The complexity of counting colourings and independent sets in sparse graphs and hypergraphs

Transcription

1 The complexity of counting colourings and independent sets in sparse graphs and hypergraphs Catherine Greenhill Abstract We consider certain counting problems involving colourings of graphs and independent sets in hypergraphs. Using polynomial interpolation techniques, we show that these problems are #P -complete. Therefore, efficient approximate counting is the most one can realistically expect to achieve. Rapidly mixing Markov chains which can be used for approximately solving some of these counting problems have been recently developed by the author and others. 1 Introduction The complexity of counting is an area which has received attention since Valiant [11] introduced the complexity class #P. However, we know much less about the complexity of counting than we do about the complexity of decision problems. As Vadhan notes [10], one area where knowledge is lacking is whether hard counting problems remain hard when extra restrictions are placed on the problem instances. A possible reason for this is that many of the reductions involve blowing up the graph. This destroys special properties such as regularity or degree-boundedness, which may be of interest. In [10], Vadhan uses reductions which preserve the properties of interest of the graphs, and enable restricted-case #P-completeness proofs to be obtained. He also gives a general-purpose lemma (stated in Lemma 1.1 below) which is of great assistance in setting up polynomial interpolation. In this paper we give three #P-completeness proofs which use the same kind of reductions as in [10]. The problems considered are those for which rapidly mixing Markov chains have recently been designed by the author and others. The existence of these Markov chains show that approximate counting is possible in polynomial time, even if exact counting is unachievable (unless P=#P). The first problem we prove to be #P-complete is that of counting all proper colourings of triangle free -regular graphs. This class of graphs includes the physically relevant class of planar grids, of particular interest to statistical physicists. Another motivation for this result is that a rapidly mixing Markov chain for approximately counting proper 7-colourings of triangle free 4-regular graphs was given in [2]. In Section 3, the methods are adapted to the situation of counting independent sets in graphs and hypergraphs. The problem of counting independent sets in graphs with School of Computer Studies, University of Leeds, Leeds LS2 9JT, UNITED KINGDOM. Supported by a Leverhulme Trust Special Research Fellowship. 1

2 maximum degree four (or higher) is known to be #P-complete, as shown by Vadhan [9]. On the other hand, the number of independent sets for graphs with maximum degree at most two can be easily calculated in polynomial time (see [8, Lemma A.4] for details). The second result of this paper closes the gap, and shows that it is #P-complete to count independent sets in graphs with maximum degree three. In fact we prove that this problem remains #P-complete when restricted to 3-regular graphs. The Markov chain described by Luby and Vigoda [6] is the first which can be used to approximately count independent sets in graphs with maximum degree at most four, in polynomial time. A new Markov chain for independent sets in graphs was described in [3], where it was also extended to hypergraphs. Like the Luby-Vigoda chain, the new chain can be used to approximately count independent sets in graphs with maximum degree at most four. In addition, it can be used to approximately count independent sets in two classes of hypergraphs, in polynomial time: (i) hypergraphs with maximum degree two, and (ii) hypergraphs with maximum degree three and maximum edge size three. Counting independent sets in hypergraphs with maximum degree two is equivalent to counting the number of edge covers in graphs, and this problem is #P -complete [1]. Thus the final question, so far as the Markov chain described in [3] is concerned, is whether it is #P-complete to count independent sets in hypergraphs with maximum degree three and maximum edge size three. The final result of this paper shows that this problem is indeed #P-complete. In fact, we prove that the problem remains #Pcomplete when restricted to hypergraphs with maximum degree at most three where every edge has size three. There are two main tools used in the proofs. As mentioned above, an important tool is polynomial interpolation, which has also been employed in [10, 11]. Here a graph (or hypergraph) is constructed by replacing certain edges by a chain of edges (or some more complicated structure) of length r, for each r in a certain range. If the chain has r edges then we call this the r-stretch of the (hyper)graph (with respect to the set of edges replaced). This allows us to relate the number N of proper colourings (or independent sets) of the original (hyper)graph to the number N r of proper colourings (or independent sets) of the r-stretch. If this relationship is well behaved then the value of N can be calculated in polynomial time from a polynomial number of values of N r. Although these interpolation proofs run along similar lines, there are subtle and nontrivial adjustments necessary to fit the argument to each new situation. In the interpolation arguments, we make repeated use of the following lemma. Lemma 1.1 ([10, Lemma A.3]) Let A, B, C, D, x 0 and y 0 be rational numbers. Define the sequences {x n } and {y n } recursively by x n+1 = A x n + B y n and y n+1 = C x n +D y n. Then the sequence {z n = x n /y n } never repeats as long as all of the following conditions hold: AD BC 0, D 2 2AD + A 2 + 4BC 0, D + A 0, D 2 + A 2 + 2BC 0, D 2 + AD + A 2 + BC 0, D 2 AD + A 2 + 3BC 0, B y 0 2 C x 0 2 (A D) x 0 y

