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@moshekaplan
Created October 14, 2012 15:37
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Efficient exponentiation
def exp(x,n):
"""Returns x^n, for integers x,n and n>=0
This is done more intelligently, as we can use the definition
that x^(a/2)^2 == x^a to make many less multiplications"""
# count is the amount of multiplications.
count = 0
result = 1
x_raised = x
# On each iteration, multiply our result by x^a if x^a is needed to compute x^n.
# For example, x^4 is in x^7, since 4 is the largest power of 2 that goes into 7.
while n>0:
if n % 2:
count += 1
result *= x_raised
x_raised = x_raised*x_raised
count +=1
n /= 2
return result, count
def naive(x,n):
"""Returns x^n, for integers x,n and n>=0
This is done by naively doing n multiplications"""
result = 1
count = 0
for i in range(n):
result *= x
count +=1
return result, count
def main():
result, count = exp(50,1024)
print count
result, count = naive(50,1024)
print count
if __name__ == '__main__':
main()
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