Integer Multiplication: $x$ and $y$ are two n-bit integers, where $n=3^k$ for some $k>0$. We break $x$ into three parts $a$, $b$, $c$, each with $n/3$ bits; and $y$ into three parts $d$, $e$, $f$, each with $n/3$ bits. The multiplication is:

$$xy = ad \cdot 2^\frac{4n}{3} + (ae + bd) \cdot 2^n + (af + cd + be) \cdot 2^\frac{2n}{3} + (bf + ce) \cdot 2^\frac{n}{3} + cf$$

The multiplication can be calculated as follows:

$r_1 = ad$
$r_2 = (a + b)(d + e)$
$r_3 = be$
$r_4 = (a + c)(d + f)$
$r_5 = cf$
$r_6 = (b + c)(e + f)$
$xy = r_1 \cdot 2^\frac{4n}{3} + (r_2 - r_1 - r_3) \cdot 2^n + (r_3 + r_4 + r_1 - r_5) \cdot 2^\frac{2n}{3} + (r_6 - r_3 - r_5) \cdot 2^\frac{n}{3} + r_5$

Write the recurrence equation representing the algorithm and find the running time of this algorithm (Some logarithms you may need: $\log{3} = 0.477$, $\log{6} = 0.778$, $\log{9} = 0.954$).

For now I am only concerned with the first part of the question (the one asking for the recurrence relation).

What I can deduce from this procedure is that each of the two integers x and y is split into three parts, the variables r1 through r6 are calculated and then summed up using the formula mentioned in the last line of the problem. Since each call results in 6 calls to the same procedure, and the number of additions made is irrelevant of the size of the integer, the recurrence relation is:

$$T(n)=6T(n/3) + \theta(1)$$

Is my reasoning correct?

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  • $\begingroup$ @D.W. I will add the text soon, just when I get the time to. $\endgroup$
    – Wais Kamal
    Jul 31, 2021 at 18:56

1 Answer 1


Close. But consider that for example a+b isn’t an n/3 bit but an n/3 + 1 bit number, and the additions take not O(1) but O(2n/3) operations.

  • $\begingroup$ It is given in the question that n is a power of 3. $\endgroup$
    – Wais Kamal
    Jul 31, 2021 at 18:56

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