Midterm Exam 2 Solutions

Transcription

1 Algorithm Design and Analysis November 12, 2010 Pennsylvania State University CSE 565, Fall 2010 Professor Adam Smith Exam 2 Solutions Problem 1 (Miscellaneous). Midterm Exam 2 Solutions (a) Your friend has found a recursive algorithm for sorting an array of length n that uses only Θ(n log n) work at the top level plus three recursive calls on subarrays of length n/3. What is the running time of your friend s algorithm? Answer: O(n log 2 n) Justification: The running time satisfies T (n) = 3T ( n/3 ) + Θ(n 1 log 1 n). Master theorem: Here a = b = 3 and the work term has the from n log b a log k n for k = 1. So the solution has the from T (n) = n log b a log k+1 n = n log 2 n. (b) Suppose you are given a directed graph G = (V, E) with lengths l : E R for each edge, a source node s and a list of candidate distances d(v) (where d(v) is the claimed to be the length of the shortest path from s to v in G). What is the running time of the fastest algorithm which checks whether the distances d are correct? Answer: O( V + E ). Describe the algorithm briefly: Similar to one iteration of Bellman-Ford, except you have to check that one of the constraints is actually tight: For each vertex v, check two conditions: (a) For every incoming edge (u, v), check that d(v) d(u) + l(u, v). (b) There exists at least one incoming neighbor u such that d(v) = d(u) + l(u, v) (that is, one of the conditions has to be met with equality otherwise d(v) doesn t actually correspond to the length of a real path. (c) Suppose you are given a directed graph G = (V, E) with capacities c : E R for each edge, two vertices s and t and a s-t flow f : E R. What is the running time of the fastest algorithm which checks whether f is a maximum s-t flow? Answer: O( V + E ). Describe the algorithm briefly: 1

2 Construct the residual graph (this takes time O( E )) and then check, using BFS or DFS, that there is no path from s to t (this takes time O( V + E ). [You might also want to check that f is really a flow by verifying that 0 f(e) c(e) for each edge and that conservation is respected at all vertices (also O( V + E time).] (d) (Extra credit) Recall that in a directed graph G, a set of paths p 1,..., p k from vertex s to vertex t is edge-disjoint if each edge in the graph is used by at most one of the paths (it could be that the paths share some vertices, but they have to use different edges). The following fact is proved in the book: if we set the capacities of all edges to be 1, then the value of the maximum s t flow in G is exactly the size of the largest collection of edge-disjoint paths from s to t. We say that s is k-connected to t if there are k or more edge-disjoint paths from s to t. Prove that if u is k-connected to v and v is k-connected to w, then u is k-connected to w. Proof by contradiction. We ll make heavy use of the min-cut/max-flow theorem. Look at the minimum u-w cut (call it (A, B)), and suppose it has value strictly less than k. The vertex v must be on one side of the cut. That means that (A, B) is either a u-v cut, or a v-w cut. Either way, it means that the maximum flow from either u to v or v to w is less than k, which leads to a contradiction. Problem 2. (Oscillating sequences) A subsequence i 1 < i 2 < < i k of an array of numbers is oscillating if A[i j ] A[i j+1 ] when j is odd and A[i j ] A[i j+1 ] when j is even. Use dynamic programming to find a polynomial time algorithm which takes an array of numbers and finds the length of the longest oscillating subsequence. Faster (correct) algorithms are worth slightly more points than slower ones. Clearly label the different parts of your solution as follows: (a) Define the subproblems your algorithm will compute. For each i = 1,.., n, we compute two quantities odd(i) and even(i). (b) Explain, in English, what the subproblems mean. odd(i) = length of the longest odd-length oscillating sequence ending at i (and using only indices less than or equal to i). even(i) = length of the longest even-length oscillating sequence ending at i (and using only indices less than or equal to i). (c) Give a recursive algorithm (or formula, if it is easier) for computing the answer to a subproblem in terms of smaller subproblems. For i = 1, define odd(1) = 1 and even(1) = 0. For i = 2,...n, define: 2

