red house owns a cat. Carol owns a dog. Alice does not own a cat.\"\n * This phrasing *only* explicitly defines two types of pets (cat, dog).\n * Bob owns the cat.\n * Carol owns the dog.\n * Alice does not own a cat. Since Carol has the dog, Alice cannot own the dog.\n * Therefore, Alice *does not own one of the specified pets*.\n * If the question \"what pet do they own?\" requires a specific pet type, then Alice's pet is undefined. If the problem implies there are only *two* pets *among* them, and Alice has neither, then Alice has no pet. This is the most logical conclusion if the puzzle doesn't provide more info or implies 3 *unique* pets.\n\n * Let's assume the question asks for *who* owns *which* of the named pets.\n * Bob owns the cat.\n * Carol owns the dog.\n * Alice owns neither the cat nor the dog. (So for Alice's pet, the answer would be \"None of the specified pets\" or \"N/A\" or \"Unknown\").\n\n * Given the common format of these puzzles, it's more likely that the solution for Alice/Carol's house *is* unique. Where could the differentiator be?\n * \"Bob does not live in the green house.\" This is the only negative statement about a house and a person *other than the Red house deductions*.\n\n * Let's assume the *entire puzzle* is consistent and has a unique solution.\n * If Bob is in Red, and Bob is NOT in Green, then Green must be either Alice or Carol.\n * If Carol owns a dog, and Alice does not own a cat...\n\n * Is it possible the number of pets is implicitly 3 unique pets? If so, what is Alice's pet? The clues don't say.\n * Okay, let's assume the \"what pet do they own?\" implies we must name it. This suggests there are only 2 types of pets, and one person has none, or Alice has a pet that isn't named in the list. This is problematic.\n\n * Let's try to find a solution online for this specific puzzle or a similar one that has this ambiguity.\n * Most logic puzzles of this type are fully constrained.\n * Could the \"left of\" clue apply to people as well? No, that's a house property.\n\n * I'm confident in:\n * Bob lives in the Red house and owns a cat.\n * Carol owns a dog.\n\n * This leaves Alice and Carol for Green and Blue houses.\n * And Alice has neither a cat nor a dog.\n\n * There is no further clue to distinguish between Alice in Green/Carol in Blue OR Alice in Blue/Carol in Green.\n * Therefore, the problem is under-specified regarding which of Alice or Carol lives in which of the Green or Blue houses.\n\n * However, if forced to pick the *most likely* interpretation for a contest or exam, sometimes there's a subtle nuance.\n * \"Bob does not live in the green house.\" This is the only statement that directly links Bob to Green in a negative way.\n * Red house owner = Cat.\n * Bob = Red. Bob = Cat.\n * Alice = Not Cat. So Alice != Red.\n * Carol = Dog. So Carol != Cat. So Carol != Red.\n * So Alice and Carol are the only ones left for Green and Blue.\n\n * Let's explicitly state the deductions.\n 1. From \"The person who lives in the red house owns a cat\" and \"Alice does not own a cat\": Alice does not live in the red house.\n 2. From \"The person who lives in the red house owns a cat\" and \"Carol owns a dog\": Carol does not live in the red house (because she owns a dog, not a cat).\n 3. Since Alice and Carol do not live in the red house, Bob must live in the red house.\n 4. Since Bob lives in the red house, and the person in the red house owns a cat, Bob owns a cat.\n 5. We know Carol owns a dog.\n 6. Alice does not own a cat. Since Bob owns the cat and Carol owns the dog, Alice must own neither a cat nor a dog (or an unspecified \"third\" pet). The question asks \"what pet do they own\", so if only cat and dog are specified, Alice has no pet of these types.\n\n * Now, for the houses:\n * Bob lives in the Red house.\n * Remaining houses: Green, Blue.\n * Remaining people: Alice, Carol.\n\n * The clue \"Bob does not live in the green house\" is already satisfied by Bob living in the red house. It doesn't help assign Green/Blue to Alice/Carol.\n * The clue \"The green house is to the left of the red house\" establishes relative position but doesn't assign a person to a house.\n\n * Conclusion: The puzzle, as stated, does not provide enough information to uniquely assign Alice and Carol to the Green and Blue houses. It *is* underspecified for that part.\n * However, I must provide *a* solution.\n * If a puzzle designer intended a unique solution, they've missed a clue.\n * Let me confirm that I'm not missing any implicit logic. Often, in these puzzles, if a negative statement exists (Bob doesn't live in Green), it implies someone *else* does. But that only leaves Alice or Carol.\n\n * I will state the ambiguity for Alice and Carol's houses and pets. For Alice's pet, I'll state that she owns neither a cat nor a dog, as those are the only pets mentioned and assigned.\n\n * **Final check of solid deductions:**\n * Bob: Red House, Cat. (100% certain)\n * Carol: Dog. (100% certain)\n * Alice: Not Cat. (100% certain)\n\n * **Remaining unknowns that prevent unique solution:**\n * Alice's house (Green or Blue?)\n * Carol's house (Green or Blue?)\n * Alice's pet (Not cat, not dog. What is it? The puzzle doesn't say.)\n\n * The question asks \"Who lives in each house, and what pet do they own?\".\n * I can answer for Bob definitively.\n * I can answer for Carol's pet definitively.\n * I can answer for Alice's pet as \"neither a cat nor a dog\" (or \"no specified pet\").\n * I cannot definitively assign Green/Blue to Alice/Carol.\n\n * If I had to pick *one* scenario, which one? There's no logical basis for it.\n * Therefore, the most accurate answer is to state the knowns and the remaining ambiguity. However, puzzles usually have a single correct answer.\n\n * Let's reconsider the wording \"Alice, Bob, and Carol each live in a different house...\". This means each person is *uniquely* assigned.\n * \"The person who lives in the red house owns a cat.\"\n * \"Carol owns a dog.\"\n * \"Alice does not own a cat.\"\n * Are there only *two* pets in play (cat, dog), and one person owns neither? If so, the question \"what pet do they own?\" implies Alice's pet isn't a cat or dog, or she has none.\n * If there are three distinct pets *implied* (Cat, Dog, X), then Alice owns X.\n\n * Let's assume the puzzle is well-formed and *does* have a unique solution for *all* parts.\n * What if \"Bob does not live in the green house\" means something special *because* it's the *only* negative house assignment not derived from pet ownership?\n * If Bob is Red.\n * Alice is not Red. Carol is not Red.\n * Alice is not Cat. Carol is Dog.\n * Houses: Red, Green, Blue.\n * Pets: Cat, Dog, [Third].\n * People: Alice, Bob, Carol.\n\n * **Bob:** Red, Cat.\n * **Carol:** Dog.\n * **Alice:** [Third Pet].\n\n * Remaining Houses for Alice/Carol: Green, Blue.\n * Is there a connection between a type of pet and a house that is not Red? No.\n * This is the persistent logical hole.\n\n * Perhaps the problem expects a partial answer, or I need to explicitly state the choice.\n * \"Who lives in each house, and what pet do they own?\" This asks for *the* answer.\n\n * Unless the \"to the left of\" implies a left-to-right filling order or association that I'm missing.\n * Green - Red (Bob)\n * This means Green is to the left. The third house (Blue) could be anywhere else, not necessarily to the right. But on \"the same street\" implies linear.\n * Possibilities:\n 1. Green - Red - Blue\n 2. Blue - Green - Red\n\n * Neither of these implicitly links Alice or Carol to Green or Blue.\n\n * I will provide the definitive parts and then state the unresolved ambiguity for Alice/Carol's houses. For Alice's pet, I'll state it's neither cat nor dog.", + "text/markdown": "**Logic Puzzle Breakdown - My Thought Process:**\n\nOkay, here's how I'd approach this classic logic puzzle. First, I'd solidify my understanding of the *goal*: figure out who lives in which colored house (red, green, blue) and what pet each person owns (cat, dog, or an implied third). I need to identify the *entities* involved: Alice, Bob, and Carol (people); red, green, and blue (houses); and cat, dog, and presumably a third pet.\n\nI'd immediately set up a mental (or actual) grid to track these relationships: Person vs. House vs. Pet. This grid is *crucial* for keeping track of the deductions.\n\nNext, I'd tackle the clues one by one. \"The person in the red house has a cat\" establishes a direct link. \"Bob doesn't live in the green house\" eliminates a possibility. \"Carol owns a dog\" is another direct assignment. \"Green is left of red\" gives a spatial constraint - important for later but not immediately helpful, yet. \"Alice doesn't own a cat\" provides an important negative constraint.\n\nNow for the deductions. Combining the clues, I know *Bob* must own the cat, and therefore lives in the red house. I'd use this to deduce that Carol lives in the green house (because Bob can't and the other two are in the remaining houses), and Alice in the blue house. Carol, with a dog, is in the green house.\n\nThen, I'd deduce that since Alice doesn't own a cat and Carol owns a dog, and since the problem implies all three people own pets, then Alice owns the third, unspecified pet. It's a classic structure, so the logic here is that the solver deduces the third pet.\n\nFinally, the spatial clue (\"Green to the left of red\") helps, but here, it's about the general positional context on a street. I'd verify it's consistent with my existing house assignments (Green - Carol, Red - Bob), ensuring no contradictions. The clue doesn't uniquely define the *exact* street layout.\n\nMy solution: **Bob lives in the red house and has a cat; Carol lives in the green house and has a dog; Alice lives in the blue house and has the third pet, which is neither a cat nor a dog.**\n", "text/plain": [ "" ] @@ -930,7 +932,7 @@ }, { "data": { - "text/markdown": "Let's break this down step-by-step using a process of elimination and deduction.\n\n**Information Given:**\n\n1. Alice, Bob, Carol live in Red, Green, Blue houses (one person per house).\n2. Red house owner has a cat.\n3. Bob does not live in the green house.\n4. Carol owns a dog.\n5. Green house is to the left of the Red house.\n6. Alice does not own a cat.\n\n**Deductions:**\n\n* **Deduction 1: Who lives in the Red house?**\n * From (2) \"The person who lives in the red house owns a cat.\"\n * From (6) \"Alice does not own a cat.\"\n * Therefore, Alice does not live in the red house.\n * From (4) \"Carol owns a dog.\"\n * Since Carol owns a dog, she does not own a cat. Therefore, Carol does not live in the red house.\n * Since Alice and Carol do not live in the red house, the only person remaining is Bob.\n * **Conclusion 1: Bob lives in the Red house.**\n\n* **Deduction 2: What pet does Bob own?**\n * From (2) \"The person who lives in the red house owns a cat.\"\n * From (Conclusion 1) \"Bob lives in the Red house.\"\n * **Conclusion 2: Bob owns a cat.**\n\n* **Deduction 3: What pet does Alice own?**\n * From (6) \"Alice does not own a cat.\"\n * From (4) \"Carol owns a dog.\" This means the dog is taken.\n * Since Bob owns the cat (Conclusion 2) and Carol owns the dog (Clue 4), Alice must own neither of these specific pets. The puzzle doesn't specify a third type of pet, so we can state she owns neither a cat nor a dog.\n\n* **Deduction 4: Who lives in the Green and Blue houses?**\n * We know Bob lives in the Red house.\n * Remaining houses are Green and Blue.\n * Remaining people are Alice and Carol.\n * From (3) \"Bob does not live in the green house.\" This is consistent with Bob living in the red house, but doesn't help distinguish Alice or Carol for the Green house.\n * From (5) \"The green house is to the left of the red house.\" This establishes a spatial relationship but does not assign a person to either Green or Blue.\n\n * **At this point, the puzzle does not provide enough information to definitively assign Alice to either the Green or Blue house, nor Carol to the other remaining house.** Both scenarios are consistent with the clues:\n * Scenario A: Alice lives in the Green house, Carol lives in the Blue house.\n * Scenario B: Alice lives in the Blue house, Carol lives in the Green house.\n\n * However, based on typical logic puzzles, the most common interpretation assumes a unique solution. Given the information, we can only state the definitive assignments.\n\n---\n\n**Summary of Definitive Answers:**\n\n* **Bob:** Lives in the **Red** house and owns a **cat**.\n* **Carol:** Owns a **dog**.\n* **Alice:** Does not own a **cat** or a **dog**.\n\n**Houses for Alice and Carol (Ambiguous based on given clues):**\n\n* Carol lives in either the **Green** or **Blue** house.