scholarly journals The problem of Euler/Tarry revisited

2020 ◽  
Author(s):  
Dieter Betten
Keyword(s):  
Pairwise Disjoint ◽  
Latin Square ◽  
Latin Squares ◽  
Main Diagonal ◽  
Euler Problem

AbstractThe problem of Euler/Tarry concerning 36 officers can be formulated in mathematical terms: Can a latin square of order 6 have an orthogonal square, or equivalently, are there 6 pairwise disjoint transversals? This was first answered (in the negative) by Tarry (1900/01). We prove the following Theorem: If a latin square of order 6 admits a reflection, i. e. an automorphism of order two which fixes the main diagonal elementwise, then it has no orthogonal square. We list the 12 isomorphism types of latin squares of order 6 and see: they all admit such a reflection. So we get a solution of the Euler problem without the tedious task of tracing the transversals.

scholarly journals Critical sets in Latin squares given that they are symmetric

2007 ◽  
pp. 38-45
Author(s):  
Ali Mojdeh ◽  
Jafari Rad
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Main Diagonal ◽  
Critical Set ◽  
Critical Sets

A uniquely completable (UC) set U is a subset of a Latin square L such that L is the only superset of U which is a Latin square. A critical set C of L is a subset of L such that C is uniquely completable and no subset of C has this property. We show that there is a symmetric Latin square with fixed main diagonal entries for each even number, and obtain a uniquely completable partial symmetric Latin square of order 2n for each n and prove that, it is critical set for n = 3, 4, 5 and 6, and make a problem.

scholarly journals On the Structure and Classification of SOMAs: Generalizations of Mutually Orthogonal Latin Squares

10.37236/1464 ◽  
1999 ◽  
Vol 6 (1) ◽  
Author(s):  
Leonard H. Soicher
Keyword(s):  
Graph Theory ◽  
Group Theory ◽  
Pairwise Disjoint ◽  
Latin Square ◽  
Latin Squares ◽  
Orthogonal Latin Squares ◽  
Mutually Orthogonal Latin Squares ◽  
Computational Group Theory

Let $k\ge0$ and $n\ge2$ be integers. A SOMA, or more specifically a SOMA$(k,n)$, is an $n\times n$ array $A$, whose entries are $k$-subsets of a $kn$-set $\Omega$, such that each element of $\Omega$ occurs exactly once in each row and exactly once in each column of $A$, and no 2-subset of $\Omega$ is contained in more than one entry of $A$. A SOMA$(k,n)$ can be constructed by superposing $k$ mutually orthogonal Latin squares of order $n$ with pairwise disjoint symbol-sets, and so a SOMA$(k,n)$ can be seen as a generalization of $k$ mutually orthogonal Latin squares of order $n$. In this paper we first study the structure of SOMAs, concentrating on how SOMAs can decompose. We then report on the use of computational group theory and graph theory in the discovery and classification of SOMAs. In particular, we discover and classify SOMA$(3,10)$s with certain properties, and discover two SOMA$(4,14)$s (SOMAs with these parameters were previously unknown to exist). Some of the newly discovered SOMA$(3,10)$s come from superposing a Latin square of order 10 on a SOMA$(2,10)$.

scholarly journals A Discussion of a Cryptographical Scheme Based in F-Critical Sets of a Latin Square

Mathematics ◽  
2021 ◽  
Vol 9 (3) ◽  
pp. 285
Author(s):  
Laura M. Johnson ◽  
Stephanie Perkins
Keyword(s):  
Necessary Conditions ◽  
Latin Square ◽  
Latin Squares ◽  
Critical Sets ◽  
Long Cycles ◽  
The Given

This communication provides a discussion of a scheme originally proposed by Falcón in a paper entitled “Latin squares associated to principal autotopisms of long cycles. Applications in cryptography”. Falcón outlines the protocol for a cryptographical scheme that uses the F-critical sets associated with a particular Latin square to generate access levels for participants of the scheme. Accompanying the scheme is an example, which applies the protocol to a particular Latin square of order six. Exploration of the example itself, revealed some interesting observations about both the structure of the Latin square itself and the autotopisms associated with the Latin square. These observations give rise to necessary conditions for the generation of the F-critical sets associated with certain autotopisms of the given Latin square. The communication culminates with a table which outlines the various access levels for the given Latin square in accordance with the scheme detailed by Falcón.