3 even(i) = max{0} {odd(j) + 1 j < i such that A[j] A[i]}. odd(i) = max{1} {even(j) + 1 j < i such that A[j] A[i]}. (d) Explain why your recursive formulation is correct. First, the base cases are correct since any single element is a valid length-1 oscillating sequence (so odd(1) = 1) and since an even length subsequence has to have length at least 2. First, note that even-length oscillating subsequences end with an increase from the second last to the last element, while odd-length oscillating subsequences end with an decrease from the second last to the last element. Consider the longest even-length sequence ending at i. The previous index in the sequence must be a number j < i such that A[j] A[i] (if no such index exists, then even(i) = 0). Any odd-length sequence ending at such an index j can be extended by 1 by adding i to the sequence. So the longest sequence ending at i will the maximum over all j (with A[j] A[i]) of even(j) + 1. The analysis for OP T 0 is symmetric, except that the minimum possible value is 1. (e) Give pseudocode for your algorithm. We describe here the simplest algorithm. There is a more complicated O(n log n) time algorithm (very similar to the one for longest increasing subsequence). Algorithm 1: Oscillating subsequence Input: Array A[1..n] Initialize arrays even, odd of length n; even[1] 0; odd[1] 1; for i = 2 to n do newevenv al 0; newoddv al 1; for j = 1 to i 1 do if A[j] A[i] then newevenv al max{odd[i] + 1, newevenv al} if A[j] A[i] then newoddv al max{even[i] + 1, newoddv al} odd[i] newoddv al; even[i] newevenv al; return max{odd[n], even[n]}; (f) Analyze its space usage and running time. Space: 2n array entries. Time: O(n 2 ) for the two nested for loops. 3

4 Problem 3. (Dining Problem) Several families are having a dinner party. To increase their social interaction, they would like to sit at tables so that no two members of the same family are at the same table. Show how to find a seating arrangement that meets this objective (or prove that no such arrangement exists) by reducing this problem to Maximum Flow. Assume that the dinner contingent has p families, that the ith family has a i members, that q tables are available and up to b j people can be seated at table j. Use the following to organize your answer: (a) Explain how to construct a graph G from the inputs a 1,..., a p and b 1,..., b q. Create G as follows: Nodes: We create a node u i for each family i {1,...p} and a node v j for each table j {1,..., q}. In addition, we add a source s and sink t. Total: V = p + q + 2. Edges: add an edge (s, u i ) with capacity a i for every family i and a node (v j, t) with capacity b j for every table j. For every pair i, j, add an edge (u i, v j ) of capacity 1. Total: E = p + q + pq. (b) Explain how to use the answer to the maximum flow problem to construct a seating assignment. Let A = p i=1 a i. If the value of the maximum s t flow in G is strictly less than A, then there is no valid seating assignment. Otherwise (that is, if there is a flow of value A), given an integral flow, we seat one member of family i at table j for every edge (u i, v j ) that carries a unit of flow. (c) Explain why your algorithm is correct. First, note that there is never a flow of value more than A since the capacities leaving s add up to A. Suppose that some seating assignment exists. Then we can construct a valid flow of value A as follows: for every member of family i seated at table j, we add one unit of flow from u i to v j. If we send a i units of flow from s to each u i, then conservation is satisfied at teach u i. For each v j, we add as many units of flow from v j to t as there are people seated at table j. Again, conservation is satisfied at v j (and capacities are respected since each table fits at most b j people). All this means that if the max flow has value less than A, then no valid seating assignment exists. Now suppose some flow of value A exists. By the properties of Ford-Fulkerson, there is an integral flow of value A. We can construct a seating assignment as describe above. The edges out of s must all be saturated to get value A. Conservation of flow at u i indicates that every member of family i gets seated. Conservation at the v j s indicates that the flow coming in to v j is at most b j, so each table gets at most b j people. Thus, the seating assignment is valid. (d) Analyze the running time of the algorithm. 4

