58th Annual IEEE Symposium on Foundations of Computer Science
The independence number of the Birkhoff polytope graph, and applications to maximally recoverable codes
Daniel Kane Shachar Lovett Sankeerth Rao Department of Computer Science University of California, San Diego
Abstract—Maximally recoverable codes are codes de- recovery from single node failure and optimal recov- signed for distributed storage which combine quick recov- ery from catastrophic failure. More precisely, they are ery from single node failure and optimal recovery from systematic linear codes which combine two types of catastrophic failure. Gopalan et al [SODA 2017] studied redundancy symbols: local parity symbols, which allow the alphabet size needed for such codes in grid topologies for fast recovery from single symbol erasure; and global and gave a combinatorial characterization for it. Consider a labeling of the edges of the complete bipartite parity symbols, which allow for recovery from the max- d graph Kn,n with labels coming from F2, that satisfies the imal information theoretic number of erasures. This was following condition: for any simple cycle, the sum of the further studied in [1], [7], [9], [11], [12]. labels over its edges is nonzero. The minimal d where this The present paper is motivated by a recent work of is possible controls the alphabet size needed for maximally Gopalan, Hu, Kopparty, Saraf, Wang and Yekhanin [5], recoverable codes in n × n grid topologies. Prior to the current work, it was known that d is between which studied the effect of the topology of the network 2 log(n) and n log n. We improve both bounds and show on the code design. Concretely, they studied grid like that d is linear in n. The upper bound is a recursive topologies. In the simplest setting, a codeword is viewed construction which beats the random construction. The as an n × n array, with entries in a finite field F2d , lower bound follows by first relating the problem to the where there is a single parity constraint for each row and independence number of the Birkhoff polytope graph, and then providing tight bounds for it using the representation each column, and an additional global parity constraint. theory of the symmetric group. More generally, a Tn×m(a, b, h) maximally recoverable d code has codewords viewed as an n×m matrix over F2, I. INTRODUCTION with a parity constraints per row, b parity constraints per The Birkhoff polytope is the convex hull of n × n column, and h additional global parity constraints. An doubly stochastic matrices. The Birkhoff polytope graph important problem in this context is, how small can we d is the graph associated with its 1-skeleton. This graph choose the alphabet size 2 and still achieve information is well studied as it plays an important role in combi- theoretical optimal resiliency against erausers. natorics and optimization, see for example the book of Gopalan et al. [5] gave a combinatorial characteriza- Barvinok [2]. For us, this graph arose naturally in the tion for this problem, in the simplest setting of m = n study of certain maximally recoverable codes. Our main and a = b = h =1. Their characterization is in terms of technical results are tight bounds on the independence labeling the edges of the complete bipartite graph Kn,n d number of the Birkhoff polytope graph, which translate by elements of F2, which satisfy the property that in to tight bounds on the alphabet size needed for maxi- every simple cycle, the sum is nonzero. d mally recoverable codes in grid topologies. Let [n]={1,...,n}. Let γ :[n] × [n] → F2 be We start by describing the coding theory question that a labeling of the edges of the complete bipartite graph motivated the current work. Kn,n by bit vectors of length d. γ n × n → Fd A. Maximally recoverable codes Definition I.1. A labeling :[ ] [ ] 2 is simple cycle free if for any simple cycle C in Kn,n it holds that Maximally recoverable codes, first introduced by Gopalan, Huang, Jenkins and Yekhanin [6], are codes γ(e) =0 . designed for distributed storage which combine quick e∈C
