CSCE 222 Lecture 25

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Divide and Conquer

Example

Multiplication of Integers

In binary form, two number $a$ and $b$ consist of $2n$ bits. Each integer can be rewritten as a sum of two $n$-bit integers that represent the most and least significant bits.

${\begin{aligned}a&=\left(a_{2n-1},\ a_{2n-2},\ \ldots ,\ a_{1},\ a_{0}\right)\\&=2^{n}A_{1}+A_{0}=\left(a_{2n-1},\ \ldots ,\ a_{n}\right)+\left(a_{n-1},\ \ldots ,\ a_{0}\right)\\b&=\left(b_{2n-1},\ b_{2n-2},\ \ldots ,\ b_{1},\ b_{0}\right)\\&=2^{n}B_{1}+B_{0}=\left(b_{2n-1},\ \ldots ,\ b_{n}\right)+\left(b_{n-1},\ \ldots ,\ b_{0}\right)\\ab&=(2^{n}A_{1}+A_{0})(2^{n}B_{1}+B_{0})=(2^{2n}+2^{n})A_{1}B_{1}+2^{n}(A_{1}-A_{0})(B_{0}-B_{1})+(2^{n}+1)A_{0}B_{0}\end{aligned}}$

Let $f(n)$ be the total number of bit operations needed to multiply two $n$-bit integers. In this case, $f(2n)=3f(n)+Cn$.

Bitwise Shift Operator

a >> n shifts the binary digits of $a$ $n$ places to the right (essentially multiplying $a$ by $2^{n}$).

Master Theorem

Let $f$ be an increasing function satisfying a recurrence relation:

$f(n)=af\left({\frac {n}{b}}\right)+cn^{d}$

Whenever

• $n=b^{k}$ for some $k\geq 1$, $a\geq 1$
• $b$ is an integer > 1
• $c$ and $d$ are real numbers ≥ 0

Then

$f(n)=af\left({\frac {n}{b}}\right)+cn^{d}={\begin{cases}O\left(n^{d}\right)&a<b^{d}\\O\left(n^{d}\log {n}\right)&a=b^{d}\\O\left(n^{\log _{b}{a}}\right)&a>b^{d}\end{cases}}$

Proof

Proposition. If $a=b^{d}$ and $n$ is a power of $b$, then a function $f$ satisfying the recurrence relation $f(n)=af(n/b)+cn^{d}$ is of the form $f(n)=f(1)n^{d}+cn^{d}\log _{b}{n}$

Proof. Let $k=\log _{b}{n}$. Then

${\begin{aligned}f(n)&=cn^{d}+ac\left({\frac {n}{b}}\right)^{d}+a^{2}c\left({\frac {n}{b^{2}}}\right)^{d}+\dots +a^{k-1}c\left({\frac {n}{b^{k-1}}}\right)^{d}+a^{k}f(1)\\&=a^{k}f(1)+\sum _{j=0}^{k-1}a^{j}c\left({\frac {n}{b^{j}}}\right)^{d}\\&=a^{k}f(1)+\sum _{j=0}^{k-1}cn^{d}\\&=a^{k}f(1)+kcn^{d}\\&=a^{\log _{b}{n}}f(1)+c(\log _{b}{n})n^{d}\\&=b^{d\log _{b}{n}}f(1)+c(\log _{b}{n})n^{d}\\&=n^{d}f(1)+c(\log _{b}{n})n^{d}\\&=O(n^{d}\log {n})\end{aligned}}$