Reconstruction of permutations distorted by single Kendall τ-errors

Abstract

Levenshtein first put forward the sequence reconstruction problem in 2001. This problem sets a model in which a sequence from some set is transmitted over multiple channels, and the decoder receives the different outputs. In this model, the sequence reconstruction problem is to find the minimum number of channels required to exactly recover the transmitted sequence. In the combinatorial context, the problem is equivalent to determining the maximum intersection between two balls of radius r, where the distance between their centers is at least d. The sequence reconstruction problem was studied for strings, permutations and so on. In this paper, we extend the study by Konstantinova et al. for reconstruction of permutations distorted by single Kendall τ-errors. While they solved the case where the transmitted permutation can be arbitrary and the erroneous patterns are distorted by at most two Kendall τ-errors, we study the setup where the transmitted permutation belongs to a permutation code of length n and the erroneous patterns are distorted by at most three Kendall τ-errors. In this scenario, it is shown that n² − n + 1 erroneous patterns are required in order to reconstruct an unknown permutation from some permutation code of minimum Kendall τ-distance 2 or an arbitrary unknown permutation for any n ≥ 3.

Appendix

In this appendix, we will discuss the size of N(n,d, 3) for all 3 ≤ n ≤ 5 and d = 1, 2 as follows. Assume that there exists some permutation $\upbeta \in \mathcal {S}_{n}$ such that $I(n,d,3)=|{B_{K}^{n}}(\upbeta ,3)\cap {B_{K}^{n}}(\epsilon _{n},3)|$ and w_K(β) = d for any 3 ≤ n ≤ 5 and d = 3, 4.

When n = 3, we can easily obtain I(3, 1, 3) = I(3, 2, 3) = I(3, 3, 3) = 6. Hence we have N(3, 1, 3) = N(3, 2, 3) = 6.

When n = 4, we have I(4, 1, 3) = I(4, 2, 3) = 12. By Lemma 4, we have ${S_{K}^{4}}(0)=1,{S_{K}^{4}}(1)=3,{S_{K}^{4}}(2)=5,~\text {and}~{S_{K}^{4}}(3)=6$. If (i,i + 1) ∈ Iv(β) for each i ∈ [3], then β = [4, 3, 2, 1] and w_K(β) = 6. Since ${S_{K}^{4}}(1)=3$, then |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}|≤ 2 for w_K(β) = 3 or 4. First, we estimate the value of I(4, 4, 3) for w_K(β) = 4. Suppose Iv(e_i) = I₁ such that I₁∉Iv(β) for some i ∈ [3]. Then, there exists a permutation α such that Iv(α) = {I₁,I₂}. Hence, |Iv(β) ∩ Iv(α)|≤ 1 and d_K(β,α) ≥ 4. Since ${S_{K}^{4}}(2)=5$, then we have |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 4 for w_K(β) = 4. Therefore, we have

$$\begin{array}{@{}rcl@{}} I(4,4,3)=\sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 2+4+6=12. \end{array}$$

Second, we estimate the value of I(4, 3, 3) for w_K(β) = 3. When w_K(γ) = 3, we have that |{γ|w_K(γ) = 3,d_K(γ,β ) ≤ 3}|≤ 4. Therefore, we have

$$\begin{array}{@{}rcl@{}} I(4,3,3)=\sum\limits_{i=0}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 1+2+5+4=12. \end{array}$$

So, we can obtain that N(4, 1, 3) = N(4, 2, 3) = 12.

