On a Property of a Three-Dimensional Matrix

Let S n be the symmetrical group acting on the set {1, 2, . . . , n} and x, y ∈ S n . Consider the set W = {(i, x(i), y(i)) | 1 ≤ i ≤ n, |i − x(i)| > 1 ∨ |i − y(i)| > 1 ∨ |x(i) − y(i)| > 1}.The main result of this paper is the following theorem. If the number ofW set entries is more than [n/3], then there exist entries (i 1 , x(i 1 ), y(i 1 )), (i 2 , x(i 2 ), y(i 2 )), (i 3 , x(i 3 ), y(i 3 )) ∈ W such that |i 1 − x(i 2 )| ≤ 1,


Introduction
Let  3   be the set of  ×  × -matrices over the field of real numbers.
Three-dimensional matrix not only is an interesting mathematical object [1][2][3], but also has applications in many fields, such as theoretical physics [4] and operational research [5,6].
Let   be the symmetrical group acting on the set {1, 2, . . ., }, ,  ∈   , ‖  ‖ ∈  3   and ( The main result of this paper is the following theorem.
Theorem 1.If ‖  ‖ ∈  3  , ,  ∈   and the number of  ∩  set entries is more than [/3], then there exist entries We give another formulation of Theorem 1.Consider the set

Proof of Theorem 1
We prove the theorem by contradiction.The set of matrix ‖  ‖ ∈  3  entries with one index fixed and the two others having values from 1 to  will be called a layer.We denote a layer by    , where  indicates the location of a fixed index and  indicates its value.For example,  2  = {  | ,  = 1, . . ., }.Furthermore, entries from  will be called basic, entries from  will be called nonbasic;  will be termed a trajectory; the layer containing a basic trajectory entry will be termed a basic layer and the layer containing a nonbasic trajectory entrywill be termed a nonbasic layer.
If  1  is a nonbasic layer, then one layer in the pair  2  −1 ,  We prove the first arrangement of item 1. Suppose that the two nonbasic layers are followed only by two basic layers, that is, . ..00110. . . .Let there be layers . . . layers are occupied, while the  1 +4 layer is nonbasic.There is a contradiction.
The other arrangements are proved in a similar way.Thus, nonbasic layers may not be arranged closer than those in the above variants.But these variants do not allow for the composition of a combination containing more than [/3] nonbasic layers.Hence, it follows that if a trajectory includes more than [/3] nonbasic entries, then one of the variants is violated and nonbasic trajectory entries can be replaced by a set of entries containing a basic entry.

Application to the Three-Dimensional Assignment Problem
The three-dimensional assignment problem (AP3) is an important combinatorial optimization problem.It is sufficient to note that the particular case of AP3, the 3dimensional matching problem, is one of the six main NPhard problems [7].The formal AP3 statement is as follows: for a matrix ‖  ‖ ∈  3  , find permutations ,  ∈   such that ∑  =1  ()() is maximized.
In this paper, the AP3 is considered for a special class of  ×  × -matrices ().A matrix One of AP3 interpretations is the following.There are  employees and two job sets of  jobs each.If the th employee performs the th job of the first set and the th job of the second set, then the effect equals   .It is required to distribute the jobs among the employees in such a way so as to maximize the total effect.
Let us describe the situation that will lead to the AP3 for matrices from ().As a rule, the employees are ordered by qualification, while the jobs are ordered by complexity.A higher effect is reached when a more qualified employee performs a more complex job, and we may arrive at the AP3 for matrices from ().
The NP-hard particular cases of the traveling salesman problem, with sets of matrices whose structures are similar to those of matrices from (), have been considered previously [15,16].
All entries that are equal to  belong to the optimal solution of AP3 for ‖  ‖ matrix.This signifies that if the optimal solution of AP3 for ‖  ‖ matrix is known, then the optimal solution of AP3 for ‖  ‖ matrix is known as well.Therefore, the AP3 for arbitrary matrices is polynomially reducible to the AP3 for matrices from ().Indeed, if for optimal T of matrix ‖  ‖ ∈ () the number of  ∩  set entries is more than [/3], then, according to Theorem 1, it is possible to make such replacement of entries such that the sum of T set entries will not change and the number of  ∩  set entries is reduced.Since  2 ≤ (1/3)⋅max   ∈   and  1 ≥ ⋅min   ∈   , then  2 / 1 ≤ 1/(3).Hence, ( 1 −   )/ 1 ≤ 1/(3), and the relative error of   taken as the AP3 approximate solution for the ‖  ‖ matrix is no more than 1/(3).Remark 6.An exact ()-algorithm of AP3 solution has been constructed [9] for  ×  × -matrices ‖  ‖ such that   ≥ 0 for   ∈  and   = 0 for   ∈  using dynamical programming.

Corollary 4 .
If ‖  ‖ ∈ (), then a set T exists on which the AP3 optimum is reached and the number of  ∩  set entries is no more than [/3].