Random sampling of minimally cyclic digraphs with given imbalance sequence

Transcription

1 Acta Univ. Sapientiae, Mathematica,? (????) xx yy Random sampling of minimally cyclic digraphs with given imbalance sequence Koko K. Kayibi University of Qatar, Doha, Qatar Shariefuddin Pirzada Department of Mathematics University of Kashmir Srinagar, India Muhammad Ali Khan King Fahd University of Petroleum and Minerals Dhahran 31261, Saudi Arabia Antal Iványi Eötvös Loránd University Faculty of Informatics H-1117, Budapest, Hungary Pázmány Péter sétány 1/C Abstract. The imbalance of a vertex v in a digraph D is defined as a(v) = d + (v) d (v), where d + (v) and d (v) respectively denote the outdegree and indegree of vertex v. The imbalance sequence of D is formed by listing vertex imbalances in nondecreasing order. We define a minimally cyclic digraph as a connected digraph which is either acyclic or has exactly one oriented cycle whose removal disconnects the digraph. In this paper we consider the problem of sampling minimally cyclic digraphs with a given imbalance sequence. We present an algorithm for generating all minimally cyclic graphs realizing an imbalance sequence with each realization produced in linear time. We then construct a Markov chain on the set M of all minimally cyclic digraphs with a given imbalance sequence and prove that this Markov chain is ergodic, reversible and rapidly mixing. Thus we can sample through M in polynomial time Mathematics Subject Classification: 05C20, 05C07, 60J10, 60J15, 65C05 Key words and phrases: imbalance sequence, minimally cyclic digraph, random walk, contraction of an imbalance sequence, ergodic Markov chain, mixing time 1

2 2 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi 1 Introduction In this paper we allow a digraph to have loops and multiple arcs. A digraph is called acyclic if it contains no oriented cycles. If d (v) and d + (v) respectively denote the indegree and outdegree of vertex v in a digraph D then the imbalance [11, 23, 2] of v is defined as a(v) = d + (v) d (v). We form the imbalance sequence of D by listing its vertex imbalances in nondecreasing order (although imbalances can be listed in non-increasing order as well). The set of distinct imbalances of a digraph is called its imbalance set [23, 2]. Mostly the literature on imbalance sequences is concerned with obtaining necessary and sufficient conditions for a sequence of integers to be an imbalance sequence of a simple (or multi) digraph [21, 23, 2]. For r 1, we define an r-digraph [2] as an orientation of a multigraph that contains at most r arcs joining any pair of vertices. An r-digraph is a special case of the (a, b)-digraph [10] in which a pair of vertices are connected by at least a and at most b arcs. In the sequel we shall refer to the following recent characterization of imbalance sequences of r-digraphs: Theorem 1 (Pirzada, Naikoo, Samee, Iványi [2]) A sequence A = [a 1,..., a n ] of integers in nondecreasing order is the imbalance sequence of an r-digraph if and only if a i rk(k n) (1) for 1 k n with equality when k = n. Proof. See [2]. If A is a sequence of integers and D is a digraph that has A as its imbalance sequence, we say that D is a realization of A. A realization is called simple if it is a simple digraph (a digraph without loops and parallel arcs [9, 11, 28]. Mubayi et al. [21] characterized imbalance sequences of simple digraphs and determined the maximum number of arcs in a simple realization. They provided a greedy algorithm for constructing a simple realization with minimum number of arcs. Wang [27] gave an asymptotic formula for the number of labeled simple realizations of an imbalance sequence.

3 Random sampling of minimally cyclic digraphs 3 We say that a realization D is minimally cyclic if D is connected and does not contain a nonempty set of arcs S such that deleting S keeps the digraph connected but preserves imbalances of all vertices. Obviously such a set S, if it exists, must add 0 to the imbalances of vertices incident to it and hence must be a union of oriented cycles. Thus a minimally cyclic digraph is either acyclic or has exactly one oriented cycle whose removal disconnects the digraph. For the sake of brevity we shall use the phrase minimal realization to refer to a minimally cyclic realization of A. We denote the set of all minimal realizations of A by M. Sampling of large sets of combinatorial structures is an often occuring problem [1, 2, 3, 6, 7, 8, 1, 15, 20, 22]. Now, suppose that A is an imbalance sequence and we want to sample all minimal realizations of A. Let D be such a realization. We can generate another realization D of A by adding an oriented cycle C to D. However, the digraph D obtained in this manner is not minimally cyclic. One way of fixing this problem would be to delete from D all the oriented 2-cycles created by the addition of C. But this may disconnect the digraph. Thus we cannot sample all minimal realizations of A by a simple addition and deletion of oriented cycles. The aim of this paper is to construct a Markov chain on the set M by a different approach. We apply a series of arithmetic operations called contractions to the imbalance sequence A. This gives us a chain of imbalance sequences. We consider the set C of all such chains and show that there is a surjection from C to M. We then construct an ergodic Markov chain on C that is reversible and rapidly mixing [18, 19], thus making it possible to sample through M at random. We structure of the paper is the following. After the introductory Section 1 in Section 2 the minimal r is computed which is sufficient to realize a potential imbalance sequence. In Section 3 we define our contraction operation and show that the contraction of an imbalance sequence produces another imbalance sequence. In Section we present an algorithm for generating minimal realizations of an imbalance sequence, with each realization produced in linear time. We also introduce the set C and obtain a surjection from C to M. We show that to every element of C we can associate a unique element of M, by constructing such a minimal realization in polynomial time. In Section 5, we define a Markov chain C mc on C and prove it to be ergodic with uniform stationary distribution. A conductance argument is used in Section 6 to show that C mc is fast mixing. Finally in Section 7 we conclude by constructing a Markov chain M mc on the set M and prove it to be ergodic, reversible and fast mixing.