3 Notice in particular that if A, B, C, D are all positive, only the first and last conditions must be checked. The second tool is the use of gadgets which enable us to reduce from regular graphs. Here we design a device which can be attached to low degree vertices and which satisfies any other conditions required. 2 Colourings First, we recall the notation used for graph colourings. Let G = (V, E) be a graph and let be the maximum degree of G. Let k be a positive integer and let C be a set of size k. A map from V to C is called a k-colouring (or simply a colouring when k is fixed). A vertex v is said to be properly coloured in the colouring X if v is coloured differently from all of its neighbours. A colouring X is called proper if every vertex is properly coloured in X. The set of all proper k-colourings of G is denoted by Ω k (G). The problem of counting proper k-colourings of graphs is known to be #P -complete for k 3, by [5]. This problem remains #P -complete when restricted to graphs with maximum degree (exactly), whenever 3. A proof of this is presented in [2], and the result is stated below. Theorem 2.1 ( [2, Theorem 6.2]) Let k, be constants such that k 3 and 3. Consider the counting problem #kcolm axdeg, which takes as an instance a graph with maximum degree (exactly), and returns the number of proper k-colourings of the graph. Then #kcolm axdeg is #P -complete. In this section we consider a more restricted versions of this counting problem, and show that it is #P -complete to count the number of proper k-colourings of triangle free -regular graphs with n vertices, when k 3 and 3. Here k and are constants. First let us describe the devices used in the proof. Let γ, γ denote two fixed, distinct colours. The chain graph C r has r + 1 vertices u 1,..., u r+1, and r edges {u i, u i+1 } for 1 i r. Define σ r, δ r by It is not difficult to prove that σ r = {X Ω k (C r ) : X(u 1 ) = γ, X(u r+1 ) = γ}, δ r = { X Ω k (C r ) : X(u 1 ) = γ, X(u r+1 ) = γ }. σ r = (k 1)r + ( 1) r (k 1), δ r = (k 1)r + ( 1) r 1 k k (1) for r 1. The full description of the rest of the devices is deferred until after the proof. For now, we describe only the critical characteristics and give notation for the number of colourings of these devices. We use a triangle free graph P which has 2 vertices of degree and two vertices, q 1, q 2 of degree one. Define α 1, β 1 by α 1 = {X Ω k (P ) : X(q 1 ) = γ, X(q 2 ) = γ}, (2) β 1 = { X Ω k (P ) : X(q 1 ) = γ, X(q 2 ) = γ }. 3

