Nonexistence of perfect permutation codes under the Kendall $\tau$-metric

Abstract

In the rank modulation scheme for flash memories, permutation codes have been studied. In this paper, we study perfect permutation codes in $S_n$, the set of all permutations on n elements, under the Kendall $\tau$-metric. We answer one open problem proposed by Buzaglo and Etzion. That is, proving the nonexistence of perfect codes in $S_n$, under the Kendall $\tau$-metric, for more values of n. Specifically, we present the polynomial representation of the size of a ball in $S_n$ under the Kendall $\tau$-metric for some radius r, and obtain some sufficient conditions of the nonexistence of perfect permutation codes. Further, we prove that there does not exist a perfect t-error-correcting code in $S_n$ under the Kendall $\tau$-metric for some n and $t=2,3,4,5,~\text {or}~\frac{5}{8}\left( {\begin{array}{c}n\\ 2\end{array}}\right) < 2t+1\le \left( {\begin{array}{c}n\\ 2\end{array}}\right)$.

Appendices

Appendix A

The purpose of this appendix is to prove Lemma 21 given in Section 5.2.

Proof

We choose any permutation $\beta \in L_i(\pi )$. Then $\beta$ is obtained by applying some t adjacent transpositions and at most $5-t$ adjacent transpositions on the former $i-1$ elements of $\pi$ and the latter $n-i$ elements of $\pi$, respectively, where $0\le t\le 5$. Thus, these operations can produce $S_{K}^{i-1}(t)B_K^{n-i}(5-t)$ permutations. If $t\ge \left( {\begin{array}{c}i-1\\ \lfloor \frac{i-1}{2}\rfloor \end{array}}\right)$, we have $S_{K}^{i-1}(t)=0$. So, the size of the first kind of permutations is $\sum _{t=0}^{5}S_{K}^{i-1}(t)B_K^{n-i}(5-t)$.

Next, we choose $\pi =[2,1,3,\ldots ,n]\in S_{n,2}$ such that $\pi (i)=i$ for all $i\ge 3$. By Lemma 10, we obtain $R_2=|R_2(\pi )|$. If $\sigma \in R_2(\pi )$, then we have $\sigma (1)=3,4,~\text {or}~5$. When $\sigma (1)=3$, we can obtain that elements 2 and 3 are exchanged and this operation needs at least 3 adjacent transpositions. Then the number of this kind of permutations in $R_2(\pi )$ is $B_K^{n-2}(2)$. When $\sigma (1)=4$, we have that elements 2 and 4 are exchanged and this operation needs at least 4 adjacent transpositions. Hence, the number of this kind of permutations in $R_2(\pi )$ is $B_K^{n-2}(1)$. When $\sigma (1)=5$, we obtain that elements 2 and 5 are exchanged and this operation needs at least 5 adjacent transpositions. Hence, the number of this kind of permutations in $R_2(\pi )$ is $B_K^{n-2}(0)$. So when $i=2$, we obtain that $R_2=B_K^{n-2}(2)+B_K^{n-2}(1)+B_K^{n-2}(0)$. By Lemma 18, when $i=n-1$, we also get $R_{n-1}=B_K^{n-2}(2)+B_K^{n-2}(1)+B_K^{n-2}(0)$.

Similarly, when $i=3$ or $n-2$, we can obtain that $R_i=B_K^{n-3}(2)+3B_K^{n-3}(1)+3$. When $4\le i\le n-3$, we can prove that $R_i=\sum _{t=0}^{2}S_{K}^{i-1}(t)\cdot B_K^{n-i}(2-t)+2\sum _{t=0}^{1}S_{K}^{i-1}(t)B_K^{n-i}(1-t)+2$.$\square$

Appendix B

The purpose of this appendix is to prove Lemma 22 given in Sect. 5.3.

Proof

In order to obtain this result, we first prove that when $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right)$, then

$$\begin{aligned} S_K^{n}(p)+S_K^{n}(q-1)<S_K^n(q)+S_K^n(p-1) \end{aligned}$$

(35)

by induction for all $n\ge 6$.

When $n=6$, we obtain $\left( {\begin{array}{c}6\\ 2\end{array}}\right) =15$. By Lemma 4, we compute $S_K^6(t)$ to obtain that $S_K^6(0)=1,S_K^6(1)=5,S_K^6(2)=14,S_K^6(3)=29,S_K^6(4)=49,S_K^6(5)=71,S_K^6(6)=90,S_K^6(7)=101$. Hence, when $n=6$, we clearly have that $S_K^{6}(p)+S_K^{6}(q-1)<S_K^6(q)+S_K^6(p-1)$ for $1\le p<q\le 7$ and $p+q\le 7$. So when $n=6$, $S_K^6(t)$ satisfies the condition in (35).