When n = 5, we have I(5, 1, 3) = I(5, 2, 3) = 20. By Lemma 4, we have ${S_{K}^{5}}(0)=1,{S_{K}^{5}}(1)=4,{S_{K}^{5}}(2)=9,~\text {and}~{S_{K}^{5}}(3)=15$. First, we estimate the value of I(5, 4, 3) for w_K(β) = 4. The inversions of all the elements of ${S_{K}^{5}}(\epsilon _{5},1)$ are (1, 2), (2, 3), (3, 4), (4, 5). Suppose (i₀,j₀) ∈ Iv(β) is an inversion with the maximum value of |i₀ − j₀|. It is easily verified that j₀ − i₀ = 2, 3, or 4. When j₀ − i₀ = 4, then (1, 5) ∈ Iv(β). Since w_K(β) = 4 and (1, 5) ∈ Iv(β), it follows that β = [5, 1, 2, 3, 4] or [2, 3, 4, 5, 1]. Without loss of generality, let β = [5, 1, 2, 3, 4]. Thus, Iv(β) = {(1, 5), (2, 5), (3, 5), (4, 5)}. let γ ∈{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}. Since w_K(β) = 4, w_K(γ) = 1, and d_K(γ,β ) ≤ 3, by Lemma 8, then Iv(γ) ⊂ Iv(β). Hence, |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}| = 1. Similarly, let γ ∈{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}, then Iv(γ) ⊂ Iv(β) and |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}| = 1. Since ${S_{K}^{5}}(3)=15$, then |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 15. So, we have

$$\begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 1+1+15=17. \end{array}$$

When j₀ − i₀ = 3, then (1, 4) or (2, 5) ∈ Iv(β). Without loss of generality, let (1, 4) ∈ Iv(β). Since w_K(β) = 4 and (1, 4) ∈ Iv(β), it follows that {(1, 4), (2, 4), (3, 4)} or {(1, 4), (1, 2), (1, 3)}⊂ Iv(β). Consider {(1, 4), (2, 4),(3, 4)}⊂ Iv(β), we easily have that |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}|≤ 2 and |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 3. So, we have

$$\begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 2+3+15=20. \end{array}$$

When j₀ − i₀ = 2, then (1, 3), (2, 4), or (3, 5) ∈ Iv(β). If Iv(β) has at least two elements of {(1, 3), (2, 4), (3, 5)}, then (1, 3), (2, 4) ∈ Iv(β) or (2, 4), (3, 5) ∈ Iv(β). Without loss of generality, consider (1, 3), (2, 4) ∈ Iv(β), it follows that |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}|≤ 2 and |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 3. So, we have

$$\begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 2+3+15=20. \end{array}$$

If Iv(β) has only an element of (1, 3), (2, 4), (3, 5), it is easily verified that β = [3, 2, 1, 5, 4] or [2, 1, 5, 4, 3]. Without loss of generality, let β = [3, 2, 1, 5, 4]. Then we have that |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}| = 3 and |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}| = 4. Obviously, when γ = [4, 1, 2, 3, 5] or [1, 5, 2, 3, 4], we have d_K(β,γ) ≥ 4. Since ${S_{K}^{5}}(3)=15$, then |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 13. So, we have

$$\begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 3+4+13=20. \end{array}$$

By the above discussion, we have that

$$\begin{array}{@{}rcl@{}} I(5,4,3)=\sum\limits_{i=1}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 20. \end{array}$$

Second, we estimate the size of I(5, 3, 3) and w_K(β) = 3. If Iv(β) contains all the inversions of any three distinct elements of ${S_{K}^{5}}(\epsilon _{5},1)$, then |Iv(β)|≥ 4. Hence, |{γ|w_K(γ) = 1,d_K(γ,β ) ≤ 3}|≤ 2. Since ${S_{K}^{5}}(2)=9$, then |{γ|w_K(γ) = 2,d_K(γ,β ) ≤ 3}|≤ 9. Therefore, by (23), we have

$$\begin{array}{@{}rcl@{}} I(5,3,3)=\sum\limits_{i=0}^{3}|\{\gamma|w_{K}(\gamma)=i,d_{K}(\gamma,\upbeta)\leq 3\}|\leq 1+2+9+7=19. \end{array}$$

So, we can get N(5, 1, 3) = N(5, 2, 3) = 20.