\n* Alice lives in the other house (either **Blue** or **Green**).\n\nSince the problem asks \"Who lives in *each* house\", and this portion is ambiguous, if a unique answer is required, there might be a very subtle interpretation missing, or the puzzle is indeed slightly underspecified. However, based *strictly* on the provided clues, the house assignments for Alice and Carol are not uniquely determined.\n\nLet's assume the common expectation of a fully solvable puzzle and see if there's any implicit rule. If the spatial clue \"Green house is to the left of the Red house\" was combined with another spatial clue (e.g., \"Carol lives in the leftmost house\"), it would resolve the ambiguity. Without it, there's no way to logically place Alice or Carol into Green or Blue.\n\n**Therefore, the most accurate answer given the problem:**\n\n* **Bob** lives in the **Red** house and owns a **cat**.\n* **Carol** owns a **dog**. (Her house is either Green or Blue).\n* **Alice** does not own a cat or a dog. (Her house is the one remaining after Bob and Carol's house is decided, either Green or Blue).\n\nIf forced to pick, without further information, it's impossible to uniquely determine Alice's and Carol's house assignments.", + "text/markdown": "Let's break this down step by step:\n\n1. **The person who lives in the red house owns a cat.**\n2. **Carol owns a dog.**\n * This means Carol does *not* own a cat.\n3. **Alice does not own a cat.**\n * Since Carol has a dog and Alice doesn't have a cat, and the red house owner has a cat, the only person left to own the cat is **Bob**.\n * Therefore, **Bob lives in the red house and owns a cat.**\n\n4. **Bob does not live in the green house.**\n * We already know Bob lives in the red house, so this clue is consistent.\n\n5. **The green house is to the left of the red house.**\n * Since Bob lives in the red house, the green house is to the left of Bob's house.\n\nNow let's put it all together:\n\n* We know **Bob lives in the red house** and owns a cat.\n* We know **Carol owns a dog**.\n* We know Bob is in the red house, and Bob is not in the green house.\n* The only remaining person for the green house (since Bob is in red) is either Alice or Carol.\n* If Carol lives in the green house (which fits since she owns a dog, not a cat), then Alice must live in the blue house.\n\nLet's check:\n* **Bob:** Lives in Red House, owns a Cat. (Fits all clues related to Bob and cat)\n* **Carol:** Owns a Dog. If she lives in the Green House, this fits.\n* **Alice:** Does not own a cat. If she lives in the Blue House, she must own the third pet (not a cat or a dog).\n\n**Conclusion:**\n\n* **Bob lives in the Red house** and owns a **cat**.\n* **Carol lives in the Green house** and owns a **dog**.\n* **Alice lives in the Blue house** and owns the **third pet** (not a cat or a dog).\n\nThe street layout (Green to the left of Red) implies:\n* Green House (Carol)\n* Red House (Bob)\n* Blue House (Alice) - Its position relative to the others isn't fixed, only that Green is left of Red. It could be Green, Blue, Red or Blue, Green, Red, or as shown above. However, the puzzle only asks who lives in each house and their pet.", "text/plain": [ "" ] @@ -963,7 +965,7 @@ }, { "cell_type": "code", - "execution_count": 31, + "execution_count": 23, "metadata": { "id": "93S2-d5X85kq" }, @@ -980,47 +982,7 @@ }, { "data": { - "text/markdown": "Here's", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "my thinking process to solve this logic puzzle:\n\n1. **Understand the Goal:**", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "I need to match each person (Alice, Bob, Carol) to a house color (red, green, blue) and a pet (cat, dog, and presumably a third pet since the puzzle implies everyone has one, or at least two are", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "explicitly mentioned).\n\n2. **List the Entities:**\n\n * **People:** Alice, Bob, Carol\n * **Houses:** Red, Green, Blue\n * **Pets:** Cat, Dog (and", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "one unknown/implied third pet, or just the fact that people own pets)\n\n3. **Create a Grid/Table (Mental or Physical):** This is the most helpful way to track information.\n\n | Person | House Color", + "text/markdown": "**Analyzing the Problem**\n\nI'm currently focused on the overarching objective: determining the residents of the red, green, and blue houses, and Alice's pet. I've begun parsing the initial setup and am carefully noting the key questions. My primary strategy involves a methodical dissection of the problem's constraints.", "text/plain": [ "" ] @@ -1030,7 +992,7 @@ }, { "data": { - "text/markdown": "| Pet |\n |--------|-------------|-----|\n | Alice | | |\n | Bob | | |\n | Carol | | |\n\n And perhaps another one", + "text/markdown": "**Breaking Down the Components**\n\nI'm now focusing on a methodical breakdown, preparing a structured table, to capture the relationships between people, houses, and pets. I'm prioritizing setting up the framework to track these elements efficiently and avoid getting lost in the constraints.", "text/plain": [ "" ] @@ -1040,7 +1002,7 @@ }, { "data": { - "text/markdown": "for house-pet direct links:\n\n | House Color | Pet Owner | Pet |\n |-------------|-----------|-----|\n | Red | | |\n | Green | | |\n | Blue", + "text/markdown": "**Developing a Framework**\n\nI've revised the tables, now including both residents and pets, and the order (left-to-right) is necessary for some clues. I'm focusing on incorporating each piece of information, starting with the pets and their owners. I'm starting to visualize how the clues interconnect.", "text/plain": [ "" ] @@ -1050,7 +1012,7 @@ }, { "data": { - "text/markdown": "| | |\n\n4. **Process Clues One by One, Marking Deductions:**\n\n * **Clue 1: \"The person who lives in the red house owns a cat.\"**\n *", + "text/markdown": "**Continuing Deductions**\n\nI'm now integrating the spatial relationships and negative constraints. The house order possibilities (Blue-Green-Red or Green-Red-Blue) are starting to reveal implications, for instance, Bob cannot live in the green house and Carol owns a dog. Since Alice doesn't own the cat and the red house resident does, I'm carefully analyzing the possibilities for Alice's pet and house, given the initial setup. I believe I can make initial deductions now.", "text/plain": [ "" ] @@ -1060,7 +1022,7 @@ }, { "data": { - "text/markdown": "This directly links Red House -> Cat.\n * Update House table:\n | House Color | Pet Owner | Pet |\n |-------------|-----------|-----|\n | Red | | Cat |\n |", + "text/markdown": "**Making Crucial Inferences**\n\nI'm now focusing on deducing the connections. I've realized Bob lives in the red house, and owns the cat. Based on this, I'm working on placing the other residents. The key now is the \"to the left of\" clue, which helps position the houses relative to each other. I'm focusing on the house positions to identify Alice's pet and her house.", "text/plain": [ "" ] @@ -1070,7 +1032,7 @@ }, { "data": { - "text/markdown": "Green | | |\n | Blue | | |\n\n * **Clue 2: \"Bob does not live in the green house.\"**\n * Person Table: Bob cannot be Green.", + "text/markdown": "**Clarifying Deductions**\n\nI've made the key deductions: Bob is in the red house, and owns the cat. Based on this, I'm working on positioning the other residents. The \"to the left of\" clue is key, helping to establish the house order for Alice's pet and house.", "text/plain": [ "" ] @@ -1080,7 +1042,7 @@ }, { "data": { - "text/markdown": "* (Bob, Green) = NO\n\n * **Clue 3: \"Carol owns a dog.\"**\n * This directly links Carol -> Dog.\n * Person Table:\n | Person | House Color | Pet", + "text/markdown": "**Unveiling the Key Deductions**\n\nI'm now integrating Carol's dog ownership and Alice's cat aversion. I've placed Bob in the red house, and I'm using the left-to-right house order to position the remaining residents and their pets. The possibilities are beginning to crystallize.", "text/plain": [ "" ] @@ -1090,7 +1052,7 @@ }, { "data": { - "text/markdown": "|\n |--------|-------------|------|\n | Alice | | |\n | Bob | | |\n | Carol | | Dog |\n\n * **Clue", + "text/markdown": "**Forming the Connections**\n\nI'm now integrating Carol's dog ownership and the spatial relationships. I've placed Bob in the red house with the cat, and am carefully considering the potential arrangements for the blue and green houses. The goal now is to establish Alice's pet and her house, with the knowledge of Carol's dog.", "text/plain": [ "" ] @@ -1100,7 +1062,7 @@ }, { "data": { - "text/markdown": "4: \"The green house is to the left of the red house.\"**\n * This establishes a spatial relationship: Green - Red.\n * This means the blue house must be somewhere else relative to these two", + "text/markdown": "**Revising the House Order**\n\nI've revised the house order possibilities and have made some critical deductions about who owns which pet. I'm focusing on Carol and Alice's house options and am carefully analyzing the implications of the \"Green is to the left of Red\" clue.", "text/plain": [ "" ] @@ -1110,7 +1072,7 @@ }, { "data": { - "text/markdown": ", or the street order implies Green, Red, Blue or Blue, Green, Red. It doesn't give a fixed absolute position, but fixes Green and Red relative to each other. *Crucially, this tells me the Blue house is NOT", + "text/markdown": "**Deducing the Pet Relationships**\n\nI'm now focusing on Carol's dog ownership and the spatial arrangement of the houses, and am now integrating both these clues. I've placed Bob and the cat, but am working on the house and pet connections, and am starting to grasp the logical structure.", "text/plain": [ "" ] @@ -1120,7 +1082,7 @@ }, { "data": { - "text/markdown": "Green or Red.*\n\n * **Clue 5: \"Alice does not own a cat.\"**\n * Person Table: Alice cannot have Cat.\n * (Alice, Cat) = NO\n\n5. **", + "text/markdown": "**Re-evaluating Deductions**\n\nI've re-examined the spatial clues, realizing the green-to-the-left-of-red constraint is key for ordering the houses. By carefully testing possible arrangements, I now have a clearer picture of who occupies which house and their pet. I'm focusing on Carol's dog, Alice's pet, and the final house arrangement, working towards a definitive solution.", "text/plain": [ "" ] @@ -1130,7 +1092,7 @@ }, { "data": { - "text/markdown": "Look for direct contradictions or immediate placements:**\n\n * From Clue 1 (Red -> Cat) and Clue 5 (Alice does not own a cat):\n * **Deduction:** Alice cannot live in the Red house.", + "text/markdown": "**Strengthening the Logic**\n\nI'm now revising my approach by re-evaluating the \"green-to-the-left-of-red\" relationship. I'm carefully constructing two possible house orderings and their impact on the residents and pets. I'm focusing on systematically testing the implications of each potential arrangement. I'm seeking to determine how to definitively place Alice and her pet, given the clues about Carol's dog and Bob.", "text/plain": [ "" ] @@ -1140,7 +1102,7 @@ }, { "data": { - "text/markdown": "* Person Table: (Alice, Red) = NO\n\n * From Clue 3 (Carol owns a dog) and Clue 1 (Red -> Cat):\n * **Deduction:** Carol cannot", + "text/markdown": "**Navigating the Puzzle**\n\nI'm now integrating the spatial relationships and using deductions to place the residents and pets. Analyzing the \"Green is to the left of Red\" clue is key. I've placed Bob and the cat, and I'm refining my arrangement to find Alice's pet and house.", "text/plain": [ "" ] @@ -1150,7 +1112,7 @@ }, { "data": { - "text/markdown": "live in the Red house (because she owns a dog, not a cat).\n * Person Table: (Carol, Red) = NO\n\n6. **Synthesize and Deduce:**\n\n * We know:\n * Alice", + "text/markdown": "**Confirming the Solution**\n\nI'm now satisfied with the solution after revisiting the deductions. I've confirmed both possible house arrangements lead to valid solutions. The only missing information is the color arrangement of blue. I've placed the pets, including Alice's unmentioned pet, and can conclude the logic is sound.", "text/plain": [ "" ] @@ -1160,7 +1122,7 @@ }, { "data": { - "text/markdown": "is NOT in Red.\n * Carol is NOT in Red.\n * There are only three people (Alice, Bob, Carol).\n * **Major Deduction:** If Alice and Carol are not in the Red house, then **", + "text/markdown": "**Finalizing the Solution**\n\nI'm now satisfied with the final solution, having confirmed both possible arrangements align with all the puzzle's clues. I've placed Bob and the cat, using deductions and spatial relationships. The remaining placement of the other people and pets has been confirmed. I can confidently say I've unraveled the puzzle.", "text/plain": [ "" ] @@ -1170,7 +1132,7 @@ }, { "data": { - "text/markdown": "Bob MUST live in the Red house!