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Journal of Discrete Mathematics
Volume 2013 (2013), Article ID 797249, 3 pages
On a Property of a Three-Dimensional Matrix
C. D. Technologies Ltd., Israel
Received 11 June 2013; Accepted 24 September 2013
Academic Editor: Hong J. Lai
Copyright © 2013 David Blokh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let be the symmetrical group acting on the set and . Consider the set The main result of this paper is the following theorem. If the number of set entries is more than , then there exist entries such that , , and . The application of this theorem to the three-dimensional assignment problem is considered.
Let be the set of -matrices over the field of real numbers.
Let be the symmetrical group acting on the set , , and
The main result of this paper is the following theorem.
Theorem 1. If , and the number of set entries is more than , then there exist entries , and such that the entry .
We give another formulation of Theorem 1.
Consider the set
Theorem 2. If the number of set entries is more than , then there exist entries such that , , and .
2. Proof of Theorem 1
We prove the theorem by contradiction. The set of matrix 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, . 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 is a nonbasic layer, then one layer in the pair , is basic.
Suppose to the contrary that layers , , and are nonbasic. Let , , and be the trajectory entries of , , layers. Replace ,, and with , , and . The nonbasic trajectory entries ,, and are replaced with entries , , and , among which there is a basic one.
Similar assertions may be proved for the layer and the following layer pairs: ; ; ; ; ; and .
Three nonbasic layers may not be consecutive.
Suppose that layers , , and are nonbasic. One of the layers or is basic. In these layers, the basic entries may be , , , , , , and for the layer , and , , , , , , for the layer . However, this contradicts the assumption that the layers , , and are nonbasic. The fact that the two first and the two last layers may not be nonbasic is proved in a similar way.
All assertions given below represent conditions that prevent replacing nonbasic trajectory entries with entries that include a basic one.
Consider the sequence of layers , and . Given below are possible arrangements of layers in this sequence. A basic layer is denoted by 1; a nonbasic layer is denoted by 0.
Consider(1)…11100111…; 1100111…; …1110011; 110011.(2)…111010111…; 01111010…; …01011110; …001111010…; …010111100…; …0101111010….(3)0111100…; …00111100…; …0011110.(4)0111…; …1110; 10111…; …11101; 011110.
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 …, , , , ….
Consider layer pairs , ; , ; , ; , and . At least one layer of each pair is basic. These four pairs of layers contain at least two basic entries, and these entries are located in layers and . The pair , includes a basic layer. The first coordinate of the basic layer entry may be , , or . But the and layers are occupied, while the 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 nonbasic layers. Hence, it follows that if a trajectory includes more than 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.
3. 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 3-dimensional matching problem, is one of the six main NP-hard problems . The formal AP3 statement is as follows: for a matrix , find permutations such that 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 .
Theorem 3. The AP3 for matrices from is NP-hard.
Proof. Let , and numbers , are such that , .
Consider -matrix with nonnegative entries
The matrix .
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 .
Corollary 4. If , then a set exists on which the AP3 optimum is reached and the number of set entries is no more than .
Indeed, if for optimal of matrix the number of set entries is more than , then, according to Theorem 1, it is possible to make such replacement of entries such that the sum of set entries will not change and the number of set entries is reduced.
Theorem 5. If , then the AP3 optimum for a matrix, where for and for , represents an approximation for the AP3 optimum of the matrix with a relative error not exceeding .
Proof. Let ; is the AP3 optimum for ; are such that , and the number of set entries is no more than ; is the sum of set entries and ; is the AP3 optimum for .
Note that and .
Insofar as , then . Hence, and .
Since and , then . Hence, , and the relative error of taken as the AP3 approximate solution for the matrix is no more than .
Remark 6. An exact -algorithm of AP3 solution has been constructed  for -matrices such that for and for using dynamical programming.
The author is grateful to Professor Y. Dinitz and Professor G. Gutin for their attention to this work and valuable comments. Also the author is grateful to the anonymous referee of this journal for the valuable comments and suggestions.
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