Update sol1.py

Author: MaximSmolskiy

Diff for: project_euler/problem_122/sol1.py

@@ -3,6 +3,30 @@
 
 Efficient Exponentiation
 
+The most naive way of computing n^15 requires fourteen multiplications:
+
+                                               n x n x ... x n = n^15.
+
+But using a "binary" method you can compute it in six multiplications:
+                                                         n x n = n^2
+                                                     n^2 x n^2 = n^4
+                                                     n^4 x n^4 = n^8
+                                                     n^8 x n^4 = n^12
+                                                    n^12 x n^2 = n^14
+                                                      n^14 x n = n^15
+
+<However it is yet possible to compute it in only five multiplications:
+
+n x n = n^2
+n^2 x n = n^3
+n^3 x n^3 = n^6
+n^6 x n^6 = n^{12}
+n^12 x n^3 = n^{15}
+
+We shall define m(k) to be the minimum number of multiplications to compute n^k; for example m(15) = 5.
+
+Find sum_{k = 1}^200 m(k).
+
 It uses the fact that for rather small n, applicable for this problem, the solution
 for each number
 can be formed by increasing the largest element.