4 We do not have closed form expressions for α 1, β 1, but show how these values can be found in constant time in Section 2.1. We do know that α 1 = (k 1)S + (k 1)(k 2)D, β 1 = (k 2)S + (k 2 3k + 3)D, (3) for some S, D such that S < D. (4) Suppose now that is odd. We use a triangle free graph Q with vertices of degree and one vertex, w, of degree one. Let ϕ be defined by ϕ = {X Ω k (Q) : X(w) = γ}, where γ is some fixed colour. Then, for our graph Q we have ϕ = (k 1) α 1 ( 1)/2, where α 1 is as defined in (2). The final device we use is denoted by U r, for r 1. This is a triangle free graph with 2r vertices of degree and two vertices, q 1, q r+1, of degree one. (When r = 1 we take U 1 = P.) Define α r, β r by α r = { X Ω k (U r ) : X(q 1 ) = γ, X(q r+1 ) = γ }, β r = {X Ω k (U r ) : X(q 1 ) = γ, X(q r+1 ) = γ} for r 1. Then, for the graph U r which we define below, (3) holds, and α r+1 = (k 1)D α r + α 1 β r, β r+1 = α 1 (k 1) α r + β 1 β r (5) for r 1 (where S, D are the quantities which appear in (4)). We can now prove that it is #P -complete to count proper k-colourings of triangle free -regular graphs with n vertices, when k 3 and 3. Theorem 2.2 Let k, be constants such that k 3 and 3. Consider the counting problem #kcolt rianglef ree Regular, which takes as an instance a triangle free -regular graph, and returns the number of proper k-colourings of the graph. Then #kcolt rianglef ree Regular is #P -complete. Proof. Let #kcolm axdeg denote the problem of counting the number of proper k-colourings of a graph with maximum degree. The problem #kcolm axdeg is #P -complete, by Theorem 2.1. We will give a polynomial-time reduction from this problem to the problem #kcolt rianglef ree Regular. This will show that the latter problem is # P-hard. Since the problem is clearly in #P, we can conclude that #kcolt rianglef ree Regular is #P-complete, as stated. We give the required polynomial-time reduction in three steps. Consider the counting problem, denoted by #kcolt rianglef reem axdeg, which takes as instance a triangle free graph with maximum degree and returns the number of proper k-colourings of the graph. This is the first intermediate problem used in the three-step reduction argument. 4

5 The second intermediate problem is denoted by #kcolt rianglef reealmost Regular. This problem takes as instance a graph where every vertex has degree or 1, and returns the number of proper k-colourings of that graph. REDUCTION 1 (#kcolm axdeg #kcolt rianglef reem axdeg ). Let G be a graph with maximum degree (exactly) and let I be a (minimal) set of edges such that deleting all edges in I makes G triangle free. For r 1 let H r be the r-stretch of G with respect to I. That is, obtain H r from G by replacing each edge {v, w} in I with the chain C r, using the identifications u 1 = v, u r+1 = w. Then H r has maximum degree for r 1, and H r is triangle free for r 2. Note that H 1 = G. Let m = I and let p s be the number of k-colourings of G such that s edges in I are monochromatic, the remaining m s edges of I are bichromatic and all other edges are bichromatic. Then Ω k (G) = p 0 and for r 1 we have Ω k (H r ) = p s σ s r δ m s m r = δ r s=0 p s x s r, where x r = σ r /δ r. These equations, for 2 r m + 2, can be expressed using the matrix equation δ m 2 Ω k (H 2 ) 1 x 2 x m 2 p 0 δ m 3 Ω k (H 3 ) 1 x 3 x m 3 p 1. δ m+2 m Ω k (H m+2 ) s=0 = x m+2 x m m+2. p m. (6) The values x r are distinct, as can be verified directly from (1). If we knew the values Ω k (H r ) for 2 r m + 2, then the values of p 0,..., p m could be found in polynomial time by inverting the Vandermonde matrix in (6). Since p 0 = Ω k (H 1 ) = Ω k (G) is the answer we seek, this completes the first reduction. REDUCTION 2 (#kcolt rianglef reem axdeg #kcolt rianglef reealmost Regular). Let G be a triangle free graph with maximum degree. Form the graph H, where every vertex has degree or 1, by attaching ( d)/2 copies of P to every vertex v with degree d < 1 (using the identifications v = q i for i = 1, 2). Suppose that R copies of P are attached to G to form the graph H. Then Ω k (H) = α 1 R Ω k (G), where α 1 is as defined in (2). We can calculate α 1 in polynomial time, as shown in the next section. Since H is triangle free, this completes the second reduction. REDUCTION 3 (#kcolt rianglef reealmost Regular #kcolt rianglef ree Regular). Let H be a triangle free graph such that every vertex has degree or 1. Let U be the set of vertices in H with degree 1. We can assume that U contains an even 5

