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On Completing Latin Squares Iman Hajirasouliha Joint work with Hossein Jowhari, Ravi Kumar, and Ravi Sundaram 1 Definitions What is a Latin Square and a Partial Latin Square (PLS)? The PLSE problem: Given a PLS, fill the maximum number of empty cells using numbers in [n] without violating the constraints. 1 2 4 2 3 1 4 4 2 3 1 3 2 4 The k-PLSE problem: How many empty cells of a PLS can be filled properly using at most k ≤ n different numbers? Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 2 Motivations and Applications Interesting object for mathematicians, Evans conjecture(1960) says that a PLS with n-1 filled cells can be completed. (Proved by Smetaniuk in 1981) Sudoku puzzles, one of the current fads, are PLSs with additional properties. The problem has application in errorcorrecting codes and recently optical networks. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 3 Previous and New Results The PLSE problem is NP-Complete (Colbourn 1984) The PLSE problem is APX-hard (This paper) 1-1/e+ε hardness for the k-PLSE problem. (This paper) Problem Approx. factor Authors, Year Technique PLSE 1/2 Kumar, Russell, Sundaram, 1995 Combinatorial ideas PLSE 1-1/e Gomes, Regis, Shmoys, LP and Randomized 2003 Rounding PLSE 2/3-ε This paper Local Search k-PLSE 1-1/e-ε This paper LP and Randomized Rounding Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 4 A problem equivalent to PLSE The 3EDM Problem: finding the number of maximum edge disjoint triangles in a tripartite simple graph. columns a a b 1 2 b 2 3 c d 4 1 c d rows 4 1 4 2 3 3 a a b b c c d d 1 numbers 2 3 4 Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 5 Local Search Algorithm for 3EDM Let G be an instance of 3EDM Fix a constant t ≥ 7. Start from any arbitrary valid solution. If possible, replace s ≤ t triangles in the current solution with s+1 edge-disjoint triangles to get another valid solution. Since the size of solution increases in each step by one, the algorithm runs in polynomial time. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 6 Local Search Analysis Let T={T1,…, Tm} be the set of edge disjoint triangles of OPT and T’={T’1,…, T’n} be the set of triangles found by the heuristic. Construct a bipartite graph H with vertex set T T’. Connect Ti and T’j in H, iff Ti and T’j share an edge in G. T1 T’1 T2 T’2 . . . Tm T’n H Optimal TrianglesLocal Search Triangles Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 7 Hurkens-Schrijver Theorem: Let H be a bipartite graph with vertex set X Y; |X|=n, |Y|=m. Let k ≥3 and assume: For each y in Y, deg (y) ≤ k. Every subset of size ≤ t of X has a system of distinct representatives in Y. Then: Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 8 PLSE and H-S Theorem H satisfies the Hurkens-Schrijver conditions. deg (T’j) ≤ 3 for each T’j. Every subset of size t in T has a System of Distinct Representation in T’ (due to local search). T1 T’1 T2 T’2 Setting k=3, we get the 2/3-ε bound. Tm . . . For t=7 we beat the previous result: m k (k 1) r k 3*16 3 1 r n 2(k 1) k 2*16 3 1 1 e T’n H Optimal Triangles Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion Local Search Triangles 9 The k-PLSE problem How many cells of a PLS can be filled using at most k ≤ n different numbers? A natural greedy algorithm: Repeat k times: Pick the number c which can fill the most cells. Fill those cells with c. The greedy algorithm is a ½ - approximation algorithm. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 10 Greedy algorithm analysis OPT solution and greedy solution are sets of triples {(i, j, k)}. To each triple y in OPT, we assign a triple x in greedy solution as accountable. Given y=(i, j, k) in OPT, we have three cases: 1) cell x=(i, j, t) is in Greedy. x is accountable for y. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 11 2) (i, j) is empty in Greedy but k has been used in Greedy. We can assign a distinct x=(i’, j’, k) in Greedy to y. Consider the iteration where Greedy chooses k. Cells with number 1 in OPT Cells with number 1 in Greedy 1 1 1 1 1 1 1 Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 12 3) cell (i, j) is empty in Greedy and number k is missing in Greedy. For each number c in OPT we can assign a number c’ in Greedy which is missing in OPT. OPT 1 2 4 1 2 Greedy 4 4 4 Red cells in OPT are mapped to Yellow cells in Greedy Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 13 LP relaxation of the problem A way to extend the PLS with a number represents a matching. Mc is the set of all matchings that extends the PLS with number c. ycM is 1 when Matching M is chosen. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 14 1-1/e-ε approximation 1. Solve the LP program. 2. Multiply the variables by 1-ε. 3. For each number pick a matching randomly according to the probability associated with the matchings. 4. If matchings intersect in a cell, choose one of them arbitrarily for the cell. Expectation of the size of solution obtained is bigger than (1-1/e-ε)LPOPT With a constant probability, at most k numbers have been picked. Introduction, 2/3 Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 15 Conclusion We defined a new and natural variation of the PLSE problem and obtained simple approximation algorithms for the PLSE and k-PLSE problems. Our results for the PLSE problem is an improvement and for the k-PLSE problem is the best possible. Introduction, 2/3 Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion 16