CS 560, HOMEWORK 4 SOLUTIONSINSTRUCTOR: HOA VUEach question is worth 25 points. The extra credit question is worth 10 points.When you are asked to design an algorithm, do the following: a) describe the algorithm,b) explain (or more rigorously prove) why it is correct, and c) provide the running time.Unless specified otherwise, the problems are from the textbook by Jeff Erickson.(1) Question 1: Consider the bubble sort algorithm: BubbleSort(A[1 . . . n])sw = f alse ;k = n ;repeatsw = f alse ;for i = 1 to k − 1 doif A[i] > A[i + 1] thenSwap A[i] and A[i + 1] ;sw = true ;endendk = k − 1;until sw = false;Algorithm 1: BubblesortIn each iteration of the outermost loop, we scan through the array, if there isa consecutive pair of entries that is in the wrong order, we swap them. We stopwhen there is no swap, i.e., the array is sorted.Part a) After the first iteration of the outermost loop, why must the largest elementends up being in A[n]? After the second iteration of the outermost loop, whymust the second largest element ends up being in A[n−1]. What is the generalizedobservation after the ith iteration of the outermost loop? Then conclude that afterthe nth iteration of the outermost loop, A is sorted.In the first iteration, the largest element will keep being swapped until it reachesA[n]. Similarly, in the second iteration, the second largest element will keep beingswapped until it reaches A[n − 2], and so on. The generalized observation is that12 INSTRUCTOR: HOA VUafter the ith iteration, the ith largest element will be at A[i]. Therefore, after then, iteration, the array is sorted.Part b) What is the running time of the algorithm. Please justify your answer.The algorithm runs in O(n2) time. Each inner loop runs at most O(n) times(since k ≤ n). After each outer loop’s iteration, another entry will be in the rightplace in A. Hence, the outer loop runs at most O(n) times. Thus the total runningtime is O(n2).(2) Question 2: Given anl array A[1 . . . n] of numbers (that are not necessarily positive).Design an algorithm that rearranges the entries inPA to maximize the sum. Make sure you prove that your algorithm maximizes the describedsum.The algorithm is to simply sort A in non-decreasing order. Claim: in the optimalarrangement A[i] ≤ A[i + 1] for all 1 ≤ i ≤ n − 1. The proof by contradiction is asfollows. Suppose A[i] > A[i + 1] for some i. Then let x = A[i] and y = A[i + 1].Note that x > y as assumed. If we swap A[i] and A[i + 1]. Then the sum changesby an amount:iy + (i + 1)x − ix − (i + 1)y > 0 ⇐⇒ x − y > 0 ⇐⇒ x > y.Hence, we get an arrangement with a larger sum and therefore, the arrangement isnot optimal. Thus, we have a contradiction. Hence, in the optimal arrangement,the order must be non-decreasing.(3) Question 3: Problem 1a page 268. For simplicity, assume all weights are distinct.To prove it, follow the following steps. Let uv be the edge with the largest weightin the cycle C. In particular, w(uv) > w(u0v0) for all other u0v0 ∈ C.Step 1) Suppose uv is in the minimum spanning tree (MST) T∗. If you delete uvfrom the MST T∗, you will have two connected components A and B where u ∈ Aand v ∈ B. Argue why there must be an edge ab 6= uv in C that has one end pointin a ∈ A and one end point b ∈ B (feel free to draw a picture to visualize).Step 2) Argue that T = T∗ − {uv} + {ab} is also a spanning tree but with asmaller weight. Deduce a contradiction and conclude that uv cannot be in an MST.Suppose uv is in the cycle C = v, u, x1, x2, . . . , xk and uv has the largest costand suppose uv ∈ T∗for some MST T∗.If you delete uv from T∗, there must be two connected trees A and B whereu ∈ A and v ∈ B. We know that u, x1, . . . , xk, v is a path from u to v. In thispath, there must be anCS 560代写、代做C/C++程序语言、代写c++、代做A edge xixi+1 between a node in A