4 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi 2 Computation of the minimal r Theorem 1 implies the following assertion. Corollary 1 If n 1 and A = (a 1,..., a n ) is an imbalance sequence of a digraph if and only if n a i = 0. (2) Proof. Substituting k = n into (1) we get that (2) is necessary. If r Let us suppose that accordig to (2) the sum of the elements of a potential imbalance sequence P = (p 1,..., p n ) holds (2) is zero and r = max a 1, a n. Then it is easy to construct such r-digraph D whose imbalance sequence is A connecting the vertices having positive imbalance with the vertices having negative imbalance using the prescribed number of arcs. Then it is a natural question the value r min (P) defined as the minimal value of r sufficient for a potential imbalance sequence P to be the imbalance sequence of some r-graph. r min (A) has the following natural bounds. Lemma 1 If A = (a 1,..., a n ) is an imbalance sequence, then an a 1 r min min( a 1, a n ). (3) n The following algorithm Rmin computes r min (A) for a sequence A = (a 1,..., a n ) satisfying (3). Rmin is based on Theorem 1, on the bounds given by Lemma 1 and on the the logarithmic search algorithm described by D. E. Knuth [16, page 10] and is similar to algorithm MinF-MaxG [10, Section.2]. The pseudocode of Rmin follows the conventions used in [5]. Input. n: the number of elements A (n 2); A = (a 1,..., a n ): a nondecreasing sequence of integers. Output. r min (A): the smallest sufficient value of r. Working variables. k: cycle variable; l: current value of the lower bound of r min (A); u: current value of the upper bound of r min (A); S: the current sum of the first k elements of A.

5 Rmin(n, A) Random sampling of minimally cyclic digraphs 5 01 l a n a 1 // line 01 02: initialization of l and u 02 u min(a n, a i ) // line 03 1: 03 while l < u 0 r = l+u 2 05 S = S 0 06 for k 1 to n 1 07 S S + a i 08 if S < rk(k n) 09 l r 10 if l = r r min = l return r 13 go to 03 1 u r 15 r min l // line 15 16: return of the computed minimal r 16 return r min The running time of Rmin is Θ(n log n) since the while cycle executes Θ(log n) times and the for cycle in it requires Θ(n) time. Lemma 2 Algorithm MinG-MaxG computes the values f and g for arbitrary sequence D = (d 1, d 2,..., d n ) in O(n log(d n /(n)) time. Proof. According to Lemma?? F is an element of the interval [ d n /(n 1), 2d n /(n 1) ] and g is an element of the interval [0, f]. Using Theorem B of [?, page 12] we get that O(log(d n /n)) calls of Interval-Test is sufficient, so the O(n) run time of Interval-Test implies the required running time of MinF-MaxG. 3 Contraction of an imbalance sequence Let D be a digraph having n vertices and m arcs. Throughout we assume that the vertices of D are labeled v 1,..., v n according to their imbalances in nondecreasing order while the arcs of D are labeled e 1,..., e m arbitrarily. Let A = [a 1,..., a n ] be the imbalance sequence of D, where a i is imbalance of vertex v i. We define an arithmetic operation, called contraction on A as follows.