6 number of vertices, as follows. If U contains an odd number of vertices then is odd. Choose a vertex v U and attach the triangle free graph Q to v using the identification v = w. This ensures that U contains an even number of vertices, and we may partition U into m subsets of size two. Let I be such a partition. For 0 s m, let p s be the number of proper k-colourings X of H such that X(v 1 ) = X(v 2 ) for exactly s pairs {v 1, v 2 } I. Then Ω k (H) = p s, so it suffices to calculate the values p 0,..., p m. Now form the triangle free -regular graph H r from H, for 1 r m+1, by attaching a copy of U r to each pair {v 1, v 2 } I, making the identifications v i = q i for i = 1, 2. Then s=0 Ω k (H r ) = p s α s r β m s m r = β r s=0 p s x s r, s=0 where x r = α r /β r for r 1. Using (3), (4) and (5), one may apply Lemma 1.1 to show that the quotients x r are distinct. Thus we can calculate the values of p 0,..., p m in polynomial time, given the values of Ω k (H r ) for 1 r m + 1. This completes the polynomial-time reduction. Hence we have proved that it is #P -complete to count proper k-colourings of triangle free -regular graphs with n vertices, when k 3 and 3. The above theorem is valid for graphs with maximum degree, where is a constant. It is probably possible to extend this result so that it holds for graphs with maximum degree (n), for some suitable function (n). But a consequence of Edwards [4] work on dense graphs implies that there must be some restriction on the function (n). Specifically, Edwards proves the following result. Suppose that δ is the minimum degree of a graph G with N vertices. Further suppose that δ αn for some rational number α such that 0 α < 1 and α = Ω(1). If 0 α < (k 2)/(k 1) then it is #P -complete to count the number of proper k-colourings of G, while if (k 2)/(k 1) < α < 1 then this problem has a polynomial time solution. For regular graphs, the maximum degree and minimum degree coincide. Thus we can conclude that we must have (n) < (k 2)n/(k 1). 2.1 The devices used We now give a full description of the devices used in the proof of Theorem 2.2. First we describe the graph F which is the building block of all our devices. Consider the complete bipartite graph K 1, 1 with vertex bipartition {u 1,..., u 1 } {w 1,..., w 1 }. Let F be the graph obtained by joining each vertex u i to a new vertex z 1 using the edge {u i, z 1 } for 1 i 1, and joining each vertex w j to a new vertex z 2 using the edge {w j, z 2 } for 1 j 1. Then F has 2( 1) vertices of degree and two vertices, 6

7 z 1 z 2 Figure 1: The graph F for = 4 z 1, z 2, of degree 1. For example, when = 4 the graph F is as shown in Figure 1. Let γ, γ be two fixed, distinct colours, and define S = {X Ω k (F ) : X(z 1 ) = γ, X(z 2 ) = γ}, D = { X Ω k (F ) : X(z 1 ) = γ, X(z 2 ) = γ }. Now define the graph U r inductively, as follows. Let U 1 be obtained from a copy of F by joining z i to the new vertex q i using the edge {z i, q i }, for i = 1, 2. Now suppose that U r has been defined with the following properties: the graph U r is triangle free and has 2r vertices of degree and two vertices q 1, q r+1 of degree one. Obtain U r+1 from U r by joining a new copy of F to U r, using the identification q r+1 = z 1 (where primed vertices belong to the new copy of F ), and then attaching a new vertex q r+2 to the remaining vertex z 2 of degree 1. Then U r+1 is triangle free and has 2(r + 1) vertices of degree and two vertices q 1, q r+2 of degree one, as required. For example, the graph U 2 is as shown in Figure 2 when = 4. q 1 q 3 Figure 2: The graph U 2 for = 4 As mentioned above, we take P to be the graph U 1, a triangle free graph with 2 vertices of degree and two vertices of degree one. If is odd, form Q by taking ( 1)/2 copies of P, identifying all vertices of degree one to give a vertex z of degree 1, then joining z to a new vertex w. Then Q has vertices of degree, and one vertex, w, of degree one. When = 5 the graph Q is as shown in Figure 3. For the proof of Theorem 2.2 to work, we must show that our assumptions hold. That is, we must prove that S < D, and that the value α 1 can be calculated in polynomial time. 7

