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On the Diameter of the Symmetric Group: Polynomial Bounds

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On the Diameter of the Symmetric Group: Polynomial Bounds On the diameter of the symmetric group: polynomial bounds L´aszl´oBabai∗, Robert Bealsy, Akos´ Seressz Abstract which has G as its vertex set; the pairs g; gs : g G; s S are the edges. Let diam(G; Sff) denoteg the2 We address the long-standing conjecture that all per- diameter2 g of Γ(G; S). mutations have polynomially bounded word length in terms of any set of generators of the symmetric group. The diameter of Cayley graphs has been studied The best available bound on the maximum required in a number of contexts, including interconnection word length is exponential in pn log n. Polynomial networks, expanders, puzzles such as Rubik's cube and bounds on the word length have previously been estab- Rubik's rings, card shuffling and rapid mixing, random lished for very special classes of generating sets only. generation of group elements, combinatorial group theory. Even and Goldreich [EG] proved that finding In this paper we give a polynomial bound on the the diameter of a Cayley graph of a permutation group word length under the sole condition that one of the is NP-hard even for the basic case when the group generators fix at least 67% of the domain. Words is an elementary abelian 2-group (every element has of the length claimed can be found in Las Vegas order 2). Jerrum [Je] proved that for directed Cayley polynomial time. graphs of permutation groups, to find the directed The proof involves a Markov chain mixing esti- distance between two permutations is PSPACE-hard. mate which permits us, apparently for the first time, No approximation algorithm is known for distance in to break the \element order bottleneck." and the diameter of Cayley graphs of permutation As a corollary, we obtain the following average- groups. Strikingly, the question of the diameter of the case result: for a 1 δ fraction of the pairs of Rubik's cube Cayley graph appears to be wide open generators for the symmetric− group, the word length (cf. [Ko]). We refer to [BHKLS] for more information is polynomially bounded. It is known that for almost about the history of the diameter problem and related all pairs of generators, the word length is less than results and to the survey [He] for applications of exp(pn ln n(1 + o(1))). Cayley graphs to interconnection networks. For G = Sn and G = An we conjecture that 1 Introduction diam(G; S) is polynomially bounded in n for all sets S of generators: Let G be a finite group and S a set of generators of G. We consider the undirected Cayley graph Γ(G; S) Conjecture 1.1. [BS1] There exists a constant C such that for all n and all sets S of generators of C G = Sn or An, diam(G; S) n . ∗Department of Computer Science, University of Chicago, ≤ 1100 East 58th Street, Chicago, IL 60637, and Mathemati- This conjecture would imply a quasipolynomial upper cal Institute of the Hungarian Academy of Science. Email: bound on the diameters of all Cayley graphs of all [email protected]. This research was partially supported by the NSF. transitive permutation groups [BS2]. yIDA Center for Communications Research - Princeton, 805 The Conjecture has been verified for very special Bunn Drive, Princeton, NJ 08540. Email: [email protected]. Department of Mathematics, Ohio State University, classes of generating sets. Driscoll and Furst [DF] z 2 231 W. 18th Avenue, Columbus, OH 43210. Email: proved an O(n ) bound for the case when all gen- [email protected]. Partially supported by the NSF. erators are cycles of bounded lengths and McKen- zie [McK] gave a polynomial bound for the case when now able to overcome this obstacle. the generators have bounded degree. (The degree of a permutation is the number of elements actually Corollary 1.1. Fix k 2 and δ > 0. Let G be ≥ moved.) We note that McKenzie's exponent is Θ(k) either Sn or An. Let S be a random subset of size k where k is the bound on the degree. in G and let G1 be the group generated by S. Then, with probability > 1 δ we have diam(G ;S) nC for We extend these results to a large class of gener- 1 a constant C = C(δ−). ≤ ating sets in the hope that this may lead to settling the Conjecture. 2 Proof of the worst-case bound: a Markov Theorem 1.1. Let be a positive constant. Let S be chain argument a set of generators for G = S or G = A : Assume n n We shall use the following fact. We explain the that S contains an element of degree n=(3+). Then concepts involved after the statement. diam(G; S) = O(nC ) where C is an≤ absolute constant (does not depend on ). Moreover, between any pair C Fact 2.1. Let X be a connected undirected graph on of elements of G, a path of length O(n ) can be found n vertices, with maximum degree ∆ and diameter d. in Las Vegas polynomial time. Then the strong mixing time for a lazy random walk on X (starting from any vertex) is O(d∆n log n). We did not attempt to optimize the exponent C; without effort, our proof yields a diameter up- A lazy random walk on the graph means a per bound of n7(log n)O(1). The bound reduces to 6 O(1) stochastic process where a particle moves from vertex n (log n) if the number of generators is bounded. to vertex; at each step, with probability 1/2 the The best upper bound known to hold for the particle does not move; the remaining probability 1/2 diameter of all Cayley graphs of An and Sn is is split evenly between the neighbors. On a connected exp(pn ln n(1 + o(1))). Except for the function im- graph, lazy random walks are ergodic and therefore the plicit in the o(1) term, this bound is the same as the distribution of the particle approaches the stationary largest order of elements of Sn [La]; indeed one of the distribution. operations used in the proof of the diameter bound is In the stationary distribution, each vertex is vis- raising an element to an arbitrary power, which makes ited with a frequency proportional to its degree. So the order of elements a bottleneck. for regular graphs (as will always be the case in our Theorem 1.1 seems to be the first result to over- applications), the stationary distribution is uniform. come the \element order bottleneck." To achieve this, The strong mixing time of a random walk we have to avoid taking powers of permutations as a means the number of steps needed for the random degree-reduction tool. Instead we increase the degree- walk to reach a distribution close to the stationary reducing power of the commutator operation by ap- 1 distribution in the following sense: each state x is plying it to carefully chosen conjugates τ and π− τπ; reached with probability p(x)e± where p(x) is the the conjugating permutation π is obtained as the re- stationary probability. Here 0 < 1=2; the influence sult of a polynomial-length random walk. of the specific value of on the≤ strong mixing time Our worst-case result brings us close to proving a estimates is a factor of log(1/) which we can ignore polynomial upper bound for the \typical case." First as long as is bounded away from zero. we need to recall Dixon's celebrated result: almost The Fact remains true if we allow loops and all pairs of permutations in Sn generate either Sn or multiple edges. The Fact follows from a result of An [Dix] (cf. [Bo, Ba2]). Landau and Odlyzko [LO] stating that the eigenvalue 1 For random pairs of generators, the best diameter gap of the transition matrix is at least (d(∆ + 1)n)− . bound to-date states that the diameter is almost al- The eigenvalue gap refers to the quantity 1 max λ ways at most nln n(1=2+o(1)) [BH]. This expression also where the maximum is taken over all eigenvalues− λ =j 1j describes the order of almost all permutations [ET], so of the transition matrix. Note that 1 is a simple6 this result again suffers from the \element order bot- eigenvalue and the lazyness of the walk guarantees tleneck." In 99% (but not in almost all) cases we are that λ < 1 for all other eigenvalues. j j By standard arguments, the Landau-Odlyzko Choose π with at most average value of A Aπ . eigenvalue gap implies an O(d∆n log n) strong mixing j \ j time. A permutation group G is t-transitive if for any The proof of Theorem 1.1 is based on Lemma 2.2 pair of ordered t-tuples of distinct elements of the below. We first state Lemma 2.1 in order to illustrate permutation domain, (x1; : : : ; xt) and (u1; : : : ; ut), σ the technique on a simple instance. there exists σ G such that for all i, x = ui. 2 i Lemma 2.1. Let G = S be a transitive permutation Lemma 2.2. Let G = S be a (t + 1)-transitive group of degree n. Leth δi > 0 be fixed. Let A [n]; permutation group of degreeh i n: Fix δ > 0. Let ⊂ A = k. Then there exists a word π of length (x1; : : : ; xt) and (u1; : : : ; ut) be t-tuples of distinct j j 2 1 O(n S log n) over S S− such that elements of [n].
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