0272-5428 2017 252 U.S. Government Work Not Protected by U.S. Copyright DOI 10.1109/FOCS.2017.31 Gopalan et al. [5] showed that the question on the Theorem I.4. Let H be an Abelian group. Let γ :[n] × minimal alphabet size needed for maximally recoverable [n] → H be simple cycle free. Then |H|≥2n/2−2. codes, reduces to the question of how small can we take As a side remark, we note that the study of graphs d = d(n) so that a simple cycle free labeling exists. with nonzero circulations was instrumental in the re- Concretely: cent construction of a deterministic quasi-polynomial • The alphabet size needed for Tn×n(1, 1, 1) codes is algorithm for perfect matching in NC [4]. However, 2d(n). beyond some superficial similarities, the setup seems • The alphabet size needed for Tn×m(a, b, h) codes inherently different than ours. For starters, they study is at least 2min(d(n−a+1),d(m−b+1))/h. general bipartite graphs, while we study the complete Before the current work, there were large gaps be- graphs. Moreover, they need to handle certain families d n tween upper and lower bounds on ( ). For upper of cycles, not necessarily simple, while in this work we K nO(n) bounds, as the number of simple cycles in n,n is , focus on simple cycles. d O n n a random construction with = ( log ) succeeds The proofs of Theorem I.3 and Theorem I.4 rely on with high probability. There are also simple explicit the study of a certain Cayley graph of the permutation constructions matching the same bounds, see e.g. [6]. In group, which encodes the property of simple cycle free d ≥ n terms of lower bounds, it is simple to see that log labeling. Surprisingly, the corresponding graph is the is necessary. The main technical lemma of Gopalan et 2 Birkhoff polytope graph. al. [5] in this context is that in fact d ≥ Ω(log n) is necessary. This implies a super-polynomial lower bound C. The Birkhoff polytope graph d n on the alphabet size 2 in terms of , which is one of Let Sn denote the symmetric group of permutations their main results. on [n]. A permutation τ ∈ Sn is said to be a cycle We improve on both upper and lower bounds and if, except for its fixed points, it contains a single non- show that d is linear in n. We note that our construction trivial cycle (in particular, the identity is not a cycle). improves upon the random construction, which for us We denote by Cn ⊂ Sn the set of cycles. The Cayley was somewhat surprising. For convenience we describe graph Bn = Cay(Sn, Cn) is a graph with vertex set Sn n it when is a power of two, but note that it holds for and edge set {(π, τπ):π ∈ Sn,τ ∈Cn}. Note that this n −1 any with minimal modifications. graph is undirected, as if τ ∈Cn then also τ ∈Cn. B Theorem I.2 (Explicit construction). Let n be a power The graph n turns out to be widely studied: it is the of two. There exists γ :[n]×[n] → Fd for d =3n which graph of the Birkhoff polytope, which is the convex hull 2 n × n is simple cycle free. of all permutation matrices. See for example [3] for a proof. Our analysis does not use this connection; Our main technical result is a nearly matching lower we use the description of the graph as a Cayley graph. bound. The following claim shows that Theorem I.4 reduces d Theorem I.3 (Lower bound). Let γ :[n] × [n] → F2 be to bounding the size of the largest independent set in the simple cycle free. Then d ≥ n/2 − 2. Birkhoff polytope graph. H B. Labeling by general Abelian groups Claim I.5. Let be an Abelian group. Assume that γ :[n]×[n] → H is simple cycle free. Then Bn contains The definition of simple cycles free labeling can be an independent set of size ≥ n!/|H|. extended to labeling by general Abelian groups, not just d F2. Let H be an Abelian group, and let γ :[n] × [n] → Proof. Define H. We say that γ is simple cycle free if for any simple n C cycle , A = π ∈ Sn : γ(i, π(i)) = h , sign(e)γ(e) =0 . i=1 e∈C where h ∈ H is chosen to maximize the size of A. Thus where sign(e) ∈{−1, 1} is an alternating sign assign- |A|≥n!/|H|. We claim that A is an independent set in ment to the edges of C (these are sometimes called Bn. circulations). We note that the analysis of Gopalan et Assume not. Then there are two permutations π, π ∈ −1 al. [5] can be extended to non-binary alphabets Fp,in A such that τ = π(π ) ∈Cn. Let Mπ = {(i, π(i)) : which case their combinatorial characterization extends i ∈ [n]} denote the matching in Kn,n associated with to the one above with H = Fp. π, and define Mπ analogously. Let C = Mπ ⊕ Mπ
