(Homework Solution): +) | 100% 1- The time to convert an array, with priorities stored at subseripts 1

+) | 100% 1- The time to convert an array, with priorities stored at subseripts 1 through n, to a minheap is in: A. e() B. o(logn) C.ol e performing mergesort on n items is: А.Ollogn) B.o(m + n) C.o(n) D, o(nlogn) 3. Which of the following is not true? 4. The cost function for the optimal matrix multiplhcation problem is: isksj lske isksi 5. The function π + 3n-log n is in which set? 6. f(n)-nlgn is in all of the following sets, except 7. Which statement is comect regarding the wweighted and weighted aetivity seheduling problems? A. O(log n) B.0(log(n)) C. Ω(n) D, dif) A. Both are casily solyed using a greedy techmique D. Weighted is solred using a greedy technique. wmw eighted is solved by dyaamie programming B. Unweighted is solved using a greedy technique, weighted is solved by dynamie programming C. Both require dynanuc programming S. What is the valne of 2

answers with explaination

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+) | 100% 1- The time to convert an array, with priorities stored at subseripts 1 through n, to a minheap is in: A. e() B. o(logn) C.ol e performing mergesort on n items is: А.Ollogn) B.o(m + n) C.o(n) D, o(nlogn) 3. Which of the following is not true? 4. The cost function for the optimal matrix multiplhcation problem is: isksj lske isksi 5. The function π + 3n-log n is in which set? 6. f(n)-nlgn is in all of the following sets, except 7. Which statement is comect regarding the wweighted and weighted aetivity seheduling problems? A. O(log n) B.0(log(n)) C. Ω(n) D, dif) A. Both are casily solyed using a greedy techmique D. Weighted is solred using a greedy technique. wmw eighted is solved by dyaamie programming B. Unweighted is solved using a greedy technique, weighted is solved by dynamie programming C. Both require dynanuc programming S. What is the valne of 2

Expert Answer

answers

1. D.Θ(nlogn)

Explanation: The algorithm to build a min heap is given below.

BUILD-HEAP(A) 
    heapsize := size(A); 
    for i := floor(heapsize/2) downto 1 
        do HEAPIFY(A, i); 
    end for 
END

A quick look over the above algorithm suggests that the running time is Θ(nlogn), since each call to Heapify costs Θ(log n) and Build-Heap makes Θ(n) such calls.

2. C. Θ(n)

Explanation:

We know that call MergeSort (log n) times. Since each recursive step is one half the length of n.

Since we know that merge sort is O(n log n) could stop here as MergeSort is called log n times, the merge must be called n times. But we can also reason that we must subdivide the n items until each input consists of one element. Clearly we must merge n such one item lists to arrive at a final out consisting of n items.

3. B. nlogn belongs to Ω(n2)

Explanation: The omega definition is given below.

Ω (g(n)) = {f(n): there exist positive constants c and
                  n0 such that 0 <= cg(n) <= f(n) for
                  all n >= n0}.

Clearly, for any positive n value, nlogn <= n2. This means that 0 <= c*n2 < nlogn is not satisfied.

4. c. C(i, j) = min i<=k<j {C(i, k) + C(k+1, j) + pi-1pkpj}

Explanation:

The most important step of the dynamic programming paradigm is to define the value of an optimal solution recursively in terms of the optimal solutions to subproblems. To help us keep track of solutions to subproblems, we will use a table, and build the table in a bottomup manner. For 1 sisjsn, let mli,jl be the minimum number of scalar multiplications needed to compute the Ai.J. The optimum cost can be described by the following recursive formulation Basis: Observe that ifi =丿then the problem s trivial, the sequence contains only one matrix, and so the cost so n other words, there is nothing o multiply) us m[i, i-Ofor i = 1, 2, , n. Step: If ij,then we are asking about the product of the subchain Ai.j and we take advantage of the structure of an optimal solution. We assume that the optimal parenthesization splits the product, Ai.j into for each value of k, 1s ks n as A.Ak1 The optimum time to compute is mi, k], and the optimum time to compute is m[k1,j]. We may assume that these values have been computed previously and stored in our array. Since Aik is a matrix, and Ak+l.j is a matrix, the time to multiply them is pi-1 Pk Pj. This suggests the following recursive rule for computing mli. if ij Sk<j To keep track of optimal subsolutions, we store the value of k in a table sli,j1. Recall, k is the place at which we split the product Ai., to get an optimal parenthesization. That is

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