8 w Figure 3: The graph Q for = 5 Lemma 2.1 With S, D as defined above, we have S < D. Proof. Let λ(s, i, t) be the number of ways to colour s unrelated vertices with exactly i colours, chosen from a set of t colours, where s, i and t are positive integers. Similarly, let λ + (s, i, t) be the number of ways to colour s unrelated vertices with exactly i colours, chosen from a set of t colours, such that a certain distinguished colour is always used. Then λ(s, i, t + 1) = λ(s, i, t) + λ + (s, i, t + 1) (7) for all positive s, i, t. It is not difficult to see that S = D = 1 i=1 1 i=1 λ( 1, i, k 1)(k i 1) 1, { λ( 1, i, k 2)(k i 1) 1 + λ + ( 1, i, k 1)(k i) 1}. Using (7), it follows that S = = < 1 i=1 1 i=1 1 i=1 = D, λ( 1, i, k 1)(k i 1) 1 ( λ( 1, i, k 2) + λ + ( 1, i, k 1) ) (k i 1) 1 { λ( 1, i, k 2)(k i 1) 1 + λ + ( 1, i, k 1)(k i) 1} 8

9 as required. Note that the quantities λ(s, i, t) are related to Stirling numbers of the second kind [12, p.105], as follows. The Stirling numbers of the second kind, S(s, i), count the number of ways of partitioning s objects into i nonempty subsets. Hence S(s, 1) = 1 and S(s, s) = 1 for s 1, while S(s, i) = i S(s 1, i) + S(s 1, i 1) (8) for 2 i s 1. In our situation, each subset corresponds to a distinct colour, and there are t(t 1) (t i + 1) ways to assign colours to these subsets. Therefore λ(s, i, t) = S(s, i) t(t 1) (t i + 1). Using this equation, it is not difficult to show that λ + (s, i, t) = i S(s, i) t(t 1) (t i + 2). We can use these facts to construct a constant time algorithm for calculating S and D, as follows. Starting with the values S(s, 1), S(s, s) for 1 s 1, use (8) to calculate the remaining values of S(s, i) for 1 i s, 1 s 1. This costs O( 2 ) operations. Then the values λ( 1, i, k 1) and λ( 1, i, k 2) can be calculated for 1 i 1 in O( ) operations. The powers (k i) 1 can be calculated for 1 i in O( log( )) operations. Finally these values can be combined to give S, D in O( ) operations. The overall cost is O( 2 ), where the most significant cost is calculating the Stirling numbers. Since is a constant, this is a constant time algorithm. Thus we can calculate α 1 in polynomial time, as claimed in Theorem 2.2. I am very grateful to a referee for suggesting an alternative formulation of the quantities S, D using exponential generating functions. Namely, S is the coefficient of x 1 y 1 in ( 1)! 2 (e x + e y 1) k 1, and D is the coefficient of x 1 y 1 in ( 1)! 2 (e x + e y 1) k 2 e x+y. This formulation provides an easy proof that S < D, since every coefficient of e x+y e x e y + 1 = (e x 1)(e y 1) is positive. However, this formulation does not appear to provide a more efficient procedure for calculating S, D. 3 Independent sets Let G = (V, E) be a hypergraph and let I(G) be the set of independent sets in G. That is, if X V then X I(G) if and only if e X for all e E. Let λ be a positive parameter and define the probability measure π on I(G) by π(x) = λ X /Z, 9

10 where Z = X I(G) λ X. The normalising factor Z is called the partition function of G. When λ = 1, evaluating Z is equivalent to calculating I(G). In [6], Luby and Vigoda described a Markov chain for independent sets in graphs which can be used to approximately count independent sets of graphs with maximum degree at most four. The problem of approximately counting independent sets of graphs with maximum degree is known to be #P -complete when 4, as follows from [9]. We prove in Section 3.2 that the problem of counting independent sets in 3-regular graphs is #P -complete. A Markov chain for independent sets in hypergraphs was described in [3]. Like the Luby-Vigoda chain, this chain can be used for approximately counting independent sets in graphs with maximum degree at most four. It can also be used for approximately counting independent sets in hypergraphs with maximum degree at most two, and hypergraphs with maximum degree at most three and maximum edge size at most three. The former problem is equivalent to the problem of counting edge coverings in graphs, and so is #P -complete (see [1]). It follows from the counting problem for graphs with maximum degree at most three that the counting problem for hypergraphs with maximum degree at most three and maximum edge size at most three is #P -complete. To disprove the conjecture that the difficulty lies in the possible presence of edges of size two, we prove in Section 3.3 that the counting problem remains #P -complete when restricted to hypergraphs with maximum degree at most three where every edge has size three. 3.1 Correspondence with SAT instances The set of independent sets in a given hypergraph corresponds in a natural way to the set of satisfying assignments of a certain SAT instance (for background on SAT see, for example, [7, Section 4.2]). This correspondence is used in Theorem 3.2 below, and the results we obtain can all be restated in terms of counting problems of SAT instances. We now give the details of this correspondence. A SAT instance is said to be monotone if no variable appears negated. If v is a variable, denote its negation by v. Using notation borrowed from Roth [8], denote by lµ-kmon the class of all monotone SAT instances where each variable appears in at most l clauses and each clause contains at most k literals. Let G be a hypergraph where each edge has at most m vertices and each vertex appears in at most edges. Let each vertex correspond to a variable (denoted by the same symbol as the vertex) and for each e E form a clause C e as follows. If e = {v 1,..., v d } then let C e = d v i. i=1 Then e E C e 10