253 denote their symmetric difference. The fact that τ ∈Cn a) Organization.: We prove Theorem I.2 in Sec- has exactly one cycle, is equivalent to C being a simple tion II and Theorem I.8 in Section III. Theorem I.7 is cycle. Let sign(·) be an alternating sign assignment to proved in Appendix A. the edges of C. Then b) Acknowledgements.: We thank Ran Gelles and Sergey Yekhanin for useful discussions on the problem sign(e)γ(e)= γ(e)− γ(e)=h−h =0. and comments on a preliminary version of this paper. e∈C e∈Mπ e∈M π We thank Igor Pak for bringing to our attention that the γ This violates the assumption that is simple cycle free. Cayley graph which we study is the Birkhoff polytope graph. The construction of a simple cycle free labeling in II. A CONSTRUCTION OF A SIMPLE CYCLE FREE Theorem I.2, combined with Claim I.5, implies that the LABELING Birkhoff polytope graph contains a large independent set. We prove Theorem I.2 in this section. We first intro- n B n n Corollary I.6. Let be a power of two. Then n duce some notation. For x ∈ [n] denote by e ∈ F ≥ n / n x 2 contains an independent set of size ! 9 . the unit vector with 1 in coordinate x and 0 in all other n ∈ Fn We also give in the appendix a construction of a larger coordinates. We let 0 2 denote the all zero vector. independent set in the Birkhoff polytope graph, not based Let n be a power of two. We define recursively a γ n × n → F3n n on a simple cycle free labeling. labeling n :[] [ ] 2 .For =2set (for example) Theorem I.7. Let n be a power of two. Then Bn contains n 6 6 an independent set of size ≥ n!/4 . γ2(0, 0) = e1,γ2(0, 1) = e2, γ , e6,γ , e6. The best previous bounds we are aware of are by 2(1 0) = 3 2(1 1) = 4 B Onn [8] who proved√ that n contains an independent n> x x n/ y y Ω( n) Assume 2. Let = mod ( 2) and = set of size ≥ n . 3n mod (n/2), where x ,y ∈ [n/2]. Define γn(x, y) ∈ F2 Our main technical result is an upper bound on recursively as the largest size of an independent set in the Birkhoff n (i) The first n bits of γn(x, y) are e if y ≤ n/2, and polytope graph. x otherwise they are 0n. B n/2 Theorem I.8. The largest independent set in n has size (ii) The next n/2 bits of γn(x, y) are ey if x ≤ n/2, (n−4)/2 ≤ n!/2 . and otherwise they are 0n/2. n/ γ x, y As a side remark, we note that general bounds on the (iii) The last 3 2 bits of n( ) are defined recur- γ x,y independence number of graphs, such as the Hoffman sively to be n/2( ). bound, give much weaker bounds. A standard application We claim that γn is indeed simple cycle free. For n = of the Hoffman bound gives a much weaker bound for 2 it is simple to verify this directly, so assume n>2. the independence number of Bn of O(n!); and if we Let C be a simple cycle in Kn,n, and assume towards restrict all permutations to have the same sign, the bound a contradiction that e∈C γn(e)=0. Assume C has improves to O((n−1)!). The reason is that the Hoffman 2k nodes, for some 2 ≤ k ≤ n, and let these be bounds (at least in its simplest form) directly relates to C =(x1,y1,x2,y2,...,xk,yk,x1). We denote X = the minimal eigenvalues of the graph. However, in our {x1,...,xk} and Y = {y1,...,yk}. Define furthermore case the eigenvalues are controlled by the irreducible L = {1,...,n/2} and U = {n/2+1,...,n}. representations of Sn, and the extreme eigenvalues are Claim II.1. Either Y ⊂ L or Y ⊂ U. given by low dimensional representations. This prohibits obtaining strong bounds on the independence number Proof. Assume that both Y ∩L and Y ∩U are nonempty. directly. Then there must exist i ∈ [k] with yi ∈ L and yi+1 ∈ U, In order to overcome this barrier, our analysis circum- where if i = k then we take the subscript modulo k. vents the effect of the low dimensional representations Recall that xi+1 is the neighbour of yi,yi+1 in C. Its n en by appealing to a structure vs. randomness dichotomy contribution to the first bits of the sum is xi+1 , since specialized for our setting. It allows us to either reduce yi ≤ n/2 and yi+1 >n/2. Note that no other edge in C the dimension of the ambient group, or restrict to pseudo- has a nonzero value in coordinate xi+1. Thus the xi+1 random assumptions about the actions of the low dimen- coordinate in the sum over C is 1, which contradicts the sional representations. assumption that the sum over C is zero.