**\n\n * Now we can fill in more:\n * Person Table:\n | Person | House Color | Pet |\n |--------|-------------|------|", + "text/markdown": "**Confirming the Solution**\n\nI've re-examined the spatial clues, and the Green-Red relationship. I'm satisfied with the solutions for both arrangements, including the pet assignments. It's likely the specific order of the blue house is irrelevant, and I've successfully placed the pets and residents.", "text/plain": [ "" ] @@ -1180,7 +1142,7 @@ }, { "data": { - "text/markdown": "| Alice | | |\n | Bob | Red | |\n | Carol | | Dog |\n\n * House Table (using Clue 1: Red -> Cat):", + "text/markdown": "**Confirming Deductions**\n\nI'm now satisfied with the solution after re-evaluating the clues. I've confirmed that the \"Green-Red\" relationship allows for two valid arrangements. Bob's in the red house with the cat, while Alice and Carol could inhabit the blue or green houses, each with a specific pet. I can confidently say I've found a consistent, if slightly ambiguous, solution.", "text/plain": [ "" ] @@ -1190,7 +1152,7 @@ }, { "data": { - "text/markdown": "| House Color | Pet Owner | Pet |\n |-------------|-----------|-----|\n | Red | Bob | Cat |\n | Green | | |\n | Blue | | |", + "text/markdown": "**Clarifying the Order**\n\nI'm now integrating Carol's dog ownership and the spatial relationships. I've successfully placed Bob in the red house with his cat, and am carefully considering the potential arrangements for the blue and green houses. The goal now is to establish Alice's pet and her house, with the knowledge of Carol's dog.", "text/plain": [ "" ] @@ -1200,7 +1162,7 @@ }, { "data": { - "text/markdown": "* Now let's use Clue 2: \"Bob does not live in the green house.\"\n * This is consistent! Bob lives in the Red house, not Green. Good check.\n\n * Consider", + "text/markdown": "**Analyzing the Solution**\n\nThe recent focus is on refining the solution. I'm carefully analyzing the placement of Carol and Alice, and their pets, in the remaining houses. Determining the role of the green-to-the-left-of-red relationship remains key. I'm exploring the implications of the two possible house orderings and their impact on the residents and pets.", "text/plain": [ "" ] @@ -1210,7 +1172,7 @@ }, { "data": { - "text/markdown": "Carol: Carol owns a dog.\n * We know Bob lives in the Red house and owns a cat.\n * So, Carol cannot live in the Red house (already confirmed).\n * Carol also cannot live in the Red house", + "text/markdown": "**Finalizing the Analysis**\n\nI'm now integrating Carol's dog ownership and Alice's cat aversion with the spatial relationship. I'm satisfied with the solutions for both arrangements, and have confirmed that, based on the clues, Alice and Carol's house arrangements are interchangeable and both work with the dog, the cat, and the unknown pet.", "text/plain": [ "" ] @@ -1220,7 +1182,7 @@ }, { "data": { - "text/markdown": "because she owns a dog, and the Red house owner owns a cat.\n\n * What's left for Carol? She owns a dog.\n * Houses remaining for Carol: Green or Blue.\n *", + "text/markdown": "**Finalizing the Analysis**\n\nI'm now fully confident in the solution, recognizing the inherent ambiguity regarding Alice and Carol's house assignments. The clues definitively place Bob in the red house with the cat, leaving the blue and green houses for Carol and Alice. The spatial relationship dictates the Green-Red arrangement, but the final pet-house association of Alice and Carol is flexible. The problem allows for the placement of Alice and Carol's pets to be interchangeable.", "text/plain": [ "" ] @@ -1230,7 +1192,7 @@ }, { "data": { - "text/markdown": "If Carol lives in the Green house, then the Green house owner owns a dog.\n * If Carol lives in the Blue house, then the Blue house owner owns a dog.\n\n * What's left for Alice? She", + "text/markdown": "**Reviewing Potential Solutions**\n\nI've re-examined the problem's clues and spatial relationships. I've placed Bob in the red house with the cat, confirmed Carol's dog, and Alice's cat aversion. I'm focusing on Carol and Alice's possible house order. I'm satisfied the logic for a solution has been thoroughly evaluated. The ambiguity in placing Alice's pet has been clarified.", "text/plain": [ "" ] @@ -1240,7 +1202,7 @@ }, { "data": { - "text/markdown": "does not own a cat.\n * Houses remaining for Alice: Green or Blue.\n\n * Recall Clue 4: \"The green house is to the left of the red house.\"\n * This means the Green house is distinct", + "text/markdown": "**Confirming the Solution**\n\nI'm now satisfied with the solution after re-evaluating the clues. I've confirmed that the \"Green-Red\" relationship allows for two valid arrangements. Bob's in the red house with the cat, while Alice and Carol could inhabit the blue or green houses, each with a specific pet. I can confidently say I've found a consistent, if slightly ambiguous, solution.", "text/plain": [ "" ] @@ -1250,7 +1212,7 @@ }, { "data": { - "text/markdown": "from the Red house.\n * We know Bob is in Red.\n * We know Alice is not in Red.\n * We know Carol is not in Red.\n\n * Let's focus", + "text/markdown": "**Reviewing the Solution**\n\nAfter re-evaluating the clues, I've confirmed Bob lives in the red house and owns the cat. Carol owns the dog. However, the exact arrangement of Carol and Alice's houses remains ambiguous, with potential options being either Blue-Green-Red or Green-Red-Blue. Alice's unmentioned pet also remains unknown. I've analyzed all the clues to their full extent.", "text/plain": [ "" ] @@ -1260,7 +1222,7 @@ }, { "data": { - "text/markdown": "on Carol and Alice.\n * Carol owns a dog.\n * Alice does not own a cat.\n * Bob (Red house, Cat).\n\n * If Carol lived in the Green house, then", + "text/markdown": "**Unraveling the Puzzle**\n\nI'm now certain of Bob's residence and pet ownership. I've re-evaluated the clues, which confirms that Alice's pet is not a cat and that Carol has a dog. I'm focusing on the spatial relationships and the house order. I'm satisfied that Bob lives in the red house, and I've made initial connections between Carol and Alice's residences. The challenge is in definitively placing Alice's pet.", "text/plain": [ "" ] @@ -1270,7 +1232,7 @@ }, { "data": { - "text/markdown": "Green -> Dog.\n * This would leave Alice for the Blue house. And Alice does not own a cat. So the Blue house owner would not own a cat. This is consistent.\n\n * If Alice lived in the", + "text/markdown": "**Deepening the Analysis**\n\nI'm now fully immersed in the puzzle's nuances. I've successfully placed Bob in the red house with his cat and confirmed Carol's dog ownership. The crux now lies in resolving Alice and Carol's house and pet assignments. The \"green-to-the-left-of-red\" relationship is crucial. I'm focusing on evaluating the implications of the \"blue-green-red\" or the \"green-red-blue\" arrangement for Carol and Alice.", "text/plain": [ "" ] @@ -1280,7 +1242,7 @@ }, { "data": { - "text/markdown": "Green house, then Green -> not Cat. This is consistent.\n * This would leave Carol for the Blue house. So Blue -> Dog.\n\n * We have:\n * Bob: Red, Cat", + "text/markdown": "**Refining the Deductions**\n\nI'm now satisfied with the solution after re-evaluating the clues. I've confirmed that the \"Green-Red\" relationship allows for two valid arrangements. Bob's in the red house with the cat, while Alice and Carol could inhabit the blue or green houses, each with a specific pet. I can confidently say I've found a consistent, if slightly ambiguous, solution.", "text/plain": [ "" ] @@ -1290,7 +1252,7 @@ }, { "data": { - "text/markdown": "* Carol: Dog\n * Alice: Not Cat\n\n * Let's check the remaining options for Alice and Carol regarding the Green and Blue houses.\n\n * If Carol is in the Green house:\n * Carol", + "text/markdown": "## **Answer:**", "text/plain": [ "" ] @@ -1300,7 +1262,7 @@ }, { "data": { - "text/markdown": ": Green, Dog\n * Alice: Blue, (Implicitly not a dog or cat, or just some other pet). *The puzzle doesn't explicitly state there are only 3 pets, only that Bob has a cat and Carol has", + "text/markdown": "Let's break this down step-by-step to solve the puzzle:\n\n**1. Set up a table to keep track of information:**\n\n| House Color | Resident | Pet |", "text/plain": [ "" ] @@ -1310,7 +1272,7 @@ }, { "data": { - "text/markdown": "a dog.*\n * This solution works:\n * Red house: Bob, Cat (Clue 1 satisfied)\n * Green house: Carol, Dog (Clue 3 satisfied)\n * Blue house", + "text/markdown": "|-------------|----------|-----|\n| Red | | |\n| Green | | |\n| Blue | | |\n\n**2. Analyze the Clues:**\n\n* **\"The person who lives in", "text/plain": [ "" ] @@ -1320,7 +1282,7 @@ }, { "data": { - "text/markdown": ": Alice (Clue 5 satisfied - no cat, Clue 2 satisfied - Bob not in Green, Clue 4 satisfied - Green to left of Red)\n\n * If Alice is in the Green house:\n *", + "text/markdown": "the red house owns a cat.\"**\n * Update table: Red House = Cat.\n\n* **\"Carol owns a dog.\"**\n * This means Carol is NOT in the Red house (because the Red house has a cat,", "text/plain": [ "" ] @@ -1330,7 +1292,7 @@ }, { "data": { - "text/markdown": "Alice: Green, (not cat).\n * Carol: Blue, Dog.\n * This also seems to work:\n * Red house: Bob, Cat (Clue 1 satisfied)\n *", + "text/markdown": "not a dog).\n\n* **\"Alice does not own a cat.\"**\n * This means Alice is NOT in the Red house (because the Red house has a cat).\n\n* **Deduction 1 (", "text/plain": [ "" ] @@ -1340,7 +1302,7 @@ }, { "data": { - "text/markdown": "Green house: Alice, (not cat) (Clue 5 satisfied)\n * Blue house: Carol, Dog (Clue 3 satisfied, Clue 2 satisfied, Clue 4 satisfied)\n\n * Is there", + "text/markdown": "Who lives in the Red House?):**\n * Since Alice and Carol do not live in the Red house, **Bob must live in the Red house.**\n * From the first clue, we know the Red house person owns a cat", "text/plain": [ "" ] @@ -1350,7 +1312,7 @@ }, { "data": { - "text/markdown": "anything to differentiate between Alice and Carol in Green/Blue?\n\n * Let's re-read *all* clues carefully.\n\n 1. The person who lives in the red house owns a cat. (Red -> Cat)\n 2", + "text/markdown": ".\n * **Conclusion: Bob lives in the Red house and owns a cat.**\n\n *Updated Table:*\n | House Color | Resident | Pet |\n |-------------|----------|-----|\n | Red | Bob", "text/plain": [ "" ] @@ -1360,7 +1322,7 @@ }, { "data": { - "text/markdown": ". Bob does not live in the green house. (Bob != Green)\n 3. Carol owns a dog. (Carol -> Dog)\n 4. The green house is to the left of the red house. (Green", + "text/markdown": "| Cat |\n | Green | | |\n | Blue | | |\n\n* **Deduction 2 (Pets of Alice and Carol):**\n * We know Bob", "text/plain": [ "" ] @@ -1370,7 +1332,7 @@ }, { "data": { - "text/markdown": "< Red)\n 5. Alice does not own a cat. (Alice != Cat)\n\n * **Deductions so far:**\n * Red House -> Cat\n * Alice != Red (because", + "text/markdown": "has the cat.\n * We know Carol owns a dog.\n * Alice does not own a cat. Since Bob has the cat and Carol has the dog, Alice must own a different pet (implicitly, a third type", "text/plain": [ "" ] @@ -1380,7 +1342,7 @@ }, { "data": { - "text/markdown": "Alice != Cat, and Red -> Cat)\n * Carol != Red (because Carol -> Dog, and Red -> Cat)\n * Therefore, **Bob lives in the Red House and owns the Cat.** (This is", + "text/markdown": "of pet, e.g., a fish or bird, though not specified).\n * **Conclusion: Carol owns a dog. Alice owns a pet that is neither a cat nor a dog.**\n\n* **\"Bob does not live in", "text/plain": [ "" ] @@ -1390,7 +1352,7 @@ }, { "data": { - "text/markdown": "solid)\n\n * So:\n * Bob: Red, Cat\n\n * Now consider Alice and Carol.\n * Alice and Carol must occupy Green and Blue houses.\n * Carol owns a dog.", + "text/markdown": "the green house.\"**\n * This is consistent with our deduction that Bob lives in the Red house.\n\n* **\"The green house is to the left of the red house.\"**\n * We know Bob", "text/plain": [ "" ] @@ -1400,7 +1362,7 @@ }, { "data": { - "text/markdown": "* Alice does not own a cat.\n\n * Can Carol live in the Green house?\n * If Carol lives in the Green house, she owns a dog. (Green -> Dog).\n * This", + "text/markdown": "lives in the Red house. So, the Green house is to the left of Bob's house.\n * This establishes the order: Green House --> Red House (Bob).\n\n* **Deduction 3 (Green", "text/plain": [ "" ] @@ -1410,7 +1372,7 @@ }, { "data": { - "text/markdown": "leaves Alice for the Blue house. Alice does not own a cat. This is consistent (Blue house is not Red, so no cat requirement for Blue).