1 On Completing Latin Squares Iman Hajirasouliha Joint work with Hossein Jowhari, Ravi Kumar, and Ravi Sundaram

2 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. 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

3 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 error- correcting codes and recently optical networks. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

4 Previous and New Results ProblemApprox. factor Authors, YearTechnique PLSE1/2Kumar, Russell, Sundaram, 1995 Combinatorial ideas PLSE1-1/eGomes, Regis, Shmoys, 2003 LP and Randomized Rounding PLSE2/3-εThis paperLocal Search k-PLSE1-1/e-εThis paperLP and Randomized Rounding 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) Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

5 A problem equivalent to PLSE  The 3EDM Problem: finding the number of maximum edge disjoint triangles in a tripartite simple graph rows columns numbers a d b 1 c d c b a Introduction, 2/3 -ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion abc d a b c d

6 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

7 Local Search Analysis Let T={T 1,…, T m } 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 T i and T’ j in H, iff T i and T’ j share an edge in G. T1T1 T2T TmTm T’ 1 T’ 2 T’ n Optimal TrianglesLocal Search Triangles H Introduction, 2/3 - ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

8 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

9 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). Setting k=3, we get the 2/3-ε bound. For t=7 we beat the previous result: T1T1 T2T TmTm T’ 1 T’ 2 T’ n Optimal Triangles Local Search Triangles H Introduction, 2/3 - ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

10 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

11 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

12 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 OPTCells with number 1 in Greedy Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

13 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 Red cells in OPT are mapped to Yellow cells in Greedy OPT Greedy Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

14 LP relaxation of the problem A way to extend the PLS with a number represents a matching. M c is the set of all matchings that extends the PLS with number c. y cM is 1 when Matching M is chosen. Introduction, 2/3-ε Approx. for PLSE, 1-1/e-ε Approx. for k-PLSE, Conclusion

15 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)LP OPT 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

16 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