6.854 Advanced Algorithms

Transcription

1 6.854 Advanced Algorithms Homework 5 Solutions 1 10 pts Define the following sets: P = positions on the results page C = clicked old results U = unclicked old results N = new results and let π : (C U) P map old results to their positions in the old list. Also let v i (N, j) be the value of placing the j th new result in position i and let v i (O, j) be the value of placing the j th old result in position i Then define the network G = (V, E) with edge capacities and costs as follows: V = {s, t, x, y, z} C U N P E = {(s, x), (s, y), (s, z), (z, y)} {x} C {y} N {z} U (C N U) P P {t} C (i, j) = (s, x) k (i, j) = (s, y) u(i, j) = n C k (i, j) = (s, z) or (z, y) 1 otherwise v i (O, j) p π(i),j (i, j) (C U) P c(i, j) = v i (N, j) (i, j) N P 0 otherwise. An integer min-cost max-flow of this graph corresponds to a solution to our problem. Grading Note: Don t be picky if they weren t very precise on their notation for v i since the original question was worded a bit poorly. 1

2 2 10 pts Assume that we are given graph G = (V, E) with source s and sink t. We will solve both parts by modifying G slightly and solving a min cost maximum flow on the modified graph. Specifically, start by computing the max-flow of G. Let M be the value of the maxflow. Construct a graph G on which a min-cost max-flow algorithm produces the required minimum-cost flow. Graph G = (V, E ) is constructed from G such that V = V {s } and E = E {(s, s), (s, t)}. The source of G is s and the sink is still t. The edge (s, s) has capacity M and cost 0, and forces the min-cost, max-flow algorithm to send M units of flow through the remaining graph. (s, t) serves as a short cut from s to t that bi-passes G. (a) 5 pts The edge (s, t) has capacity M/10 and cost 0 and therefore allows at most 10% of the flow to bypass the original graph. Recall that s is the source of G and allows at most M units of flow to s. The max-flow value of G is therefore M, but the flow in its subgraph G can be in the range [9M/10, M]. We can thus establish a one-to-one mapping from a feasible 90% flow in G to a flow in G with same cost. So we compute the min-cost max-flow of G and look for the required minimum cost flow in its subgraph G. (b) 5 pts In this case, the edge (s, t) has capacity M and cost K. Again, we have a one-to-one mapping from a feasible flow in G to a flow in G with same cost. We compute the min-cost flow of G and extract the required flow from subgraph G. Our algorithms on require a single call to a max flow algorithm on G and a single call to min cost max flow algorithm on G, which will run asymptotically as fast as a min cost max flow algorithm on G. For (b) there s the minor caveat that we might incur a logarithmic cost on K if we apply the weakly polynomial time analysis of Goldberg-Tarjan to G. However, it s easy to check that, if K > nc, then we won t use the auxiliary edge at all in a min cost max flow. Accordingly, the number of iterations in Goldberg-Tarjan is at most O (mn log((nc) 2 )) = O (mn log(nc)). Grading Note: Don t take off points if they don t mention this issue. It s enough to just assert that min cost max flow runs quickly on G (especially considering that Goldberg-Tarjan can be shown to actually run in strongly polynomial time). 2

3 3 10 pts (a) 2 pts e C e C 1/u(e) = = H(C) e C H (C) e C 1/u(e) H (C) e C 1 + H (C) e C,e bottleneck edge 1/u(e) H (C) e C The last step follows from noting that H (C) e C is negative and that e C,e bottleneck edge 1/u(e) is positive. (b) 5 pts We can obtain f by adding negative cost cycles C 1, C 2,... C l to f with (positive) weights w 1, w 2,..., w l. Then we can write: Cost(f ) Cost(f) = w w l. e C 1 e C l Additionally, w 1 1/u(e) w l 1/u(e) u(e)/u(e) = m e C 1 e C l e since the total weight of the cycles that an edge is included in cannot exceed that edges capacity. We combine both equations to see that, Cost(f ) Cost(f) w 1 e C w l e C l m w 1 e C 1 1/u(e) w l e C l 1/u(e) Cost(f) Cost(f ) m w 1 e C w l e C l w 1 e C 1 1/u(e) w l e C l 1/u(e) w i e C i w i e C i 1/u(e) for some i in 1,..., l. This follows from the fact that for positive n 1,..., n l and positive d 1,..., d l, there is always some i such that n i d i n 1+n n l d 1 +d d l. Grading Note: They can also use directly that, when the n 1,..., n l s are negative, there is always some i such that n i d i n 1+n n l d 1 +d d l. Both of these facts were posted as hints on Piazza so don t need to be proven. We conclude that e C i e C i 1/u(e) Cost(f ) Cost(f) m for C i. The same must be true for our chosen cycle since it is optimal by definition. Combining with the inequality from part (a) gives the result. 3

