Kehao ch10problems(2 of them)

1_beginner/chapter5/practice/alternating.py

@@ -1,12 +1,12 @@
-""" 
+"""
+Alternating
 
 Ask the user for an integer. The print the numbers from 1 to that number,
 but alternating in sign. For example, if the input was 5, what would be printed
 is 1, -1, 2, -2, 3, -3, 4, -4, 5. (Note, DO NOT include the last negative
 number).
 
 Do this with a for loop
-
 """
 
 # Write code here.

2_intermediate/chapter10/practice/img_avg.py

@@ -1,27 +1,31 @@
-""" Here is the challenge problem for 2d loops>
-Images are often represented as 3d arrays,
+"""
+Image Average
+
+Here is the challenge problem for 2D loops:
+Images are often represented as 3D arrays,
 where the rows and columns are the pixels in the image,
-and each pixel has an r, g, and b value.
+and each pixel has an RGB (red, green, blue) value
+which determines the color of the pixel.
 
 The interesting thing is that we can iterate over images.
 The challenge is, given an image, create a program that
 will return a different image where each pixel is the average
 of the pixels surrounding it in the original image.
 
-The neighbors of an image are all the pixels that surroun it,
-1 on each side, and 4 on the diagonals, for 8 in total. 
-Each pixel doesn't necessarily have 8 neighbors, though (think about why)
+The neighbors of an image are all the pixels that surround it,
+1 on each side, and 4 on the diagonals, for 8 in total. Each
+pixel doesn't necessarily have 8 neighbors, though (think about why).
 
 The code to grab an image from the internet and make it
 into an array is given to you. The code also displays the new image
 you create in the end.
 
-NOTE: The image is 3 dimensional because each pixel has rgb values.
+NOTE: The image is 3 dimensional because each pixel has RGB values.
 To find the average value of all of a pixels neighbors, you must
 change the average of the red value to the red value, blue to blue, etc.
-For example, if the neighbors of a pixel with value [1,2,3]
-were [20,30,40] and [10,120,30], the new pixel that would replace the original one would be
-[15,75,35]
+For example, if the neighbors of a pixel with value [1, 2, 3]
+were [20, 30, 40] and [10, 120, 30], the new pixel that would replace the
+original one would be [15, 75, 35]
 """
 
 from PIL import Image

2_intermediate/chapter10/practice/odd_sum.py

@@ -0,0 +1,11 @@
+# Odd Sum
+# Given a 2D list, find the sum of all elements at odd indexes for all
+# the lists at odd indexes. Print this sum times the sum of all
+# first element of all the 1D lists in the 2D list.
+#
+# Ex:[[1,2,3,6],[2,41,2,1]]should have print 42 after the program runs.
+#
+# Write the code below.
+
+two_d_list = [[1, 2, 3, 5, 2], [2, 3, 1, 4], [2, 3, 1, 2, 21], [21, 3, 1, 41]]
+# two_d_list should print 51 after the program runs.

2_intermediate/chapter10/practice/smooth_max.py

@@ -0,0 +1,9 @@
+# Given a 2D list, let's call a element "smooth" if index of the
+# element in its 1D list plus the element is even. For example,
+# given the 2D list [[0,4][2,6]], the 1st element of each of the
+# 1D list is considered "smooth" because 0 + 0 is 0 and 0 + 2 is 2
+# (both are even numbers). Find the maximum "smooth" element and
+# print it. Using the example [[0,4][2,6]] again, the maximum
+# "smooth" element is 2 because 2 is bigger than 0.
+
+two_d_list = [[425, 214, 412, 123], [312, 214, 123, 343]]

2_intermediate/chapter10/solutions/img_avg.py

@@ -1,27 +1,31 @@
-""" Here is the challenge problem for 2d loops>
-Images are often represented as 3d arrays,
+"""
+Image Average
+
+Here is the challenge problem for 2D loops:
+Images are often represented as 3D arrays,
 where the rows and columns are the pixels in the image,
-and each pixel has an r, g, and b value.
+and each pixel has an RGB (red, green, blue) value
+which determines the color of the pixel.
 
