CSE 551 TEST 2 SOLUTIONS IN CLASS 03/27/18, 1 HOURS 15 MINUTES

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1 CSE 551 TEST 2 SOLUTIONS IN CLASS 03/27/18, 1 HOURS 15 MINUTES Regrading of Midterms If you believe that your grade has not been recorded correctly, return the entire paper to the instructor with a short note indicating what you believe to be the error. Other than for that reason, test grades are almost never changed. If you believe that you did not receive the proper credit, first read these sample solutions carefully to see if you can understand the answer to your concern. If that does not resolve it, write a clear explanation of why you believe the grade is in error and submit that, along with the entire test paper, to the instructor. Please do not discuss in your explanation how your solution is like that of another student, as FERPA legislation makes it impossible for me to discuss one student s work with another. Please take into account that many papers were graded, and it is impossible to change the grade on one paper without giving every other student the same opportunity. If you nevertheless want the paper regraded, be advised that the entire paper will be regraded and the grade may go up, stay the same, or go down. The new grade will be final. It is a violation of the Academic Integrity Policy to request a grade change simply because you need or want a higher grade. If you require a clarification of the sample solutions (not a grade change or review as discussed above), ask in the office hours of the TA or instructor. You will be asked whether you have read the sample solution and to indicate what precisely is unclear to you about it, so read these sample solutions carefully first. Under no circumstances can anyone change a grade other than the instructor, so do not ask the TA to do so he is not able to. Grade review requests, whether submitted as described above or not, will not be considered if received after 10 April Name ASU ID Ima SAMPLE Answer every question, whether or not you think your answer is complete or correct. This is a closed book/closed notes examination. Ensure that no papers or electronic devices can be seen by any class member. Part of the exercise of doing the 1

2 questions is understanding the questions. Read each carefully so as not to waste time. All questions are equally weighted (but not necessarily equally difficult.) Please do not explain a general algorithm by giving examples on specific inputs. Describe the algorithm for the general case. Question Mark Out Of Total 40 Question 1. A 2-tree is an undirected graph, defined recursively as follows. A triangle (i.e. a simple graph on three nodes and three edges) is a 2-tree. When G 1 = (V 1, E 1 ) is a 2-tree, G 2 = (V 2, E 2 ) is a 2-tree, V 1 V 2 = {x, y}, and e = {x, y} E 1 E 2, the graph G = (V 1 V 2, E 1 E 2 ) is a2-tree. No graph is a 2-tree unless it is obtained in this way. In a node-weighted 2-tree G = (V, E), there is a weight function w : V Z (that is every node v V has an integer weight w(v)). Devise the most efficient algorithm that you can for finding a maximum weight independent set of nodes in a node-weighted 2-tree. If you define OPT in some way in a dynamic programming solution, each subproblem should be of the same type (i.e., a 2-tree). If it is something else you must say what to do in that case. Although not all 2-trees are maximal outerplanar graphs, it remains true that every 2-tree has at least two nodes of degree 2 (note that two nodes of degree 2 cannot be adjacent in a 2-tree with four or more nodes.) We describe an algorithm, reversing the construction that must have built the 2-tree from triangles. Define three functions on vertices corresponding to the edges of the 2-tree: f(x, y), s(x, y), n(x, y). Note that f(x, y) = s(y, x), s(x, y) = f(y, x), and n(x, y) = n(y, x). Initially for every edge (x, y) set all the f, s, n values to 0. While there are at least three vertices, pick a vertex v with degree 2 (which always exists). Let x, y be the

3 neighbors of v ({x, y} is an edge). Set f(x, y) f(x, y) + f(x, v) + n(v, y) s(x, y) s(x, y) + n(x, v) + s(v, y) n(x, y) n(x, y) + s(x, v) + f(v, y) + w(v) Now delete v. When only two vertices x, y remain, the final answer is max(f(x, y) + w(x), s(x, y) + w(y), n(x, y)). (This can be done in linear time, by maintaining a list of degree 2 nodes.) Although not every 2-tree is a mop, this is a small adaptation of the solution given on Homework 3 s solutions. Question 2. A tree G = (V, E) is given. Devise the most efficient algorithm that you can to find the number of edges in a maximum cardinality matching of G (i.e. a largest set of edges from E so that no node in V appears in more than one of the chosen edges). Explain.. Recall that a tree is a connected gtaph with no cycles. Unless you modify the tree. it does not have a root or children or parent nodes. If you do choose a root, there is no limit on the number of children. If you define OPT in some way in a dynamic programming solution, each subproblem should be of the same type (i.e., a tree). If it is a forest (or something else), you must say what to do in that case. Good answer: A tree is a bipartite graph. Colour the nodes black and white by first choosing any node to be black. Then as long as there is an uncoloured node x with a coloured neighbour y, colour x the opposite from y. Now direct all edges from black nodes to white ones and give each capacity 1. Add a new node s and an edge from s to each black node, with capacity 1. Then add a new node t and an edge from each white node to t with capacity 1. Then the value of the s, t-maxflow is the cardinality of the largest matching. So use Ford-Fulkerson. Excellent answer (a linear time algorithm): (This is like the answer to Question 1.) For each node v, we will keep track of two values out(v) and in(v); the first will record information when v is available to be matched, the second when v is not available to be matched. Here is the algorithm: Input: Tree G = (V, E) Initialize out(v) = in(v) = 0 for all v V. Compute the degree d(v) of each v V. Place all nodes v V with d(v) = 1 in L while L 2, choose an x L and let y be the only neighbour of x

