Why is squaring a number faster than multiplying two random numbers?

Question

Multiplying two binary numbers takes n^2 time, yet squaring a number can be done more efficiently somehow. (with n being the number of bits) How could that be?

Or is i

Answer 1

First of all great question! I wish there were more questions like this.

So it turns out that the method I came up with is O(n log n) for general multiplication in the arithmetic complexity only. You can represent any number X as

X = x_{n-1} 2^{n-1} + ... + x_1 2^1 + x_0 2^0
Y = y_{m-1} 2^{m-1} + ... + y_1 2^1 + y_0 2^0

where

x_i, y_i \in {0,1}

then

XY = sum _ {k=0} ^ m+n r_k 2^k

where

r_k = sum _ {i=0} ^ k x_i y_{k-i}

which is just a straight forward application of FFT to find the values of r_k for each k in (n +m) log( n + m) time.

Then for each r_k you must determine how big the overflow is and add it up accordingly. For squaring a number this means O(n log n) arithmetic operations.

You can add up the r_k values more efficiently using the Schönhage–Strassen algorithm to obtain a O(n log n log log n) bit operation bound.

The exact answer to your question is already posted by Eric Bainville.

However, you can get a much better bound than O(n^2) for squaring a number simply because there exist much better bounds for multiplying integers!