6 6 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi Choose any ordered pair (a, a j ) with i j from A. Replace a i by a i + a j, delete a j and sort nondecreasinly the received sequence. Note that we do not require a i and a j to be consecutive terms of A. The new sequence has one less element and we denote it by A/(i, j). We refer to A/(i, j) as a minor of A. Our terminology is inspired by the concept of minor and edge contraction from graph theory. The proof of Theorem 3 explains our choice of terminology. Example 1 Let A = [ 3, 0, 1, 2] where a 1 = 3, a 2 = 0, a 3 = 1 and a = 2. Then A/(1, 2) = [ 3, 1, 2], A/(3, 1) = [ 2, 0, 2]. For computational purposes, we preserve the labeling of of the elements from A into its minors. Thus for A/(1, 2) = [ 3, 1, 2] we have a 1 = 3, a 2 is deleted, a 3 = 1 and a = 2, while for A/(3, 1) = [ 2, 0, 2] element a 1 is deleted, a 2 = 0, a 3 = 2 and a = 2. An imbalance sequence A is an r-imbalance sequence if at least one of its realizations is an r-digraph. We also observe that an r-imbalance sequence is also an (r + 1)-imbalance sequence. The next result allows us to construct imbalance sequences and their realizations recursively. It also establishes a relation between the arithmetic operations of contraction discussed above and the edge contraction operations of graphs. Theorem 2 If A is an r-imbalance sequence, then all its minors are 2rimbalance sequences. Proof. Let A be an imbalance sequence of an r-digraph. Suppose that B = A/(p, q) and let a p and a q be both negative with a p a q. Then b l = a p + a q so that b l < a p. Thus, for all k q, we have b i b i + (q k)a q, q a i, (since all these elements are negative) (since A is a non decreasing sequence).

7 Random sampling of minimally cyclic digraphs 7 Therefore b i q a i (q k)a q a i rk(k n), (since A is an imbalance sequence) (2r)k(k n + 1) (since n k + 2), For q < k n 1, we have b i = a i rk(k n), (since A is an imbalance sequence) (2r)k(k n + 1) for n k = 2 and equality holds when k = n 1. Thus in either case B is an imbalance sequence of a 2r-digraph by Theorem??. By symmetry, we have that Theorem 2 holds if a p and a q are both positive. Now suppose that a p 0 and a q 0 with a p a q. If b l = a p + a q, then b l 0. For all k p, we have b i a i rk(k n), (since A is an imbalance sequence) (2r)k(k n + 1), (since n k + 2). For all p < k l, we have b i a i a p a i rk(k n), (since A is an imbalance sequence) (2r)k(k n + 1), (since n k + 2). For all k > l, we have b i a i + a q a i rk(k n), (since A is an imbalance sequence) (2r)k(k n + 1).

8 8 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi The last inequality holds if n k = 2, with equality when k = n 1. Thus once again B is an imbalance sequence of a 2r-digraph by Theorem??. By symmetry, Theorem 2 holds if a p a q. Suppose that D is a realization of A/(i, j), for some 1 i, j n. We construct a digraph D that is a realization of A from D. Now, if D is a realization of A/(i, j), there is a vertex v i in D having imbalance a i, where a i = a i + a j for a i and a j in A. First, we define an available arc as an arc incident to v i. Suppose that there are p i available arcs directed towards v i and q i available arcs directed away from v i. Let p i, q i, p j and q j be non-negative integers such that p i q i = a i, p i + p j = p i, q i + q j = q i, The input of the following Algorithm 1 are??? p i q i + 1 = a i, () p j q j 1 = a j. (5) Algorithm 1: AddVertex(). Inputs: The imbalances B/(i, j) and B, and a realization of B/(i, j), denoted by D. Output: a realization of B, denoted by D. Let p i be the number of available arcs directed towards v i and q i be the number of available arcs directed away from v i. Step 1. To D add a isolated vertex labelled v j and labelled the new digraph D. Step 2 if a j 0 and p i < a, choose a vertex v of D at random, then add two j new arcs between v and v i such that one is directed towards v and the other is directed towards v i if v v i or add a loop at v if v = v i. Go to Step 2. if a j 0 and p i a, choose an integer p j j with a j p j p i and concede p j available arcs directed towards v i to v j and concede a j p j available arcs directed away from v i to v j. if a j < 0 and q i < a, choose a vertex v of D at random, then add two j new arcs between v and v i such that one is directed towards v and the other is directed towards v i if v v i or add a loop at v if v = v i. Go to Step 2. if a j < 0 and q i a, choose an integer q j j with a j q j q i and concede q j available arcs directed away from v i to v j and concede a j q j available edges directed towards v i to v j

9 Random sampling of minimally cyclic digraphs 9 v i v i v j D D Figure 1: Digraph D constructed from D Step 3. Output D. The new digraph D, thus obtained is a realization of B as adding an oriented cycle in Step 2 preserves the imbalances. We now show that Algorithm 1 gives a minimally cyclic realization of A. Theorem 3 The realization D obtained from Algorithm 1 is a minimally cyclic digraph. Moreover D is obtained in linear time. Proof. Indeed, we have that a i = p i q i (by construction) = a i + a j (by definition of contraction of imbalance sequence). Now, let a and b be imbalances of v i and v j respectively. Then a = p i q i + 1 (by construction) = (p i p j) (q i q j) + 1 = (p i q i ) (p j q j 1) = a i + a j a j = a i. Moreover, by construction, we have that b = a j and all the other vertices v k for k ij, have imbalances a k in D and D. Hence D is a realization of A. Now, for the minimality of D, we first note that equations () and (5) are not satisfied only if a i and a j have opposite sign. So without loss of generality,

