DC3: Integer Multiplication

San Skulrattanakulchai

December 6, 2018

Integer Multiplication

DPV Ch 2.1
Let two $n$-bit positive integers $x$ and $y$ be given. We wish to find their product $xy$.
Standard multiplication algorithm takes $\Theta(n^2)$ bit operations. We will present an algorithm that beats this bound.
Gauss once made the following observation: Computing the product of two complex numbers, \[ (a+bi)(c+di) = (ac - bd) + (bc + ad)i, \] seems to involve four real-number multiplications: $ac$, $bd$, $bc$, and $ad$. Nevertheless, it can be achieved with three: $ac$, $bd$, and $(a+b)(c+d)$. This is because \[ bc+ad = (a+b)(c+d) - ac - bd. \]

Divide-and-conquer algorithm

Gauss' observation is the basis for the following divide-and-conquer integer multiplication algorithm.

  if n = 1 return x*y
  x1 ← leftmost ⌊n/2⌋ bits of x
  x0 ← rightmost ⌈n/2⌉ bits of x
  y1 ← leftmost ⌊n/2⌋ bits of y
  y0 ← rightmost ⌈n/2⌉ bits of y
  p1 ← multiply(x1, y1)
  p2 ← multiply(x0, y0)
  p3 ← multiply(x1 + x0,  y1 + y0)
  return p1*2^{2⌈n/2⌉}
      + (p3-p1-p2)*2^{⌈n/2⌉}
      + p2

Running time

The procedure's worst-case running time $T(n)$ satisfies the recurrence \[ T(n) = \left\{ \begin{array}{ll} 1 & \mbox{if $n = 1$} \\ n + 3T(n/2) & \mbox{if $n > 1$.} \end{array} \right. \] By the Master Theorem, $T(n) = \Theta(n^{\log_2 3})=o(n^{1.59})$, which is $o(n^2)$.

Remarks

Multiplying an $n$-bit number $z$ by a perfect power of two, say $2^k$, where $k$ is $O(n)$, takes time $O(n)$ since it can be done simply by shifting the bits of $z$.
In practice, instead of $n=1$, use the base case that corresponds to the natural word size of the processor (say $n=64$).
An even faster multiplication algorithm exists and is based on the Fast Fourier Transform (FFT).