CSE 101: Design and Analysis of Algorithms Homework 5 Winter 2016 Due: March 10, 2016


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1 CSE 101: Design and Analysis of Algorithms Homework 5 Winter 2016 Due: March 10, You are given a directed graph G = (V, E) with unit edge capacities (c e = 1, for all e E). You also given nodes s, t V and a positive integer k. The goal is to delete at most k edges in order reduce the maximum st flow in G by as much as possible. In other words, give an efficient algorithm to find a set of edges E E where E k such that the maximum st flow of G = (V, E \ E ) is minimized. Solution: Since all edges have unit capacity, removing k edges can reduce max flow by at most k. By the maxflow/mincut theorem, removing k edges from the min st cut will reduce the flow by k. Thus choosing any k edges from the min st cut (if there are fewer than k edges, remove them all) will be an optimal solution. procedure maxflowreduction(g, c, s, t, k) f = fordfulkerson(g, c, s, t) G f residual graph on G with respect to f for all v V : visited(v) = false explore(g f, s) for all (u, v) E: if E < k and visited(u) and not visited(v): add (u, v) to E return E Here f denotes the flow on each edge found by the maxflow algorithm. Knowing that, we can construct a residual graph G with respect to f and use the residual graph to find the min st cut. The running time is dominated by the time needed to find the max flow.
2 2. DPV 8.3: STINGY SAT is the following problem: given a set of clauses (each a disjunction of literals) and an integer k, find a satisfying assignment in which at most k variables are true, if such an assignment exists. Prove that STINGY SAT is NPcomplete. Solution: NPcompleteness is proved with 2 steps: (a) Stingy SAT is in NP. Given an arbitrary solution to Stingy SAT we can check whether the solution is a satisfying assignment by evaluating the formula. Furthermore, we can check that fewer than k literals are assigned the true value by inspecting each one once. (b) Reduce a known NPcomplete problem to Stingy SAT. To complete our NPcompleteness proof we reduce SAT to Stingy SAT. i. Transformation: Given a SAT formula f with n variables we simply chose (f, n) as an instance of Stingy SAT. ii. Proof: We have now to show that: f is a yesinstance of SAT if and only if (f, n) is a yesinstance of Stingy SAT: ( ) Suppose that f is a yesinstance of SAT. No more than n variables can be true, because there are a total of n variables. So any satisfying assignment of instance f for SAT will be a satisfying assignment of instance (f, n) for Stingy SAT. So (f, n) is a yesinstance of Stingy SAT. ( ) Suppose that (f, n) is a yesinstance of Stingy SAT. Any satisfying assignment of that instance will be also a satisfying assignment of instance f for SAT. So f is a yesinstance of SAT.
3 3. DPV 8.13: Determine which of the following problems are NPcomplete and which are solvable in polynomial time. In each problem you are given an undirected graph G = (V, E), along with: (a) A set of nodes L V, and you must find a spanning tree such that its set of leaves includes the set L. Solution: This can be solved in polynomial time. Algorithm. Delete all the vertices in the set L for the given graph and find a spanning tree of the remaining graph. Now, for each vertex v L, add it to any of its neighbor present in the tree. Proof of correctness. If the graph becomes unconnected after removing L, or some vertex in L has no neighbors in G \ L, then no spanning tree exists having all the vertices in L as leaves. It is clear that such a tree, if it is possible to construct one, must have all the vertices in L as leaves. Complexity. The algorithm is linear. Therefore the problem is P. (b) A set of nodes L V, and you must find a spanning tree such that its set of leaves is precisely the set L. Solution: i. (b) is in NP. (b) is a decision problem and you can quickly (polynomial time) test whether a solution is correct by going through the leaves of the tree. ii. Reduce a known NPcomplete problem to (b). (b) is a generalization of Rudrata (s, t)path. So we are going to reduce Rudrata (s, t) Path to (b). A. Transformation Given a graph G and 2 vertices s and t as an instance of Rudrata (s, t)path, we choose the same graph G and L = s, t as an instance of (b). B. Proof: We have now to show that: G, s and t is a yesinstance of Rudrata (s, t)path if and only if G and L = s, t is a yesinstance of (b): ( ) Suppose that G, s and t is a yesinstance of Rudrata (s, t)path. It means that there exists an undirected Rudrata path going from s to t. Such a path is a spanning tree of the G with only 2 leaves: s and t. Therefore G and L = s, t is a yesinstance of (b). ( ) Suppose G and L = s, t is a yesinstance of (b). It means that it exists a spanning tree with only 2 leaves: s and t. Such a tree is an undirected Rudrata path going from s to t. Therefore G, s and t is a yesinstance of Rudrata (s, t)path. (c) A set of nodes L V, and you must find a spanning tree such that its set of leaves is included in the set L.
4 Solution: i. (c) is in NP. Same as previous. ii. Reduce a known NPcomplete problem to (c). Like (b), (c) is also a generalization of Rudrata (s, t)path. We are going to reduce Rudrata (s, t)path to (c) using the same reduction as for (b). So we just have to prove the answer mapping. Proof: We have now to show that: G, s and t is a yesinstance of Rudrata (s, t)path if and only if G and L = s, t is a yesinstance of (c): ( ) Suppose that G, s and t is a yesinstance of Rudrata (s, t)path. It means that there exists an undirected Rudrata path going from s to t. Such a path is a spanning tree of the G with only 2 leaves: s and t. So the set of leaves is included in L. (because equal to L) Therefore G and L = s, t is a yesinstance of (c). ( ) Suppose G and L = s, t is a yesinstance of (b). It means that it exists a spanning tree with only 2 leaves: s and t. Such a tree is an undirected Rudrata path going from s to t. Therefore G, s and t is a yesinstance of Rudrata (s, t)path.
