On Solving Covering Problems
•Download as PPT, PDF•
2 likes•461 views
The set covering problem and the minimum cost assignment problem (respectively known as unate and binate covering problem) are well-known NP-complete problems.We investigate the complexity and approximation ratio of two lower bound computation algorithms from both a theoretical and practical point of view, and produce an algorithm that is 2-3 order of magnitude faster.
Recommended
More Related Content
What's hot
What's hot (19)
Similar to On Solving Covering Problems
Similar to On Solving Covering Problems (20)
More from Olivier Coudert
Recently uploaded
Recently uploaded (20)
On Solving Covering Problems
- 1. On Solving Covering Problems 33rd DAC, June 5, 1996 Olivier Coudert Synopsys, Inc.
- 3. Binate Covering Problem Input: Output: A Boolean function f(y1, ..., yn) A cost function Cost on the assignment of the yk’s A minimum cost n-tuple satisfying f
- 4. Binate Covering Problem f = y2y3 + (y1 y4)⊕ Cost(yk = 0) = 0 Cost(yk = 1) = 1 y1 + y3 + y4 y2 + y3 + y4 y2 + y3 + y4 y1 + y2 + y4 y3y1 y2 y4 1 1 0 1 0 0 1 1 1 0 1 0 Minimum solutions: y1 = 1, y2 = y3 = y4 = 0 y1 = y2 = y3 = 0, y4 = 1
- 5. Binate Covering Problem 2-level Boolean function and relation minimization State minimization DAG covering Exact encoding 3-level NAND minimization ....
- 7. Branch-and-bound Resolution y1 y2 y3 y4 0 1 1 0 1 1 0 1 x1 x2 x3 y1 0 1 y3 y4 1 0x1 x2 y2=0 y3 1x2 y1=0 y4 0x1 y1=1 y1 x3 1 y3 y4 1 1 1 x2 y2=1 y1 x3 1 y3 1 1 x2 y4=0 y1 1 y3 1x2 y4=1
- 8. Branch-and-bound Resolution Pruning technique Lower bound computation
- 10. Independent Set Based Lower Bound x4x1 x5 x2 x3 y1 y2 y3 y4 x1 x2 x3 x4 x5 1 1 1 0 0 1 10 0 1 1 1 0 1 Cost(yk = 0) = 2 Cost(yk = 1) = 3
- 11. Optimum Independent Set x1 x2 x3 x4 x5 x6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x1 x2 x3 x4 x5 x6 Solution is n/2, while the lower bound is only 1
- 12. Optimum Ind. Set Based Lower Bound The optimum independent set based lower bound is NP-complete … and can be arbitrary far from the minimum covering solution
- 13. Greedy Independent Set Select the minimum degree vertex
- 14. Greedy Independent Set 2n+1 vertices 1 n n Independent set (n) (n+1) (2n-1) Greedy: 2 Maximum: n
- 15. Greedy Ind. Set Based Lower Bound The greedy independent set based lower bound can be arbitrary far from the optimum independent set based lower bound ..., which itself can be arbitrary far from the minimum covering solution
- 16. 1 x7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x1 x2 x3 x4 x5 x6 1 1 1 1 1 1 1 11 x7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x1 x2 x3 x4 x5 x6 1 1 1 1 1 1 1 1 Greedy Log-approximation of Covering Optimal≤ log(|X|) ∗≥Optimal Greedy log(|X|)
- 17. Which Lower Bound is Better? Greedy Independent set based Greedy solution based arbitrary far from the optimal solution within log(|X|) of the optimal solution can take advantage of limit lower bound
- 30. further prune Limit Lower Bound Takes advantage of an independent set based lower bound to the search spacefurther prune Can reduce the search space by 1000
- 31. Which Lower Bound is Better? Greedy Independent set based Greedy solution based arbitrary far from the optimal solution within log(|X|) of the optimal solution can take advantage of limit lower bound
Editor's Notes
- - What’s the problem? Why is it useful? - An approach: Branch-and-bound resolution - Depend on the quality of a lower bound
- - A function - A cost function - Decompose the function as a product of sum - Put in regards the variables of the function - Build the covering matrix - Solve the covering matrix
- - Exact method are important to evaluate the performances of polynomial heuristics algorithms
- - Choose a branching column y - Assume that y=0 or y=1, and consequently simplify the matrix - Exponential search space
- - Prevent from exploring unsuccessful branches that do not yield a better solution - Relies on lower bound computation
- - From the matrix, one build a graph such that two rows are connected by an edge iff there is an assignment of a variable that covers both of these rows. - An independent set of this graph produces a lower bound of the optimal solution
- -How far the optimum independent set based lower bound can be from the cost of the optimal solution? - Well, as far as one wants… - If the rows are connected to all (clique) or most (dense graph) of the other, the independent set is small, and thus the lower bound is of poor quality.
- - Not only the maximum independent set is NP-complete, but the lower bound it produces can be arbitrary far from the optimal solution.
- - Need a polynomial independent set algorithm - Greedy: select a node according to some criterion, put it in the independent set, remove all edges and nodes connected to the chosen node, and iterate. - Simple criterion: take the node of minimal degree - Here, it yields the optimal independent set - But in general, what can we expect from such a greedy algorithm?
- - Here, the greedy produces an independent set that is arbitrary far from the maximum independent set - One can change the criterion selection, but likely, if the criterion is polynomial, one can build a counter expample that still produces an arbitrary large gap between the greedy and the maximum independet sets.
- - General status of the Greedy independent set based lower bound: there are 2 arbitrary large gap between the computed lower bound and the cost of the optimal solution.
- - Consider a unate covering matrix (I.e., a set covering problem) - It is log-approximable by a greedy algorithm - Indeed, the log-ratio is the best possible one (except if P = NP) - Only the multiplicative constant may be improved - The upper bound on the solution yields a lower bound as well…. -…and this lower bound is no further than log(|X|) from the cost of the optimal solution.
- - So the greedy solution based lower bound looks much better than the (greedy or maximum) independent set lower bound. - But the independent set lower bound can take advantage of the limit lower bound to further prune the search.
- - This set covering problem has several optimum solution of cost 6.
- - This set covering problem has several optimum solution of cost 6.
- - This set covering problem has several optimum solution of cost 6.
- - This set covering problem has several optimum solution of cost 6.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - Let’s build an independent set. It consists in selecting an element, putting it in the independent set, and deleting the sets covering this element as well as the elements covered by these sets. - At the end, we got an independent set of cost 5. - Assuming that we already found a solution 6 (which is optimal), it’s impossible to prune the recursion since the lower bound is only 5.
- - But the sets that do not contain any elements selected for the independent set cannot be used to build a better solution. So we can delete them...
- -And if we do so, we create essential sets, I.e., sets that must be taken as belonging to the minimum solution since they are the only one to cover some elements. - Consequenly simplifying the matrix yield a unique solution, whose cost is 6 - So it is impossible to improve upon the solution, and the search is complete
- - Thus the limit lower bound uses an independent set based lower bound that is close enough to the current best solution to further prune the search spaceee - Indeed, in examples coming from logic minimizatiuon, the reduction of the search space range from 50 to 1000.
- - Thus the independent set based lower bound is, from the practical point of view, more effective than the theoritically better greedy solution based lower bound. - But for high density covering matrix (e.g., some examples coming from operating research), the second can be more effective than the first one.