254 Thus we can assume from now on that either Y ⊂ L b) Non-Uniform Action on Tuples.: Let [n]m = or Y ⊂ U. {(i1,...,im):i1,...,im ∈ [n] distinct} denote the family of ordered m-tuples of distinct elements of [n]. Its Claim II.2. Either X ⊂ L or X ⊂ U. size is (n)m = n(n − 1) ···(n − m +1). A permutation Proof. Assume that Y ⊂ L, and the case of Y ⊂ U is π ∈ Sn acts on [n]m by sending I =(i1,...,im) identical. Assume that both X ∩ L and X ∩ U are both to π(I)=(π(i1),...,π(im)). Below when we write · nonempty. Then there must exist i ∈ [k] with xi ∈ L and Prπ∈A[ ] we always mean the probability of an event π ∈ A xi+1 ∈ U. Recall that yi is the neighbour of xi,xi+1 under a uniform choice of . m in C. Its contribution to the 2nd batch (of n/2 bits) Notice that if Prπ∈A[π(I)=J] ≥ c /(n)m for some n/2 I,J ∈ n of the sum is e , since xi ≤ n/2 and xi+1 >n/2. pair [ ]m, this will allow us to reduce to a yi Note that no other edge in C has a nonzero value in lower dimensional version of the problem. In particular, A {π ∈ A π I J} coordinate n+yi, where we here we need the assumption if we let = : ( )= , we note that |A|≤|A| n /cm that Y ⊂ L or Y ⊂ U. Thus the n + yi coordinate in ( )m . On the other hand, after multiplying the sum over C is 1, which contradicts the assumption on the left and right by appropriate permutations (an that the sum over C is zero. operation which doesn’t impact our final problem), we can assume that I = J = {n − m +1,...,n}. Then, if A B A Thus we have that X ⊂ U or X ⊂ L, and similarly were an independent set for n, would correspond Y ⊂ U Y ⊂ L C K to an independent set for Cay(Sn−m, Cn−m). Then, if or . Thus, is a simple cycle in n/2,n/2 a we could prove the bound that |A |≤ n−m n − m , embedded in Kn,n in one of four disjoint ways: L × L, c ( )! |A|≤ a n L × U, U × L or U × U. Observe that in each of these we could inductively prove that cn !. copies, the last 3n/2 coordinates of the sum are precisely c) Uniform Action on Tuples.: When the action of A m m γn/2, so by induction C cannot have zero sum. on -tuples is near uniform for all , we will attempt to show that two elements of A differ by a simple cycle using techniques from the Fourier analysis of Sn. In fact, III. THE INDEPENDENCE NUMBER OF THE BIRKHOFF we will show the stronger statement that some pair of POLYTOPE GRAPH elements of A differ by a single cycle of length n. We prove Theorem I.8 in this section. Let A be an Some slight complications arise here when parity of n independent set in Bn. We prove an upper bound on the the permutations here is considered. In particular, all - a size of A. Concretely, we will show that |A|≤ cn n! for cycles have the same parity. This is actually a problem n some absolute constants a, c >√1. As we will see at the for even, as all such cycles will be odd, and thus our end, the choice of a =4,c= 2 works. statement will fail if A consists only of permutations The proof relies on representation theory, in particular with the same parity. Thus, we will have to consider our n representation theory of the symmetric group. We refer statement only in the case of odd. Even in this case readers to the excellent book of Sagan [10], which though, parity will still be relevant. In particular, note A provides a thorough introduction to the topic. We will try that the difference between two permutations in can be n to adhere to the notations in that book whenever possible. a cycle of length only if the