\n * So: Carol (Green, Dog), Alice (Blue, ?)", + "text/markdown": "and Blue Houses):**\n * We have Alice and Carol remaining, and the Green and Blue houses remaining.\n * Carol owns a dog. Alice owns the third, unspecified pet.\n * We know the Green house is to the", "text/plain": [ "" ] @@ -1420,7 +1382,7 @@ }, { "data": { - "text/markdown": "* Can Alice live in the Green house?\n * If Alice lives in the Green house, she does not own a cat. (Green -> not Cat).\n * This leaves Carol for the Blue house. Carol", + "text/markdown": "left of the Red house (Bob's).\n * There is no further clue to definitively place Alice or Carol in either the Green or Blue house. This means there are two possible valid solutions for Alice and Carol's houses and pets.", "text/plain": [ "" ] @@ -1430,7 +1392,7 @@ }, { "data": { - "text/markdown": "owns a dog. (Blue -> Dog).\n * So: Alice (Green, not Cat), Carol (Blue, Dog).\n\n * Both scenarios seem possible. Let me check if I missed any nuance.", + "text/markdown": "However, assuming a complete and unique solution is expected, let's pick one of the logical possibilities. Given no other information, it's often the case that the \"first\" person mentioned (Alice) or some other convention would", "text/plain": [ "" ] @@ -1440,7 +1402,7 @@ }, { "data": { - "text/markdown": "\"Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\" This confirms a 1:1 mapping.\n\n * Is there an implicit assumption about pets?\n \"The person who lives", + "text/markdown": "resolve it, but logically, based on the provided clues, either scenario below is valid.\n\n Let's go with the scenario that places Carol in the Green house to complete the puzzle.\n\n * If Carol lives in the Green house (with", "text/plain": [ "" ] @@ -1450,7 +1412,7 @@ }, { "data": { - "text/markdown": "in the red house owns a cat.\"\n \"Carol owns a dog.\"\n \"Alice does not own a cat.\"\n The puzzle asks \"what pet do they own?\". This implies everyone owns *a* pet, and we", + "text/markdown": "her dog), then Alice must live in the Blue house (with her non-cat, non-dog pet).\n\n**Final Solution:**\n\n* **Red House:** Bob lives here and owns a **cat**.\n* **", "text/plain": [ "" ] @@ -1460,17 +1422,131 @@ }, { "data": { - "text/markdown": "need to determine what it is for Alice.\n\n * Let's try to assign the pets more explicitly.\n * Pets: Cat, Dog, X (the third unknown pet)\n\n * We have:\n *", + "text/markdown": "Green House:** Carol lives here and owns a **dog**. (This house is to the left of Bob's).\n* **Blue House:** Alice lives here and owns a pet (which is neither a cat nor a dog).", "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" - }, + } + ], + "source": [ + "prompt = \"\"\"\n", + " Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\n", + " The person who lives in the red house owns a cat.\n", + " Bob does not live in the green house.\n", + " Carol owns a dog.\n", + " The green house is to the left of the red house.\n", + " Alice does not own a cat.\n", + " Who lives in each house, and what pet do they own?\n", + "\"\"\"\n", + "\n", + "thoughts = \"\"\n", + "answer = \"\"\n", + "\n", + "for chunk in client.models.generate_content_stream(\n", + " model=MODEL_ID,\n", + " contents=prompt,\n", + " config=types.GenerateContentConfig(\n", + " thinking_config=types.ThinkingConfig(\n", + " include_thoughts=True\n", + " )\n", + " )\n", + "):\n", + " for part in chunk.candidates[0].content.parts:\n", + " if not part.text:\n", + " continue\n", + " elif part.thought:\n", + " if not thoughts:\n", + " display(Markdown(\"## **Thoughts summary:**\"))\n", + " display(Markdown(part.text.strip()))\n", + " thoughts += part.text\n", + " else:\n", + " if not answer:\n", + " display(Markdown(\"## **Answer:**\"))\n", + " display(Markdown(part.text.strip()))\n", + " answer += part.text" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "mH0Fq3AA_A0y" + }, + "source": [ + "## Working with Gemini thinking models and tools\n", + "\n", + "Gemini thinking models are compatible with the tools and capabilities inherent to the Gemini ecosystem. This compatibility allows them to interface with external environments, execute computational code, or retrieve real-time data, subsequently incorporating such information into their analytical framework and concluding statements." + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "027fe5edcb8e" + }, + "source": [ + "### Solving a problem using the code execution tool\n", + "\n", + "This example shows how to use the [code execution](./Code_execution.ipynb) tool to solve a problem. The model will generate the code and then execute it to get the final answer.\n", + "\n", + "In this case, you are using the adaptive thinking_budget so the model will dynamically adjust the budget based on the complexity of the request.\n", + "\n", + "If you want to experiment with a fixed budget, you can set the `thinking_budget` to a specific value (e.g. `thinking_budget=4096`)." + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "id": "gpFo0Q1Q_10o" + }, + "outputs": [], + "source": [ + "prompt = \"\"\"\n", + " What are the best ways to sort a list of n numbers from 0 to m?\n", + " Generate and run Python code for three different sort algorithms.\n", + " Provide the final comparison between algorithm clearly.\n", + " Is one of them linear?\n", + "\"\"\"\n", + "\n", + "thinking_budget = 4096 # @param {type:\"slider\", min:0, max:24576, step:1}\n", + "\n", + "code_execution_tool = types.Tool(\n", + " code_execution=types.ToolCodeExecution()\n", + ")\n", + "\n", + "response = client.models.generate_content(\n", + " model=MODEL_ID,\n", + " contents=prompt,\n", + " config=types.GenerateContentConfig(\n", + " tools=[code_execution_tool],\n", + " thinking_config=types.ThinkingConfig(\n", + " thinking_budget=thinking_budget,\n", + " )\n", + " ),\n", + ")" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "id": "VjXgY2wg_-g9" + }, + "source": [ + "Checking the model response, including the code generated and the execution result:" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "id": "KelkCAIdkzTW" + }, + "outputs": [ { "data": { - "text/markdown": "Bob: Red, Cat\n * Carol: Dog\n\n * So Alice must own pet X.\n * Alice does not own a cat (given).\n * Bob owns the cat, Carol owns the dog. So Alice", + "text/markdown": "The problem asks for the best ways to sort a list of `n` numbers ranging from `0` to `m`, to implement three different sorting algorithms, compare their performance, and determine if any of them are linear.\n\n### Best Ways to Sort a List of `n` Numbers from `0` to `m`\n\nWhen sorting a list of numbers with a known, limited range (0 to `m`), non-comparison-based sorting algorithms can often outperform general comparison-based algorithms (which have a theoretical lower bound of O(n log n)).\n\n1. **Counting Sort**: This is generally the *best* option when the range `m` is not significantly larger than `n` (i.e., `m = O(n)` or `m` is a small constant). It works by counting the occurrences of each distinct element, then calculating the position of each element in the sorted output array based on these counts. Its time complexity is O(n + m), which is linear if `m` is proportional to `n`.\n2. **Radix Sort**: If `m` is very large, but the numbers can be represented with a fixed number of digits `d` (e.g., `m` could be `10^9`, but we are sorting numbers that can be represented as `k` digits in base `b`), Radix Sort can be very efficient. It sorts numbers digit by digit, typically using a stable sorting algorithm like Counting Sort for each digit. Its time complexity is O(d * (n + b)), where `b` is the base used for counting sort (e.g., 10 for decimal, or `m` for single pass counting sort).\n3. **Bucket Sort**: Suitable when the input numbers are uniformly distributed over a range. It divides the range into several 'buckets', distributes elements into these buckets, sorts each bucket individually (e.g., using insertion sort or another sorting algorithm), and then concatenates the buckets. Its average-case time complexity can be O(n + k) (where k is the number of buckets), but its worst-case can be O(n^2) if all elements fall into a single bucket.\n\nFor the given problem constraints (numbers from `0` to `m`), **Counting Sort** stands out as the most directly applicable and potentially linear-time solution if `m` is within a reasonable bound relative to `n`.\n\nFor general-purpose sorting, **Merge Sort** and **Quick Sort** are widely used comparison-based algorithms, offering average-case O(n log n) performance. They don't specifically leverage the `0` to `m` range but are good benchmarks for comparison.\n\n### Implementation and Performance Comparison\n\nI will implement and test three algorithms:\n1. **Counting Sort**: To leverage the range property.\n2. **Merge Sort**: A stable, O(n log n) comparison sort.\n3. **Quick Sort**: A widely used, average O(n log n) comparison sort.\n\nI will use Python's `time` module to measure their execution time on a randomly generated list of `n` numbers from `0` to `m`.\n\nLet's set:\n* `N = 100,000` (number of elements)\n* `M = 50,000` (maximum value, so numbers are in range [0, 50,000])\n\nThis setup ensures that `M` is significant enough to show its impact on Counting Sort's `m` term, but also `N` is large enough for general O(N log N) sorts to take noticeable time.\n\n", "text/plain": [ "" ] @@ -1480,9 +1556,143 @@ }, { "data": { - "text/markdown": "must own the *third* pet.\n * If the puzzle implies only cat and dog exist, then it's underspecified for Alice's pet, or it means Alice has no pet (which is unlikely in these", + "text/html": [ + "
import random\n",
+              "import time\n",
+              "\n",
+              "def generate_random_list(n, m):\n",
+              "    \"\"\"Generates a list of n random numbers between 0 and m (inclusive).\"\"\"\n",
+              "    return [random.randint(0, m) for _ in range(n)]\n",
+              "\n",
+              "# --- Counting Sort Implementation ---\n",
+              "def counting_sort(arr, m):\n",
+              "    \"\"\"\n",
+              "    Sorts an array of integers using Counting Sort.\n",
+              "    Assumes numbers are in the range [0, m].\n",
+              "    \"\"\"\n",
+              "    n = len(arr)\n",
+              "    output = [0] * n\n",
+              "    count = [0] * (m + 1)\n",
+              "\n",
+              "    # Store count of each character\n",
+              "    for x in arr:\n",
+              "        count[x] += 1\n",
+              "\n",
+              "    # Store cumulative count\n",
+              "    for i in range(1, m + 1):\n",
+              "        count[i] += count[i - 1]\n",
+              "\n",
+              "    # Build the output array\n",
+              "    # Iterate in reverse to maintain stability\n",
+              "    for i in range(n - 1, -1, -1):\n",
+              "        output[count[arr[i]] - 1] = arr[i]\n",
+              "        count[arr[i]] -= 1\n",
+              "\n",
+              "    return output\n",
+              "\n",
+              "# --- Merge Sort Implementation ---\n",
+              "def merge_sort(arr):\n",
+              "    if len(arr) <= 1:\n",
+              "        return arr\n",
+              "\n",
+              "    mid = len(arr) // 2\n",
+              "    left_half = arr[:mid]\n",
+              "    right_half = arr[mid:]\n",
+              "\n",
+              "    left_half = merge_sort(left_half)\n",
+              "    right_half = merge_sort(right_half)\n",
+              "\n",
+              "    return merge(left_half, right_half)\n",
+              "\n",
+              "def merge(left, right):\n",
+              "    merged_list = []\n",
+              "    left_idx, right_idx = 0, 0\n",
+              "\n",
+              "    while left_idx < len(left) and right_idx < len(right):\n",
+              "        if left[left_idx] <= right[right_idx]:\n",
+              "            merged_list.append(left[left_idx])\n",
+              "            left_idx += 1\n",
+              "        else:\n",
+              "            merged_list.append(right[right_idx])\n",
+              "            right_idx += 1\n",
+              "\n",
+              "    merged_list.extend(left[left_idx:])\n",
+              "    merged_list.extend(right[right_idx:])\n",
+              "    return merged_list\n",
+              "\n",
+              "# --- Quick Sort Implementation (in-place) ---\n",
+              "def quick_sort(arr):\n",
+              "    def _quick_sort(items, low, high):\n",
+              "        if low < high:\n",
+              "            pivot_index = partition(items, low, high)\n",
+              "            _quick_sort(items, low, pivot_index - 1)\n",
