**solve the following recurrence relations a. π(π) = π( π/4) + π( π/2 ) + π^2 b. T(n) = T(n/5) + T(4n/5) + n c. π(π) = 3π( n/4 ) + ππ^2 f. π(π) = (π/πβ5) * π(π β 1) + 1 g. π(π) = π(log π) + log π h. π(π) = π (π^ 1/ 4) + 1 i. π(π) = π + 7 βπ β π(βπ) j. π(π) = π ( 3π/4 ) + 1/root(n)**

The **Answer to the Question**

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**Here's the Solution to this Question**

a.

so, we have geometric series:

$T(n)=n^2+5n^2/16+25n^2/256+...=n^2/(1-5/16)$

$T(n)=\Theta(n^2)$

c.

Comparing withΒ $T(n)=aT(n/b)+f(n)$Β :

$a=3,b=4$

so, we have:

$f(n)=\Omega(n^{log_ba})=\Omega(n^{log_43})$

by Master Theorem:

$T(n)=\Theta(f(n))=\Theta(n^2)$

b.

additive term is exactly the current subproblem size, so T(n) is simply the sum of all numbers that appear in this tree. the:

$T(n)\ge nlog_5n$Β andΒ $T(n)\le nlog_{5/2}n$

so,

$T(n)=\Theta(nlogn)$

g.

for recursion tree:

there areΒ $log^*(n)$Β Β levels, and each subproblem at level i has one problem of sizeΒ $log^{(i)}n$

whereΒ $log^{(i+1)}n=log(log^{(i)}n)$Β andΒ $log^{(i)}n=logn$

So, the upper bound we have:

$T(n)=logn+T(logn)=O(\displaystyle \sum^{log^*(n)}_{i=0}log^{(i+1)}n)=O(logn)$

This is true becauseΒ $logn<1/2$Β for large n, andΒ $log^{(i+1)}n<\frac{1}{2^i}logn$Β and use geometric series.Β

Also,Β $T(n)=\Omega (logn)$

so,

$T(n)=\Theta(logn)$

j.

Comparing withΒ $T(n)=aT(n/b)+f(n)$Β :

$a=1,b=4/3$

$f(n)=1/\sqrt n$

so, we have:

$log_ba=log_{4/3}1=0$

$n^{log_ba}=n^{log_{4/3}1}=1$

$f(n)=O(n^{log_ba})=O(n^{log_{4/3}1})=O(1)$

then, by Master Theorem:

$T(n)=\Theta(n^{log_{4/3}1})=\Theta(1)$

h.

$π(π) = π (π^{ 1/ 4}) + 1=π (π^{ 1/ 16}) + 2=π (π^{ 1/ 64}) + 3=π (π^{ 1/ 4^n}) + n$

$T(n)=O(n)$

i.

$π(π) = π + 7 π(\sqrtπ)\sqrt n=7\sqrt n(7T(n^{1/4})n^{1/4}+n^{1/2})+n=$

$=7^kn^{1/2+1/4+1/8+...+1/2^k}T(n^{1/2^k})+...+n$

$1/2+1/4+1/8+...+1/2^k=1-1/2^k$

letΒ $n^{1/2^k}=2$Β , then:

$π(π)=7^nT(2)n/2+...+n$

$π(π)=O(7^nn)$

f.

$π(π) = (π/(πβ5)) π(π β 1) + 1$

$T(n)=(\frac{n}{n-5})^{n-1}T(1)+(\frac{n}{n-5})^{n-2}+...+1$

$T(n)=O(1)$