initial permutations had the a) Overall Strategy.: Our basic plan will be to same parity. Thus, we lose very little by restricting our A break our analysis into two cases based on whether or attention to only elements of with the more common A not the action of A on m-tuples is nearly uniform for parity. This will lose us a factor of 2 in the size of , all m. This will be in analogy with standard structure but will make our analysis somewhat easier. We are now vs. randomness arguments. If the action on m-tuples is prepared to state our main technical proposition: highly non-uniform, this will allow us to take advantage Proposition III.1. Let n be an odd integer and let c>1 of this non-uniformity to reduce to a lower-dimensional be a sufficiently small constant. Let A ⊂ Sn be a set of case. On the other hand, if A acts nearly uniformly permutations satisfying: on m-tuples, this suggests that it behaves somewhat A randomly. This intuition can be cashed out usefully by (i) All elements of are of the same sign. (ii) For any even m 255 d) Remark.: In the second condition above, we hm =(n − m, 1, 1,...,1) for 0 ≤ m ≤ n − 1 denote consider only even m. This is because if this condition the partition corresponding to a hook. Then fails, we are going to use our other analysis to recursively m λ (−1) if λ = hm consider permutations of [n − m], and would like n − m χ ((n)) = . (3) 0 otherwise to also be odd. We prove Proposition III.1 below, and then show that Thus we can simplify Equation (2) to it implies Theorem I.8. n −1 m h ϕ, ψ = (−1) χ m (ϕ). (4) Proof. First, note that by replacing all π ∈ A by πσ m=0 σ for some odd permutation if necessary, it suffices to g) Bounding the characters on ϕ.: The character assume that all π ∈ A are even. We will assume this h0 corresponds to the trivial representation, and by our henceforth. definition of ϕ it equals χh0 (ϕ)=1. Observe that we e) Rephrasing the problem using class functions.: χλ ϕ can simplify ( ) as Let Cn denote the set of n-cycles in Sn. Define two class λ 1 λ −1 −1 functions ϕ, ψ ∈ R[Sn] as χ ϕ χ σπ π σ ( )=|A|2|S | ( ( ) ) (5) n π,π ∈A,σ∈Sn ϕ 1 σπ π −1σ−1 = |S ||A|2 ( ) 1 λ −1 n = χ (π(π ) ). (6) σ∈Sn,π,π ∈A |A|2 π,π ∈A ψ 1 τ. = |C | First, we argue that the evaluation of characters on ϕ n τ∈C n is always nonnegative. It is easy to see that our conclusion is equivalent to Claim III.2. χλ(ϕ) ≥ 0 for all λ n. showing that ϕ, ψ > 0. ζ ∈ R S ζ 1 π Let λ n denote a partition of n, namely λ = Proof. Let [ n] be given by = |A| π∈A . (λ1,...,λk) where λ1 ≥ ... ≥ λk ≥ 1 and λi = n. Then S The irreducible representations of n are the Specht χλ ϕ 1 Sλ π Sλ π −1 {Sλ λ n} ( )=|A|2 Tr ( ) (( ) ) modules, which are indexed by partitions : . λ π,π ∈A Let χ : Sn → R denote their corresponding charac- λ λ T λ 2 = Tr S (ζ)S (ζ) = S (ζ)F , ters. Their action extends linearly to R[Sn]. Namely, if ζ ∈ R Sn ζ ζππ ∈ R Sn M [ ] is given by = π∈Sn [ ] where where for a matrix its Frobenius norm is given ζ ∈ R χλ ζ ζ χλ π M2 |M |2 π then ( )= π∈Sn π ( ). by F = i,j . In particular it is always As ϕ, ψ ∈ R[Sn] are class functions, their inner nonnegative. product equals h The following lemma bounds χ m (ϕ). Observe that λ λ hm ϕ, ψ = χ (ϕ)χ (ψ). (1) in particular for c =1it gives χ (ϕ)=0. However, λn we would use it to obtain effective bounds when c>1. Let (n) ∈Cn be a fixed cycle of length n. As all elements Lemma III.3. Let m ∈{1,...,n− 1}. For any even k λ λ hm c −1 in ψ are conjugate to (n),wehaveχ (ψ)=χ ((n)) k ∈{m,...,n} it holds that χ (ϕ) ≤ k . (m) and we can simplify Equation (1) to Proof. Let M μ denote the (not irreducible) Young mod- λ λ ϕ, ψ = χ (ϕ)χ ((n)). (2) ule associated with a partition μ n. In the case of λn μ = hk it corresponds to the action of Sn on [n]k. That π ∈ S M hk π Thus, we are lead to explore the action of the irreducible is, for any n we have that ( ) is a matrix I,J ∈ n characters on the full cycle (n). whose rows and columns are indexed by [ ]k M hk π f) Characters action on the full cycle.: The respectively, where ( )I,J =1π(I)=J . Observe that h −1 h T Murnaghan-Nakayama rule is a combinatorial method M k (π )= M k (π) . We extend this action to λ to compute the value of a character χ on a conjugacy R[Sn] linearly. ζ 1 π ∈ R S class, which in our case is (n). In this special case it Recall that = |A| π∈A [ n]. By assumption is very simple. It equals zero unless λ is a hook, e.g. (ii) in Proposition III.1 we have its corresponding tableaux has only one row and one ck M hk ζ π I J ≤ . column, and otherwise its either −1 or 1. Concretely, let ( ) I,J =Pr[ ( )= ] π∈A (n)k 256 Thus, we can bound the Frobenius norm of M hk (ζ) by A are even permutations, it holds by the definition of ϕ that hk 2 hk 2 M (ζ)F = | M (ζ) | I,J λ 1 λ −1 I,J χ (ϕ)= χ (π(π ) ) |A|2 ck π,π ∈A ≤ | M hk ζ | ck. ( ) I,J = 1 λ −1 λ (n)k χ π π χ ϕ . I,J = |A|2 ( ( ) )= ( ) π,π ∈A This is useful as In particular if λ = hm then λ = hn−1−m. h h h T Tr(M k (ϕ)) = Tr M k (ζ) M k (ζ) Next, we lower bound ϕ, ψ as follows. The dom- M hk ζ 2 ≤ ck. = ( ) F inant terms are χh0 (ϕ)=χhn−1 (ϕ)=1. For any 1 ≤ m ≤ (n − 1)/2 − 1, the corresponding term in The Kostka numbers Kλ,μ denote the multiplicity of m h λ μ Equation (4) appears twice, once as − χ m ϕ and the Specht module S in the Young module M .We ( 1) ( ) once as − n−1−mχhn−1−m ϕ − mχhm ϕ . The can thus decompose ( 1) ( )=( 1) ( ) term for m =(n − 1)/2 appears once. μ λ χhm ϕ ≥ m Tr(M (ϕ)) = Kλ,μχ (ϕ). Furthermore, as ( ) 0 for all by Claim III.2, λ the only negative terms correspond to odd 1 ≤ m ≤ n − / λ ( 1) 2. Thus we can lower bound We saw that χ (ϕ) ≥ 0 for all λ. By Young’s rule, Kλ,μ 2m equals the number of semistandard tableaux of shape λ 1 c − 1 ϕ, ψ ≥1 − . (7) and content μ. In particular, it is always a nonnegative 2 2m m≥1,modd m integer. In the special case of λ = hm and μ = hk for k ≥ m, Young’s rule is simple to compute and gives It is not hard to show that this√ is positive if c>1 is c k small enough. If we take = 2, the right hand side Kh ,h . of Equation (7) is slightly negative for large enough m m k = m (the limit as m →∞is −0.02451...). However, when 8 h c −1 χ 0 n ≥ 8, the second term can be replaced by 8 rather Recall that is the trivial representation, for which (3) K χh0 ϕ c6−1 1 h0,hk =1and ( )=1. Thus ϕ, ψ than 6 , making our lower bound on 2 at least (3) k 0.057. This completes our proof. χhm ϕ ≤ K χλ ϕ 1+ m ( ) λ,hk ( ) λ h k = Tr(M k (ϕ)) ≤ c . We are now prepared to prove Theorem I.8. Proof. We first prove that if n is odd and if all permu- n! tations in A have the same sign, then |A|≤ (n−1)/2 . We next apply Lemma III.3 to bound χhm (ϕ) for 2 We proceed by induction on n. Firstly, we note that all 1 ≤ m ≤ n − 1.Ifm ≤ n/2 then we can apply if n =1, the bound follows trivially. Lemma III.3 for k =2m and obtain the bound For odd n>1, we note that unless there is some even c2m − 1 m 257 n A m−i m−i We now need to reduce to the case of odd and Let Ti,j = {2 (j − 1) + 1,...,2 j} for 0 ≤ i i consisting only of permutations of the same sign. First, i ≤ m, 1 ≤ j ≤ 2 . Note that {Ti,j : j ∈ [2 ]} is a A m−i restricting to only permutations of the most common partition of [n] for every i, that |Ti,j| =2 and that A sign, we can assume that all permutations in have Ti,2j−1 ∪ Ti,2j is a partition of Ti−1,j . |A| the same sign, losing only a factor of 2 in .Now, We define a sequence of subsets of Sn.For1 ≤ i ≤ m if n is odd, we are done. otherwise, let j be the most 2m−i+1 m−i+1 let Mi = m−i . For any set R of size |R| =2 π n π A 2 likely value of ( ) for taken from . We have that let indi(R, ·) be a bijection between subsets of R of size Prπ∈A[π(n)=j] ≥ 1/n. Without loss of generality, m−i 2 and ZMi . Define A0 = Sn and j = n and we can let A = {π ∈ A : π(n)=n}. Since ⎧ ⎫ A is an independent set in Cay(Sn−1, Cn−1), and since ⎨ 2 i−1 ⎬ n − A π ∈ A π T ,π T ≡ M . 1 is odd, we have i = ⎩ i−1 : indi( ( i−1,j ) ( i,2j−1)) 0mod i⎭ j=1 (n−2)/2 n/2−1 |A|≤n|A |≤n(n − 1)!/2 = n!/2 . 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Bipartite perfect matching 2) ( )= for all i,j = i. is in quasi-nc. In Proceedings of the 48th Annual ACM SIGACT i Symposium on Theory of Computing, pages 754–763. ACM, Proof. We prove the claim by induction on . The case of 2016. i =1follows from the definition of A1. By assumption −1 [5] P. Gopalan, G. Hu, S. Kopparty, S. Saraf, C. Wang, and π, π fix both T1,1 and T1,2. However, as τ = π(π ) is S. Yekhanin. Maximally recoverable codes for grid-like topolo- T T gies. In Proceedings of the Twenty-Eighth Annual ACM-SIAM a cycle, it must be contained in either 1,1 or 1,2. This Symposium on Discrete Algorithms, pages 2092–2108. SIAM, implies that τ(x)=x for all x ∈ T1,1 or all x ∈ T1,2. 2017. Consider next the case of i>1. By induction [6] P. Gopalan, C. Huang, B. Jenkins, and S. Yekhanin. Explicit i−1 π(Ti−1,j )=π (Ti−1,j ) for all j ∈ [2 ]. Moreover, maximally recoverable codes with locality. 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Springer ind ( ( ) ( )) 0mod Science & Business Media, 2013. j=1 [11] I. Tamo and A. Barg. Bounds on locally recoverable codes with multiple recovering sets. In Information Theory (ISIT), 2014 For any j = j we know that IEEE International Symposium on, pages 691–695. IEEE, 2014. π(Ti−1,j)=π (Ti−1,j ) and π(Ti,2j−1)= [12] I. Tamo and A. Barg. A family of optimal locally recoverable π (Ti,2j−1), so indi(π(Ti−1,j ),π(Ti,2j−1)) = codes. IEEE Transactions on Information Theory, 60(8):4661– 4676, 2014. indi(π (Ti−1,j ),π (Ti,2j−1)). Thus we obtain that also indi(π(Ti−1,j ),π(Ti,2j −1)) = APPENDIX indi(π (Ti−1,j ),π (Ti,2j −1)). Moreover, as we We prove Theorem I.7 in this section. Assume that also know that π(Ti−1,j )=π (Ti−1,j ) and that m n n A ⊂ Sn |A|≥n / π T , · Z =2 . We construct of size ! 4 , indi( ( i−1,j ) ) is a bijection to Mi , it must be A B such that is an independent set in n. the case that π(Ti,2j −1)=π (Ti,2j −1) and hence 258 also π(Ti,2j )=π (Ti,2j ). Thus we conclude that i π(Ti,j)=π (Ti,j) for all j ∈ [2 ]. To conclude, as τ = π(π)−1 is a cycle, it must be contained in either Ti,2j −1 or Ti,2j . Thus, τ must fix all points in Ti,2j −1 or all points in Ti,2j .Wesetji ∈ {2j − 1, 2j} accordingly. A |A | Finally, we compute the size of .As i = 2m−i+1 2m−i+1 |Ai−1|/Mi and Mi = 2m−i ≤ 2 we obtain that n n n |A|≥ ! ≥ ! ! . m 2i 2m+1 = n i=1 2 2 4 259