scholarly journals Atomic Latin Squares based on Cyclotomic Orthomorphisms

10.37236/1919 ◽  
2005 ◽  
Vol 12 (1) ◽  
Author(s):  
Ian M. Wanless
Keyword(s):  
Finite Fields ◽  
Complete Graph ◽  
Prime Order ◽  
Automorphism Groups ◽  
Latin Square ◽  
Latin Squares ◽  
Established Method ◽  
Cyclic Groups ◽  
Complete Bipartite Graphs ◽  
First Time

Atomic latin squares have indivisible structure which mimics that of the cyclic groups of prime order. They are related to perfect $1$-factorisations of complete bipartite graphs. Only one example of an atomic latin square of a composite order (namely 27) was previously known. We show that this one example can be generated by an established method of constructing latin squares using cyclotomic orthomorphisms in finite fields. The same method is used in this paper to construct atomic latin squares of composite orders 25, 49, 121, 125, 289, 361, 625, 841, 1369, 1849, 2809, 4489, 24649 and 39601. It is also used to construct many new atomic latin squares of prime order and perfect $1$-factorisations of the complete graph $K_{q+1}$ for many prime powers $q$. As a result, existence of such a factorisation is shown for the first time for $q$ in $\big\{$529, 2809, 4489, 6889, 11449, 11881, 15625, 22201, 24389, 24649, 26569, 29929, 32041, 38809, 44521, 50653, 51529, 52441, 63001, 72361, 76729, 78125, 79507, 103823, 148877, 161051, 205379, 226981, 300763, 357911, 371293, 493039, 571787$\big\}$. We show that latin squares built by the 'orthomorphism method' have large automorphism groups and we discuss conditions under which different orthomorphisms produce isomorphic latin squares. We also introduce an invariant called the train of a latin square, which proves to be useful for distinguishing non-isomorphic examples.

Pairs of Latin Squares to Counterbalance Sequential Effects and Pairing of Conditions and Stimuli

1989 ◽  
Vol 33 (18) ◽  
pp. 1223-1227 ◽  
Author(s):  
James R. Lewis
Keyword(s):  
Human Factors ◽  
Repeated Measures ◽  
Random Sequence ◽  
Latin Square ◽  
Latin Squares ◽  
Sequential Effects ◽  
Long Run ◽  
Within Subjects ◽  
Human Factors Research

This paper discusses methods with which one can simultaneously counterbalance immediate sequential effects and pairing of conditions and stimuli in a within-subjects design using pairs of Latin squares. Within-subjects (repeated measures) experiments are common in human factors research. The designer of such an experiment must develop a scheme to ensure that the conditions and stimuli are not confounded, or randomly order stimuli and conditions. While randomization ensures balance in the long run, it is possible that a specific random sequence may not be acceptable. An alternative to randomization is to use Latin squares. The usual Latin square design ensures that each condition appears an equal number of times in each column of the square. Latin squares have been described which have the effect of counterbalancing immediate sequential effects. The objective of this work was to extend these earlier efforts by developing procedures for designing pairs of Latin squares which ensure complete counterbalancing of immediate sequential effects for both conditions and stimuli, and also ensure that conditions and stimuli are paired in the squares an equal number of times.

Complete Diagonals of Latin Squares

1979 ◽  
Vol 22 (4) ◽  
pp. 477-481 ◽  
Author(s):  
Gerard J. Chang
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Special Case

AbstractJ. Marica and J. Schönhein [4], using a theorem of M. Hall, Jr. [3], see below, proved that if any n − 1 arbitrarily chosen elements of the diagonal of an n × n array are prescribed, it is possible to complete the array to form an n × n latin square. This result answers affirmatively a special case of a conjecture of T. Evans [2], to the effect that an n × n incomplete latin square with at most n − 1 places occupied can be completed to an n × n latin square. When the complete diagonal is prescribed, it is easy to see that a counterexample is provided by the case that one letter appears n − 1 times on the diagonal and a second letter appears once. In the present paper, we prove that except in this case the completion to a full latin square is always possible. Completion to a symmetric latin square is also discussed.

