On a Property of a Three-Dimensional Matrix

Abstract

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 of 𝑊 set entries is more than [𝑛/3], then there exist entries (𝑖₁,𝑥(𝑖₁),𝑦(𝑖₁)),(𝑖₂,𝑥(𝑖₂),𝑦(𝑖₂)),(𝑖₃,𝑥(𝑖₃),𝑦(𝑖₃))∈𝑊 such that |𝑖₁−𝑥(𝑖₂)|≤1, |𝑖₁−𝑦(𝑖₃)|≤1, and |𝑥(𝑖₂)−𝑦(𝑖₃)|≤1. The application of this theorem to the three-dimensional assignment problem is considered.

1. Introduction

Let 𝑅³_𝑛 be the set of 𝑛×𝑛×𝑛-matrices over the field of real numbers.

Three-dimensional matrix not only is an interesting mathematical object [1–3], but also has applications in many fields, such as theoretical physics [4] and operational research [5, 6].

Let 𝑆_𝑛 be the symmetrical group acting on the set {1,2,…,𝑛}, 𝑥,𝑦∈𝑆_𝑛, ‖𝑎_𝑖𝑗𝑘‖∈𝑅³_𝑛 and 𝐺=𝑎_𝑖𝑗𝑘∣||,𝑎𝑖−𝑗≤1∧𝑖−𝑘≤1∧𝑗−𝑘≤1𝐻=_𝑖𝑗𝑘∣||,𝑎𝑖−𝑗>1∨𝑖−𝑘>1∨𝑗−𝑘>1𝑇=_{𝑖𝑥(𝑖)𝑦(𝑖)}∣1≤𝑖≤𝑛. (1)

The main result of this paper is the following theorem.

Theorem 1. If ‖𝑎_𝑖𝑗𝑘‖∈𝑅³_𝑛, 𝑥,𝑦∈𝑆_𝑛 and the number of 𝑇∩𝐻 set entries is more than [𝑛/3], then there exist entries 𝑎_{𝑖1𝑗1𝑘1},𝑎_{𝑖2𝑗2𝑘2}, and 𝑎_{𝑖3𝑗3𝑘3}∈𝑇∩𝐻 such that the entry 𝑎_{𝑖1𝑗2𝑘3}∈𝐺.

We give another formulation of Theorem 1.

Consider the set (𝑖)(𝑖)|(𝑖(𝑖𝑥(𝑖)(𝑖.𝑊=𝑖,𝑥,𝑦∣1≤𝑖≤𝑛,𝑖−𝑥)|>1∨𝑖−𝑦)>1∨−𝑦)>1 (2)

Theorem 2. If the number of 𝑊 set entries is more than [𝑛/3], then there exist entries (𝑖₁,𝑥(𝑖₁),𝑦(𝑖₁)),(𝑖₂,𝑥(𝑖₂),𝑦(𝑖₂)),(𝑖₃,𝑥(𝑖₃),𝑦(𝑖₃))∈𝑊 such that |𝑖₁−𝑥(𝑖₂)|≤1, |𝑖₁−𝑦(𝑖₃)|≤1, and |𝑥(𝑖₂)−𝑦(𝑖₃)|≤1.

2. Proof of Theorem 1

We prove the theorem by contradiction. The set of matrix ‖𝑎_𝑖𝑗𝑘‖∈𝑅³_𝑛 entries with one index fixed and the two others having values from 1 to 𝑛 will be called a layer. We denote a layer by 𝐿^𝑟_𝑠, where 𝑟 indicates the location of a fixed index and 𝑠 indicates its value. For example, 𝐿²_𝑠={𝑎_𝑖𝑠𝑘∣𝑖,𝑘=1,…,𝑛}. Furthermore, entries from 𝐺 will be called basic, entries from 𝐻 will be called nonbasic; 𝑇 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 𝐿¹_𝑠 is a nonbasic layer, then one layer in the pair 𝐿²_𝑠−1,𝐿³_𝑠−1 is basic.

