Proof of a conjecture of Kl{\o}ve on permutation codes under the Chebychev distance

Let $d$ be a positive integer and $x$ a real number. Let $A_{d, x}$ be a $d\times 2d$ matrix with its entries $$ a_{i,j}=\left\{ \begin{array}{ll} x\ \&\mbox{for} \ 1\leqslant j\leqslant d+1-i, 1\ \&\mbox{for} \ d+2-i\leqslant j\leqslant d+i, 0\ \&\mbox{for} \ d+1+i\leqslant j\leqslant 2d. \end{array} \right. $$ Further, let $R_d$ be a set of sequences of integers as follows: $$R_d=\{(\rho_1, \rho_2,\ldots, \rho_d)|1\leqslant \rho_i\leqslant d+i, 1\leqslant i \leqslant d,\ \mbox{and}\ \rho_r\neq \rho_s\ \mbox{for}\ r\neq s\}.$$ and define $$\Omega_d(x)=\sum_{\rho\in R_d}a_{1,\rho_1}a_{2, \rho_2}\ldots a_{d,\rho_d}.$$ In order to give a better bound on the size of spheres of permutation codes under the Chebychev distance, Kl{\o}ve introduced the above function and conjectured that $$\Omega_d(x)=\sum_{m=0}^d{d\choose m}(m+1)^d(x-1)^{d-m}.$$ In this paper, we settle down this conjecture positively.


Introduction
A permutation code is a subset of the symmetric group S n , equipped with a distance metric. Permutation codes are of potential use in various applications such as power-line communications and coding for flash memories used with rank modulation [6,7]. Permutation codes were extensively studied over the last decade. Hamming metric is naturally the first to be considered. Later, Ulam metric [4] and Kendall τ -metric [2] were introduced and are now the two most investigated metrics. However in [9], a new metric named the Chebyshev metric was proposed by Kløve et al., when they were studying the multi-level flash memory model. A combinatorial survey on metrics related to permutations was given in [3].
The two main questions in coding theory are fundamental limits on the parameters of the code (information rate versus minimum distance) and constructions of codes that attain these limits. It turns out that both topics are difficult for permutation codes. Few explicit constructions were obtained for various metrics and no general bounds better than the GV-bound and Sphere packing bounds were found in [1,2,4,6,9] except for the Hamming metric [5]. Both the GV-bound and the Sphere packing bound depends on the volume (V (n, d)) of a typical "ball" which consists of permutations in S n at distance at most d from the identity permutation. The calculation of the volume of that ball becomes a crucial problem.
The Chebychev distance d(p, q) between two permutations p = (p 1 , p 2 , . . . , p n ) and q = (q 1 , q 2 , . . . , q n ) is defined by Let ι be the identity permutation in S n . Since d(ι, q) = d(p, pq), we have V (d, n) = |T d,n |. The permanent of an n × n matrix A is defined by perA = p∈Sn a 1,p 1 . . . a n,pn .
Let A (d,n) be the n × n matrix with a d,n i,j = 1 if |i − j| d and a d,n i,j = 0 otherwise. Clearly, V (d, n) = perA (d,n) . Although the permanent looks similar to the determinant of a matrix, it is a difficult problem to compute the permanent for general matrices. The celebrated van der Waerden theorem gives a lower bound for the permanent of the so called doubly stochastic n × n matrix. Here doubly stochastic means that all the elements are non-negative and that the sum of the elements in any row or column is 1. Thus, if A is an n × n matrix where the sum of the elements in any row or column is a constant k, then van der Waerden theorem gives a lower bound on the permanent of A.
By noticing that most rows and columns of A (d,n) have the sum 2d + 1, Kløve defined a closely related matrix B (d,n) with row sum and column sum 2d + 1 so that van der Waerden's theorem can be applied. The matrix B (d,n) is defined as follows: otherwise.
With this new defined matrix B (d,n) , Kløve [10] gives a lower bound on V (d, n) as follows: For example, Let R d be a set of sequences of integers as follows: Define Let Kløve [9] also gave the following lower bound on V (n, d): In particular, Ω d (2) = Ω d . Kløve gave the following generalized conjecture and verified it for d 9.
In this paper, we shall prove that Conjecture 2 is true. Actually, for any m × n matrix A = (a i,j ) with m n, the permanent function of A is already defined as follows (see, for example, [11]): where P (n, m) denotes the set of all m-permutations of the n-set {1, 2, . . . , n}.
In fact, by the definition of R d , we know that R d is exactly the subset of all dpermutations of the 2d-set {1, 2, . . . , 2d} such that σ ∈ R d if and only if a 1,σ 1 a 2,σ 2 · · · a d,σ d = 0. Hence we have Ω d (x) = per(A d,x ).
In order to prove Theorem 3, we first give a related combinatorial identity.
Lemma 4. Let m and n be positive integers. Then where k 0 = 1 and k m+1 = n.
For example, we have Proof of Lemma 4. We compute the multiple sum in the order from k m to k 1 . It can be proved by induction on k m−1 , k m−2 , . . . , k m−i−1 respectively that n km=k m−1 k m (n + 1) km−k m−1 n n−km = (n − k m−1 + 1)n n−k m−1 +1 , By choosing i = m − 1 in (2.2), we complete the proof of (2.1).

Proof of Theorem 3. It is clear that (1.3) is equivalent to
By the definition of Ω d (x + 1), we know that each x comes from the first term in x + 1.
To compute b m , we first choose m x's from m (x + 1)'s which are not in the same row nor in the same column of the matrix A d,x+1 , and then choose (d − m) 1's in the other d − m rows so that no 1's are in the same column. Suppose that the m x's are chosen from the rows indexed by d+1−i 1 , d+1−i 2 , . . . , d+1−i m with i 1 < i 2 < · · · < i m , respectively. By noticing that the d + 1 − i-th row has i (x + 1)'s and all the x's we choose must be in different columns, we have i 1 (i 2 − 1)(i 3 − 2) · · · (i m − m + 1) ways to do this. As for the number of ways to choose 1's in the remaining rows, we notice that the i-th row has d + i 1's including those 1's in (x + 1)'s and all these 1's forms several right trapezoids in the matrix A d,x+1 . Therefore, there are (d + 1) i 1 −1 d i 2 −i 1 −1 (d − 1) i 3 −i 2 −1 · · · (d − m + 1) d−im ways to choose the remaining 1's. It follows that b m = 1 i 1 <i 2 <···<im d i 1 (i 2 − 1)(i 3 − 2) · · · (i m − m + 1) where k s = i s − s + 1 (s = 1, . . . , m), k 0 = 1, and k m+1 = d − m + 1. By replacing n by d − m + 1 in (2.1), we obtain b m = d m (d − m + 1) d . This completes the proof.