5 4. For a graph G = (V, E), a set of nodes S V is independent if no two nodes in S are joined by an edge. That is, u, v S, (u, v) / E. Consider the following problems. Independent Set Input: An undirected graph G = (V, E) and a nonnegative integer k. Decision Problem: Does G contain an independent set of size k? Bigger Independent Set Input: An undirected graph G = (V, E) and nodes S V where S is an independent set. Decision Problem: Does G contain an independent set larger than S? Show that the Bigger Independent Set decision problem is NPcomplete if the Independent Set decision problem is NPcomplete. Solution: BIS NP. An instance of Bigger Independent Set (BIS) consists of a graph G and a set S. A certificate for BIS is a candidate set S V. First, check that S > S by counting the nodes. This takes O( V ). Next check that S is an independent set, by iterating over E and ensuring that each edge contains at most one node in S. This takes O( E ). Thus, BIS can be verified in polynomial time and so BIS NP. BIS NPhard. To show this, we give a reduction from Independent Set (IS) to BIS. An instance of IS consists of a graph G and an integer k. We create an instance of BIS as follows. First make k 1 dummy nodes S = {v 1,..., v k 1 } and let V = V S. Next, create a set of edges E = E {(u, v) : u V, v S}. That is, E contains all the edges from the original graph plus an edge from every original node (u V ) to every dummy node (v S). Now, G has an independent set of size k iff G = (V, E ) has an independent set bigger than S. If G has an independent set S of size k then (1) k = S > S = k 1, (2) S V because we didn t delete any nodes, and (3) S is still an independent set in G because we didn t add any edges between nodes in V. Thus if G has an independent set S of size k then G has an independent set of size k, bigger than the size of S. Every dummy node in V has an edge to every node in V so if an independent set in G contains a single dummy node, it cannot contain any nodes in V. Thus, because there are only S dummy nodes, if there is a a bigger independent set S in G then S does not contain any dummy nodes meaning S V. Further, S is also an independent set in G since E E. Lastly, because removing nodes from an independent set still results in an independent set, any bigger independent set S in G can be made into an independent set of size k in G.
6 5. Suppose you wish to pack a list L of of n items each with some weight 0 < w i 1. You are packing the items into bins, each of which has as maximum weight capacity of 1. You are allowed to use any number of bins and the goal is to pack all the items using as few bins as possible. In the online version of this problem, your task is to pack items into bins one by one, as they arrive. You don t know what weights are coming later and you are not permitted to change the bin assignment of an object after it has been packed. For a particular assignment of items to bins B 1,..., B m, the sum of the w i values packed into a bin B j is called the content of B j, and is denoted by c(b j ). Note that 1 c(b j ) is the empty space that the bin B j has left. We call 1 c(b j ) the gap of B j. Given a list L = (w 1, w 2,..., w n ), define the NextFit heuristic algorithm as follows. Pack w 1, w 2,..., w i into B 1 until w i+1 > 1 c(b 1 ), then w i+1, w i+2,..., w j into B 2 until w j+1 > 1 c(b 2 ), etc. In other words, NextFit keeps packing items into the current bin until the next item doesn t fit at which point it moves on to the next bin. Once a bin has no space for the next item, that bin is never used again. Let NF (L) be the number of bins used by the NextFit algorithm, and let L be the number of bins needed by an optimal packing. Show that as L, the ratio NF (L)/L is upperbounded by 2. Solution: Suppose we are currently placing items into B i. Consider when we first add an item to bin B i+1. If w j is the last item added to bin B i then we move to the next bin when the weight of item w j+1 > 1 B i (the gap). Next, consider ordered pairs of bins: (B i, B i+1 ), etc. Since we only place items in the next bin once we have exceeded the gap of the previous bin, every pair of bins must follow c(b i ) + c(b i+1 ) 1. Let m be the number of bins used by the greedy algorithm. Suppose we temporarily ignore capacities of bins. Then, we could put the items from each pair of bins into a single super bin Bi s. We would then use ms = m/2 bins exactly for our solution but c(bi s ) 1 for all i. Due to the capacity constraint, the optimal solution must use at least as many bins than the super solution. Let m be the number of bins used by the optimal solution. In conclusion: m = 2 m s 2 m
7 6. DPV 9.7: In the MULTIWAY CUT problem, the input is an undirected graph G = (V, E) and a set of terminal nodes s 1, s 2,..., s k V. The goal is to find the minimum set of edges in E whose removal leaves all terminals in different components. (a) Show that this problem can be solved exactly in polynomial time when k = 2. (b) Give an approximation algorithm with ratio at most 2 for the case k = 3. Solution: (a) When k = 2, this problem becomes exactly the (s,t)minimum cut problem, which we know can be solved efficiently using a variety of maxflow algorithms. (b) Imagine trying to find the minimum cut that separates s 1, from the rest of the terminals. We could create a dummy vertex t and connect s 2, s 3 to t with edge weights that are infinite. Letting s = s 1, we could then find the minimum (s,t)cut in this graph, which would be the minimum cut to separate s 1 from the other terminals. Let E 1 be the set of edges in this cut, and let δ(e 1 ) be the value of the cut. We could repeat this procedure to find the minimum cut to separate s 2 from the other terminals, and s 3 from the others, giving us E 2 and E 3 (and δ(e 2 ), δ(e 3 )). Let OPT be the value of the best 3Way Cut. We know that OP T δ(e i ), i = 1, 2, 3, since we must at least separate one of the terminals from the rest, and δ(e i ) is the minimum way to do such. Let C be the union of the two E i that have the smallest cut values δ(e i ) of the three. Without loss of generality let the two smallest be E 1, E 2. We know δ(c) δ(e 1 )+δ(e 2 ) 2OP T, which shows that this is a 2approximation.
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