10 10 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi we assume that a i 0. Thus it is required that there are at least a i 1 arcs directed towards v i and a j + 1 arcs directed away from v i for equations () and (5) to be satisfied. That is, p i a i 1 and q j a j + 1. Therefore, in Algorithm 1, the loop consisting of checking equations () and (5), and adding a pair of opposite arcs until these equations are satisfied is exited exactly after max{a i 1 p i, a j + 1 q j } iterations (O(n) steps). That is, when equations () and (5) are minimally satisfied. Hence D is a minimal realization. Chain of imbalance sequences A chain of an imbalance sequence A is a sequence of imbalance sequences A ( n), A ( n 1),..., A ( 1) with A ( n) = A and A ( i 1) being the minor of A ( i) for every 1 i n 1. ConstructRecursionChain() gives an algorithm from constructing a recursion chain of A. Algorithm 2: ConstructRecursionChain(). Input: the imbalance sequence A Output: A recursion chain of A. Step 1. Let i = n and let A n = A. Step 2. If i = 1, stop, return (A ( n), A (n 1),..., A ( 1)). Else Step 3. Choose two integers k and j, with k j and 1 k, j i. Let A (i 1) = A ( i)/(j, k). Go back to Step 2. Now, since each contraction in Step 3 of ConstructRecursionChain() reduces the number of entries of an imbalance sequence by 1, the last element A ( 1) of the chain contains exactly one element and so the length of the chain is equal to the number of elements n of the imbalance sequence A. Thus for all 1 i n the sequence A ( i) contains i elements. To every chain of an imbalance sequence A of length n we can associate bijectively a chain of n 1 ordered pairs with i th element equal to (j, k), where A (n i) = A (n i+1) /(j, k). That is (v j, v k ) is contracted to obtain A (n i) from A (n i+1). This bijection allows us to represent every chain of imbalance sequences by the sequence of pairs (j, k). Figure 2 illustrates the procedure for the imbalance sequence A = [2, 1, 0, 3] (arranged in non-increasing order), where the pair (j, k) by which an imbalance sequence is contracted is written next to the arrow. Hence this chain of

11 Random sampling of minimally cyclic digraphs 11 imbalance sequence A can be written as ((1, 2), (, 3), (, 1)). [2,1,0, 3] ( a, a ) 1 2 [3,0, 3] ( a, a ) 3 ( a, a ) [3, 3] 1 [0] Figure 2: A chain of the imbalance sequence A = [2, 1, 0, 3] Heneceforth we will not make any distinction between a chain of an imbalance sequence and the chain of ordered pairs that it corresponds to. Furthermore, we will use the word chain to mean a chain of an imbalance sequence. By using the definition of chain, it is easy to prove the next result, which gives a characterization for a sequence of ordered pairs to be a chain. We say that a label i is terminated if it is the second element of an ordered pair. Lemma 3 Let A be an imbalance sequence having n elements. A string of ordered pairs (i, j) with i j and i, j {1, 2,, n} is a chain of A if and only if the following hold. (i) For all labels i {1, 2,, n} there is at least one ordered pair containing i. (ii) If label i appears as the second element of an ordered pair, then it can not be contained in any subsequent ordered pair. (iii) For all labels i, except exactly one, there is an ordered pair where i is terminated.

