[SOLVED] CS计算机代考程序代写 AI algorithm x

x
CS570
Analysis of Algorithms Fall 2009
Exam I
Name: _____________________ Student ID: _________________
____Monday ____Friday 2-5 ____Friday 5-8 ____DEN
Maximum
Received
Problem 1
20
Problem 2
15
Problem 3
15
Problem 4
15
Problem 5
15
Problem 6
20
Total
100
2 hr exam
Close book and notes
If a description of an algorithm is required, please limit your description within 200 words, anything beyond 200 words will not be considered. Proof/justification or complexity analysis will only be considered if the algorithm is either correct or provided by the problem.

1) 20 pts
Mark the following statements as TRUE or FALSE. No need to provide any justification.
[TRUE]
Given a min-heap with n elements, the time complexity of select the smallest element from the n elements is O(1).
[FALSE] But we give credit to both true and false
Given a min-heap with n elements, the time complexity of select the second smallest element from the n elements is O(1).
[FALSE] Algorithms (e.g., bubble sort) with O(n2) time complexity could cost O(n) in some instances.
Given a problem with input of size n, a solution with O(n) time complexity always costs less in computing time than a solution with O(n2) time complexity.
[FALSE]
By using a heap, we can sort any array with n integers in O(n) time. [TRUE] m is at most n(n-1)
For any graph with n vertices and m edges we can say that m = O(n2). [FALSE] m could be n(n-1)
For any graph with n vertices and m edges we can say that O(m+n) = O(m).
[FALSE] Consider the following counter-example:
G(V, E). V={A,B,C}, (A,B)=2, (A,C)=1, (B,C)=1. P=AB is the shortest path between A and B, but it is not in the MST.
Given the shortest path P between two nodes A and B in graph G(V,E), then there exists a minimum spanning tree T of G, such that all edges of path P is contained in T.
[FALSE] Consider the following counter-example:
w1 prefers m1 to m2; w2 prefers m2 to m1; m1 prefer w1 to w2; m2 prefer w2 to w1
Consider an instance of the Stable Matching Problem in which there exists a man m and a woman w such that m is ranked last on the preference list of w and w is ranked last on the preference list of m, then in every stable matching S for this instance, the pair (w, m) belongs to S.
[FALSE] A graph could have multiple MSTs.
While there are many algorithms to find the minimum spanning tree in a graph, they all produce the same minimum spanning tree.
[TRUE]
A BFS tree is a spanning tree

2) 15 pts
Here is a divide-and-conquer algorithm that aims at finding a minimum spanning tree. Given a graph G = (V, E), partition the set V of vertices into two sets V1 and V2 such that |V1| and |V2| differ by at most 1. Let E1 be the set of edges that are incident only on vertices in V1, and let E2 be the set of edges that are incident only on vertices in V2. Recursively solve a minimum spanning tree problem on each of the two subgraphs G1 = (V1, E1) and G2 = (V2, E2). Finally, select the minimum-weight edge in E that crosses the cut (V1, V2), and use this edge to unite the resulting two minimum spanning trees into a single spanning tree.
Either argue that the algorithm correctly computes a minimum spanning tree of G, or provide an example for which the algorithm fails.
This algorithm fails. Take the following simple graph:
Never mind how the algorithm first divides the graph, the MST in one subgraph (the one containing u) will be one edge of cost 1, and in the other subgraph, it will be one edge of cost 8. Adding another edge of cost 1 will give a tree of cost 10. But clearly, it would be better to connect everyone to u via edges of cost 1.

