HOMEWORK III SOLUTIONS

Transcription

1 HOMEWORK III SOLUTIONS CSE 552 RANDOMIED AND APPROIMATION ALGORITHMS. Question. In a Steiner tree of a graph with metric distances, one can always bypass any Steiner vertex of degree 2. If you do this, every Steiner point must have degree at least three. Such bypassing would change the quality of the approximation. In the solutions given here, such bypassing is not done. Note that when an edge is not shown in the diagrams, its cost is that obtained by metric closure. a) Consider the following example: c a b Where the gray vertices are in the set R, and the white vertices are in set V R. The following three cases are possible: I) MST = (c, b): In this case the result for our algorithm discards vertex as it is a leaf on the MST and outputs path c as the answer. This case happens when c + b c + a, implying a b. Using the triangle inequality we also know that a + b c, therefore 2a > c, ensuring that edge a is not selected. In this case we get the optimal result. II) MST = (c, a): Again, the answer from the algorithm will be c and it is the optimal solution with a similar argument as in the previous case. III) MST = (a, b): In this case the result involves vertex and the algorithm returns a, b. This implies that a + b a + c, and a + b b + c, which implies

2 2 CSE 552 RANDOMIED AND APPROIMATION ALGORITHMS that a c, and b c. And therefore a + b 2c. The solution S produced by our algorithm satisfies the inequality c S 2c. In the worst case, the algorithm may always be forced to choose S > c for connecting two points in R. If OP T = x c, for some constant x R {v}, our approximate solution AP P > x c and we also know that AP P x 2c and therefore AP P OP T 2. When R = 3, the only case when we may not get the optimal solution is when the Steiner point v is not a leaf node in the MST of R {v} and has degree 2 connecting to two nodes of R. Consider the following example: c d W a b In this case the solution returned is {a, b, d}. Since {a, b} is chosen over {a, c} or {b, c} that implies that a c and b c. Also, a + b 2c. Since a + b > c, we know that OP T = c + d. The algorithm solution is AP P = a + b + d or AP P 2c + d. Which implies that AP P OP T 2. b) Consider the following example: x + ɛ ɛ W x ɛ x ɛ Here the length for OP T = x + 2 ɛ and the length of the solution returned by the algorithm is AP P = 2x + ɛ. The approximation ratio is 2 as we make ɛ smaller. This is a tight example for the bound when R = 3 but can be extended for arbitrary values for R if remains the connecting vertex for only two points in R and values for x and ɛ are substituted accordingly.

3 HOMEWORK III SOLUTIONS 3 c) The approximation ratio does not change. For any two vertices {v i, v k } V R in the generated MST of R {v i, v k } the adjacent vertices of v i and v k are disjoint, and {v i, v k } together can do no worse than when they individually remain within twice the optimal solution of the length required to cover the vertices of R that they are adjacent to. Since we take the minimum spanning tree of all R + 2 possible vertices that contain R, the resultant spanning tree, can be viewed as a MST of R {v i } union MST of R 2 {v k }, where R R 2 = R and R R 2 =. The MST of R {v i } is at most two times the optimal using the same analysis as in part a), the same holds for R 2 {v k }. Taken together they cannot do worse than twice the optimal for R. 2. Question 2. a) We form a MST on G. When l V, OP T = MST. When l = 2, OP T 2 MST. It reduces to the cheapest Hamiltonian path. We shortcut by connecting 2 leaves with the edge between them which has the minimum weight among all the shortcuts. Each time we add an edge in the increasing order until the graph has only l leaves. Every time we do the shortcutting, we delete at least leaf, so after at most S l steps, we are able to reach MST l, where S is the number of leaves in the graph. Consider the process from MST to 2MST, we can delete the leaves from S to 2, so each step the average cost increased is 2 MST MST, since l 2, we will stop at S l steps. The actual increased cost COST l = AV G l s l, where AV G l is the average cost we need to add at each step for transforming from MST to MST l and AV G is the average cost we need to add at each step for transforming from MST to 2MST, so AV G l AV G. Thus, Cost l Avg S l = The total cost is at most MST + S l most OP T + S l OP T. b) S n. When l = 3/ n, we have 2 MST MST S l = S l MST MST. Since OP T MST, our cost is at MST MST + S l l 2 MST = MST + ( + S 2 S 2 ) MST 3n MST + ( + 2 n 3 ) MST = n MST + ( n 3 ) MST = MST + ( n 3 ) MST < 5 MST 5 OP T

4 CSE 552 RANDOMIED AND APPROIMATION ALGORITHMS Thus, the approximation ratio is Question 3. Given n items, let the profit for each item be defined as p i, and the size of each item as s i, where i =..n and both p and s are non-negative. Sort the objects by ratio of profit to size. Let a i = p i s i, then we have: a a 2... a n. Let m = {a, a 2,..., a k } and n = {a k }, thus the algorithms selects max(profit(m), profit(n)). Suppose it is possible to pick a fraction of an item to store in the knapsack. Our selection of items will be {a, a 2,..., a k } with a selection vector of {,..., x}, where x is the selected size of the k th item, and x [0, ]. Let this selection be denoted as OP T F. In this case the knapsack is completely full. Back to our case, where we disallow fractional values, we know that, profit(m) profit(op T F ) profit(op T F ) = profit(m) + x p k profit(m) + x p k profit(m) + profit(n) 2 max(profit (m), profit(n)) Therefore this approach yields a profit that is at least half of the optimal profit. b) Let us consider the following cases: ) l = k: In this case the exact analysis as done in the previous section holds. 2) l > k: This case cannot happen because /2 K has to be exceeded first before K is exceeded. 3) l < k: Here we have two sub cases: I) profit(a,.., a k ) > profit(a k ): We know that a l {a,..., a k } and our algorithm makes a selection of max({a,..., a l }, {a l }), and the second selection fills the remaining space in the knapsack. In this case, we are sure that this approach does as good as the first approach. II) profit(a,.., a k ) < profit(a k ): In this case, we cannot guarantee that we will perform better than the previous algorithm. It is possible to have a final selection of {a,..., a k } which we know has a lesser profit than a k. Although we cannot guarantee that this algorithm will outperform the algorithm in part a, we can still provide the same general guarantees, similarly as we did in part a.

5 HOMEWORK III SOLUTIONS 5. Question. The weighted vertex cover problem is a special case of the set cover problem. From the analysis of the set cover problem in the textbook, the worst case performance of the greedy algorithm gives a solution that is could be a H n factor from the optimal solution. Consider the following example: n + ɛ n In this case, the greedy algorithm will output a vertex cover consisting of all the nodes except the center one with cost = H n n n when in fact, if you include only the center node you have a vertex cover with cost + ɛ. The ratio of the vertex cover found and the optimal vertex cover is Ω(log(n)) Consider the following algorithm: 5. Question 5. Given a directed graph G = (V, E), where V = {v, v 2,..., v n }. For every edge e E, where e = (v i, v j ): if i < j, E = E e, else E 2 = E 2 e. Output E if E > E 2, otherwise output E 2. Claim: The generated graph does not contain any cycles. Proof: Assume set E is generated and there exists a cycle v i v j v k v i. But that would imply that i < j < k < i which is a contradiction. Therefore, there are no cycles in E. Similarly we can prove the case when E 2 is generated. Since E = E E 2, then E E or E 2 2 E. Therefore this gives a factor 2 2 approximation for the acyclic subgraph problem.