Now we assume that $S_K^m(t)$ satisfies the condition in (35) for some integers $m\ge 6$, that is, if $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right)$, then

$$\begin{aligned} S_K^{m}(p)+S_K^{m}(q-1)<S_K^m(q)+S_K^m(p-1). \end{aligned}$$

(36)

When $n=m+1$, by Lemma 4, we obtain

$$\begin{aligned} S_K^{m+1}(q)=S_K^{m+1}(q-1)+S_K^m(q)-S_K^m(q-m-1)\\ S_K^{m+1}(p)=S_K^{m+1}(p-1)+S_K^m(p)-S_K^m(p-m-1) \end{aligned}$$

for $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$. Hence, we have

$$\begin{aligned}&S_K^{m+1}(q)+S_K^{m+1}(p-1)+S_K^m(p)+S_K^m(q-m-1)=S_K^{m+1}(q-1)\nonumber \\&\qquad +S_K^{m+1}(p)+S_K^m(q)+S_K^m(p-m-1). \end{aligned}$$

(37)

Using the induction hypothesis on $S_K^m(q)$, we can obtain some results as follows. When $p\ge m+1$, then $1\le p-m<q$ and $p+q-m-1<\frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right) -\frac{m}{2}-1$. Thus, by (36), we have

$$\begin{aligned} S_K^m(p-m)+S_K^m(q-1)<S_K^m(q)+S_K^m(p-m-1). \end{aligned}$$

(38)

Since $q-1\ge \max \{p,q-m-1\}$, by (36), then $S_K^m(p)+S_K^m(q-m-1)<S_{K}^m(p-m)+S_K^m(q-1)$. Hence, by (38), we obtain $S_K^m(p)+S_K^m(q-m-1)<S_{K}^m(q)+S_K^m(p-m-1)$.

When $p<m+1$, then $S_K^m(p-m-1)=0$. If $q<m+1$, we also have $S_K^m(q-m-1)=0$. Since $p<q$, then $S_K^m(p)<S_K^m(q)$. If $q\ge m+1$, then $1\le p,q-m$ and $p+q-m<\frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right) -\frac{m}{2}$. Hence, by (36), we obtain $S_K^m(p)+S_K^m(q-m)\le S_K^m(p+q-m)$. Assume $q\le \frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right)$, then $p+q-m\le q$. Since $S_K^m(t)$ is an increasing sequence for all $0\le t\le \frac{n}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right)$, then $S_K^m(p+q-m)\le S_K^m(q)$. Assume $\frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right) <q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$, then $\left( {\begin{array}{c}m\\ 2\end{array}}\right) -q<\frac{1}{2}\left( {\begin{array}{c}m\\ 2\end{array}}\right)$. By Lemma 2, $S_K^m(q)=S_{K}^{m}(\left( {\begin{array}{c}m\\ 2\end{array}}\right) -q)$. Since $p+q-m<\left( {\begin{array}{c}m\\ 2\end{array}}\right) -q<\left( {\begin{array}{c}m\\ 2\end{array}}\right)$, we also have $S_K^m(p)+S_K^m(q-m-1)<S_K^m(p)+S_K^m(q-m)\le S_K^m(p+q-m)<S_K^m(\left( {\begin{array}{c}m\\ 2\end{array}}\right) -q)=S_K^m(q)$. By the above discussion, we always have $S_K^m(p)+S_K^m(q-m-1)<S_K^{m}(q)+S_K^m(p-m-1)$ for $0\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$. By (37), we obtain that

$$\begin{aligned} S_K^{m+1}(q-1)+S_K^{m+1}(p)<S_K^{m+1}(q)+S_K^{m+1}(p-1), \end{aligned}$$

for $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}m+1\\ 2\end{array}}\right)$. So, by induction, if $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right)$, then

$$\begin{aligned} S_K^{n}(p)+S_K^{n}(q-1)<S_K^n(q)+S_K^n(p-1). \end{aligned}$$

(39)

By (39), we obtain

$$\begin{aligned} S_K^{n}(p+q-t)+S_K^{n}(t)<S_K^{n}(p+q-t+1)+S_K^{n}(t-1) \end{aligned}$$

for all $1\le t\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right)$. Hence, we can get

$$\begin{aligned} S_K^{n}(p)+S_K^{n}(q)+p-1<S_K^{n}(p+q)+S_K^{n}(0) \end{aligned}$$

for $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right)$. Since $S_K^{n}(0)=1$, then we have

$$\begin{aligned} S_K^{n}(p)+S_K^{n}(q)+p-2<S_K^{n}(p+q) \end{aligned}$$

for $1\le p<q$ and $p+q\le \frac{1}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right)$. $\square$