 The interesting thing is that we can iterate over images.
 The challenge is, given an image, create a program that
 will return a different image where each pixel is the average
 of the pixels surrounding it in the original image.
 
-The neighbors of an image are all the pixels that surroun it,
-1 on each side, and 4 on the diagonals, for 8 in total. 
-Each pixel doesn't necessarily have 8 neighbors, though (think about why)
+The neighbors of an image are all the pixels that surround it,
+1 on each side, and 4 on the diagonals, for 8 in total. Each
+pixel doesn't necessarily have 8 neighbors, though (think about why).
 
 The code to grab an image from the internet and make it
 into an array is given to you. The code also displays the new image
 you create in the end.
 
-NOTE: The image is 3 dimensional because each pixel has rgb values.
+NOTE: The image is 3 dimensional because each pixel has RGB values.
 To find the average value of all of a pixels neighbors, you must
 change the average of the red value to the red value, blue to blue, etc.
-For example, if the neighbors of a pixel with value [1,2,3]
-were [20,30,40] and [10,120,30], the new pixel that would replace the original one would be
-[15,75,35]
+For example, if the neighbors of a pixel with value [1, 2, 3]
+were [20, 30, 40] and [10, 120, 30], the new pixel that would replace the
+original one would be [15, 75, 35]
 """
 
 from PIL import Image
@@ -37,9 +41,12 @@
 
 # write code to create newimg here
 def solution1():
-    """Iterating over the image here. i is a variable from 0 to the width of the image.
-	j is a variable that ranges from 0 to the height of the image. i is associated with
-	values"""
+    """
+    Iterating over the image here. i is a variable from
+    0 to the width of the image.
+    j is a variable that ranges from 0 to the height of the image.
+    i is associated with values
+    """
     for i in range(len(img)):
         for j in range(len(img[0])):
             x_n = [0]

2_intermediate/chapter10/solutions/odd_sum.py

@@ -0,0 +1,22 @@
+# Odd Sum
+# Given a 2D list, find the sum of all elements at odd indexes for all
+# the lists at odd indexes. Print this sum times the sum of all
+# first element of all the 1D lists in the 2D list.
+#
+# Ex:[[1,2,3,6],[2,41,2,1]]should have print 42 after the program runs.
+#
+# Write the code below.
+
+two_d_list = [[1, 2, 3, 5, 2], [2, 3, 1, 4], [2, 3, 1, 2, 21], [21, 3, 1, 41]]
+# two_d_list should print 51 after the program runs.
+
+odd_sum = 0
+for outer_idx in range(1, len(two_d_list), 2):
+    for inner_idx in range(1, len(two_d_list[outer_idx]), 2):
+        odd_sum += two_d_list[outer_idx][inner_idx]
+
+first_sum = 0
+for inner_list in range(len(two_d_list)):
+    first_sum += two_d_list[inner_list][0]
+
+print(odd_sum * first_sum)

2_intermediate/chapter10/solutions/smooth_max.py

@@ -0,0 +1,20 @@
+# Given a 2D list, let's call a element "smooth" if index of the
+# element in its 1D list plus the element is even. For example,
+# given the 2D list [[0,4][2,6]], the 1st element of each of the
+# 1D list is considered "smooth" because 0 + 0 is 0 and 0 + 2 is 2
+# (both are even numbers). Find the maximum "smooth" element and
+# print it. Using the example [[0,4][2,6]] again, the maximum
+# "smooth" element is 2 because 2 is bigger than 0.
+
+two_d_list = [[425, 214, 412, 123], [312, 214, 123, 343]]
+curr_max = None
+
+for outer_idx in range(len(two_d_list)):
+    for inner_idx in range(len(two_d_list[outer_idx])):
+        curr_elem = two_d_list[outer_idx][inner_idx]
+        to_check = curr_elem + inner_idx
+        if to_check % 2 == 0:
+            if curr_max is None or curr_elem > curr_max:
+                curr_max = curr_elem
+
+print(curr_max)