Suppose to the contrary that layers 𝐿¹_𝑠, 𝐿²_𝑠−1, and 𝐿³_𝑠−1 are nonbasic. Let 𝑎_{𝑠𝑗1𝑘1}, 𝑎_{𝑖2,𝑠−1,𝑘2}, and 𝑎_{𝑖3,𝑗3,𝑠−1} be the trajectory entries of 𝐿¹_𝑠, 𝐿²_𝑠−1, 𝐿³_𝑠−1 layers. Replace 𝑎_{𝑠𝑗1𝑘1},𝑎_{𝑖2,𝑠−1,𝑘2}, and 𝑎_{𝑖3,𝑗3,𝑠−1} with 𝑎_{𝑠,𝑠−1,𝑠−1}, 𝑎_{𝑖2𝑗1𝑘1}, and 𝑎_{𝑖3𝑗3𝑘2}. The nonbasic trajectory entries 𝑎_{𝑠𝑗1𝑘1},𝑎_{𝑖2,𝑠−1,𝑘2}, and 𝑎_{𝑖3,𝑗3,𝑠−1} are replaced with entries 𝑎_{𝑠,𝑠−1,𝑠−1}, 𝑎_{𝑖2𝑗1𝑘1}, and 𝑎_{𝑖3𝑗3𝑘2}, among which there is a basic one.

Similar assertions may be proved for the 𝐿¹_𝑠 layer and the following layer pairs: 𝐿²_𝑠−1,𝐿³_𝑠; 𝐿²_𝑠,𝐿³_𝑠−1; 𝐿²_𝑠,𝐿³_𝑠; 𝐿²_𝑠,𝐿³_𝑠+1; 𝐿²_𝑠+1,𝐿³_𝑠; and 𝐿²_𝑠+1,𝐿³_𝑠+1.

Three nonbasic layers may not be consecutive.

Suppose that layers 𝐿¹_𝑠−1, 𝐿¹_𝑠, and 𝐿¹_𝑠+1 are nonbasic. One of the layers 𝐿²_𝑠 or 𝐿³_𝑠 is basic. In these layers, the basic entries may be 𝑎_{𝑠−1,𝑠,𝑠−1}, 𝑎_{𝑠−1,𝑠,𝑠}, 𝑎_{𝑠,𝑠,𝑠−1}, 𝑎_𝑠𝑠𝑠, 𝑎_{𝑠,𝑠,𝑠+1}, 𝑎_{𝑠+1,𝑠,𝑠}, and 𝑎_{𝑠+1,𝑠,𝑠+1} for the layer 𝐿²_𝑠, and 𝑎_{𝑠−1,𝑠−1,𝑠}, 𝑎_{𝑠−1,𝑠,𝑠}, 𝑎_{𝑠,𝑠−1,𝑠}, 𝑎_𝑠𝑠𝑠, 𝑎_{𝑠+1,𝑠,𝑠}, 𝑎_{𝑠,𝑠+1,𝑠}, 𝑎_{𝑠+1,𝑠+1,𝑠} for the layer 𝐿³_𝑠. However, this contradicts the assumption that the layers 𝐿¹_𝑠−1, 𝐿¹_𝑠, and 𝐿¹_𝑠+1 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 𝐿¹₁,𝐿¹₂,…,𝐿¹_𝑠−1,𝐿¹_𝑠,𝐿¹_𝑠+1,…,𝐿¹_𝑛−1, and 𝐿¹_𝑛. Given below are possible arrangements of layers in this sequence. A basic layer 𝐿¹_𝑠 is denoted by 1; a nonbasic layer 𝐿¹_𝑠 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 …𝐿¹_𝑠, 𝐿¹_𝑠+1, 𝐿¹_𝑠+2, 𝐿¹_𝑠+3, 𝐿¹_𝑠+4….

