n/a

n/a


1. Consider the following iterative function: int square(int n) { int result = 0; for (int i = 1; i <= n;="" i++)="" result="" +="2" *="" i="" -="" 1;="" return="" result;="" }="" rewrite="" the="" function="" square="" using="" recursion="" and="" add="" preconditions="" and="" postconditions="" as="" comments.="" then="" prove="" by="" induction="" that="" the="" recursive="" function="" you="" wrote="" is="" correct.="" 2.="" suppose="" the="" number="" of="" steps="" required="" in="" the="" worst="" case="" for="" two="" algorithms="" are="" as="" follows:="" ="" algorithm="" 1:="" f(n)="3n2" +="" 9="" ="" algorithm="" 2:="" g(n)="51n" +="" 17="" determine="" at="" what="" integer="" value="" of="" n,="" algorithm="" 2="" becomes="" more="" efficient="" than="" algorithm="" 1.="" 3.="" given="" the="" following="" function="" that="" sorts="" an="" array="" of="" values:="" void="" bubblesort(double[]="" array)="" {="" for="" (int="" i="0;" i="">< array.length;="" i++)="" for="" (int="" j="array.length" -="" 1;="" j=""> i; j--) if (array[j] < array[j="" -="" 1])="" swap(array,="" j,="" j="" -="" 1);="" }="" let="" n="" be="" the="" length="" of="" the="" array.="" using="" summation="" evaluation,="" determine="" the="" number="" of="" swaps="" that="" are="" performed="" in="" the="" worst="" case="" as="" a="" function="" of="" n.="" 4.="" given="" the="" following="" recursive="" function="" and="" it="" corresponding="" helper="" function="" that="" returns="" the="" sum="" of="" all="" the="" elements="" of="" an="" array="" that="" are="" located="" at="" even="" subscripts:="" int="" sumevenelements(int="" array[],="" int="" i)="" {="" if="" (i="">= array.length) return 0; return array[i] + sumEvenElements(array, i+2); } int sumEvenElements(int array[]) { return sumEvenElments(array, 0) } Assume n is the length of the array. Find the initial condition and recurrence equation that expresses the execution time for the worst case of the recursive function and then solve that recurrence.