CODEWITHSUNDEEP
performancealgorithmruntime
Steven
Steven

Reputation: 199

exponential multiplication algorithm that runs in O(n) time?

I am reading an algorithms textbook and I am stumped by this question:

Suppose we want to compute the value x^y, where x and y are positive integers with m and n bits, respectively. One way to solve the problem is to perform y - 1 multiplications by x. Can you give a more efficient algorithm that uses only O(n) multiplication steps?

Would this be a divide and conquer algorithm? y-1 multiplications by x would run in theta(n) right? .. I don't know where to start with this question

Upvotes: 3

Views: 214

Answers (3)

user1196549
user1196549

Reputation:

The y-1 multiplications solution is based on the identity x^y = x * x^(y-1). By repeated application of the identity, you know that you will decrease y down to 1 in y-1 steps.

A better idea is to decrease y more "energically". Assuming an even y, we have x^y = x^(2*y/2) = (x^2)^(y/2). Assuming an odd y, we have x^y = x^(2*y/2+1) = x * (x^2)^(y/2).

You see that you can halve y, provided you continue the power computation with x^2 instead of x.

Recursively:

Power(x, y)=
    1 if y = 0
    x if y = 1
    Power(x * x, y / 2) if y even
    x * Power(x * x, y / 2) if y odd

Another way to view it is to read y as a sum of weighted bits. y = b0 + 2.b1 + 4.b2 + 8.b3...

The properties of exponentiation imply: 

x^y = x^b0 . x^(2.b1) . x^(4.b2) . x^(8.b2)... 
    = x^b0 . (x^2)^b1 . (x^4)^b2 . (x^8)^b3...

You can obtain the desired powers of x by squaring, and the binary decomposition of y tells you which powers to multiply.

Upvotes: 0

zvisofer
zvisofer

Reputation: 1368

I understand this better in an iterative way:

You can compute x^z for all powers of two: z = (2^0, 2^1, 2^2, ... ,2^(n-1))

Simply by going from 1 to n and applying x^(2^(i+1)) = x^(2^i) * x^(2^i).

Now you can use these n values to compute x^y:

result = 1
for i=0 to n-1:
    if the i'th bit in y is on:
        result *= x^(2^i)
return result

All is done in O(n)

Upvotes: 2

user2897690
user2897690

Reputation:

Apply a simple recursion for divide and conquer. Here i am posting a more like a pseudo code.

x^y :=
    base case: if y==1 return x;
        if y%2==0:  
            then (x^2)^(y/2;
        else 
            x.(x^2)^((y-1)/2);

Upvotes: 1

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