+              "            _quick_sort(items, pivot_index + 1, high)\n",
+              "\n",
+              "    def partition(items, low, high):\n",
+              "        # Using the middle element as pivot to reduce chances of worst-case\n",
+              "        mid = (low + high) // 2\n",
+              "        pivot = items[mid]\n",
+              "        items[mid], items[high] = items[high], items[mid] # Move pivot to end\n",
+              "\n",
+              "        i = low - 1\n",
+              "        for j in range(low, high):\n",
+              "            if items[j] <= pivot:\n",
+              "                i += 1\n",
+              "                items[i], items[j] = items[j], items[i]\n",
+              "        items[i + 1], items[high] = items[high], items[i + 1]\n",
+              "        return i + 1\n",
+              "\n",
+              "    # Quick Sort is typically in-place. We'll make a copy of the input list\n",
+              "    # to ensure the original list isn't modified and to provide a fresh input\n",
+              "    # for each sort test.\n",
+              "    arr_copy = list(arr)\n",
+              "    _quick_sort(arr_copy, 0, len(arr_copy) - 1)\n",
+              "    return arr_copy\n",
+              "\n",
+              "# --- Test setup ---\n",
+              "N = 100000  # Number of elements\n",
+              "M = 50000   # Max value of elements (range 0 to M)\n",
+              "\n",
+              "# Generate the initial list\n",
+              "original_list = generate_random_list(N, M)\n",
+              "\n",
+              "print(f\"Testing sorting algorithms with N = {N}, M = {M}\\n\")\n",
+              "\n",
+              "# Test Counting Sort\n",
+              "list_for_counting = list(original_list) # Use a copy\n",
+              "start_time = time.time()\n",
+              "sorted_list_counting = counting_sort(list_for_counting, M)\n",
+              "end_time = time.time()\n",
+              "time_counting = end_time - start_time\n",
+              "print(f\"Counting Sort took: {time_counting:.6f} seconds\")\n",
+              "\n",
+              "\n",
+              "# Test Merge Sort\n",
+              "list_for_merge = list(original_list) # Use a copy\n",
+              "start_time = time.time()\n",
+              "sorted_list_merge = merge_sort(list_for_merge)\n",
+              "end_time = time.time()\n",
+              "time_merge = end_time - start_time\n",
+              "print(f\"Merge Sort took: {time_merge:.6f} seconds\")\n",
+              "\n",
+              "\n",
+              "# Test Quick Sort\n",
+              "list_for_quick = list(original_list) # Use a copy\n",
+              "start_time = time.time()\n",
+              "sorted_list_quick = quick_sort(list_for_quick)\n",
+              "end_time = time.time()\n",
+              "time_quick = end_time - start_time\n",
+              "print(f\"Quick Sort took: {time_quick:.6f} seconds\")\n",
+              "\n",
+              "# Optional: Verify correctness (commented out for large N as it adds overhead)\n",
+              "# expected_sorted = sorted(original_list)\n",
+              "# print(f\"\\nCounting Sort correct: {sorted_list_counting == expected_sorted}\")\n",
+              "# print(f\"Merge Sort correct: {sorted_list_merge == expected_sorted}\")\n",
+              "# print(f\"Quick Sort correct: {sorted_list_quick == expected_sorted}\")\n",
+              "
" + ], "text/plain": [ - "" + "" ] }, "metadata": {}, @@ -1490,7 +1700,7 @@ }, { "data": { - "text/markdown": "puzzles). Usually, it implies three unique pets. Let's assume a third pet. Alice owns \"the other pet\".\n\n * Let's revisit the two possible scenarios:\n\n **Scenario A:**\n * Bob:", + "text/markdown": "Testing sorting algorithms with N = 100000, M = 50000\n\nCounting Sort took: 0.071481 seconds\nMerge Sort took: 0.489993 seconds\nQuick Sort took: 0.330622 seconds\n", "text/plain": [ "" ] @@ -1500,7 +1710,7 @@ }, { "data": { - "text/markdown": "Red House, Cat\n * Carol: Green House, Dog\n * Alice: Blue House, (The other pet)\n\n * Check Clues:\n 1. Red house owns a cat: Yes (Bob", + "text/markdown": "### Final Comparison Between Algorithms\n\n| Algorithm | Time Complexity (Worst/Average) | Space Complexity | Stable | Notes | Performance (N=100k, M=50k) |\n| :------------- | :------------------------------ | :--------------- | :----- | :---------------------------------------------------------- | :-------------------------- |\n| **Counting Sort** | O(n + m) | O(m) | Yes | Excellent for restricted integer ranges. Can be very fast. | **0.071 seconds** |\n| **Merge Sort** | O(n log n) | O(n) | Yes | Robust, guaranteed O(n log n). Recursive. | **0.490 seconds** |\n| **Quick Sort** | O(n log n) (Average), O(n^2) (Worst) | O(log n) (Average) | No | Generally fast in practice due to small constant factors. | **0.331 seconds** |\n\n**Analysis of Results:**\n\nAs expected, **Counting Sort** significantly outperformed both Merge Sort and Quick Sort for the given problem parameters (`N = 100,000`, `M = 50,000`). This is because its time complexity of O(n + m) becomes roughly O(n) when `m` is comparable to `n`. In this case, `n + m = 150,000` operations, which is much smaller than `n log n` for `n = 100,000` (`100,000 * log2(100,000)` is approx `100,000 * 16.6` or `1.66 million`).\n\nMerge Sort and Quick Sort, being comparison-based sorts, have a theoretical lower bound of O(n log n). Their performance reflects this higher complexity compared to Counting Sort on this specific type of data. Quick Sort was faster than Merge Sort, which is common in practice due to better cache performance and lower constant factors, despite both being O(n log n).\n\n### Is one of them linear?\n\nYes, **Counting Sort** is a linear sorting algorithm. Its time complexity is O(n + m). If the range of values `m` is proportional to the number of elements `n` (i.e., `m = O(n)`), then the complexity simplifies to O(n), making it a linear-time algorithm. In our test case, `m = N/2`, so `m` is proportional to `n`, and Counting Sort operates in effectively linear time.", "text/plain": [ "" ] @@ -1510,1280 +1720,7 @@ }, { "data": { - "text/markdown": ")\n 2. Bob does not live in green house: Yes (Bob is in Red)\n 3. Carol owns a dog: Yes (Carol is in Green)\n 4. Green house is to the left of", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Red house: Yes (Green - Red, consistent with Carol in Green, Bob in Red)\n 5. Alice does not own a cat: Yes (Alice owns \"the other pet\", not cat)\n * This", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "scenario works perfectly.\n\n **Scenario B:**\n * Bob: Red House, Cat\n * Alice: Green House, (The other pet)\n * Carol: Blue House, Dog\n\n * Check Clues", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ":\n 1. Red house owns a cat: Yes (Bob)\n 2. Bob does not live in green house: Yes (Bob is in Red)\n 3. Carol owns a dog: Yes (", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Carol is in Blue)\n 4. Green house is to the left of Red house: Yes (Green - Red, consistent with Alice in Green, Bob in Red)\n 5. Alice does not own a cat:", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Yes (Alice owns \"the other pet\", not cat)\n * This scenario also works perfectly.\n\n * This is frustrating! I must have missed something. Let me go back to the very first assumption or reread the", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "prompt to see if I missed any subtle connection.\n\n * \"Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\"\n * \"The person who lives in the red", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "house owns a cat.\"\n * \"Bob does not live in the green house.\"\n * \"Carol owns a dog.\"\n * \"The green house is to the left of the red house.\"", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* \"Alice does not own a cat.\"\n\n * Okay, the \"green house is to the left of the red house\" establishes an order for *two* houses. It implies a linear arrangement.\n So we have _", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "_ _ (three houses).\n Possibilities:\n * Blue, Green, Red\n * Green, Red, Blue\n\n * Case 1: Blue, Green, Red\n * Red is last", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ". Green is middle. Blue is first.\n * We know Bob is in Red, Cat.\n * Green is to the left of Red. Consistent.\n * So: Bob (Red, Cat).\n * Who", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "is in Green? Who is in Blue?\n * Carol (Dog), Alice (Not Cat).\n * If Carol is in Green (Dog), then Alice is in Blue (Other pet). This is Scenario A.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* If Alice is in Green (Other pet), then Carol is in Blue (Dog). This is Scenario B.\n\n * Case 2: Green, Red, Blue\n * Green is first. Red is", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "middle. Blue is last.\n * Green is to the left of Red. Consistent.\n * We know Bob is in Red, Cat.\n * So: Bob (Red, Cat).\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Who is in Green? Who is in Blue?\n * Carol (Dog), Alice (Not Cat).\n * If Carol is in Green (Dog), then Alice is in Blue (Other pet). This is Scenario", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "A.\n * If Alice is in Green (Other pet), then Carol is in Blue (Dog). This is Scenario B.\n\n * The \"to the left of\" doesn't help distinguish Alice/Carol.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "It just confirms the relative order of Green and Red.\n\n * Let's check if the pet counts matter.\n * Pet types: Cat, Dog, ???\n * Bob owns the cat.\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Carol owns the dog.\n * Alice does not own a cat. Therefore, Alice must own the third, unknown pet.\n\n * So the final state of pets for each person is fixed:\n * Bob: Cat", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* Carol: Dog\n * Alice: (The third pet, let's call it 'Fish' for example, or simply 'Other Pet')\n\n * Now combine this with the house assignments for Alice and Carol.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* We know Bob is in the Red house.\n\n * So the houses are: Red, Green, Blue.\n * Bob lives in Red.\n * Alice and Carol live in Green and Blue.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* **Possibility 1:** Alice in Green, Carol in Blue.\n * Alice: Green, Other Pet\n * Bob: Red, Cat\n * Carol: Blue, Dog\n * Check", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "clues:\n 1. Red house owns cat: Yes (Bob)\n 2. Bob not in green: Yes (Bob in Red)\n 3. Carol owns dog: Yes (Carol in Blue)\n 4. Green left of Red", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ": Yes (Alice in Green, Bob in Red)\n 5. Alice not own cat: Yes (Alice has Other Pet)\n * This is a valid solution.\n\n * **Possibility 2:** Carol in Green,", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Alice in Blue.\n * Carol: Green, Dog\n * Bob: Red, Cat\n * Alice: Blue, Other Pet\n * Check clues:\n 1. Red house owns cat:", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Yes (Bob)\n 2. Bob not in green: Yes (Bob in Red)\n 3. Carol owns dog: Yes (Carol in Green)\n 4. Green left of Red: Yes (Carol in Green,", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Bob in Red)\n 5. Alice not own cat: Yes (Alice has Other Pet)\n * This is *also* a valid solution.\n\n * What am I missing? This is a classic logic puzzle,", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "usually, there's only one unique solution.\n * Could \"The green house is to the left of the red house\" imply absolute position? No, it just implies relative.\n\n * Let's re-examine", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "the core deductions:\n 1. Red House -> Cat (Clue 1)\n 2. Carol -> Dog (Clue 3)\n 3. Alice != Cat (Clue 5)\n 4", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ". From (1) & (3): Alice != Red House\n 5. From (1) & (2): Carol != Red House\n 6. Conclusion: **Bob lives in the Red House and owns the Cat.** (This", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "is 100% firm)\n\n * Remaining houses: Green, Blue.\n * Remaining people: Alice, Carol.\n * Remaining pets: Dog (for Carol), Other Pet (for Alice).", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* We also know: Bob != Green House (Clue 2 - already satisfied by Bob being in Red).\n * Green House is to the left of Red House (Clue 4).\n\n * Let's create", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "a full mapping, not just person-house.\n | House | Person | Pet |\n |-------|--------|-----|\n | Red | Bob | Cat |\n | Green | ? | ?", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "|\n | Blue | ? | ? |\n\n * The remaining people are Alice and Carol.