Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 9.1, Problem 2E
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To prove the lower bounds in the worst case for both the maximum and minimum of n numbers is
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Design a transform-and-conquer algorithm for finding the minimum and the maximum element of n numbers using no more than 3n/2 comparisons.
Justify the number of comparisons of your algorithm.
Let’s assume that you want to find both the minimum and maximum of m numbers. Design an algorithm that achieves the goal in [3n/2]−2 comparisons in the worst case (you can write a pseudo code or a paragraph,don’t write code). Explain it.
Solve the recurrence by using repeated substitution. Show the work.
T(n) = T(n-1) + n
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Introduction to Algorithms
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- Running Time Analysis: Give the tightest possible upper bound for the worstcase running time for each of the following in terms of N. You MUST choose youranswer from the following (not given in any particular order), each of which could be reused (could be the answer for more than one of a) – f)):O(N2), O(N½)O(N3 log N), O(N log N), O(N), O(N2 log N), O(N5), O(2N), O(N3),O(log N), O(1), O(N4), O(N12)O(NN), O(N6), O(N8), O(N9)**For any credit, you must explain your answer. Assume that the most time-efficientimplementation is used. Assume no duplicate values and that you can implement theoperation as a member function of the class – with access to the underlying data structure. a) Pop a value off a stack containing N elements implemented as an array.Explanation: b) Printing out all the odd values stored in a binary search tree containing N positiveintegers in ascending order. Explanation: c) Finding the minimum value in a binary search tree of size N. Explanation:arrow_forwardSecond order Spada numbers are well established in the insurance industry. Formally they are defined by the recurrence: Si={Si−1 %( Si−2+1) :i≥ 2 , 1:i=0 ,7 :i=1} Give an O(n) time algorithm to compute the n-th Second order Spada numberarrow_forwardSolve the following recurrence by giving tight -notation bound. Assume that for n ≤ 10, T(n) = O(1). (Hint: Let n = 2m) T(n) = 9T(n¹/3) + (log(n)arrow_forward
- Running Time Analysis: Give the tightest possible upper bound for the worst case running time for each of the following in terms of N. You MUST choose your answer from the following (not given in any particular order), each of which could be reused (could be the answer for more than one of a) – f)): O(N2), O(N3 log N), O(N log N), O(N), O(N2 log N), O(N5), O(2N), O(N3), O(log N), O(1), O(N4), O(NN), O(N6)arrow_forwardShow that if n is an integer and n3 +5 is odd, the n is even using a proof by contradiction.arrow_forwardIf set A has size m, and set B has size n, then what is the size of the cross product of A and B? 2m+n Om + n 2mn mn (m + n)² (mn)² None of the above is correct.arrow_forward
- Compute the following via the division algorithm (i.e.n=dq+rwith0≤r < d). Show your work. (a)8675 mod 18 (b)309 mod 75arrow_forward3. Prove by induction that T(n) = 2T (n/2) + cn is O(n logn).arrow_forwardSolve each recurrence, using O, N, or 0, whichever is appropriate K(n) = K(n – 2/n + 1) + n. -arrow_forward
- Use substitution to show T(n) = 4T(n/4)+n is T(n) is both in O(n²) and O(n log n)- note for O(n?) will need to subtract off a lower-order term (e.g. n² is a lower-order term to n³) to make it work. %3D Explain why it can be in both.arrow_forwardShow that: logn = O(n)arrow_forward7. For n 2 1, in how many out of the n! permutations T = (T(1), 7(2),..., 7 (n)) of the numbers {1, 2, ..., n} the value of 7(i) is either i – 1, or i, or i +1 for all 1 < i < n? Example: The permutation (21354) follows the rules while the permutation (21534) does not because 7(3) = 5. Hint: Find the answer for small n by checking all the permutations and then find the recursive formula depending on the possible values for 1(n).arrow_forward
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