3. †(n) = 6+ (n-1) + 4+ (n-2)+4(3")
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- int n = 1; int k - 2; int r = n; if (k < n) { r - kFind the result : X= (3*9*(3+(9*3/(3))))sum= 0; for (int i = 0; i 1) { sum++; i= 1/2; } = 2*log2 (n) We denote by Ta(n), Tb (n), Te(n) the running time of the three fragments. 1. Give evaluations for Ta(n), Tb (n), Te(n). 2. Is T(n) = O(Ta(n)) ? Answer YES or NO and justify your answer. 3. Is Te(n) = (Ta(n)) ? Answer YES or NO and justify your answer.
- n=0; for del = 0.0 :0.4 : pi n=n+1; pe(n)=1.2*sin(del); end disp(pe);Bum 0; for (int i = 1; iConstruct a DFA A so that L(A) = L(N) where N is the following NFA:int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;ST(1/2) - +n T(n) = Answer: Display response n> 1 n<1del(int n){ j=(n/3)-1 while (n>0) for(i=n to i=j*3){ print "i"; 5. n--; }//end for 7 j=j/3; }//endwhile }//del() what is the asymptotic run time of del? anyalse each line and what is teh final run time is teh answer O(n^2/3)?SEE MORE QUESTIONS