4 compute in(y) max(in(y) + out(x), in(y) + in(x), out(y) + out(x) + 1) compute out(y) out(y) + max(in(x), out(x)) delete node x, set d(y) d(y) 1 if d(y) = 1, then add y to L. Let z be the only node in L The maximum cardinality matching has size max(in(z), out(z)). Question 3. A pipeline network has been constructed by providing a directed graph G, a source s, a terminal t, and a nonnegative capacity on every directed edge. We want it to have an s, t-flow maximum value of v, but it does not. (1) When our network has a maxflow of value v > v, Melvin says that we can reduce the capacity on a single directed edge to reduce the maxflow value. Is Melvin always right? (If yes, explain how you would find such an edge. If no such edge need exist, give an example. In either case, explain!) Melvin has been misquoted many times. Melvin says that for every flow network, there exists an edge on which you can reduce the capacity in order to reduce the maxflow. He does NOT say that you can pick any edge, or a random edge, or that he will pick an edge, to do this. Yes, Melvin is always right. By the maxflow-mincut theorem there is a cut of capacity v, say (A, B). This can be found using the Ford-Fulkerson algorithm. Choose any directed edge going from a node in A to a node in B, and reduce its capacity. Then (A, B) has capacity less than v, and so the maxflow is reduced. You cannot just pick a saturated edge in some maxflow and reduce its capacity, in general. (2) When our network has a maxflow of value v < v, Elmer says that we can increase the capacity on a single directed edge to increase the maxflow value. Is Elmer always right? (If yes, explain how you would find such an edge. If no such edge need exist, give an example. In either case, explain!) Elmer has also been misquoted many times. No, Elmer can be wrong. Construct a graph with three nodes {s, x, t}; directed edge (s, x) has capacity 1, and directed edge (x, t) has capacity 1. The maxflow value is 1. Suppose that v = 2. Increasing the capacity of either one (but not both) of the edges leaves the maxflow value unchanged.

5 The question says give an example. Saying how you might make an example does not actually give it, and indeed takes more space than simply giving one. Question 4. Here is the input to our happy worker problem: n workers W = {w 1,..., w n } and n jobs J = {j 1,... j n } δ days in a work schedule, where δ n. For each worker w W, there is a set T (w) J with T (w) = δ (these are the jobs to which w can be assigned). For each job j J, there is a set U(j) W with U(j) = δ (these are the workers who can be assigned to job j). We want to make a happy work schedule for δ days of work, so that (1) every worker is assigned exactly one job on each day, and does different jobs on different days, and (2) every job is done by exactly one worker each day. Devise the most efficient algorithm that you can to determine, given the input described, whether or not there is a happy work schedule. When it is possible, your algorithm should output the schedule determined. Hint: Can you figure out an assignment for the first day? There is always a happy work schedule. A schedule for the first day would be a perfect matching between workers and jobs. To find one, whenever worker w can do job j, make an edge from w to j with capacity 1. Add node s with edges to all workers, each of capacity 1. Add node t with edges from all jobs, each of capacity 1. Then when there is an integer-valued flow of value n, there is a perfect matching (and hence a schedule for one day). The assignment is precisely the worker-job edges with flow value equal to 1. Ford-Fulkerson does this. Having found a schedule for one day, for each edge (w, j) in the matching found remove w from U(j) and j from T (w). Then we have a problem of identical type for δ 1 days. So repeat until δ = 0 and then stop, outputting the schedules for each day. You are not asked about correctness of the method. But why does it work? You can appeal to the Philip Hall theorem, or directly consider what a minimum cut would look like if an iteration hypothetically failed for find the flow of value n.