On Completing Latin Squares

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-e Approx. for PLSE, 1-1/e-e
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-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
4
Previous and New Results

• The PLSE problem is NP-Complete (Colbourn 1984)
• The PLSE problem is APX-hard (This paper)
• 1-1/ee hardness for the k-PLSE problem. (This
  paper)

Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
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.

columns
a
b
c
d
a
a
rows
b
a
b
c
c
b
d
d
c
d
1
3
2
4
numbers
Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
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 s1 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-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
7
Local Search Analysis

• Let TT1,, Tm be the set of edge disjoint
  triangles of OPT and TT1,, Tn be the set
  of triangles found by the heuristic.
• Construct a bipartite graph H with
• vertex set T ? T.
• Connect Ti and Tj in H, iff Ti and Tj share an
  edge in G.

T1
T1
T2
T2
. . .
Tm
Tn
H
Optimal Triangles
Local Search Triangles
Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
8
Hurkens-Schrijver Theorem

• Let H be a bipartite graph with vertex set X ? Y
  Xn, Ym.
• 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-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
9
PLSE and H-S Theorem

• H satisfies the Hurkens-Schrijver conditions.
• deg (Tj) 3 for each Tj.
• Every subset of size t in T has a System of
  Distinct Representation in T (due to local
  search).
• Setting k3, we get the 2/3-e bound. For t7 we
  beat the previous result

T1
T1
T2
T2
. . .
Tm
Tn
H
Optimal Triangles
Local Search Triangles
Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
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-e Approx. for PLSE, 1-1/e-e
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-e Approx. for PLSE, 1-1/e-e
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 OPT

• Cells with number 1 in Greedy

Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
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.

OPT
Greedy
Red cells in OPT are mapped to Yellow cells in
Greedy
Introduction, 2/3-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
14
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-e Approx. for PLSE, 1-1/e-e
Approx. for k-PLSE, Conclusion
15
1-1/e-e approximation

• 1. Solve the LP program.
• 2. Multiply the variables by 1-e.
• 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-e)LPOPT
• With a constant probability, at most k numbers
  have been picked.

Introduction, 2/3 Approx. for PLSE, 1-1/e-e
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-e
Approx. for k-PLSE, Conclusion