A Partial Generalization of Mann's Theorem Concerning Orthogonal Latin Squares

1988 ◽  
Vol 31 (4) ◽  
pp. 409-413 ◽  
Author(s):  
E. T. Parker ◽  
Lawrence Somer
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Necessary Condition ◽  
Orthogonal Latin Squares ◽  
Mutually Orthogonal Latin Squares ◽  
Complete Set

AbstractLetn = 4t+- 2, where the integert ≧ 2. A necessary condition is given for a particular Latin squareLof ordernto have a complete set ofn — 2mutually orthogonal Latin squares, each orthogonal toL.This condition extends constraints due to Mann concerning the existence of a Latin square orthogonal to a given Latin square.

Two remarks about Sudoku squares

2006 ◽  
Vol 90 (519) ◽  
pp. 425-430 ◽  
Author(s):  
A. D. Keedwell
Keyword(s):  
Current Problem ◽  
Latin Square ◽  
Latin Squares ◽  
Critical Set ◽  
Critical Sets ◽  
The Given ◽  
A Current ◽  
Defining Sets

Smallest defining setsA standard Sudoku square is a 9 × 9 latin square in which each of the nine 3 × 3 subsquares into which it can be separated contains each of the integers 1 to 9 exactly once.A current problem is to complete such a square when only some of the cells have been filled. These cells are often called ‘givens’. (Such problems are currently (2005) published daily in British newspapers.) In more mathematical terms, the given filled cells constitute a defining set or uniquely completable set for the square if they lead to a unique completion of the square. If, after deletion of any one of these givens, the square can no longer be completed uniquely, the givens form a critical set. The investigation of critical sets for ‘ordinary’ latin squares is a topic of current mathematical interest. (See [1] for more details.)

scholarly journals An Efficient Algorithm for Eigenvalue Problem of Latin Squares in a Bipartite Min-Max-Plus System

Symmetry ◽  
2020 ◽  
Vol 12 (2) ◽  
pp. 311 ◽  
Author(s):  
Mubasher Umer ◽  
Umar Hayat ◽  
Fazal Abbas ◽  
Anurag Agarwal ◽  
Petko Kitanov
Keyword(s):  
Eigenvalue Problem ◽  
Efficient Algorithm ◽  
Time Complexity ◽  
Latin Square ◽  
Latin Squares ◽  
Clear Advantage

In this paper, we consider the eigenproblems for Latin squares in a bipartite min-max-plus system. The focus is upon developing a new algorithm to compute the eigenvalue and eigenvectors (trivial and non-trivial) for Latin squares in a bipartite min-max-plus system. We illustrate the algorithm using some examples. The proposed algorithm is implemented in MATLAB, using max-plus algebra toolbox. Computationally speaking, our algorithm has a clear advantage over the power algorithm presented by Subiono and van der Woude. Because our algorithm takes 0 . 088783 sec to solve the eigenvalue problem for Latin square presented in Example 2, while the compared one takes 1 . 718662 sec for the same problem. Furthermore, a time complexity comparison is presented, which reveals that the proposed algorithm is less time consuming when compared with some of the existing algorithms.

scholarly journals Hownotto prove the Alon-Tarsi conjecture

2012 ◽  
Vol 205 ◽  
pp. 1-24 ◽  
Author(s):  
Douglas S. Stones ◽  
Ian M. Wanless
Keyword(s):  
Latin Square ◽  
Latin Squares ◽  
Exact Computation

AbstractThe sign of a Latin square is −1 if it has an odd number of rows and columns that are odd permutations; otherwise, it is +1. LetLEnandLonbe, respectively, the number of Latin squares of ordernwith sign +1 and −1. The Alon-Tarsi conjecture asserts thatLEn≠Lonwhennis even. Drisko showed thatLEp+1≢Lop+1(modp3) for primep≥ 3 and asked if similar congruences hold for orders of the formpk+ 1,p+ 3, orpq+ 1. In this article we show that ift≤n, thenLEn+1≢L0n+1(modt3) only ift = nandnis an odd prime, thereby showing that Drisko’s method cannot be extended to encompass any of the three suggested cases. We also extend exact computation ton≤ 9, discuss asymptotics forLo/LE, and propose a generalization of the Alon-Tarsi conjecture.

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