12 12 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi Proof. The necessity of conditions (i), (ii) and (iii) is obvious from definition. To prove sufficiency consider a string C of ordered pairs (i, j) satisfying conditions (i), (ii) and (iii). From condition (i) and (iii) C consists of at least n 1 ordered pairs. Condition (iii) ensures that a terminated label does not appear again in C. Thus C contains exactly n 1 ordered pairs and we can think of each ordered pair (i, j) appearing in C as the (i, j)-minor of its predecessor. We now present an algorithm for associating a minimal realization D to any chain C. Let D i denote a realization of the element A i of the chain. Algorithm 2: BuildOneRealization(). Input: Given the recursion chain of A denoted (A ( n), A ( n 1),..., A ( 1)), Output: One realization of A, denoted by D. Step 1. Let i = 1. Let a single isolated vertex, denoted by D ( 1), be the realization of the entry A ( 1). Step 2. If i = n, stop, return D ( n). Else Step 3. Use A ( i + 1) and D ( i) in Algorithm AddVertex() to construct D ( i + 1). Go back to Step 2. The next result shows that BuildOneRealization() is correct and constructs a minimal realization of an imbalance sequence in polynomial time. Theorem Let A be an imbalance sequence having n entries and let C = ((a, b)(c, d) (y, z)) be a chain of an imbalance sequence A. Then there is a minimally cyclic digraph D having n vertices and m arcs with m n such that D is reconstructible from C and D is a realization of A. Moreover, this reconstruction is done in time polynomial in n. Proof. By AddVertex(), the digraph D n which is the output of BuildOne- Realization(), is assured to be a minimal realization of A. Now, AddVertex() constructs D i from D (i 1) in O(n) time and there are n 1 such constructions. Thus BuildOneRealization() runs in O(n 2 ) time... Given an imbalance sequence A/(i, j) and its realization D, there are at least ( p j )( qj ) p i q different digraphs D that can be obtained by this construction i as in Step 3 of Algorithm 1, there are ( p j p i )( qj q i ) ways of distributing arcs of v i to v i and v j. Moreover, if D has i vertices, then there are i 1 ways of choosing a vertex at random in Step 2 of Algorithm 1. We observe that the realization D of imbalance sequence A constructed from realization D of A/(i, j) is not unique in general. Now, suppose that we define a total ordering ϕ on the arcs

13 Random sampling of minimally cyclic digraphs 13 of D and that among all the possible realizations D in Step 3 of Algorithm 1, we choose the one such that the labels of the arcs adjacent to v i in D are the smallest, then such a realization is unique. Suppose for example that we define the ordering a < b < c < d < e < f, then Figure 3 illustrates how two different labeling of edges yield two different realizations. Now, let C = ((a, b)(c, d) (y, z)) be a chain of an imbalance sequence A having n entries. We associate to C a sequence of orderings ϕ = (ϕ (2),..., ϕ (n 1) ) where ϕ (i) is the total ordering on the edge set of D (i). Then, to every pair (C, ϕ) is associated a single minimal realization D of A. From now on, when we refer to a chain C, we mean the pair (C, ϕ). Therefore, we have the following observation. Lemma For any imbalance sequence A, there is a function ψ from the set of chains C of A to the set of minimal realizations M of A. The next result shows that this function is a surjection. Theorem 5 Let D be a minimal realization of the imbalance sequence A. Then there exists a chain (C, ϕ) of A reconstructing D. Proof. Let C be any sequence of pairs (k, l) satisfying the conditions of Lemma 3, where every pair corresponds to an arc of D. For i = 2, 3,..., n 1, let A (i) = A (i+1) /(r, s). Define a total order µ (i) on the vertices of D (i) such that the vertices of D (i) that are adjacent to v r in D are assigned the smallest labels. This induces a total order ϕ (i) on the edges of D (i) (with parallel edges ordered arbitrarily). Let ϕ = (ϕ (n 1), ϕ (n 2),..., ϕ (1) ). Then the chain (C, ϕ) reconstructs D, since Step 1 of Algorithm 1 attaches the edges with smallest labels to v r. The function ψ ensures that, given an imbalance sequence A, there is an ergodic Markov chain on the set M provided there is an ergodic Markov chain on the set C. Thus we have the following result. Proposition 1 If we can sample at random through C then we can sample at random through M. Proof. Suppose we can sample at random a chain (C, ϕ) from C. We can now construct a minimally cyclic digraph D that is reconstructible from (C, ϕ) in polynomial time.

14 1 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi 2 3 a d c f 2 3 b a 1 e [2,1,0, 3] b 1 3 ( a, a ) a f [3,0, 3] e f 1 b c ( a, a ) 3 1 b a [3, 3] 1 1 ( a, a ) 1 [0] Figure 3: Two realizations of A = [2, 1, 0, 3] from two different arc labeling 5 The Markov chain C mc First we define an ergodic Markov chain on the set C. The definitions and notations given here conform to [12] and [13].

