
Efficient exponentiation
Problem 122
Published on Friday, 2nd June 2006, 06:00 pm; Solved by 3681The most naive way of computing n15 requires fourteen multiplications:
n
n
...
n = n15
But using a "binary" method you can compute it in six multiplications:
n
n = n2
n2
n2 = n4
n4
n4 = n8
n8
n4 = n12
n12
n2 = n14
n14
n = n15
However it is yet possible to compute it in only five multiplications:
n
n = n2
n2
n = n3
n3
n3 = n6
n6
n6 = n12
n12
n3 = n15
We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
For 1
k
200, find
m(k).