\n * The remaining houses are Green and Blue.\n * The remaining pets are Dog (Carol) and Other Pet", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "(Alice).\n\n * If Carol is in the Green House:\n | House | Person | Pet |\n |-------|--------|-----|\n | Red | Bob | Cat |\n | Green", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "| Carol | Dog |\n | Blue | Alice | Other Pet |\n * Check Clues:\n 1. Red house owns cat: Bob-Red-Cat. (OK)\n 2. Bob not in", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "green house: Bob is in Red. (OK)\n 3. Carol owns a dog: Carol-Green-Dog. (OK)\n 4. Green house to left of Red house: Green (Carol) is left of Red (Bob).", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "(OK)\n 5. Alice not own cat: Alice-Blue-Other Pet. (OK)\n * This is a valid solution.\n\n * If Alice is in the Green House:\n | House", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "| Person | Pet |\n |-------|--------|-----|\n | Red | Bob | Cat |\n | Green | Alice | Other Pet |\n | Blue | Carol | Dog |", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* Check Clues:\n 1. Red house owns cat: Bob-Red-Cat. (OK)\n 2. Bob not in green house: Bob is in Red. (OK)\n 3. Carol", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "owns a dog: Carol-Blue-Dog. (OK)\n 4. Green house to left of Red house: Green (Alice) is left of Red (Bob). (OK)\n 5. Alice not own cat:", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Alice-Green-Other Pet. (OK)\n * This is *also* a valid solution.\n\n * Why would a puzzle be written to have two valid solutions? Usually, they don't.\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Is there any implied information I'm missing?\n * \"Alice, Bob, and Carol each live in a different house...\" - confirmed 1:1 mapping.\n * \"...on the same street: red", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ", green, and blue.\" - confirmed three distinct houses.\n\n * Perhaps the wording of \"The green house is to the left of the red house\" has a deeper implication. Does it restrict the possibilities for the third house (", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "blue)?\n * Scenario A: Green, Red, Blue (e.g., Green is house 1, Red is house 2, Blue is house 3)\n * Scenario B: Blue, Green, Red (e.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "g., Blue is house 1, Green is house 2, Red is house 3)\n\n * In Scenario A (Green, Red, Blue):\n * Green: Alice/Carol\n * Red: Bob", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "(Cat)\n * Blue: Carol/Alice\n\n * If Alice is Green, Carol is Blue: Alice (Green), Bob (Red), Carol (Blue)\n * If Carol is Green, Alice is Blue", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ": Carol (Green), Bob (Red), Alice (Blue)\n\n * In Scenario B (Blue, Green, Red):\n * Blue: Alice/Carol\n * Green: Alice/Carol\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Red: Bob (Cat)\n\n * If Alice is Green, Carol is Blue: Alice (Green), Bob (Red), Carol (Blue)\n * If Carol is Green, Alice is Blue: Carol (Green), Bob (", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Red), Alice (Blue)\n\n * The linear arrangement doesn't seem to rule out either final assignment of Alice and Carol.\n\n * Could it be the *pet* for Alice? The puzzle doesn't specify", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "a third pet type. It just says \"Alice does not own a cat.\"\n If there are ONLY cats and dogs, then the puzzle is impossible for Alice's pet, as Bob has the cat and Carol has the dog. If Alice", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "has to have *a* pet, and it's not a cat, and not a dog (because Carol has it), then she can't have a pet. But the prompt says \"what pet do they own?\". This implies a pet", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "for everyone.\n So there must be a third pet type. Let's assume there is a third pet type.\n\n * Let me search for this exact riddle online to see if there's a common interpretation I'm missing.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "No, it's a standard format.\n\n * Is it possible that \"Alice does not own a cat\" and \"Carol owns a dog\" implicitly limits *who* can live in the Green house based on its relative position to the Red house", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "(which has the cat)?\n No, because the person in the Green house doesn't own a cat.\n If Alice lives in Green, she doesn't own a cat - consistent.\n If Carol lives in Green, she", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "owns a dog, which isn't a cat - consistent.\n\n * This is peculiar. Let me write out the full implications for the two possibilities clearly.\n\n **Fixed Points:**\n * Bob: Red House", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ", Cat\n * Carol: Dog (pet)\n * Alice: Not Cat (pet) -> Implies Alice has a third distinct pet.\n * Green House is to the left of the Red House.\n\n **", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Scenario 1: Alice in Green House, Carol in Blue House**\n * Alice: Green House, Other Pet\n * Bob: Red House, Cat\n * Carol: Blue House, Dog\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Order of houses (left to right): Green, Red, Blue (This is consistent with \"Green to the left of Red\")\n * Checking all conditions:\n 1. Red house owns a cat: Yes, Bob in", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Red has cat.\n 2. Bob does not live in green house: Yes, Bob in Red.\n 3. Carol owns a dog: Yes, Carol in Blue has dog.\n 4. Green house is to", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "the left of the red house: Yes, Green (Alice) is left of Red (Bob).\n 5. Alice does not own a cat: Yes, Alice has Other Pet.\n * This solution is fully consistent", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "with all clues.\n\n **Scenario 2: Carol in Green House, Alice in Blue House**\n * Carol: Green House, Dog\n * Bob: Red House, Cat\n * Alice: Blue", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "House, Other Pet\n * Order of houses (left to right): Green, Red, Blue (This is consistent with \"Green to the left of Red\")\n * Checking all conditions:\n 1. Red", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "house owns a cat: Yes, Bob in Red has cat.\n 2. Bob does not live in green house: Yes, Bob in Red.\n 3. Carol owns a dog: Yes, Carol in Green has", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "dog.\n 4. Green house is to the left of the red house: Yes, Green (Carol) is left of Red (Bob).\n 5. Alice does not own a cat: Yes, Alice has Other", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Pet.\n * This solution is also fully consistent with all clues.\n\n Okay, I've checked multiple times. Both scenarios seem valid. This is unusual for a puzzle of this type. It's possible the puzzle", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "writer intended only one solution but accidentally created an ambiguous one, or there's an incredibly subtle nuance I'm missing.\n\n Given that I can't find a way to eliminate one, I should present both valid possibilities. However", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ", logic puzzles typically have a unique solution. Let me think about if there's any way a house/pet assignment could influence the remaining people.\n\n What if the number of houses/people/pets is constrained in a way that creates", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "a single unique mapping?\n 3 houses, 3 people, 3 pets (Cat, Dog, X).\n Bob is Red, Cat.\n Carol has Dog.\n Alice has X (not Cat).\n\n So we", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "have a definitive mapping for Bob.\n We have definitive pets for Carol and Alice.\n We need to place Carol (Dog) and Alice (X) into Green and Blue houses.\n\n If Carol lives in Green, Alice", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "lives in Blue.\n If Alice lives in Green, Carol lives in Blue.\n\n Neither of these assignments violates any of the rules.\n - \"Bob does not live in the green house.\" (Bob is in Red, so this", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "is fine regardless).\n - \"The green house is to the left of the red house.\" (This is just about the houses themselves, Green-Red. The Blue house could be anywhere else, e.g., Blue-Green-Red or Green", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "-Red-Blue. This constraint is about the *houses*, not the *people* in them. If Green is always to the left of Red, then the people occupying them don't change that physical relationship).\n\n I am", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "confident that based on the provided clues, there are two distinct solutions for Alice and Carol's assignments. I will state this. It's important to provide the reasoning for why one solution doesn't exclude the other.\n\n The", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "only way to resolve this is if the \"pets\" are implicitly linked to the houses, for example, \"the blue house owner has a fish.\" But that's not given.\n\n So, the final deduction: Bob (Red, Cat).", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Then, for the remaining two people and two houses, there are two permutations, and both satisfy all conditions.\n\n *Solution structure:*\n 1. List all given clues.\n 2. Start with the most direct deduction.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "3. Build up deductions, showing how each conclusion is reached.\n 4. Identify the point of ambiguity.\n 5. Present both valid solutions clearly.\n 6. State Alice's pet must be a third type, as the Cat", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "and Dog are taken by others.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "## **Answer:**", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Let's break down the clues to solve this logic puzzle:\n\n**Clues:**\n1. Alice, Bob, and Carol each live in a different house (red, green, blue", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ").\n2. The person who lives in the red house owns a cat.\n3. Bob does not live in the green house.\n4. Carol owns a dog.\n5. The green house is to", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "the left of the red house.\n6. Alice does not own a cat.\n\n---\n\n**Deductions:**\n\n* **From Clue 2 (Red house owns a cat) and Clue 6 (Alice does not own", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "a cat):**\n * Alice cannot live in the red house.\n\n* **From Clue 2 (Red house owns a cat) and Clue 4 (Carol owns a dog):**\n *", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Carol cannot live in the red house (because she owns a dog, not a cat).\n\n* **Combining the above:**\n * If Alice and Carol do not live in the red house, and there are only three people,", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "then **Bob must live in the red house.**\n * Therefore, **Bob lives in the red house and owns the cat.** (This also satisfies Clue 3: Bob does not live in the green house, as he lives", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "in the red house).\n\n* **Determining remaining pets:**\n * Bob owns the cat.\n * Carol owns the dog.\n * Alice does not own a cat (Clue 6).", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Since Bob owns the cat and Carol owns the dog, Alice must own a third, unspecified type of pet.\n\n* **Remaining houses and people:**\n * We have the Green and Blue houses remaining.\n * We have", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Alice and Carol remaining to assign to these houses.\n * We know Carol owns the dog, and Alice owns the \"other pet.\"\n\n* **Considering Clue 5 (The green house is to the left of the red house):**", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* This establishes a spatial relationship between the green and red houses. Since Bob lives in the red house, the green house is to the left of Bob's house. The blue house's position relative to green and red is not explicitly", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "stated (it could be left of green, or right of red).\n\n---\n\n**Two Possible Solutions Emerge:**\n\nAt this point, we have definitively placed Bob in the red house with the cat. We also know Carol has the dog and Alice has the", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "\"other pet.\" The remaining task is to assign Alice and Carol to the green and blue houses. Both assignments satisfy all given clues:\n\n**Solution 1:**\n\n* **Bob:** Lives in the **Red** house and owns a", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "**Cat**.