15 Random sampling of minimally cyclic digraphs 15 Definition 1 (Elementary change of a chain) Let (C, ϕ) be a chain of a degree sequence A. An elementary change of (C, ϕ) consists of exactly one of the following. (I) If C = ( (i, j) (k, l) ), then, (i) swap i and j and change all occurrences of i into j in all pairs coming after (i, j), (ii) swap k and i and change all occurrences of i into k and all occurences of k into i in all pairs coming after (i, j), (iii) swap k and j and change all occurrences of k into j in all pairs coming after (i, j), (iv) swap i and l and change all occurrences of i into l in all pairs coming after (i, j), (v) swap l and j and change all occurrences of l into j in all pairs coming after (i, j). (II) Choose i with 2 i n 1 and perform an inversion in the canonical ordering ϕ (i). Remark 1 Note that every elementary change is an inversion, and hence is reversible. Definition 2 (Random walk on the set of chains of A) The random walk consists of the following steps. Step 1. Pick any chain (C s, ϕ) of A. Step 2. Toss an unbiased coin. If it lands on head, then stay at C s. If tail is obtained, then choose randomly any elementary change. Step 3. Move to C t if C t is obtained from C s by the elementary change chosen in Step 2. Step. Go back to Step 2. Theorem 6 Let C be the set of all chains of an imbalance sequence A containing n elements. Then the random walk given in Definition 2 is an ergodic Markov chain running on C with uniform stationary distribution π. Proof. Let G be the graph whose vertices are chains of A and two vertices C i and C j are connected by an edge if and only if there is an elementary change transforming C i into C j. It suffices to show that G is a connected non-bipartite graph. Indeed, every chain C i is joined to itself by a loop since we stay at C i with probability 1 2. Hence G is not bipartite. For connectivity, suppose that C i and C j are two chains having the same canonical ordering ϕ but they differ in the string of ordered pairs. We show by induction, that there is a path P

16 16 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi in G connecting C i and C j, where P consists of changing the ordered pairs of C i into ordered pairs of C j one at a time, starting from the first pair, then the second, and so on, until the last pair is reached. We change an ordered pair by swapping its entries with other entries taken from subsequent ordered pairs. First, we say that two ordered pairs are in agreement if they are equal component-wise. Now, suppose, without loss of generality, that C i and C j are not in agreement for the first ordered pair and let C i = ((a, b) ) and C j = ((c, d) ). Then by Lemma 3(i), the vertex C i contains c and d in some subsequent ordered pairs. So swapping a and c and swapping b and c make the first ordered pairs to be in agreement. We then make the second pair to be in agreement, then the third, and so on. Suppose now that we have reached a chain C, where all the first k 1 ordered pairs are in agreement with C j but that the k th are not. Let these pairs be (c, d) in C j and (a, b) in C. Suppose that there is neither c nor d in a subsequent pair of C. Then by Lemma 3(ii), c and d are terminated in one of the k previous ordered pairs of C. This contradicts the inductive hypothesis. Therefore the k th ordered pairs can be made to be in agreement by swapping a and c, then b and d in C. Hence by induction, we can make all the ordered pairs, except the last ones to be in agreement. Assume that we have reached the last pairs. If they are in agreement, then we are done. If not, then unless the inductive hypothesis or Lemma 3 (iii) are contradicted, if the last pair is (a, b) in C, then it is (b, a) in C j. Hence swapping a and b makes it to be in agreement in both strings of pairs. Now, suppose that C i and C j are in agreement for all their ordered pairs, but they differ on the canonical ordering ϕ, then a sequence of inversions (Definition 1(II)) connect them. Therefore the path P exists and hence, the graph G is connected. Moreover, by Remark 1, the stationary distribution π is uniform since every elementary change is reversible. We denote the Markov chain given in Theorem 6 by C mc. 6 Mixing time of C mc Now, we view the Markov chain C mc as a network G whose vertices are all the chains of A and there is a directed edge e = (x, y) if there is an elementary transition from x to y, i.e., the transition probability from x to y, denoted P(x, y) is positive. Intuitively, the aim of this section is to show that the network G has no bottleneck in the sense that there are no cuts between any

17 Random sampling of minimally cyclic digraphs 17 set of vertices S to its complement that block the flow of the Markov chain and thus prevent the Markov chain from mixing rapidly. To make this more precise we need some preliminary definitions which are conform to [13, 17, 26]. If we denote the stationary distribution by π, the capacity c(e) of e = (x, y) is given by c(e) = π(x)p(x, y). Since C mc is reversible (as every elementary transition has a reverse), to every directed edge e = (x, y), there corresponds another edge e 1 = (y, x) such that c(e) = c(e 1 ). Let P denote the set of all simple paths from x to y and P be the set of all simple paths containing edge e. A flow in G is a function F, from the set of simple paths to the real numbers such that p P xy F(p) = π(x)π(y) for all vertices x, y of G such that x y. A flow along an edge e is then defined as F(e) = p P e F(p). For a flow F, a measure of existence of an overload along an edge is given by the quantity ρ(e) where ρ(e) = F(e) c(e), and the cost of the flow F, denoted ρ(f) is given by ρ(f) = max ρ(e). e If a network G, representing a Markov chain, can support a flow of low cost, then it can not have any bottlenecks, and hence its mixing time should be small. This intuition is confirmed by the following Theorem from [26]. Theorem 7 (Sinclair [26]) Let M be an ergodic reversible Markov chain with holding probabilities P(x, x) 1 2 at all states x. The mixing time of M satisfies ( τ x (ϵ) 8(ρ(F)) 2 ln 1 π(x) + ln 1 ). ϵ

