In the code below, we define random$(n)$ which returns iin$[n]$ with probability $\frac{1}{n},$ in import random

import math

def f$2($xs$)$:

s $=0$

n $=$ len$($xs$)$

k $=$ random$($n$)$

$($b$)$

m $=$ int$($math$.$ log$2($n$))$

if k $$ n$**0\mathrm{.}25$:

m$\}=\backslash$textrm$\{$n$\}//$

for i in range$($n$//2)$:

for j in range$($m$)$:

s $=$ s $*$ xs$[$j$]+$ xs$[$i$]$

return s

$($c$)$

import random

def f$3()$:

s $=0$

while coinflip$()==0$:

s $+=1$

return s

import random

def f$4($xs$)$:

i $=$ len$($xs$)$

$($d$)5$ while i $>0$:

if random$(10000)==1$:

return $-1$

i $-=1$

return i $($e$)$

import random

def f$5($xs$)$:

s $=0$

n $=$ len$($xs$)$

if n :

return xs$[1]$

k $=$ random$(100)$

if random$($n$**$k$)==1$:

for j in range$($n$**$k$)$:

s $=$ s $*$ xs$[$j$]+$ xs$[$j$]$

return s

$($f$)$

import random

def f$6($xs$)$:

s $=0$

n $=$ len$($xs$)$

while random$($n$)!=$ random$($n$)$:

s $+=$ xs$[$random$($n$)-1]$

return s

import random

def f$7($xs$)$:

s $=0$

if n : $\backslash$ len$($xs$)$

if n : $\backslash$ len$($xs$)$

if n : $\backslash$ len$($xs$)$

$($g$)$

while random$($n$)!=$ random$($n$)$:

s $+=$ xs$[$random$($n$)-1]$

if random.randint$(1,$ nn$/3$:

return f$7($xs$[$:n$//2])$

return s

$(1)-$time, and CoinFlip simply returns $0$ or $1,$ each with probability $\frac{1}{2}.$

import math

import randomdef random$($n$)$:

return random.randint$(1,$ n$)$def coinflip$()$:

return random$(2)-1$

For each function that follows, give each function's best case, worst case, and average

case asymptotic running times. Justify your answer.

import random

import rando

def fl$($xs$)$:

def fl$($xs$)$:

def fl$($xs$)$:

def fl$($xs$)$:

$($a$)$

m $=$ int$($math$.$sqrt$($n$))$

if k $==1$:

m$\}=\backslash$textrm$\{$n$\}//$

for j in range$($m$)$:

s $*=$ xs$[$j$]$

return s