On the finite completion of partial latin cubes

A classical question in combinatorics is the following: given a partial Latin square P, when can we complete P to a Latin square L? In this paper, we investigate the class of ε‐dense partial Latin squares: partial Latin squares in which each symbol, row, and column contains no more than ‐many nonblank cells. Based on a conjecture of Nash‐Williams, Daykin and Häggkvist conjectured that all ‐dense partial Latin squares are completable. In this paper, we will discuss the proof methods and results used in previous attempts to resolve this conjecture, introduce a novel technique derived from a paper by Jacobson and Matthews on generating random Latin squares, and use this technique to study ε‐dense partial Latin squares that contain no more than filled cells in total. In this paper, we construct completions for all ε‐dense partial Latin squares containing no more than filled cells in total, given that . In particular, we show that all ‐dense partial Latin squares are completable. These results improve prior work by Gustavsson, which required , as well as Chetwynd and Häggkvist, which required , n even and greater than 107.

Avoiding partial Latin squares and intricacy

Embedding partial Latin squares in Latin squares with many mutually orthogonal mates

In this paper we develop two methods for completing partial latin squares and prove the following. Let $$A$$A be a partial latin square of order $$nr$$nr in which all non-empty cells occur in at most $$n-1$$n-1$$r\times r$$r×r squares. If $$t_1,\ldots , t_m$$t1,?,tm are positive integers for which $$n\geqslant t_1^2+t_2^2+\cdots +t_m^2+1$$n?t12+t22+?+tm2+1 and if $$A$$A is the union of $$m$$m subsquares each with order $$rt_i$$rti, then $$A$$A can be completed. We additionally show that if $$n\geqslant r+1$$n?r+1 and $$A$$A is the union of $$n$$n identical $$r\times r$$r×r squares with disjoint rows and columns, then $$A$$A can be completed. For smaller values of $$n$$n we show that a completion does not always exist.

Suppose that L is a latin square of order m and P ⊑ L is a partial latin square. If L is the only latin square of order m which contains P, and no proper subset of P has this property, then P is a critical set of L. The critical set spectrum problem is to determine, for a given m, the set of integers t for which there exists a latin square of order m with a critical set of size t. We outline a partial solution to the critical set spectrum problem for latin squares of order 2n. The back circulant latin square of even order m has a well‐known critical set of size m2/4, and this is the smallest known critical set for a latin square of order m. The abelian 2‐group of order 2n has a critical set of size 4n‐3n, and this is the largest known critical set for a latin square of order 2n. We construct a set of latin squares with associated critical sets which are intermediate between the back circulant latin square of order 2n and the abelian 2‐group of order 2n. © 2007 Wiley Periodicals, Inc. J Combin Designs 16: 25–43, 2008

A massively parallel evolutionary algorithm for the partial Latin square extension problem

3-Homogeneous latin trades

We show that two partial latin squares of order mk are simultaneously avoidable if m > 4 and $${k>\frac{m^3(m^2-1)}{2}}$$ . If m = 4, we show the same conclusion when k > 56.

A partial Latin square (PLS) is a partial assignment of n symbols to an $$n\times n$$n×n grid such that, in each row and in each column, each symbol appears at most once. The PLS extension problem is an NP-hard problem that asks for a largest extension of a given PLS. We consider the local search such that the neighborhood is defined by (p, q)-swap , i.e., the operation of dropping exactly p symbols and then assigning symbols to at most q empty cells. As a fundamental result, we provide an efficient $$(p,\infty )$$(p,ź)-neighborhood search algorithm that finds an improved solution or concludes that no such solution exists for $$p\in \{1,2,3\}$$pź{1,2,3}. The running time of the algorithm is $$O(n^{p+1})$$O(np+1). We then propose a novel swap operation, Trellis-swap, which is a generalization of (p, q)-swap with $$p\le 2$$p≤2. The proposed Trellis-neighborhood search algorithm runs in $$O(n^{3.5})$$O(n3.5) time. The iterated local search (ILS) algorithm with Trellis-neighborhood is more likely to deliver a high-quality solution than not only ILSs with $$(p,\infty )$$(p,ź)-neighborhood but also state-of-the-art optimization solvers such as IBM ILOG CPLEX and LocalSolver.

An r×s partial Latin rectangle (lij) is an r×s matrix containing elements of {1,2,…,n}∪{⋅} such that each row and each column contain at most one copy of any symbol in {1,2,…,n}. An entry is a triple (i,j,lij) with lij≠⋅. Partial Latin rectangles are operated on by permuting the rows, columns, and symbols, and by uniformly permuting the coordinates of the set of entries. The stabilizers under these operations are called the autotopism group and the autoparatopism group, respectively.We develop the theory of symmetries of partial Latin rectangles, introducing the concept of a partial Latin rectangle graph. We give constructions of m-entry partial Latin rectangles with trivial autotopism groups for all possible autoparatopism groups (up to isomorphism) when: (a) r=s=n, i.e., partial Latin squares, (b) r=2 and s=n, and (c) r=2 and s≠n.

An improved approximation algorithm for the partial Latin square extension problem

Outlines of Latin Squares

Given enough data, Deep Neural Networks (DNNs) are capable of learning complex input-output relations with high accuracy. In several domains, however, data is scarce or expensive to retrieve, while a substantial amount of expert knowledge is available. It seems reasonable that if we can inject this additional information in the DNN, we could ease the learning process. One such case is that of Constraint Problems, for which declarative approaches exists and pure ML solutions have obtained mixed success. Using a classical constrained problem as a case study, we perform controlled experiments to probe the impact of progressively adding domain and empirical knowledge in the DNN. Our results are very encouraging, showing that (at least in our setup) embedding domain knowledge at training time can have a considerable effect and that a small amount of empirical knowledge is sufficient to obtain practically useful results.

We prove that a countable group G can be embedded into a two-generator simple group S which is an amalgamated free product of groups G * F and F, where F and F are free groups on two generators.S is also the product of two commuting free subgroups.If G has solvable word problem, then we can construct a recursive presentation for S.1. Introduction.It is well known that every countable group is a subgroup of a simple two-generator group (see [3] and [7]).The main result of our paper is Theorem A. (a) A countable group G can be embedded into a simple two-generator group S which is an amalgamated free product of G * Fx and F, where F and Fx are both free groups on two generators.S is also the product of two commuting free subgroups.(b) If G has solvable word problem, we can construct the group S of (a) with a recursive presentation.It was first shown by R. J. Thompson [8] by a construction completely different from ours that a countable group with solvable word problem is a subgroup of a recursively presented finitely generated simple group.Our construction of the group S is similar to the construction of the simple free products of R. Camm [1], or see [6] for a slightly simplified version.2. The construction of the group S. Let 77 be a countable infinite group with an element x of order at least 5, and assume that the elements of 77 are numbered by the integers, 77 = {g" i G Z}, such that g0 = 1, gx = x, g2 = x2, g3 = x3, g4 = x4.Let A = (a) and F = (b,c) be free groups on a and b, c, respectively.Then the subgroup U of A * H generated by the elements of the form ", = aW *'eZ-{0}, where p is a permutation of Z -{0}, is a free group freely generated by the elements u" / g Z -{0}, with transversal AinA*H.Similarly, the elements Vj = bjcu\ 7GZ-{0), of F, where a is a second permutation of Z -{0}, freely generate a free subgroup V of F with transversal B = (b).

Autotopism stabilized colouring games on rook’s graphs