Using the following code, determine the average case time complexity for the algorithm. function hanoi(numofdisks, start, dest, spare) { if (numofdisks > 0) { hanoi(numofdisks - 1,start,spare,dest); printf("move top of disk from peg %i to peg %i \n", start, dest);
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Using the following code, determine the average case time complexity for the
function hanoi(numofdisks, start, dest, spare)
{
if (numofdisks > 0)
{
hanoi(numofdisks - 1,start,spare,dest);
printf("move top of disk from peg %i to peg %i \n", start, dest);
hanoi(numofdisks - 1,spare,dest,start);
}
}
Step by step
Solved in 2 steps
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- Modiflow y the beprogram given to include response time program;ROUND ROBIN CPU SCHEDULING ALGORITHM #include<stdio.h> #include<conio.h> using namespace std; int main() { int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max; float awt=0,att=0,temp=0; printf("Enter the no of processes--"); scanf("%d",&n); for(i=0;i<n;i++) { printf("\nEnter Burst Time for process %d--", i+1); scanf("%d",&bu[i]); ct[i]=bu[i]; } printf("\nEnter the size of time slice--"); scanf("%d",&t); max=bu[0]; for(i=1;i<n;i++) if(max<bu[i]) max=bu[i]; for(j=0;j<(max/t)+1;j++) for(i=0;i<n;i++) if(bu[i]!=0) if(bu[i]<=t) { tat[i]=temp+bu[i];…What is the worse-case complexity scenario for the following snippets of code? Please show the breakdown of your calculations.Two loops in a row: A nested loop in which the number of times the inner loop executes depends on the value of the Outer loop index. For ( I = 0; I < N; i++) { For ( j = N; j > I; j- - ) { Sequence of statements } }Modify below program to include response time Program:SJF CPU SCHEDULING ALGORITHM: #include<stdio.h>#include<conio.h>using namespace std; int main(){int p[20], bt[20], wt[20], tat[20], i, k, n, temp; float wtavg, tatavg;printf("\nEnter the number of processes--"); scanf("%d", &n);for(i=0;i<n;i++){p[i]=i;printf("Enter Burst Time for Process %d--", i); scanf("%d", &bt[i]);}for(i=0;i<n;i++)for(k=i+1;k<n;k++)if(bt[i]>bt[k]){temp=bt[i]; bt[i]=bt[k];bt[k]=temp;temp=p[i];p[i]=p[k];p[k]=temp;}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++){wt[i] =wt[i-1]+bt[i-1];tat[i] =tat[i-1]+bt[i];wtavg = wtavg + wt[i];tatavg = tatavg + tat[i];}printf("\n\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n"); for(i=0;i<n;i++)printf("\n\t P%d \t\t %d \t\t %d \t\t %d", p[i], bt[i], wt[i], tat[i]);printf("\nAverage Waiting Time--%f", wtavg/n);printf("\nAverage Turnaround Time--%f", tatavg/n); getch();}