3) 15 pts
Suppose you are given two sets A and B, each containing n positive integers. You can choose to reorder each set however you like. After reordering, let ai be the ith element of set A, and let bi be the ith element of set B. You then receive a payoff of . Give an algorithm that will maximize your payoff. Prove that your algorithm maximizes the payoff, and state its running time.
Solution:
Algorithm:
Sort A and B into monotonically decreasing order.
Proof:
Consider any indices i, j, p, q such that i < j and p < q, and consider the terms aibp and ajbq. We want to show that it is no worse to include these terms in the payoff than to include aibq and ajbp , that is, aibp ajbq ≥ aibq ajbp .Since A and B are sorted into monotonically decreasing order and i < j, p < q, we have ai≥ aj and bp ≥ bq . Since ai and aj are positive and bp − bq is nonnegative, we have aibp−bq ≥ ajbp−bq .Multiplying both sides by aibq ajbq yields aibp ajbq ≥ aibq ajbp .Since the order of multiplication does not matter, sorting A and B into monotonically increasing order works as well.Complexity:Sorting 2 sets of n positive integers is O(nlogn).4) 15 ptsIndicate, for each pair of expressions (A, B) in the table below, whether A is O, Ω or Θ of B. Assume that k 1, and ε > 0 are constants. Answer “yes” or “no” in each box in the following form. No explanations required.
ABO
Yes No No
No No No Yes Yes Yes
The explanation is not required:
(1) log(logn)=O(logn) => klog(logn)=O(εlogn) => log(logkn)=O(lognε) =>
logkn=O(nε)
(2) -1<=sinn <=1, so we can not determine which one is larger;(3) nlogm =Θ(nlgm), mlogn =Θ(mlgn), let m=10x, and n = 10y, then nlgm = mlgn = 2xy.ΩΘ5) 15 ptsSuppose you are given a number x and an array A with n entries, each being a distinct number. Also it is known that the sequence of values A[1];A[2]; … ;A[n] is unimodal. In other words for some unknown index p between 1 and n, we haveA[1] < A[2] < … < A[p] and A[p] > A[p + 1] > … > A[n].
Give an algorithm with running time O(log n) to find out if x belongs to A, if yes the algorithm should return the index j such that A[j] = x. You should justify both your algorithm and the running time.
Solution: The idea is to first find out p and then break A into two separated sorted arrays, then use binary search on these two arrays to check if x is belong to A.
Let FindPeak() be the function of finding the peak in A . Then FindPeak(A [1 : n ]) works as follows:
Look at A [n / 2], there are 3 cases:
(1) IfA[n/2-1] A[n/2]>A[n/2+1],thenthepeakmustcomestrictlybeforen/2. Run FindPeak(A [1 : n / 2]).
(3) IfA[n/2-1] A[n/2+1],thenthepeakisA[n/2],returnn/2.
Now we know the peak index(p value). Then we can run binary search on A [1 : p ] and A [p +1 : n ] to see if x belong to them because they are both sorted.
In the procedure of finding p, we halve the size of the array in each recurrence. The running time T(n) satisfies T (n ) = T (n/2) + O(1) . Thus T (n) = O (log n). Also both binary search has running time at most O (log n), so total running time is O (log n ).
Explanations:
Note that trying to get x in one shot (in one divide and conquer recursion) usually does not work. Binary search certainly cannot work because the array is not sorted. Modified binary search can hardly work either. The problem is that in each round you need to abandon half the array. However, this decision is hard to made. For example, the array is [1 2 3 4 5 -1], x=-1. When divide we have mid=3. We find A[mid-1] S[i+1], then we delete S[i]; if such i does not exist, which means the digits are in increasing order, then we simply delete the last digit. (*) We call the position i “peak”.
Proof by induction
(1) We first prove (*) can achieve minimal N when n = 1. We have
S’ = a1a2…ai-1ai+1…an-1
Without loss of generality, suppose we delete S[j] rather than S[i], the we have S1 = a1a2…aj-1aj+1…an-1 (j!=i), and N1 is the number represented by S1. Ifj i,sinceai+1 pk+1, similar to (1), we can prove N’ > N; Therefore, we show that N<=N’.(3) With (1) and (2), we prove the correctness of the algorithm.Complexity:O(n), but O(mn) is also acceptable.