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The blocks in the Karatsuba three-product identity are integers, not single digits: $p_0$ can need $2m$ positions and $h$ can exceed $q-1$. Concatenating the displayed blocks therefore fails. Accumulate their shifted values with ordinary addition, normalizing each digit position by Euclidean division: a coefficient $v$ becomes the digit $v\mathrm{mod}B$, with carry $\lfloor v/B\rfloor$ passed to the next position. Subtraction likewise propagates borrows. Although $h\ge 0$ for these nonnegative integers, intermediate digitwise subtraction can still borrow.

In the four-digit Karatsuba worked multiplication, grouping by base $q=100$ gives unnormalized coefficients $[672,2840,2652]$, written from high to low. Normalize from the right: $2652=26\cdot 100+52$; $2840+26=2866=28\cdot 100+66$; $672+28=700=7\cdot 100+0$. The final groups are $[7,00,66,52]$, yielding $7,006,652$. The zero group must retain its two positions.

Since $a,b,c,d<B^m$, their sums are less than $2B^m\le B^{m+1}$ and can need $m+1$ digits. Allocate that extra digit before the third recursive product; discarding it can change the answer. The GMP manual describes this sum growth and uses a difference-based form to avoid it.^[ref]

Karatsuba multiplication is a divide-and-conquer algorithm for multiplication of integers. Splitting each operand into two blocks seems to require four smaller products; the Karatsuba three-product identity recovers the two cross terms together from just three products. Repeating that saving through recursion changes the time complexity from quadratic to $O(n^{{\log }_{2}3})$, where $n$ counts digits in a fixed positional notation. The exponent is about $1.585$: the benefit comes from the algorithm, even though each splitting step needs more addition and subtraction.

The Karatsuba worked multiplication below computes $1234\cdot 5678=7,006,652$ using $12\cdot 56$, $34\cdot 78$, and $46\cdot 134$. The last operand already shows why “half-size” needs care: Karatsuba carry handling and Karatsuba recursion termination account for the extra digit. Karatsuba digit-operation cost states the operation model; Karatsuba practical crossover explains why the asymptotic saving alone does not select a useful threshold.

The original 1962 paper presents a binary numeral system automaton construction: its Theorem 2 and proof reduce multiplication to taking squares of numbers, then reduce a larger square of a number to three smaller squares of numbers. Its footnote explicitly addresses the extra digit in a sum of halves. The sequential time complexity analysis below uses a different operation model from that paper's automaton size and depth bounds, including its Boolean circuit realization without feedback.^[ref]

For the sum-based Karatsuba three-product identity, a safe fixed base-case threshold is three digits or more. Return zero directly if either operand is zero; at or below the threshold, use ordinary digit-by-digit multiplication. Otherwise split by the maximum current operand length and invoke Karatsuba multiplication on $(a,c)$, $(b,d)$, and $(a+b,c+d)$.

Each recursive operand has at most $\lceil n/2\rceil +1$ digits. This is strictly less than $n$ for $n\ge 4$, proving termination. A rule stopping only at one digit does not supply this guarantee: in base two, $3=11_2$ splits into $1,1$, whose sum is $2=10_2$, still two digits. An implementation must choose a sufficient threshold or explicitly handle such nonshrinking calls. Leading zero padding must not be repeatedly added in a way that defeats the decreasing size measure.

For signed integers, multiply absolute values with the same nonnegative procedure and restore the sign afterwards. This changes the representation bookkeeping, rather than the three-product identity.

Use decimal expansion, $B=10$, $n=4$, $m=2$, and $q=100$. The Karatsuba operand splitting is $1234=12\cdot 100+34$, $5678=56\cdot 100+78$. The Karatsuba three-product identity givesThe third product has a three-digit operand. A small-input base case can evaluate these three products directly, or a smaller threshold can continue recursion. Either choice preserves the identity; only the operation count changes.

The same Karatsuba operand splitting handles $123\cdot 45$: choose $n=3$, $m=2$, and write $123=1\cdot 100+23$, $45=0\cdot 100+45$. Then $p_2=0$, $p_0=1035$, $p_s=24\cdot 45=1080$, and $h=45$, giving $4500+1035=5535$. A zero high block and unequal lengths require no new algebraic rule.