Derive a good upper bound for recurrence T(n) = 5T(n/3)+ n, with base case T(n) = c when n ≤ 3, for some constant c > 0.
Derive a good upper bound for recurrence T(n) = 5T(n/3)+ n, with base case T(n) =...
3. (20 points) Consider the recurrence To - 316-3 +770-2 where T = r = 2. Use Constructive Mathematical Induction to derive an upper bound for Tr. Assume that r. saba Primarily upper bound b as tightly as possible, and secondarily upper bound a as tightly as possible (a) What do you learn from the Base Case? (b) State the Inductive Hypothesis. (e) Show the Inductive Step. (d) Derive the constants. (e) State the final result.
(5 pts.) (b) Use a recursion tree to determine a good asymptotic upper bound on the recurrence T(n) = 6T ([n/4]) + 11n. Verify your bound by the substitution method.
Consider the recurrence T (n) = 3 · T (n/2) + n. • Use the recursion tree method to guess an asymptotic upper bound for T(n). Show your work. • Prove the correctness of your guess by induction. Assume that values of n are powers of 2.
Prove that the solution of the recurrence T(n) = T(n/2) +6(logk n) with T(1-6(1), for any integer k 2 0, is T(n) = Θ(logk+1 n) (Hint: the upper bound T(n) = O(logk+1 n) is easy; the lower bound T(n) = Ω(logk +1 n) is harder.) Prove that the solution of the recurrence T(n) = T(n/2) +6(logk n) with T(1-6(1), for any integer k 2 0, is T(n) = Θ(logk+1 n) (Hint: the upper bound T(n) = O(logk+1 n) is easy;...
Solve the recurrence relation T(n)=T(n1/2)+1 and give a Θ bound. Assume that T (n) is constant for sufficiently small n. Can you show a verification of the recurrence relation? I've not been able to solve the verification part so far note: n1/2 is square root(n)
pleas answer asap 3. (20 points) Algorithm Analysis and Recurrence There is a mystery function called Mystery(n) and the pseudocode of the algorithm own as below. Assume that n 3* for some positive integer k21. Mystery (n) if n<4 3 for i1 to 9 5 for i-1 to n 2 return 1 Mystery (n/3) Print "hello" 6 (1) (5 points) Please analyze the worst-case asymptotic execution time of this algorithm. Express the execution time as a function of the input...
please derive the binary recurrence equation ie t(n) = t(n/2) + 1, t(1)=1 given that n is not restricted to be power of two by considering the case that n can either be an odd or even number.
Find the best big O bound you can on T(n) if it satisfies the recurrence T(n) ≤ T(n/4) + T(n/2) + n, with T(n) = 1 if n < 4.
FOR ALGORITHM A WORST CASE TIME COMPLEXITY IS DESCRIBED BY RECURRENCE FORMULA T(n)= n/ T (n )thi T (c)=1 if c < 100 FOR ALGORITHM B WORST TIME COMPLEXITY IS DESCRIBED BY RECURRENCE FORMULA T(n) = 2T (2/2) + n/logn ; (c) = 1 fc 2100 WHICH ALGORITHM IS ASYMPTOTICALLY FASTER? WHY?
Compute the recurrence relation, T(n), for the following function, solve it, and give a e bound. Justify your answer public static double myPower(double r, int n) if (n1){ return 1 } else if (n % 2 == 0) { double tmp myPower (r, n/2); return tmp tmp; } else{ myPower (r, (n 1)/2); return }