Chapter 2, Problem 3E

Program Plan Intro

Asymptotic Notations:

For asymptotic analysis of algorithms, some mathematical tools are used to represent time complexity of algorithm, such mathematical tools are known as Asymptotic notations. Some of the asymptotic notations are described below:

Big-O:

A function “g(n)” is $O\left(f\left(n\right)\right)$ if there exists a constant $c>0$ such that “cf(n)” grows faster than $"g\left(n\right)"$ for all $n>=n_0$ , that is, “g(n)” is $O\left(f\left(n\right)\right)$ iff for some constants “c” and “$n_0$”, $g(n)<=cf(n)$ for all $n>=n_0$. Here, “f(n)” denotes upper bound for “g(n)”. The important points for above statements is as follows :

• “n” is the data set size.
• “f(n)” denotes a function that is calculated while taking n as the parameter.
• $O\left(f\left(n\right)\right)$ means that curve denoted by “f(n)” is an upper bound for a function’s need for resources.

Big Theta:

A function “g(n)” is $\Theta (f(n))$ iff two positive constants “c₁” and “c₂” and a positive integer $n_0$ exists such that $c_{1}f\left(n\right)\le g\left(n\right)\le c_{2}f\left(n\right)$ for all $n>=n_0$, that is, “f(n)” denotes the exact bound for “g(n)”, which denotes that growth rate of “g(n)” and “f(n)” are equal. The important points for above statements is as follows :

• “n” is the data set size.
• “f(n)” denotes a function that is evaluated taking “n” as the parameter.
• $\Theta (f(n))$ means that curve denoted by “f(n)” is an exact bound for the resource needs of a function.

Refer question 3 from chapter 2 for given statements.

Explanation of Solution

 (a).

The statement ${\displaystyle {\sum }_{i=1}^n{i}^2}$ is $O(n^3)$ if a constant “c” exists such that ${\displaystyle {\sum }_{i=1}^n{i}^2}\le c{n}^3$.

• Here, ${\displaystyle {\sum }_{i=1}^n{i}^2}$ denotes sum of squares of natural numbers, that is

  $\begin{array}{l}{\displaystyle {\sum }_{i=1}^n{i}^2}=1^2+2^2+\mathrm{...}{n}^2\\ =\frac{n(n+1)(2n+1)}{6}\\ =\frac{2n^3+3n^2+n}{6}\end{array}$

• So, ${\displaystyle {\sum }_{i=1}^n{i}^2}\le n.n^2=n^3$ , thus $c=1$ for any “n”.

Hence, ${\displaystyle {\sum }_{i=1}^n{i}^2}$ is $O(n^3)$, and in general ${\displaystyle {\sum }_{i=1}^n{i}^k}$ is $O(n^{k+1})$.

Explanation of Solution

(b).

• Function $\frac{a{n}^k}{logn}$ is $O\left(n^k\right)$ if there is a “c” and “N” for which $\frac{a{n}^k}{logn}\le c{n}^k$ for all $n\ge N$
• The inequality becomes $a/logn\le c$ for $N>1$, and since $\frac{a}{logn}\to 0$, $c=a$ can be substituted.
• But, no value for “c” is possible for which $c\le a/logn$.

Thus, $\frac{a{n}^k}{logn}$ is $not\theta (n^k)$. Hence, the function is $O\left(n^k\right)$.

Explanation of Solution

(c).

• The function $n^{1.1}+nlogn$ is $\theta \left(n^{1.1}\right)$ if two constants exists such that $c1n^{1.1}\le n^{1.1}+nlogn\le c2n^{1.1}$.
• These inequalities could be transformed into $c1\le 1+n^{-1}logn\le c2$ by rule of L'Hospital.
• $n^{-1}logn=(logn)/n^1$ has same limit as $(loge)/(1.n^1)=(10.loge)/n^1$ that is 0.
• Therefore, $c1=1,c2=1+10loge$ .

Hence, the given function is $\theta \left(n^{1.1}\right)$.

Explanation of Solution

(d).

• The function $2^n$ is $O(n!)$ if there are “c” and “N” for which $2^n\le cn!$ for all $n\ge N$ . If N = 1, then $2^n\le cn!$ implies that $2^n\le c(1*2*3\mathrm{...}*n)$, that is , $\frac{2}{1}*\frac{2}{2}*\frac{2}{3}\mathrm{...}*\frac{2}{n}\le 2=c$ , that is, value of constant $c=2$ .
• Hence, function $2^n$ is $O(n!)$.
• n! is O(2ⁿ)if there exists “c” and “N” such that $n!\le c{2}^n$ for $n\ge N$ .
• The inequality $n!\le c{2}^n$ implies $1*2*3\mathrm{....}*n\le c(2^n)$ , that is , $\frac{1}{2}*\frac{2}{2}*\frac{3}{2}\mathrm{...}*\frac{n}{2}\le c$ for all “n’s” .
• In this case, such a constant could not be found

Hence, function $n!$ is $notO(2^n)$.

Explanation of Solution

(e)

• A constant “c” exists such that for some “N” and all $n\ge N,2^{n+a}\le c{2}^n$.
• If $c\ge 2^{n+a}/2^n=2^a$, that is, if the value of constant $c\ge 2^a$

Hence, function $2^{n+a}$ is $O(2^n)$.

Explanation of Solution

(f).

• A constant “c” doesnot exists such that for some “N” and all $n\ge N,2^{2n+a}\le c{2}^n$.
• There exists no constant “c” such that $c\ge 2^{2n+a}/2^n=2^{n+a}$.

Hence, the function $2^{n+a}$ is $notO(2^n)$.

Explanation of Solution

(g).

• As $n=2^{logn}$, $n^a=2^{alogn}$ , therefore, if $2^{alogn}>2^{\sqrt{logn}}$, then $n^a>2^{\sqrt{logn}}$
• A constant “c” has to be found such that for some “N” and all $n\ge N,2^{\sqrt{logn}}\le c{n}^a$ or $2^{\sqrt{logn}}\le c{2}^{alogn}$.
• That is, $c\ge 2^{\sqrt{logn}}/2^{alogn}$ , which is possible because the function $2^{\sqrt{logn}}/2^{alogn}$ is decreasing.

Hence, function $2^{\sqrt{logn}}$ is $O(n^a)$.