Reconstructing Permutations from Cycle Minors

The ith cycle minor of a permutation p of the set {1, 2, . . . , n} is the permutation formed by deleting an entry i from the decomposition of p into disjoint cycles and reducing each remaining entry larger than i by 1. In this paper, we show that any permutation of {1, 2, . . . , n} can be reconstructed from its set of cycle minors if and only if n ≥ 6. We then use this to provide an alternate proof of a known result on a related reconstruction problem. 1 Background and Notation For any positive integer n, let [n] denote the set {1, 2, 3, . . . , n}. Let Sn be the set of all permutations of [n]. Consider a permutation p ∈ Sn and the corresponding sequence p(1), p(2), . . . , p(n), which we abbreviate p1p2 . . . pn. Definition. Let n ≥ 2, p ∈ Sn and i ∈ [n]. The ith sequence reduction of p, denoted p ↓ i, is the permutation of [n− 1] formed by first deleting pk = i from the p1p2 . . . pn and then decreasing any number greater than i in the resulting sequence by 1. For instance, 13425 ↓ 3 = 1324, because we first remove the 3 from 13425, leaving 1425, after which we decrease the 4 and the 5 by 1. We denote by R(p) the set of all sequence reductions of p and by M(p) the multiset of all sequence reductions of p. For example, R(13425) = {2314, 1234, 1324, 1342} and M(13425) = {2314, 1234, 1324, 1324, 1342}. Several reconstruction problems related to sequence reductions have been formulated. One such problem asks for which n can any permutation of length n be uniquely reconstructed from its set of sequence reductions. The analogous reconstruction problem for multisets of sequence reductions has also been investigated. Formally, these problems are equivalent to asking for which n is the restriction of R (or M , respectively) to Sn an injective map. the electronic journal of combinatorics 16 (2009), #R19 1 These problems are motivated by the famous Ulam Conjecture [7], which states that a graph with n ≥ 3 vertices can be reconstructed from its multiset of induced (n−1)-vertex subgraphs. Harary [2] conjectured further that if n ≥ 4, we can reconstruct a graph with n nodes from its set (ignoring multiplicity) of induced (n − 1)-vertex subgraphs. The problem of reconstructing from sequence reductions is a natural analogue of the Ulam conjecture for permutations in the following sense. The inversion graph of a permutation p ∈ Sn is the graph with vertices labeled 1, 2, . . . , n and with an edge between vertices i and j with i < j if and only if i is to the right of j in the sequence p1p2 . . . pn. The (n−1)vertex subgraphs of the inversion graph of a permutation p are isomorphic (ignoring the labels) to the inversion graphs of the corresponding sequence reductions of p. The following theorem has been proven independently by Ginsburg [1], Ince [3], Raykova [5], and Smith [6]. Theorem 1.1. Let n ≥ 5 be a positive integer, and let p and q be two permutations in Sn. Then R(p) = R(q) implies that p = q (and thus M(p) = M(q) implies that p = q). In addition, there are counterexamples for n = 2, 3, and 4. We have M(3142) = M(2413), M(312) = M(231), and M(12) = M(21), and the same counterexamples hold for R. In this paper we solve a natural variant on the problem of reconstructing permutations from their sequence reductions. We also use this variant to provide an alternate proof of Theorem 1.1. 1.1 Cycle Minors Rather than considering a permutation p ∈ Sn as a sequence consisting of the numbers in [n], we consider the decomposition of p into disjoint cycles, i.e. a composition of disjoint cycles of the form (i, p(i), p(p(i)), . . .). Definition. Let n ≥ 2, p ∈ Sn and i ∈ [n]. Then the ith cycle minor of p, denoted p ⇓ i, is the permutation of [n − 1] formed by first deleting the entry i from the decomposition of p into disjoint cycles, and then subtracting 1 from any number greater than i. For example, suppose n = 9 and p = (1546)(279)(3)(8). The permutation p can be represented by the directed graph shown in Figure 1. Figure 1: The directed graph associated with the permutation (1546)(279)(3)(8). Then p ⇓ 5 is the permutation (145)(268)(3)(7), which has the directed graph shown in Figure 2, where new edges and labels are shown in red. the electronic journal of combinatorics 16 (2009), #R19 2 Figure 2: The cycle minor (1546)(279)(3)(8) ⇓ 5 = (145)(268)(3)(7). Note that there are multiple ways of writing a given permutation as a product of disjoint cycles. For instance, the permutation (21)(3) can also be written (3)(12). By considering the representation of a permutation as a directed graph, it is clear that the definition of cycle minor is independent of such choices. Definition. We denote the set of cycle minors of p by C(p) and the multiset of cycle minors of p by M(p). We will also use the following conventions throughout the next section. If p(a) = b, we write a 7→ b in p. We say that a and b are adjacent in p if either a 7→ b or b 7→ a in p. We also say that a permutation p “contains” the k-cycle (a1a2 . . . ak), or the k-cycle is “in” p, if it appears in the decomposition of p into disjoint cycles. 2 Reconstruction from Cycle Minors We now state our main result. Theorem 2.1. Suppose n ≥ 6 and p, q ∈ Sn such that C(p) = C(q). Then p = q. Furthermore, there are counterexamples for n = 2, 3, 4, and 5: C((12)) = C((1)(2)) = {(1)}, C((123)) = C((132)) = {(12)}, C((13)(24)) = C((14)(23)) = {(2)(13), (1)(23), (3)(12)}, C((14253)) = C((13524)) = {(1423), (1342), (1324), (1243)}. The counterexample for n = 5 is the only pair of permutations in S5 with the same set of cycle minors. To prove Theorem 2.1, we first provide several preliminary lemmas. Lemma 2.2. Let n ≥ 3. Suppose p, q ∈ Sn and C(p) = C(q). Then C(p ⇓ n) = C(q ⇓ n) and C(p ⇓ 1) = C(q ⇓ 1). Proof. First, notice that for any p ∈ Sn and any 1 ≤ i < j ≤ n, the permutations p ⇓ j ⇓ i and p ⇓ i ⇓ j − 1 are each formed by deleting i and j simultaneously from the the electronic journal of combinatorics 16 (2009), #R19 3 cycle representation of p and subsequently subtracting 1 from any number that is between i and j and subtracting 2 from any number that is greater than j. Hence, we obtain p ⇓ i ⇓ (j − 1) = p ⇓ j ⇓ i (2.1) for any 1 ≤ i < j ≤ n. From the definition of C(p ⇓ n) and using (2.1) repeatedly, we find C(p ⇓ n) = {p ⇓ n ⇓ i | 1 ≤ i ≤ n − 1} = {p ⇓ i ⇓ (n − 1) | 1 ≤ i ≤ n − 1} = {p ⇓ i ⇓ (n − 1) | 1 ≤ i ≤ n} = {r ⇓ (n − 1) | r ∈ C(p)} Similarly, we have C(q ⇓ n) = {t ⇓ (n − 1) | t ∈ C(q)} Since C(p) = C(q) by assumption, it follows that C(p ⇓ n) = C(q ⇓ n). A similar argument shows that C(p ⇓ 1) = C(q ⇓ 1). Lemma 2.3. Suppose n ≥ 2 and p, q ∈ Sn such that p ⇓ 1 = q ⇓ 1 and p ⇓ n = q ⇓ n. Then one of the following is true: