Improve 66 in Python based on the idea that we can search the Stern-Brocot tree for a solution.

python/e066.py

@@ -62,7 +62,7 @@ def get_fractions(n, x):
     return (n, d)
 
 
-def get_minimal_solution(d):
+def get_minimal_solution_old(d):
     for i in range(0, 100):
         x, y = get_fractions(d, i)
         if x * x - d * y * y == 1:
@@ -73,8 +73,32 @@ def is_square(n):
     return math.sqrt(n).is_integer()
 
 
+def get_minimal_solution(d, d1=0, n1=1, d2=1, n2=1):
+    """
+    This is an alternative to the alternate fraction approach explained here
+    [1]. Instead of calculating the continued fraction we search the
+    Stern-Brocot tree for fractions (x / y) that satisfy the pell equation.
+
+    If the solution does not satisfy the equation we can searh the respective
+    branch by calculating the mediant between the current fraction and one of
+    the previous fractions.
+
+    [1] https://en.wikipedia.org/wiki/Pell's_equation#Fundamental_solution_via_continued_fractions
+    [2] https://en.wikipedia.org/wiki/Stern–Brocot_tree
+    """
+
+    x = n1 + n2
+    y = d1 + d2
+    p = x * x - d * y * y
+    if p == 1:
+        return (x, y)
+    if p < 1:
+        return get_minimal_solution(d, d1, n1, y, x)
+    else:
+        return get_minimal_solution(d, y, x, d2, n2)
+
+
 def euler_066():
-    # XXX: This seems really long and complicated. We can do better.
     x_max = 0
     d_max = 0
 
@@ -84,8 +108,7 @@ def euler_066():
 
         x, y = get_minimal_solution(d)
         if x > x_max:
-            x_max = x
-            d_max = d
+            x_max, d_max = x, d
 
     print("d: {} x: {}".format(d_max, x_max))
     s = d_max