Merge pull request #118 from ahmedHathout/master

Implementing Power

Author: danielabrozzoni

Algorithms/Divide & Conquer/Compute Power/Python/Compute Power.py

@@ -0,0 +1,37 @@
+def getPower(base, exponent):
+    return _getPower(base, exponent) if (exponent >= 0) else 1. / _getPower(base, exponent * -1);
+
+def _getPower(base, exponent):
+    """
+    Given 2 numbers x and y, this function computes x raised to the power of y.
+    this is a divide and conquer algorithm which means that this function does
+    not multiply x times itself y times, it takes into consideration that if y is
+    an even number then x^y = x^(y/2) * x^(y/2). If y is an odd number then
+    x^y = x^((y-1) / 2) * x^((y-1) / 2) * x. This of course saves a lot of work.
+    this algorithm runs in O(log y) instead of O(y)
+    """
+    # The exponent is 0 so the answer is 1
+    if (exponent == 0):
+        return 1;
+
+    # The exponent is even so base ^ exponent = base ^ (exponent / 2) * base ^ (exponent / 2)
+    if (exponent % 2 == 0):
+        root = _getPower(base, exponent / 2);
+        return root * root;
+
+    # The exponent is not even so base ^ exponent = base ^ ((exponent - 1) / 2) * base ^ ((exponent - 1) / 2)
+    semiRoot = _getPower(base, (exponent - 1) / 2); # Sorry for that silly name but could not find a better one :)
+    return semiRoot * semiRoot * base;
+
+while(True):
+    try:
+        base = input("Enter the base: ");
+        exponent = input("Enter the exponent: ");
+
+        result = getPower(base, exponent);
+        print("{0} ^ {1} = {2}".format(base, exponent, result));
+        break;
+
+    except Exception as e:
+        print("------------------------------------------------")
+        print("Please Enter both a base and an integer exponent");