Question 5 we have 4 different algorithms and all of them are exactly the same but except one for loop, if they all result to the correct answer which one is faster (big-O is better) 1- for(i = n; i > -1; i /= 2) 2-for(i = 0; i < n; i++) 3-for(i = 0; i < n; i += 2) 4-for(i = 1; i < n; i *= 2)  Full explain this question and text typing work only thanks

Perform an experimental analysis on the two algorithms & Visualize their running times as a function of the input size with a log-log chart. 1. prefixAverage12. prefixAverage2 note: I need with an answer with example to understand it better. Please help me out to complete this question been struggling.

220  CMPG 221 1-1 2021 Time Remaining: 00:04:31 ▲Hide Time Remaining▲ I_Know 6: SU4 All Part 5 of 5 - Run Time Question 5 of 53 Points Study the two methods below both will do the same work. If you use your algorithm analysis skills to analize these two algorithms. Will you use the first method or the second method? Motivate your answer.  public MyLinkedList split() {     int nodec=0 ;     int index=0;     Node ptr=head;          MyLinkedList newlist = new MyLinkedList<>();     if((ptr==null)||(ptr==tail))      return this;          for(;ptr!=null;ptr=ptr.next)      {         nodec++;      }           ptr = head;     if(nodec%2==0)       index = nodec/2;     else       index = nodec/2+1;     int i =0;       for(;ptr!=null;ptr=ptr.next)      {         if(i>=index)       {        newlist.append(ptr.element);           this.delete(ptr.element);       }        i++;     }        return newlist;   }   public MyLinkedList split() {     int…

Transcribed Image Text The Big-Oh of an algorithm is derived from the T(n) of the algorithm. Suppose an algorithm has T(n) = 3n^3 + 5n^2. What is/are the Big-Oh of the said algorithm? For the purpose of this item, the ^ symbol is used to denote exponentiation (i.e. "raised to"). * O(n) O(5n^2) O(log n) O(3n) O(n*2) O(n*4) O(5n) O(n*3)

Calculate the running time of the algorithms using big-O notation: a) for (i = 1; i*i*i<n; i++) printf(“%d\n”, i)   b) for (i = n; i > 1; i = ceil(i/10)) printf(“%f\n”, i);

Generate a graph of performance improvement for code below. Tip: Figure 5.17 in the textbook provides an example of a graph depicting performance improvements associated with loop unrolling.  code: //normal code int main() {          int n, sum=0;     n=50;     int a[50];     {         for(int i=0; i<n; i++)         {             sum= sum +a[i];         }         printf("sum is %d", sum);         return 0;     } //with unrolling int main() { int n, sum=0; n=50; int a[50] for(int i=0; i<n; i+=10) { scanf("%d", &a[i]); } sum= sum + a[0]; sum= sum + a[1]; sum= sum + a[2]; sum= sum + a[3]; sum= sum + a[4]; sum= sum + a[5]; sum= sum + a[6]; sum= sum + a[7]; sum= sum + a[8]; sum= sum + a[9];   printf("sum is %d", sum); return 0; }

Which of these two algorithms has a better big-O run-time? For each, list the big-O run-time notation for each and how you determined it. def gnomesort(seq):    i=0    while i<len(seq):        if i==0 or seq[i-1]<=seq[i]:            i+=1        else:            seq[i], seq[i-1] = seq[i-1], seq[i]            i-=1 and def mergesort(seq):    mid=len(seq)//2    l,r=seq[:mid],seq[mid:]    if len(l)>1: l=mergesort(l)    if len(r)>1: r=mergesort(r)    res=[]    while l and r:        if l[-1] >= r[-1]:            res.append(l.pop())        else:            res.append(r.pop())    res.reverse()    return (l or r) + res

Design a type of sieve algorithm that would (1) Create a list of consecutive integers from 1 to n. (2) Initially, let k=2. (3) Consider each of the multiples of k less than or equal to n and either place or remove a mark on it as follows:  - If you encounter an unmarked multiple, mark it with a zero. - If you encounter a multiple that has already been marked, remove the mark by restoring to its original value. (4) Increase k by one. (5) Repeat steps (3) and (4) as long as k <= n.

Consider the recursive algorithm below for computing the sum of the first n cubes: S(n) = 13 + 23 + … + n3 ALGORITHM S(n) //n is a positive integer if (n==1) return 1 else return S(n – 1) + (n*n*n);     Solve the recurrence relation in your answer for the previous question. The runtime efficiency of the algorithm is?    A.  O(n)  B.  O(n2)  C.  O(n3)  D.  None of the above

Question 6 sm.Is 2^(n+1) = O(2^n )? Is 2^(2n) = O(2^n ) ?  Suppose that on a particular computer, an Algorithm A, with inputs of size n, runs in 10n^2 steps, and another Algorithm B runs in 60n lg n steps [Note: lg x = log2(x)]. For what values of positive integers n is Algorithm A better than Algorithm B? Full explain this question and text typing work only     We should answer our question within 2 hours takes more time then we will reduce Rating Dont ignore this line