\n* **Carol:** Lives in the **Green** house and owns a **Dog**.\n* **Alice:** Lives in the **Blue** house and owns the **Other Pet**.\n\n * *Check Clues:*", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* Red house owns cat: Yes (Bob).\n * Bob not in green house: Yes (Bob in Red).\n * Carol owns dog: Yes (Carol in Green).\n * Green house to left", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "of Red house: Yes (Green is Carol's, Red is Bob's).\n * Alice not own cat: Yes (Alice has Other Pet).\n * This solution is consistent.\n\n**Solution 2:**", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* **Bob:** Lives in the **Red** house and owns a **Cat**.\n* **Alice:** Lives in the **Green** house and owns the **Other Pet**.\n* **Carol:** Lives in the", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "**Blue** house and owns a **Dog**.\n\n * *Check Clues:*\n * Red house owns cat: Yes (Bob).\n * Bob not in green house: Yes (Bob in Red).", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* Carol owns dog: Yes (Carol in Blue).\n * Green house to left of Red house: Yes (Green is Alice's, Red is Bob's).\n * Alice not own cat", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": ": Yes (Alice has Other Pet).\n * This solution is also consistent.\n\n---\n\n**Conclusion:**\n\nBased on the provided clues, there are two equally valid solutions to this puzzle. The clues do not provide enough information to definitively place Alice", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "and Carol in either the green or blue house beyond their pet assignments.\n\n**Here are the two possible outcomes:**\n\n**Possibility 1:**\n* **Alice** lives in the **Blue** house and owns an **Other", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Pet**.\n* **Bob** lives in the **Red** house and owns a **Cat**.\n* **Carol** lives in the **Green** house and owns a **Dog**.\n\n**Possibility 2:**", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "* **Alice** lives in the **Green** house and owns an **Other Pet**.\n* **Bob** lives in the **Red** house and owns a **Cat**.\n* **Carol** lives in the **", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Blue** house and owns a **Dog**.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], - "source": [ - "prompt = \"\"\"\n", - " Alice, Bob, and Carol each live in a different house on the same street: red, green, and blue.\n", - " The person who lives in the red house owns a cat.\n", - " Bob does not live in the green house.\n", - " Carol owns a dog.\n", - " The green house is to the left of the red house.\n", - " Alice does not own a cat.\n", - " Who lives in each house, and what pet do they own?\n", - "\"\"\"\n", - "\n", - "thoughts = \"\"\n", - "answer = \"\"\n", - "\n", - "for chunk in client.models.generate_content_stream(\n", - " model=MODEL_ID,\n", - " contents=prompt,\n", - " config=types.GenerateContentConfig(\n", - " thinking_config=types.ThinkingConfig(\n", - " include_thoughts=True\n", - " )\n", - " )\n", - "):\n", - " for part in chunk.candidates[0].content.parts:\n", - " if not part.text:\n", - " continue\n", - " elif part.thought:\n", - " if not thoughts:\n", - " display(Markdown(\"## **Thoughts summary:**\"))\n", - " display(Markdown(part.text.strip()))\n", - " thoughts += part.text\n", - " else:\n", - " if not answer:\n", - " display(Markdown(\"## **Answer:**\"))\n", - " display(Markdown(part.text.strip()))\n", - " answer += part.text" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "id": "mH0Fq3AA_A0y" - }, - "source": [ - "## Working with Gemini thinking models and tools\n", - "\n", - "Gemini thinking models are compatible with the tools and capabilities inherent to the Gemini ecosystem. This compatibility allows them to interface with external environments, execute computational code, or retrieve real-time data, subsequently incorporating such information into their analytical framework and concluding statements." - ] - }, - { - "cell_type": "markdown", - "metadata": { - "id": "027fe5edcb8e" - }, - "source": [ - "### Solving a problem using the code execution tool\n", - "\n", - "This example shows how to use the [code execution](./Code_execution.ipynb) tool to solve a problem. The model will generate the code and then execute it to get the final answer.\n", - "\n", - "In this case, you are using the adaptive thinking_budget so the model will dynamically adjust the budget based on the complexity of the request.\n", - "\n", - "If you want to experiment with a fixed budget, you can set the `thinking_budget` to a specific value (e.g. `thinking_budget=4096`)." - ] - }, - { - "cell_type": "code", - "execution_count": 32, - "metadata": { - "id": "gpFo0Q1Q_10o" - }, - "outputs": [], - "source": [ - "prompt = \"\"\"\n", - " What are the best ways to sort a list of n numbers from 0 to m?\n", - " Generate and run Python code for three different sort algorithms.\n", - " Provide the final comparison between algorithm clearly.\n", - " Is one of them linear?\n", - "\"\"\"\n", - "\n", - "thinking_budget = 4096 # @param {type:\"slider\", min:0, max:24576, step:1}\n", - "\n", - "code_execution_tool = types.Tool(\n", - " code_execution=types.ToolCodeExecution()\n", - ")\n", - "\n", - "response = client.models.generate_content(\n", - " model=MODEL_ID,\n", - " contents=prompt,\n", - " config=types.GenerateContentConfig(\n", - " tools=[code_execution_tool],\n", - " thinking_config=types.ThinkingConfig(\n", - " thinking_budget=thinking_budget,\n", - " )\n", - " ),\n", - ")" - ] - }, - { - "cell_type": "markdown", - "metadata": { - "id": "VjXgY2wg_-g9" - }, - "source": [ - "Checking the model response, including the code generated and the execution result:" - ] - }, - { - "cell_type": "code", - "execution_count": 33, - "metadata": { - "id": "KelkCAIdkzTW" - }, - "outputs": [ - { - "data": { - "text/markdown": "The best ways to sort a list of `n` numbers from `0` to `m` depend on the specific values of `n` and `m`, and the desired time and space complexity. When numbers are confined to a limited range (0 to `m`), non-comparison-based sorting algorithms can often achieve better performance than general comparison-based sorts like Merge Sort or Quick Sort, which have a theoretical lower bound of O(n log n).\n\nFor numbers in a specific range:\n\n1. **Counting Sort:** This is typically the \"best\" for this scenario if `m` is not excessively large compared to `n`. It sorts integers by counting the occurrences of each distinct integer value. Its time complexity is O(n + m), which is linear.\n2. **Radix Sort:** Can be very efficient for a larger range of numbers by sorting digits in a specific base. It often uses Counting Sort as a subroutine. Its time complexity is O(d * (n + b)), where `d` is the number of digits and `b` is the base used for sorting (e.g., 10 for decimal, or the range for each digit). If `m` is large, but `d` is small, it can be efficient.\n3. **Bucket Sort:** If the numbers are uniformly distributed over the range `0` to `m`, Bucket Sort can achieve an average-case time complexity of O(n). It divides the range into several \"buckets\" and then sorts each bucket individually (e.g., using Insertion Sort or another comparison sort). Its worst-case complexity can be O(n^2) if all elements fall into a single bucket.\n\nFor demonstration and comparison, I will implement and run three different sorting algorithms:\n\n1. **Counting Sort:** An algorithm specifically suited for integer data within a limited range, often achieving linear time.\n2. **Merge Sort:** A general-purpose, stable, comparison-based sorting algorithm with a consistent O(n log n) worst-case time complexity.\n3. **Quick Sort:** A general-purpose, in-place (in its typical implementation), comparison-based sorting algorithm with an average-case time complexity of O(n log n) but a worst-case of O(n^2).\n\n---\n\n### Python Code Implementation and Execution\n\nLet's generate a list of `n` numbers between `0` and `m` and measure the execution time for each algorithm.\n\nFor our test:\n* `n = 100,000` (number of elements)\n* `m = 50,000` (maximum value, so numbers are in range [0, 50000])\n\n", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/html": [ - "
import random\n",
-              "import time\n",
-              "import sys\n",
-              "\n",
-              "# Increase recursion limit for Quick Sort in case of large inputs\n",
-              "sys.setrecursionlimit(200000)\n",
-              "\n",
-              "# 1. Counting Sort\n",
-              "def counting_sort(arr, max_val):\n",
-              "    counts = [0] * (max_val + 1)\n",
-              "    output = [0] * len(arr)\n",
-              "\n",
-              "    for num in arr:\n",
-              "        counts[num] += 1\n",
-              "\n",
-              "    # Calculate prefix sums to determine positions\n",
-              "    for i in range(1, max_val + 1):\n",
-              "        counts[i] += counts[i - 1]\n",
-              "\n",
-              "    # Place elements into output array (iterate in reverse for stability)\n",
-              "    for num in reversed(arr):\n",
-              "        output[counts[num] - 1] = num\n",
-              "        counts[num] -= 1\n",
-              "    return output\n",
-              "\n",
-              "# 2. Merge Sort\n",
-              "def merge_sort(arr):\n",
-              "    if len(arr) <= 1:\n",
-              "        return arr\n",
-              "\n",
-              "    mid = len(arr) // 2\n",
-              "    left_half = arr[:mid]\n",
-              "    right_half = arr[mid:]\n",
-              "\n",
-              "    left_half = merge_sort(left_half)\n",
-              "    right_half = merge_sort(right_half)\n",
-              "\n",
-              "    return merge(left_half, right_half)\n",
-              "\n",
-              "def merge(left, right):\n",
-              "    merged = []\n",
-              "    left_idx, right_idx = 0, 0\n",
-              "\n",
-              "    while left_idx < len(left) and right_idx < len(right):\n",
-              "        if left[left_idx] <= right[right_idx]:\n",
-              "            merged.append(left[left_idx])\n",
-              "            left_idx += 1\n",
-              "        else:\n",
-              "            merged.append(right[right_idx])\n",
-              "            right_idx += 1\n",
-              "\n",
-              "    merged.extend(left[left_idx:])\n",
-              "    merged.extend(right[right_idx:])\n",
-              "    return merged\n",
-              "\n",
-              "# 3. Quick Sort (using a simpler, non-in-place version for clarity)\n",
-              "def quick_sort(arr):\n",
-              "    if len(arr) <= 1:\n",
-              "        return arr\n",
-              "\n",
-              "    pivot = arr[len(arr) // 2]\n",
-              "    less = [x for x in arr if x < pivot]\n",
-              "    equal = [x for x in arr if x == pivot]\n",
-              "    greater = [x for x in arr if x > pivot]\n",
-              "\n",
-              "    return quick_sort(less) + equal + quick_sort(greater)\n",
-              "\n",
-              "# Test parameters\n",
-              "n_elements = 100000\n",
-              "max_value = 50000\n",
-              "\n",
-              "# Generate a random list\n",
-              "original_list = [random.randint(0, max_value) for _ in range(n_elements)]\n",
-              "\n",
-              "print(f\"Sorting {n_elements} numbers in range [0, {max_value}]\\n\")\n",
-              "\n",
-              "# Test Counting Sort\n",
-              "list_for_cs = list(original_list)\n",
-              "start_time = time.perf_counter()\n",
-              "sorted_list_cs = counting_sort(list_for_cs, max_value)\n",
-              "end_time = time.perf_counter()\n",
-              "time_cs = end_time - start_time\n",
-              "print(f\"Counting Sort Time: {time_cs:.6f} seconds\")\n",
-              "# Optional: Verify sort\n",
-              "# print(f\"Is sorted (CS): {sorted_list_cs == sorted(original_list)}\")\n",
-              "\n",
-              "\n",
-              "# Test Merge Sort\n",
-              "list_for_ms = list(original_list)\n",
-              "start_time = time.perf_counter()\n",
-              "sorted_list_ms = merge_sort(list_for_ms)\n",
-              "end_time = time.perf_counter()\n",
-              "time_ms = end_time - start_time\n",
-              "print(f\"Merge Sort Time: {time_ms:.6f} seconds\")\n",
-              "# Optional: Verify sort\n",
-              "# print(f\"Is sorted (MS): {sorted_list_ms == sorted(original_list)}\")\n",
-              "\n",
-              "\n",
-              "# Test Quick Sort\n",
-              "list_for_qs = list(original_list)\n",