18 18 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi In order to make use of Theorem 7 we have to define a flow of low cost for C mc. We do this by first assuming that for all pairs (x, y) of vertices of G we channel a particular commodity, called {}}{ xy, say, and we want the commodity {}}{ xy to be different from commodities for other pairs so that commodities do not mix up. We also like to channel all the π(x)π(y) units of commodity {}}{ xy along a special path ^p, called the canonical path between x and y so that and F(^p) = π(x)π(y) F(p) = 0 for all other paths p between x and y such that p ^p. Now assume that G is a network having N vertices whose degrees are at least L 1 and at most L 2. Also assume that the length of any canonical path is at most M. With these assumptions we have the following observations. e π(x)π(y) 1, i.e., the total quantity of commodities flowing through x y the network is at most one unit. F(e) M, i.e., the total flow along all the edges is at most M. e Since the total number of edges in G is at least 1 2 L N, we have, for all edges 1 F(e) 2M (L 1 n)n, where n, the number of entries in A, is also an upper bound of the number of parallel edges between two vertices of G. Lemma 5 ρ(f) 2nM. Proof. For all edges e = (x, y), we have c(e) π(x) P(x,y) n 1 1 N nl and thus 2 ρ(e) = F(e) c(e) 2nM, since, for any edge e = (x, y), we have that P(x, y) n L 2 as there are a maximum of n multiple edges between x and y. Finally, in order to prove that the Markov chain C mc is fast mixing, we need to show that M is a polynomial in n. From the definition of canonical path used in the proof of Theorem 6, only 2n 2 swaps are required to change one chain into another through the canonical path as there are exactly n 1 pairs. Thus we have the following observation.

19 Random sampling of minimally cyclic digraphs 19 Lemma 6 Let M denote the length of a canonical path. Then M 2n 2. Theorem 8 The mixing time of C mc satisfies ( τ x (ϵ) 128mn 5 ln m + 2 ln n m ln ϵ ), mn where m is the number of edges in any realization of A. Proof. By Theorem 7 and Lemmas 5 and 6, we have ( τ x (ϵ) 32n 2 (2n 2) 2 ln 1 π(x) + ln 1 ). ϵ Now, C mc is uniform, hence ln 1 π(x) = ln N, where N is the number of vertices in G. Further, it is routine to check that N m mn n 2n, where m, the number of edges in any realization, is an upper bound of the number of parallel edges in any realization of A, m mn is an upper bound of the number of different linear ordering of edges of the realizations G (i) of A (i) for 1 i n 1 and n 2n ( ( n ) 2) n 1 is an upper bound for the number of different chains for a given linear ordering ϕ and the proof is complete. 7 Sampling minimal realizations Let π be a probability distribution over a set Ω. A Markov chain M on set π(x) Ω is said to be reversible with respect to π if for all x, y Ω, π(y) = P(y x) P(x y), where P(i j) denotes the probability of moving from state i to state j. The Markov chain M is then said to obey the detailed balance condition. Let A be an imbalance sequence and let G denote the graph whose vertices are all the chains of A. That is, G is the graph where the random walk C mc is performed. We define a Markov chain, denoted by M mc, on the set of realizations, denoted by M, as follows. Let G denote a graph whose vertices are all realizations of A and let G 1 and G 2 be two realizations of A that are reconstructible from the chains C 1 and C 2 respectively. Then, there is an edge from G 1 to G 2 if and only if there is an edge in G from C 1 to C 2.

20 20 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi Theorem 9 The Markov chain M mc is ergodic, reversible and fast mixing. Proof. By definition the Markov chain M mc is ergodic. It is also fast mixing as reconstructing a realization from a chain is done in polynomial time by Theorem. Now, to prove that M mc is reversible, we only have to prove that it satisfies the detailed balance condition. Indeed let t G1 and t G2 denote the number of chains reconstructing G 1 and G 2 respectively. Then, since the Markov chain C mc has uniform stationary distribution, we have Moreover, it is easy to see that π(g 1 ) π(g 2 ) = t G 1 t G2. P(G 1 G 2 ) P(G 2 G 1 ) = t G 2 t G1. Acknowledgements. The authors thank professor Zoltán Kása (Sapientia Hungarian University of Transylvania) and the unknown referee for their useful corrections. References [1] M. Bayati, J. H. Kim, A. Saberi, A sequential algorithm to generate random graphs, arxiv 2012, arxiv:cs/070212v5, [2] A. Berger, M. Mller-Hannemann, Uniform sampling of digraphs with a fixed degree sequence, in (ed. D. M. Thilikos) 36th Int. Workshop on Graph Theoretic Concepts in Computer Science (June 28-30, 2010, Zarós, Crete, Greece), LNCS 610 (2010) [3] J. K. Blitzstein, P. Diaconis, A sequential importance sampling algorithm for generating random graphs with prescribed degrees. Internet Mathematics 6 (2011), [] S. Chien, L. Rasmussen, A. Sinclair, Clifford algebras and approximating the permanent. J. Computer Systems Sciences 67 (2003) [5] T. H. Cormen, Ch. E. Leiserson, R. L. Rivest, C. Stein, Introduction to Algorithms Third edition, The MIT Press/McGraw Hill, Cambridge/New York, 2009.

21 Random sampling of minimally cyclic digraphs 21 [6] C. I. Del Genio, H. Kim, Z. Toroczkai, K. E. Bassler, Efficient and exact sampling of simple graphs with given arbitrary degree sequence, PLoS ONE 5 e10012 (2010). [7] C. Efthymiou, P. G. Spirakis, Random sampling of colourings of sparse random graphs with a constant number of colours. Theoret. Comput. Sci. 07 (2008) [8] É. Fusy, Uniform random sampling of planar graphs in linear time. Random Structures Algorithms 35 (2009) [9] J. L. Gross, J. Yellen, Handbook of Graph Theory, CRC Press, Boca Raton, 200. [10] A. Iványi, Reconstruction of complete interval tournaments. II. Acta Univ. Sapientiae, Math., 2 (2010), [11] A. Iványi, S. Pirzada, Comparison based ranking. In Algorithms of Informatics (ed by. A. Iványi). AnTonCom, Budapest, tony/oktatas/secondexpert/chapter25comparisons- 10Sept.pdf [12] M. Jerrum, A. Sinclair, Approximating the permanent, SIAM J. Comput. 18 (1989) [13] M. Jerrum, A. Sinclair, The Markov Chain Monte Carlo method: an approach to approximate counting and integration, in Approximation algorithms for NP-hard problems (ed. D. S. Hochbaum), PWS Publishing House, 1995, [1] H. Kim, C. I. Del Genio, K. E. Bassler, Z. Toroczkai, Constructing and sampling directed graphs with given degree sequences, arxiv, 2011, [15] Z. Király, Recognizing graphic degree sequences and generating all realizations, Technical Report of Egerváry Research Group, TR , Budapest. Last modification 23 April, [16] D. E. Knuth, The Art of Computer Programming. Volume 3. Sorting and Searching (second edition). Addison Wesley, Upper Saddle River, [17] M. Luby, D. Randall, A. Sinclair, Markov chain algorithms for planar lattice structures, SIAM J. Comput. 31 (2001)

22 22 K. K. Kayibi, M. A. Khan, S. Pirzada, A. Iványi [18] F. Martinelli, A. Sinclair, Mixing time for the Solid-to-Solid Model. In Proc. of ACM STOC, pp [19] F. Martinelli, A. Sinclair, D. Weitz, Fast mixing for independent sets, colorings, and other models on trees. Random Structures Algorithms 31 (2007) [20] I. Miklós, P. L. Erdős, L. Soukup, Towards random uniform sampling of bipartite graphs with given degree sequence. arxiv 2010, arxiv: v2. [21] D. Mubayi, T. G. Will, D. B. West, Realising degree imbalances in directed graphs, Discrete Math., 239 (2001) [22] A. Noga, W. Fernandez de la Vega, R. Kannan, M. Karpinski, Random sampling and approximation of MAX-CSPs (Special issue on STOC2002, Montreal, QC). J. Comput. System Sci. 67 (2003) [23] S. Pirzada, On imbalances in digraphs, Kragujevac J. Math., 31 (2008) [2] S. Pirzada, T. A. Naikoo, U. Samee, A. Iványi, Imbalances in directed multigraphs, Acta Univ. Sapientiae, Math., 2 (2010) [25] S. Pirzada, T. A. Naikoo, N. A. Shah, On imbalances in oriented tripartite graphs, Acta Mathematica Sinica, English Series, 27 (2011), [26] A. Sinclair, Convergence rates of Monte Carlo experiments (ed. S. G. Whittington), in: Numerical methods for polymeric systems, IMA volumes, 1997, [27] X. Wang, Asymptotic enumeration of digraphs by excess sequence, Graph theory, combinatorics, and algorithms, Vol. 1, 2 (Kalamazoo, MI, 1992), Wiley-Interscience Publ., Wiley, New York, 1995, pp [28] E. W. Weisstein, Simple directed graph. In MathWorld A Wolfram Web Resource. Received: