Karatsuba algorithm

The basic principle of Karatsuba's algorithm is divide-and-conquer, using a formula that allows one to compute the product of two large numbers ${\displaystyle x}$ and ${\displaystyle y}$ using three multiplications of smaller numbers, each with about half as many digits as ${\displaystyle x}$ or ${\displaystyle y}$, plus some additions and digit shifts. This basic step is, in fact, a generalization of a similar complex multiplication algorithm, where the imaginary unit i is replaced by a power of the base.

Let ${\displaystyle x}$ and ${\displaystyle y}$ be represented as ${\displaystyle n}$-digit strings in some base ${\displaystyle B}$. For any positive integer ${\displaystyle m}$ less than ${\displaystyle n}$, one can write the two given numbers as

${\displaystyle x=x_{1}B^{m}+x_{0},}$

${\displaystyle y=y_{1}B^{m}+y_{0},}$

where ${\displaystyle x_{0}}$ and ${\displaystyle y_{0}}$ are less than ${\displaystyle B^{m}}$. The product is then

${\displaystyle {\begin{aligned}xy&=(x_{1}B^{m}+x_{0})(y_{1}B^{m}+y_{0})\\&=x_{1}y_{1}B^{2m}+(x_{1}y_{0}+x_{0}y_{1})B^{m}+x_{0}y_{0}\\&=z_{2}B^{2m}+z_{1}B^{m}+z_{0},\\\end{aligned}}}$

where

${\displaystyle z_{2}=x_{1}y_{1},}$

${\displaystyle z_{1}=x_{1}y_{0}+x_{0}y_{1},}$

${\displaystyle z_{0}=x_{0}y_{0}.}$

These formulae require four multiplications and were known to Charles Babbage.[4] Karatsuba observed that ${\displaystyle xy}$ can be computed in only three multiplications, at the cost of a few extra additions. With ${\displaystyle z_{0}}$ and ${\displaystyle z_{2}}$ as before one can observe that

${\displaystyle {\begin{aligned}z_{1}&=x_{1}y_{0}+x_{0}y_{1}\\&=x_{1}y_{0}+x_{0}y_{1}+x_{1}y_{1}-x_{1}y_{1}+x_{0}y_{0}-x_{0}y_{0}\\&=x_{1}y_{0}+x_{0}y_{0}+x_{0}y_{1}+x_{1}y_{1}-x_{1}y_{1}-x_{0}y_{0}\\&=(x_{1}+x_{0})y_{0}+(x_{0}+x_{1})y_{1}-x_{1}y_{1}-x_{0}y_{0}\\&=(x_{1}+x_{0})(y_{0}+y_{1})-x_{1}y_{1}-x_{0}y_{0}\\&=(x_{1}+x_{0})(y_{1}+y_{0})-z_{2}-z_{0}.\\\end{aligned}}}$

To compute the product of 12345 and 6789, where B = 10, choose m = 3. We use m right shifts for decomposing the input operands using the resulting base (B^m = 1000), as:

12345 = 12 · 1000 + 345

6789 = 6 · 1000 + 789

Only three multiplications, which operate on smaller integers, are used to compute three partial results:

z₂ = 12 × 6 = 72

z₀ = 345 × 789 = 272205

z₁ = (12 + 345) × (6 + 789) − z₂ − z₀ = 357 × 795 − 72 − 272205 = 283815 − 72 − 272205 = 11538

We get the result by just adding these three partial results, shifted accordingly (and then taking carries into account by decomposing these three inputs in base 1000 like for the input operands):

result = z₂ · (B^m)² + z₁ · (B^m)¹ + z₀ · (B^m)⁰, i.e.

result = 72 · 1000² + 11538 · 1000 + 272205 = 83810205.

Note that the intermediate third multiplication operates on an input domain which is less than two times larger than for the two first multiplications, its output domain is less than four times larger, and base-1000 carries computed from the first two multiplications must be taken into account when computing these two subtractions.

If n is four or more, the three multiplications in Karatsuba's basic step involve operands with fewer than n digits. Therefore, those products can be computed by recursive calls of the Karatsuba algorithm. The recursion can be applied until the numbers are so small that they can (or must) be computed directly.

In a computer with a full 32-bit by 32-bit multiplier, for example, one could choose B = 2³¹ and store each digit as a separate 32-bit binary word. Then the sums x₁ + x₀ and y₁ + y₀ will not need an extra binary word for storing the carry-over digit (as in carry-save adder), and the Karatsuba recursion can be applied until the numbers to multiply are only one digit long.

Karatsuba's basic step works for any base B and any m, but the recursive algorithm is most efficient when m is equal to n/2, rounded up. In particular, if n is 2^k, for some integer k, and the recursion stops only when n is 1, then the number of single-digit multiplications is 3^k, which is n^c where c = log₂3.

Since one can extend any inputs with zero digits until their length is a power of two, it follows that the number of elementary multiplications, for any n, is at most ${\displaystyle 3^{\lceil \log _{2}n\rceil }\leq 3n^{\log _{2}3}\,\!}$.

Since the additions, subtractions, and digit shifts (multiplications by powers of B) in Karatsuba's basic step take time proportional to n, their cost becomes negligible as n increases. More precisely, if T(n) denotes the total number of elementary operations that the algorithm performs when multiplying two n-digit numbers, then

${\displaystyle T(n)=3T(\lceil n/2\rceil )+cn+d}$

for some constants c and d. For this recurrence relation, the master theorem for divide-and-conquer recurrences gives the asymptotic bound ${\displaystyle T(n)=\Theta (n^{\log _{2}3})\,\!}$.

It follows that, for sufficiently large n, Karatsuba's algorithm will perform fewer shifts and single-digit additions than longhand multiplication, even though its basic step uses more additions and shifts than the straightforward formula. For small values of n, however, the extra shift and add operations may make it run slower than the longhand method.