4 (c) 3 pts We want to keep canceling cycles until Cost(f) Cost(f ) = 1. After that, we ll finish on the next cycle cancel. By part (b), each time we cancel a cycle, Cost(f) Cost(f ) reduces to ( 1 m) 1 (Cost(f) Cost(f )), so after q cycle cancels, Cost(f) Cost(f ) has reduced by a factor of ( q. 1 m) 1 Let U be the maximum capacity in in our graph and let Z be the maximum cost. Setting q = O(m log(uzm)) ensures that ( 1 m) 1 q 1. UZm Since Cost(f) Cost(f ) is at most UZm to begin with, it follows that after q iterations Cost(f) Cost(f ) < 1, as required. 4

5 4 10 pts (a) 5 pts Let us write R = P Q as a set of inequalities: ( A R = {x : D ) x ( ) b } e If R is empty, by Farkas s lemma we have that there exists a w 0 such that w ( T A D) = 0 but w ( ) ( T b e < 0. Write w = y ) z with y having as many entries as the number of columns in A. Then the statement follows because w ( ) T A D = y T A + z T D and w ( ) T b e = y T b + z T e. (b) 5 pts Let c T = y T A. Then, for any vector x in P : since Ax b and z is non-negative. For any vector w in Q: c T x = y T Ax y T b c T w = y T Aw = z T Dw z T e since Dw e and y is non-negative. From part (a) we know that y T b < z T e, so piecing everything together: c T x y T b < z T e c T w. 5

6 5 10 pts (a) 5 pts Let o be the minimum mean cost of a cycle in H. We first claim that o min(minmeancost). Suppose o is the cost of a cycle C. Set x(u, v) = 1 for every (u, v) C and 0 otherwise. C The constraints of the LP are clearly satisfied. Specifically, circles conserve flow (constraint one) and x(u, v) = C 1 = 1. Furthermore, the objective value of the solution based on C C is: c(u, v)x(u, v) = c(u, v) 1 C = o u,v (u,v) C So, for every cycle, there is a solution to the LP with objective value equal to the cycle s mean cost. Accordingly, the optimal solution to the LP can only be less than or equal to the mean cost of any cycle i.e. o. Next we claim that min(minmeancost) o. Specifically, given any solution to our LP with cost t we will show how to exhibit a cycle C with minimum cost t. For a given solution to the LP, consider the circulation Cir with flow x(u, v) on edge (u, v). We know that Cir is a valid circulation because the first constraint of the linear program enforces flow conservation at every node v. Decompose the circulation into cycles, C 1,..., C l with corresponding flows w 1,..., w l. x(u, v) = i:(u,v) C i w i. Each w i here is a positive number. Accordingly, x(u, v) = u,v l (u,v) C i w i = l C i w i By our constraints, u,v x(u, v) = 1 so it must be that l C i w i is equal 1. Additionally, we can write the objective value of our solution as: l c(u, v)w i = (u,v) C i = l w i c(u, v) (u,v) C i l C i w i c(u, v) 1 C i. (u,v) C i The final form is simply a weighted average of the mean costs for C 1,..., C l since l C i w i = 1. If this average has value t, it must be that one of the cycles has mean cost t, which gives the result. (b) 3 pts Grading note: There are a number of ways to find the dual of MinMeanCost. I ll give one approach, but try to be fairly loose on the grading here if they get the correct answer 6

7 but don t show much work. There are readily available rule books for converting to dual form and it s fine if something like that was used black-box. I m showing way more work here than necessary. Let s start by writing the LP in matrix form. Specifically, let c be a vector containing each c(u, v), x be a vector containing each x(u, v), and b be a vector with its first V positions set to 0 and its last position set to 1. A is a matrix with width E it has a column for every edge in H. For i V, the i th row of A encodes the flow conservation constraint for node v. Specifically, it contains a 1 in the position corresponding to each edge (u, v) into v and a 1 in the position for each edge (v, u) out of v. The last row of A encodes the u,v x(u, v) = 1 constraint and so is simply the all ones vector. So, our program can be written as: min c T x such that: Ax = b x 0 Grading note: Notice that this is a minimization problem so this isn t exactly the version of standard form we saw in class. However, solving this program is equivalent to solving max c T x such that: Ax = b x 0 although our final optimum value with be negated. So we can just apply the usual dual formulation (given in class): which is equivalent to solving Final Dual: This is in turn equivalent to: min y T b such that: y T A c T min y T b such that: y T A c T. max y T b such that: y T A c T which is the dual formulation I expect a lot of students to get, especially if they used rule books. Now, note that y T b is simply equal to the final value of y because only the last entry of b is nonzero. Lets call this value s. 7

8 Consider the column of A corresponding to edge (u, v). In this column, the u th entry is equal to 1, the v th entry is equal to 1 and the final entry is equal to 1. Everything else is 0. So, to ensure that y T A c T for this column specifically, it must be that: y u y v + s c(u, v) c(u, v) + y v y u s c(u, v) ( y v ) + ( y u ) s where y u and y v denote the u th and v th entries of y. If we let y v and y u correspond to p(v) and p(u) from our formulation of Goldberg- Tarjan, the dual linear program is asking us to find the maximum value of y T b = s such that, for every edge, c(u, v) p(v) + p(u) s. By definition ɛ is exactly the solution to this optimization problem. (c) 2 pts This just follows from strong duality. From part (a) we know that the primal is feasible whenever there s a remaining negative cost cycle in H. Primal feasibility is all that s needed for strong duality to hold. 8