and a node in B. Now,T − {uv} + {xixi+1} is also a spanning tree (since A and B are trees that are notconnected to each other, by adding xixi+1, you connect A and B and thereforehave a spanning tree) with a smaller weight since w(xixi+1) have a spanning tree with a smaller weight than T∗ which contradicts with theassumption that T∗is an MST. Thus, uv must not be in an MST.CS 560, HOMEWORK 4 SOLUTIONS 3(4) Question 4: Suppose a graph G has n nodes where n is even. Furthermore, everynode in G has degree (degree of a node is the number of edges that node is incidenton) at least n/2. Prove that G is connected or give a counter-example to disprovethe claim.Suppose for contradiction that G has the described property and is disconnected.Then, pick a vertex v in G. Let S be the set of nodes reachable from v. We knowthat V \ S 6= ∅ (in other words, there are nodes not in S), otherwise, the graphis connected. Now, since the total number of nodes is n, either S or V \ S has atmost n/2 nodes.In the case that S has at most n/2 nodes, consider v, by our assumption, it isconnected to at least n/2 edges but it can have at most n/2 − 1 neighbors in S.Hence, it must be connected to another node in V \S. This leads to a contradiction,since then there is a path from v to another node in V \ S contradicting the factthat S is the set of all reachable nodes from v.In the case that V \S has at most n/2 nodes, again, consider any node u ∈ V \S.By our assumption, it is connected to at least n/2 edges but it can have at mostn/2 − 1 neighbors in V \ S. Hence, it must be connected to another node w in S.This leads to a contradiction, since then there is a path from v to w and since wuis an edge, there is a path from v to u contradicting the fact that S is the set of allreachable nodes from v.(5) Extra credits: Problem 3 page 178.The (greedy) algorithm is as follows. Assume X = [a, b]. Originally, we pick theset that starts at a and covers as much to the right as possible. Call this set S1.Let the sets we picked so far be S1, . . . , Si. We pick Si+1 that starts before Si endsand goes furthest to the right. We can implement the algorithm in O(n log n) timeby first sorting the sets according to the ending point. To add Si+1, we can binary4 INSTRUCTOR: HOA VUsearch among the remaining sets for the set that starts at a point before Si endsand goes furthest to the right. In particular, it takes O(log n) time to figure outSi+1.Proof of correctness: We prove by induction that after the ith step, the solution{S1, . . . , Si} is part of some optimal solution. Clearly, after the 0th step, the solutionis an empty set which is a subset of the optimal solution. Induction hypothesis:suppose {S1, . . . , Si} is part of some optimal solution T = {T1, T2, . . .}. Let the orderingof Ti’s be such that if Ti+1 = [a, b] and Ti = [c, d] then b ≥ d. Observe thatthen T1 = S1, T2 = S2, . . . , Ti = Si. Now, if we have covered the entire desiredinterval X then we are done since we know that S1, . . . , Siis an optimal solution.Otherwise, the algorithm then picks Si+1 that goes furthest to the right. Therefore,for any remaining set (not yet picked by the algorithm) R:(S1 ∪ . . . Si) ∪ R ⊆ (S1 ∪ . . . Si) ∪ Si+1. (∗)We need to show that {S1, . . . , Si, Si+1} is also part of some optimal solution.Suppose not. Then, consider Ti+1 in T. Ti+1 must start no later than where Ti ends.Note that the algorithm considered both Ti+1 and Si+1 at the (i+ 1)th step. Basedon (∗), if we remove Ti+1 from T and add Si+1 to T, we still have a collection of thesame number of sets that covers the entire X. Hence, {S1, . . . , Si, Si+1} is also partof some optimal solution which is a contradiction. Therefore, {S1, . . . , Si, Si+1} isalso part of some optimal solution.转自：http://www.daixie0.com/contents/13/4340.html