scholarly journals An Efficient Local Search for Partial Latin Square Extension Problem

2015 ◽  
pp. 182-198 ◽  
Author(s):  
Kazuya Haraguchi
Keyword(s):  
Local Search ◽  
Latin Square ◽  
Extension Problem ◽  
Partial Latin Square

An improved approximation algorithm for the partial Latin square extension problem

2004 ◽  
Vol 32 (5) ◽  
pp. 479-484 ◽  
Author(s):  
Carla P. Gomes ◽  
Rommel G. Regis ◽  
David B. Shmoys
Keyword(s):  
Approximation Algorithm ◽  
Latin Square ◽  
Extension Problem ◽  
Partial Latin Square

scholarly journals When is a partial Latin square uniquely completable, but not its completable product?

2008 ◽  
Vol 308 (13) ◽  
pp. 2830-2843
Author(s):  
Nicholas Cavenagh ◽  
Diane Donovan ◽  
Abdollah Khodkar ◽  
John Van Rees
Keyword(s):  
Latin Square ◽  
Partial Latin Square

scholarly journals The number of solutions to an equation arising from a problem on latin squares

1983 ◽  
Vol 34 (1) ◽  
pp. 138-142 ◽  
Author(s):  
I. P. Goulden ◽  
S. A. Vanstone
Keyword(s):  
Abelian Group ◽  
Recent Article ◽  
Latin Square ◽  
Latin Squares ◽  
Number Of Solutions ◽  
Graph Theoretic ◽  
Partial Latin Square

AbstractA recent article of G. Chang shows that an n × n partial latin square with prescribed diagonal can always be embedded in an n × n latin square except in one obvious case where it cannot be done. Chang's proof is to show that the symbols of the partial latin square can be assigned the elements of the additive abelian group Zn so that the diagonal elements of the square sum to zero. A theorem of M. Halls then shows this to be embeddable in the operation table of the group. In this paper, we show that when n is a prime one can determine exactly the number of distinct ways in which this assignment can be made. The proof uses some graph theoretic techniques.

scholarly journals Completing Partial Latin Squares with One Nonempty Row, Column, and Symbol

10.37236/5675 ◽  
2016 ◽  
Vol 23 (2) ◽  
Author(s):  
Jaromy Kuhl ◽  
Michael W. Schroeder
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Partial Latin Square ◽  
Partial Latin Squares

Let $r,c,s\in\{1,2,\ldots,n\}$ and let $P$ be a partial latin square of order $n$ in which each nonempty cell lies in row $r$, column $c$, or contains symbol $s$. We show that if $n\notin\{3,4,5\}$ and row $r$, column $c$, and symbol $s$ can be completed in $P$, then a completion of $P$ exists. As a consequence, this proves a conjecture made by Casselgren and Häggkvist. Furthermore, we show exactly when row $r$, column $c$, and symbol $s$ can be completed.

scholarly journals Restricted completion of sparse partial Latin squares

2019 ◽  
Vol 28 (5) ◽  
pp. 675-695 ◽  
Author(s):  
Lina J. Andrén ◽  
Carl Johan Casselgren ◽  
Klas Markström
Keyword(s):  
Positive Integer ◽  
Latin Square ◽  
Latin Squares ◽  
Empty Cell ◽  
Partial Latin Square ◽  
Partial Latin Squares

AbstractAnn×npartial Latin squarePis calledα-dense if each row and column has at mostαnnon-empty cells and each symbol occurs at mostαntimes inP. Ann×narrayAwhere each cell contains a subset of {1,…,n} is a (βn,βn, βn)-array if each symbol occurs at mostβntimes in each row and column and each cell contains a set of size at mostβn. Combining the notions of completing partial Latin squares and avoiding arrays, we prove that there are constantsα,β> 0 such that, for every positive integern, ifPis anα-densen×npartial Latin square,Ais ann×n (βn, βn, βn)-array, and no cell ofPcontains a symbol that appears in the corresponding cell ofA, then there is a completion ofPthat avoidsA; that is, there is a Latin squareLthat agrees withPon every non-empty cell ofP, and, for eachi,jsatisfying 1 ≤i,j≤n, the symbol in position (i,j) inLdoes not appear in the corresponding cell ofA.

scholarly journals A Generalisation of Transversals for Latin Squares

10.37236/1629 ◽  
2002 ◽  
Vol 9 (1) ◽  
Author(s):  
Ian M. Wanless
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Soluble Groups ◽  
Partial Latin Square ◽  
Cayley Tables

We define a $k$-plex to be a partial latin square of order $n$ containing $kn$ entries such that exactly $k$ entries lie in each row and column and each of $n$ symbols occurs exactly $k$ times. A transversal of a latin square corresponds to the case $k=1$. For $k>n/4$ we prove that not all $k$-plexes are completable to latin squares. Certain latin squares, including the Cayley tables of many groups, are shown to contain no $(2c+1)$-plex for any integer $c$. However, Cayley tables of soluble groups have a $2c$-plex for each possible $c$. We conjecture that this is true for all latin squares and confirm this for orders $n\leq8$. Finally, we demonstrate the existence of indivisible $k$-plexes, meaning that they contain no $c$-plex for $1\leq c < k$.

Comment on a Note by J. Marica and J. Schönheim

1970 ◽  
Vol 13 (4) ◽  
pp. 539-539 ◽  
Author(s):  
Charles C. Lindner
Keyword(s):  
Latin Square ◽  
Main Diagonal ◽  
Partial Latin Square

In [2] it is shown that an n × n partial latin square with n — 1 cells occupied on the main diagonal can be completed to a latin square. We can use the technique in [2] to prove the following result.An n × n partial latin square with n — 1 cells occupied with n — 1 distinct symbols can be completed to a latin square if the occupied cells are in different rows or different columns.Let P be an n × n partial latin square based on 0,1, 2, …, n — 1 satisfying the above conditions, and let (x0, y0), (x1, y1), …, (xn-2, yn-2) De the occupied cells where y0, y1, … yn-2 are distinct.

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