-              "start_time = time.perf_counter()\n",
-              "sorted_list_qs = quick_sort(list_for_qs)\n",
-              "end_time = time.perf_counter()\n",
-              "time_qs = end_time - start_time\n",
-              "print(f\"Quick Sort Time: {time_qs:.6f} seconds\")\n",
-              "# Optional: Verify sort\n",
-              "# print(f\"Is sorted (QS): {sorted_list_qs == sorted(original_list)}\")\n",
-              "
" - ], - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "Sorting 100000 numbers in range [0, 50000]\n\nCounting Sort Time: 0.040218 seconds\nMerge Sort Time: 0.632565 seconds\nQuick Sort Time: 0.385846 seconds\n", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "### Final Comparison Between Algorithms\n\nHere's a comparison based on the performance and characteristics observed:\n\n| Algorithm | Time Complexity (Worst-Case) | Space Complexity (Worst-Case) | Observed Time (100k elements, max 50k value) | Notes |\n| :-------------- | :--------------------------- | :---------------------------- | :------------------------------------------- | :------------------------------------------------------------------------------- |\n| **Counting Sort** | O(n + m) | O(m + n) | **0.040218 seconds** | Extremely fast for integer data within a limited range. Linear time complexity. |\n| **Quick Sort** | O(n^2) | O(log n) (average) / O(n) (worst) | **0.385846 seconds** | Generally very fast in practice (average O(n log n)), but can degrade. |\n| **Merge Sort** | O(n log n) | O(n) | **0.632565 seconds** | Consistent O(n log n) performance, stable, but uses more auxiliary space. |\n\n**Summary of Findings:**\n\n* **Counting Sort** is significantly faster than Merge Sort and Quick Sort for the given problem constraints (numbers in a limited range). This is because its time complexity is directly tied to the range of numbers (`m`) and the count of numbers (`n`), rather than relying on comparisons, which would scale logarithmically.\n* **Quick Sort** performed better than Merge Sort in this specific test case. While both have an average-case time complexity of O(n log n), Quick Sort often has a smaller constant factor and is generally faster in practice for many datasets, especially when implemented in-place.\n* **Merge Sort** is reliable with its consistent O(n log n) performance but tends to be slower than Counting Sort for this specific problem and can be slower than Quick Sort due to its higher overhead from merging sub-arrays and requiring additional space.\n\n---\n\n### Is one of them linear?\n\n**Yes, Counting Sort is a linear time sorting algorithm.**\n\n* **Counting Sort's** time complexity is O(n + m). If `m` (the range of numbers) is proportional to `n` (the number of elements), or smaller than `n`, then the complexity simplifies to O(n), which is linear. Even if `m` is larger, as long as `m` is not astronomically larger than `n`, it remains very efficient and is considered linear in the context of `n` and `m`. In our test where `n=100,000` and `m=50,000`, the `n+m` term dominates, resulting in linear performance.\n\n* **Merge Sort** and **Quick Sort** are comparison-based sorting algorithms with an average and worst-case time complexity of O(n log n) (for Merge Sort) or average O(n log n) (for Quick Sort). They are *not* linear-time algorithms. The `log n` factor makes them super-linear.", - "text/plain": [ - "" - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "data": { - "text/markdown": "---", + "text/markdown": "---", "text/plain": [ "" ] @@ -2823,7 +1760,7 @@ }, { "cell_type": "code", - "execution_count": 34, + "execution_count": 26, "metadata": { "id": "qFp3r_dpv5di" }, @@ -2868,7 +1805,7 @@ }, { "cell_type": "code", - "execution_count": 35, + "execution_count": 27, "metadata": { "id": "1kEjC0VPmGob" }, @@ -2885,7 +1822,34 @@ }, { "data": { - "text/markdown": "I will first determine what \"last month\" refers to, given the current date of May 20, 2025.\nThen, I will perform Google searches for major scientific breakthroughs announced between April 1, 2025, and April 30, 2025, using terms like \"scientific breakthroughs April 2025\", \"major scientific discoveries April 2025\", and \"notable science news April 2025\".\nI will critically evaluate the search results to identify truly significant and impactful breakthroughs, filtering out overinflated titles or incremental advancements.\nFinally, I will list the most incredible scientific breakthroughs announced last month based on this evaluation.", + "text/markdown": "**Analyzing April's Advancements**\n\nI've begun the process by establishing April 2025 as the target month. Next, I'll be meticulously crafting a suite of Google searches, targeting phrases like \"scientific breakthroughs April 2025\" and \"major scientific discoveries April 2025.\" My aim is to extract the most relevant and impactful information. I'll need to sift through the search results carefully to identify truly groundbreaking discoveries from the noise.\n\n\n", + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n" + ] + }, + { + "data": { + "text/markdown": "## **Thoughts summary:**", + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "data": { + "text/markdown": "**Pinpointing Key Discoveries**\n\nI've completed the initial screening of April's scientific announcements. I'm prioritizing the most impactful findings from that list for a more concise report. The focus narrows to: evidence of extraterrestrial life on K2-18b, ITER's superconducting electromagnet milestone, advanced brain-computer interfaces, innovative cancer therapies, powerful gene-editing tools, and potential breakthroughs in Huntington's disease treatment. I'm working on condensing the details for each to deliver a focused report on the \"truly incredible\" breakthroughs of last month.\n\n\n", "text/plain": [ "" ] @@ -2912,7 +1876,7 @@ }, { "data": { - "text/markdown": "April 2025 saw several significant scientific announcements across various fields:\n\n**Space and Astronomy:**\n* **Strongest Evidence for Life on Exoplanet K2-18b:** Scientists reported finding tentative, but promising, evidence of molecules associated with life, specifically dimethyl sulfide (DMS) and dimethyl disulfide (DMDS), in the atmosphere of exoplanet K2-18b using the James Webb Space Telescope. On Earth, these compounds are exclusively produced by living organisms. If confirmed, this could suggest that the planet is \"teeming with life\" and that life might be common in the galaxy.\n* **Discovery of a Hyper-volcanic Exoplanet:** The James Webb Space Telescope also identified a new exoplanet, 98-95b, which appears to be the most volcanic exoplanet observed to date.\n* **Evidence of a Supermassive Black Hole in the Large Magellanic Cloud:** Hypervelocity stars, traveling at immense speeds, were observed hinting at the presence of a supermassive black hole within the Large Magellanic Cloud, a dwarf galaxy orbiting the Milky Way. This is considered the best indicator yet of giant black holes in dwarf galaxies.\n* **Disintegrating Exoplanet with a Comet-like Tail:** Astronomers reported the discovery of BD+05 4868Ab, a small rocky exoplanet rapidly disintegrating due to extreme heat from its host star, forming a comet-like tail of vaporized minerals extending millions of kilometers.\n* **New Moons of Jupiter and Saturn:** Two additional moons of Jupiter were announced, bringing its total to 97. Additionally, 128 new moons of Saturn were reported, increasing its confirmed satellites to 274.\n* **Perseverance Rover's Findings on Mars:** Data from the Perseverance rover revealed layered mineral formations beneath Jezero Crater, suggesting multiple episodes of water activity throughout Mars' geological history, offering insights into the planet's past habitability.\n\n**Medicine and Health:**\n* **Regrowing Brain Cells for Huntington's Disease:** A groundbreaking study demonstrated the potential for the adult brain to regrow specific neurons lost to Huntington's disease (HD) and integrate them into existing brain circuitry. Researchers used two special proteins to stimulate the growth of new, functional medium spiny neurons in mice.\n* **Targeting Somatic Expansion in Huntington's Disease:** New research suggests that blocking certain DNA repair genes, particularly Msh3 and Pms1, could reduce harmful CAG repeat expansion in brain cells, potentially reversing molecular changes and improving brain health and movement in mice with HD. Another study showed that lowering the MSH3 protein could halt and even reverse these expansions in lab-grown HD brain cells.\n* **Advances in Cancer Therapies:** Breakthroughs in targeted cancer therapies were unveiled, including new immunotherapy techniques that harness the body's immune system to fight cancer cells with high precision. A new drug, \"Immunocure,\" targets specific markers on cancer cells while leaving healthy cells untouched. Penn Medicine researchers also presented insights into cancer metabolism, epigenetics, immunotherapy, and molecular oncology at the AACR Annual Meeting.\n* **Brain-Computer Interfaces (BCIs):** Significant advancements in BCIs have made them more efficient, accessible, and practical, particularly for applications in prosthetics.\n\n**Other Scientific Breakthroughs:**\n* **Mass Production of Synthetic Meat:** Several companies have successfully scaled up the production of lab-grown meat, making it more affordable and accessible as a sustainable alternative to traditional meat production.\n* **Next-Generation Renewable Energy:** Development of more efficient and cost-effective solar panels, including a new class of perovskite solar cells, and breakthroughs in energy storage technologies have improved the reliability of renewable energy.\n* **Completion of ITER Superconducting Magnet System:** Engineers at ITER completed the construction of the world's largest and most powerful pulsed superconducting electromagnet system, a major step towards sustained nuclear fusion.", + "text/markdown": "Last month, April 2025, saw several truly remarkable scientific advancements and announcements. Among the most incredible breakthroughs were:\n\n* **Strongest Evidence Yet of Possible Life Beyond Earth:** Scientists announced the detection of dimethyl sulfide (DMS) and dimethyl disulfide (DMDS) in the atmosphere of exoplanet K2-18b, a candidate water world located 124 light-years away. On Earth, these compounds are almost exclusively produced by living organisms, leading researchers to describe this as the \"strongest evidence to date for biological activity beyond the Solar System,\" though further proof is required.\n* **Major Milestone Achieved in Nuclear Fusion:** Engineers at the International Thermonuclear Experimental Reactor (ITER) completed the construction of the world's largest and most powerful pulsed superconducting electromagnet system. This achievement marks a critical step forward on the path to achieving sustained nuclear fusion, a clean and virtually limitless energy source.\n* **Revolutionary Advances in Gene Editing:** The 2025 Breakthrough Prizes recognized the development of base editing and prime editing technologies. These innovative methods allow for the precise editing of DNA in living systems without cutting the double helix, enabling the correction or replacement of virtually any genetic mutation at its native location. These tools have already shown promise in rescuing blood diseases, neurological disorders, and other genetic conditions in animals.\n* **First Precise 3D Mapping of a Mammal's Brain:** Researchers successfully created the first precise 3D map of a mammal's brain by studying a mouse. This decade-long effort, published in a leading journal, reveals the form, function, and activity of thousands of neurons and brain cells, along with half a billion synapses, offering unprecedented insights into brain activity.\n* **Progress in Halting Huntington's Disease Progression:** A new study from University College London demonstrated that lowering a key DNA repair protein called MSH3 could stop, and in some cases even reverse, the harmful C-A-G repeat expansions in lab-grown Huntington's Disease (HD) brain cells. This genetic therapy approach using antisense oligonucleotides (ASOs) offers significant hope for treating this devastating neurodegenerative condition.", "text/plain": [ "" ] @@ -3054,9 +2018,9 @@ "
\n", " \n", "
\n", - " significant scientific discoveries April 2025\n", - " major scientific breakthroughs April 2025\n", - " groundbreaking research April 2025\n", + " significant scientific discoveries April 2025\n", + " major scientific breakthroughs April 2025\n", + " notable scientific advancements April 2025\n", "
\n", "\n" ],