Consider layer pairs 𝐿²_𝑠+1, 𝐿³_𝑠+1; 𝐿²_𝑠+1, 𝐿³_𝑠+2; 𝐿²_𝑠+2, 𝐿³_𝑠+1; 𝐿²_𝑠+2, and 𝐿³_𝑠+2. 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 𝐿¹_𝑠+2 and 𝐿¹_𝑠+3. The pair 𝐿²_𝑠+3, 𝐿³_𝑠+3 includes a basic layer. The first coordinate of the basic layer entry may be 𝑠+2, 𝑠+3, or 𝑠+4. But the 𝐿¹_𝑠+2 and 𝐿¹_𝑠+3 layers are occupied, while the 𝐿¹_𝑠+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 [𝑛/3] nonbasic layers. Hence, it follows that if a trajectory includes more than [𝑛/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.

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 [7]. The formal AP3 statement is as follows: for a matrix ‖𝑎_𝑖𝑗𝑘‖∈𝑅³_𝑛, find permutations 𝑥,𝑦∈𝑆_𝑛 such that ∑^𝑛_𝑖=1𝑎_{𝑖𝑥(𝑖)𝑦(𝑖)} is maximized.

In this paper, the AP3 is considered for a special class of 𝑛×𝑛×𝑛-matrices 𝑈(𝑛). A matrix 𝑎_𝑖𝑗𝑘∈𝑈(𝑛),if𝑎_𝑖𝑗𝑘≥0,min_{𝑎𝑖𝑗𝑘∈𝐺}𝑎_𝑖𝑗𝑘≥𝐶⋅max_{𝑎𝑖𝑗𝑘∈𝐻}𝑎_𝑖𝑗𝑘,𝐶≥3. (3)

One of AP3 interpretations is the following. There are 𝑛 employees and two job sets of 𝑛 jobs each. If the 𝑖th employee performs the 𝑗th job of the first set and the 𝑘th job of the second set, then the effect equals 𝑎_𝑖𝑗𝑘. 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 𝑈(𝑛). 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 𝑈(𝑛).

The particular cases of AP3 [5, 8–11], the nonpolynomial exact algorithms for the AP3 [6, 12], and heuristics for the AP3 [13, 14] were considered.

The NP-hard particular cases of the traveling salesman problem, with sets of matrices whose structures are similar to those of matrices from 𝑈(𝑛), have been considered previously [15, 16].

Theorem 3. The AP3 for matrices from 𝑈(𝑛) is NP-hard.

Proof. Let ‖𝑏_𝑖𝑗𝑘‖∈𝑅³_𝑛, 𝑏_𝑖𝑗𝑘≥0 and numbers 𝐴, 𝐵 are such that 𝐴=3⋅max𝑏_𝑖𝑗𝑘, 𝐵=𝑛𝐴.
Consider 4𝑛×4𝑛×4𝑛-matrix ‖𝑎_𝑖𝑗𝑘‖ with nonnegative entries 𝑎_{4𝑠−3,4𝑠−2,4𝑠−2}𝑎=𝐵𝑠=1,…,𝑛;_{4𝑠−2,4𝑠−1,4𝑠−1}𝑎=𝐵𝑠=1,…,𝑛;_{4𝑠,4𝑠,4𝑠}𝑎=𝐵𝑠=1,…,𝑛;_𝑟𝑢𝑣||||||=𝐴𝑟,𝑢,𝑣=1,…,𝑛;if𝑟−𝑢≤1,𝑟−𝑣≤1,𝑢−𝑣≤1,𝑎_𝑟𝑢𝑣𝑎≠𝐵;_{4𝑖−1,4𝑗−3,4𝑘−3}=𝑏_𝑖𝑗𝑘𝑎𝑖,𝑗,𝑘=1,…,𝑛;_𝑖𝑗𝑘=0inothercases. (4)
The matrix ‖𝑎_𝑖𝑗𝑘‖∈𝑈(4𝑛).
All entries that are equal to 𝐵 belong to the optimal solution of AP3 for ‖𝑎_𝑖𝑗𝑘‖ matrix. This signifies that if the optimal solution of AP3 for ‖𝑎_𝑖𝑗𝑘‖ matrix is known, then the optimal solution of AP3 for ‖𝑏_𝑖𝑗𝑘‖ matrix is known as well. Therefore, the AP3 for arbitrary matrices is polynomially reducible to the AP3 for matrices from 𝑈(𝑛).