5 The time required to construct the graph G is O( V + E ) = O(pq). Running Ford- Fulkerson would take most A iterations of time O( V + E ) each, for a total of O(pqA). The capacity scaling algorithm runs in time O(pq log(a)). Problem 4. (Earthmover distance.) Currently, there are n piles of dirt scattered over a large construction site. Pile number i weighs w i tons and is at position p i in the plane. The dirt in these piles has to be moved into k new piles at positions q j (for j = 1,..., k) with weights v j (also in tons). The good news: You have a bulldozer. The bad news: You have to figure out the cheapest way to move the dirt. For any non-negative real number t, the cost of moving t tons of dirt from position p to positions q is t dist(p, q) where dist(p, q) is the distance from p to q. Note that t need not be an integer. Your goal is to minimize the total cost of moving the dirt from its current locations to the new ones. The input consists of the points p 1,..., p n, q 1,..., q k R 2 and weights w 1,..., w n, v 1,..., v k N. You may ignore the cost of driving around the construction site when not carrying dirt; you only need to account for the cost of moving dirt from pile to pile. Also, because no nuclear reactions will be involved, you may assume that mass is conserved, that is, n i=1 w i = k j=1 v j. Note: In what follows, the different parts of the problem are independent: you do not need to solve one to solve the others. (a) Give a linear program that captures this problem. Variables: For every pair i, j, variable x ij indicates the weight of the dirt moved from position p i to q j. Constraints: For each pair i, j, we have x ij 0. For each i, we have k j=1 x ij w j (since we can use at most w j tons from location p i ). For each j, we have n i=1 x ij = v j (since we need v j tons at location q j ). Objective function to be minimized: i,j x ij dist(p i, q j ) (b) Prove or disprove the following conjecture: In an optimal solution, the paths taken by the bulldozer when moving dirt will never cross. [Hint: Think of an exchange argument.] We will show that in any solutions with two crossed paths, we can improve the cost of the solution by rerouting some dirt. Suppose there are two pairs (p i, q j ) and (p l, q m ) such that the corresponding line segments cross, and such that some positive amount of dirt is currently planned to be pushed from p i to q j and p l to q m. Let x ij be the amount of dirt that will be moved directly from p i to q j, and let u = min(x ij < x lm. 5

6 Consider rerouting a little bit of dirt by uncrossing the two paths. That is, we decrease x ij and x lm by u, and we increase x im and x lj by u. The new plan still respects all the constraints. The change in cost is u (dist(p i, q m ) + dist(p l, q j ) dist(p i, q j ) dist(p l, q m ). Here is the key fact: uncrossing two line segments reduces the sum of the lengths of the segments. This follows by applying the triangle inequality to the two triangles in the following picture: p p l i c a b qj a < b + c q m (c) Consider the following greedy algorithm: Repeat the following basic step until all the dirt has been moved: find the closest pair (p i, q j ) such that p i still has dirt and q j still needs dirt, and move a ton of dirt from p i to q j. Is this greedy algorithm correct? Justify your answer. No. Suppose that p 1 = ( 2, 0), p 2 = (0, 1), q 1 = (0, 1) and q 2 = (2, 0) and suppose that all weights are 1. The greedy algorithm will move 1 ton from p 2 to q 1 (distance: 2) and from p 1 to q 2 (distance: 4). Total cost = 6. However, dist(p 1, q 1 ) = dist(p 2, q 2 ) = 5. So uncrossing the bulldozer paths improves total cost to 2 5 < 6. (d) (Extra credit) Prove or disprove: if the weights of the initial and final piles are integers, there exists an optimal solution to the problem in which the bulldozer only moves integer amounts of dirt. [not yet written.] 6