From b18a92c9fd006bc8b1086a4050baf9ac88a5c112 Mon Sep 17 00:00:00 2001
From: Albert Hu <alberthu@berkeley.edu>
Date: Sun, 30 Apr 2017 23:54:18 -0700
Subject: [PATCH 1/2] Day 21: helpingStephan codefights

---
 day21/README.md         | 50 +++++++++++++++++++++++++++++++++++++++++
 day21/helpingStephan.py | 40 +++++++++++++++++++++++++++++++++
 2 files changed, 90 insertions(+)
 create mode 100644 day21/README.md
 create mode 100644 day21/helpingStephan.py

diff --git a/day21/README.md b/day21/README.md
new file mode 100644
index 0000000..c32b785
--- /dev/null
+++ b/day21/README.md
@@ -0,0 +1,50 @@
+Question of the day: https://codefights.com/challenge/hT9JyzWh77Y6NR2Qd
+
+Stephan is a little boy who loves math. It came as a revelation to him that
+one can define the greatest common divisor for rational numbers, not just
+for integers. He's decided to dig into the subject and analyze the properties
+of this new-to-him GCD.
+
+Stephan needs lots of GCDs to gather all possible information about them. But
+Stephan is still a little boy and he can't calculate the GCD of two numbers as
+quickly as he'd like. So he asks for your help: Given float numbers x and y,
+calculate their greatest common divisor.
+
+It is guaranteed that both x and y will have at most 5 digits after the decimal
+point.
+
+Example:
+
+For `x = 2.4` and `y = 4.8`, the output should be
+`helpingStephan(x, y) = 2.4`.
+
+## Ideas
+
+How do I even find GCD without decimals? uhhhhh....
+
+Prime factorization? That seems complicated but doable for small numbers. 
+I can get the prime factorization of each number by repeatedly dividing
+primes in increasing magnitude up to the value of the number, and keeping
+the number of times each prime divides into the number in a hash. Then get
+the primes that each number has in common and get the `min` between them.
+
+Let me just google this.. 
+
+Aha! [Euclid's Algorithm](https://en.wikipedia.org/wiki/Greatest_common_divisor#Using_Euclid.27s_algorithm)
+I totally learned this in Algorithms class in college but forgot it existed.
+
+Not sure what the runtime is for this algorithm, but this stack overflow post
+kinda explains it: http://stackoverflow.com/questions/3980416/time-complexity-of-euclids-algorithm
+
+So I could just change the floats that I'm given into ints by multiplying by
+a factor of 10, apply Euclid's Algorithm in `O(n)` runtime where `n` is
+proportional to the value of the input numbers, and then divide by the earlier
+factor of 10 I used to multiply.
+
+## Code
+
+[Python](helpingStephan.py)
+
+## Follow up
+
+Floats.. https://docs.python.org/2/tutorial/floatingpoint.html#tut-fp-issues
diff --git a/day21/helpingStephan.py b/day21/helpingStephan.py
new file mode 100644
index 0000000..f9d5ff3
--- /dev/null
+++ b/day21/helpingStephan.py
@@ -0,0 +1,40 @@
+def helpingStephan(x, y):
+    xSplit = str(x).split(".")
+    ySplit = str(y).split(".")
+    
+    XdigitsAfterDecimal = "0"
+    if len(xSplit) > 1:
+        XdigitsAfterDecimal = xSplit[1]
+        
+    YdigitsAfterDecimal = "0"
+    if len(ySplit) > 1:
+        YdigitsAfterDecimal = ySplit[1]
+                                      
+                                  
+    digitsAfterDecimal = max(len(XdigitsAfterDecimal), len(YdigitsAfterDecimal))
+                
+    x *= 10**digitsAfterDecimal
+    y *= 10**digitsAfterDecimal
+
+    a = max(x, y)
+    b = min(x, y)
+     
+    r = -1
+    while r != 0:
+        r = round(a % b, 6) * 1.0
+        a = b
+        b = r
+        
+    return a / 10**digitsAfterDecimal
+
+def main():
+    assert helpingStephan(2.4, 4.8) == 2.4
+    assert helpingStephan(36.21, 4.19) == 0.01
+    assert helpingStephan(4, 6) == 2
+    assert helpingStephan(2.43, 4.899) == 0.003
+    assert helpingStephan(47.804, 11.8683) == 0.0001
+    assert helpingStephan(17.27568, 49.90752) == 1.91952
+    assert helpingStephan(22.14728, 11.07364) == 11.07364
+
+if __name__ == "__main__":
+    main()

From 88ea6b1ad0f4c19b21a1401332284eb8263c1494 Mon Sep 17 00:00:00 2001
From: Albert Hu <alberthu@berkeley.edu>
Date: Wed, 3 May 2017 21:26:54 -0700
Subject: [PATCH 2/2] Day 21: Assume 5 digit max after decimal point

---
 day21/helpingStephan.py | 17 +++--------------
 1 file changed, 3 insertions(+), 14 deletions(-)

diff --git a/day21/helpingStephan.py b/day21/helpingStephan.py
index f9d5ff3..e4d33ac 100644
--- a/day21/helpingStephan.py
+++ b/day21/helpingStephan.py
@@ -1,17 +1,5 @@
 def helpingStephan(x, y):
-    xSplit = str(x).split(".")
-    ySplit = str(y).split(".")
-    
-    XdigitsAfterDecimal = "0"
-    if len(xSplit) > 1:
-        XdigitsAfterDecimal = xSplit[1]
-        
-    YdigitsAfterDecimal = "0"
-    if len(ySplit) > 1:
-        YdigitsAfterDecimal = ySplit[1]
-                                      
-                                  
-    digitsAfterDecimal = max(len(XdigitsAfterDecimal), len(YdigitsAfterDecimal))
+    digitsAfterDecimal = 5
                 
     x *= 10**digitsAfterDecimal
     y *= 10**digitsAfterDecimal
@@ -21,13 +9,14 @@ def helpingStephan(x, y):
      
     r = -1
     while r != 0:
-        r = round(a % b, 6) * 1.0
+        r = round(a % b, 6)
         a = b
         b = r
         
     return a / 10**digitsAfterDecimal
 
 def main():
+    assert helpingStephan(1, 1) == 1
     assert helpingStephan(2.4, 4.8) == 2.4
     assert helpingStephan(36.21, 4.19) == 0.01
     assert helpingStephan(4, 6) == 2