Corollary 4. If ‖𝑎_𝑖𝑗𝑘‖∈𝑈(𝑛), then a set 𝑇 exists on which the AP3 optimum is reached and the number of 𝑇∩𝐻 set entries is no more than [𝑛/3].
Indeed, if for optimal 𝑇 of matrix ‖𝑎_𝑖𝑗𝑘‖∈𝑈(𝑛) the number of 𝑇∩𝐻 set entries is more than [𝑛/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 𝑇∩𝐻 set entries is reduced.

Theorem 5. If ‖𝑎_𝑖𝑗𝑘‖∈𝑈(𝑛), then the AP3 optimum for a ‖𝑎^′_𝑖𝑗𝑘‖ matrix, where 𝑎^′_𝑖𝑗𝑘=𝑎_𝑖𝑗𝑘 for 𝑎^′_𝑖𝑗𝑘∈𝐺 and 𝑎^′_𝑖𝑗𝑘=0 for 𝑎^′_𝑖𝑗𝑘∈𝐻, represents an approximation for the AP3 optimum of the ‖𝑎_𝑖𝑗𝑘‖ matrix with a relative error not exceeding 1/(3𝐶).

Proof. Let ‖𝑎_𝑖𝑗𝑘‖∈𝑈(𝑛); 𝑚₁ is the AP3 optimum for ‖𝑎_𝑖𝑗𝑘‖; 𝑓,𝑔∈𝑆_𝑛 are such that 𝑚₁=∑^𝑛_𝑖=1𝑎_{𝑖𝑓(𝑖)𝑔(𝑖)}, and the number of {𝑎_{𝑖𝑓(𝑖)𝑔(𝑖)}∣𝑖=1,…,𝑛}∩𝐻 set entries is no more than [𝑛/3]; 𝑚₂ is the sum of {𝑎_{𝑖𝑓(𝑖)𝑔(𝑖)}∣𝑖=1,…,𝑛}∩𝐻 set entries and 𝑚₃=𝑚₁−𝑚₂; 𝑚^′ is the AP3 optimum for ‖𝑎^′_𝑖𝑗𝑘‖.
Note that 𝑚₃=∑^𝑛_𝑖=1𝑎^′_{𝑖𝑓(𝑖)𝑔(𝑖)} and 𝑚^′≥𝑚₃.
Insofar as 𝑎_𝑖𝑗𝑘≥𝑎^′_𝑖𝑗𝑘,𝑖,𝑗,𝑘=1,…,𝑛, then 𝑚₁≥𝑚^′. Hence, 𝑚₃≤𝑚^′≤𝑚₁ and (𝑚₁−𝑚^′)/𝑚₁≤(𝑚₁−𝑚₃)/𝑚₁=𝑚₂/𝑚₁.
Since 𝑚₂≤(1/3)𝑛⋅max_{𝑎𝑖𝑗𝑘∈𝐻}𝑎_𝑖𝑗𝑘 and 𝑚₁≥𝑛⋅min_{𝑎𝑖𝑗𝑘∈𝐺}𝑎_𝑖𝑗𝑘, then 𝑚₂/𝑚₁≤1/(3𝐶). Hence, (𝑚₁−𝑚^′)/𝑚₁≤1/(3𝐶), and the relative error of 𝑚^′ taken as the AP3 approximate solution for the ‖𝑎_𝑖𝑗𝑘‖ matrix is no more than 1/(3𝐶).

Remark 6. An exact 𝑂(𝑛)-algorithm of AP3 solution has been constructed [9] for 𝑛×𝑛×𝑛-matrices ‖𝑎_𝑖𝑗𝑘‖ such that 𝑎_𝑖𝑗𝑘≥0 for 𝑎_𝑖𝑗𝑘∈𝐺 and 𝑎_𝑖𝑗𝑘=0 for 𝑎_𝑖𝑗𝑘∈𝐻 using dynamical programming.

Acknowledgments

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.