11 is a SAT instance with I(G) satisfying assignments. Replace the variable v i by w i for each v i V. This gives an instance of µ-mmon. The above construction can be reversed, showing that the problem of finding the number of independent sets in hypergraphs with maximum degree at most and maximum edge size at most m is equivalent to the counting problem # µ-mmon. 3.2 Three-regular graphs First let us describe the device Q r which we use in the next theorem. The device Q r, defined for r 0, has 4(r + 1) vertices of degree three and one vertex, u 2r of degree two. The device is depicted in Figure 4. Let M r = I(Q r ), and let M r (0) be defined by u u 0 1 u 2 u 2r- 1 u 2r Figure 4: The graph Q r M r (0) = {X I(Q r ) : u 2r X}. Then M 0 = 9, M 0 (0) = 6 and, for r 0, M r+1 = 4M r + 2M r (0), M r+1 (0) = 3M r + M r (0). (9) We now prove that the problem of counting independent sets in 3-regular graphs is #P -complete. This proof has two steps, and the first step follows the method of [2, Theorem 6.2]. The interpolation is slightly more complicated than in the colourings situation, since we must distinguish between chains with both endpoints included in the independent set, and chains with neither endpoint included. Theorem 3.1 Let #REGULAR-IND-SETS(3) denote the following counting problem. An instance is a 3-regular graph and the problem is to calculate the number of independent sets in that graph. Then #REGULAR-IND-SETS(3) is #P -complete. Proof. Let #IND-SETS( ) denote the counting problem which takes a graph G of maximum degree at most, and returns the number of independent sets in G. The problem #IND-SETS(4) is #P -complete, as follows from [9]. We give a polynomialtime reduction from this problem to #REGULAR-IND-SETS(3). This shows that the latter problem is #P-hard. Since the problem is clearly also in #P, we can conclude that #REGULAR-IND-SETS(3) is #P-complete, as stated. 11

12 We give the required polynomial-time reduction in two steps, with the intermediate problem #IND-SETS(3). REDUCTION 1 (#IND-SETS(4) #IND-SETS(3)) Let G be a graph with maximum degree at most 4. We construct a graph H = (V, E) with maximum degree at most three and edge bipartion E = E E as follows. Replace each vertex v of degree four by two vertices v 1, v 2 of degree three. The edge {v 1, v 2 } is placed in E, while two neighbours of v in G are joined to v 1 using an edge in E, and the other two neighbours of v are joined to v 2 using an edge in E. All other edges in G become edges in E. Let m = E. For 0 t m, 0 i t define sets S t,i by S t,i = { X V : e E e X, { e E : e X = } = i, { e E : e X } = t i. }. (10) Say that an edge e E is well behaved in X if e X = or e X. If X S t,i then t edges in E are well behaved. Let n t,i = S t,i for 1 t m, 1 i t. Then I(G) = n m,i. (11) i=0 This follows since the set I(G) is in one-one correspondence with the set of subsets X V such that all edges in E are well behaved. Let f r denote the rth Fibonacci number, defined inductively by f 1 = 1, f 2 = 1 and f r+2 = f r+1 + f r for r 1. We extend the definition, defining f 0 = 0 and f 1 = 1. Let C r be the chain graph with r + 1 vertices u 1,..., u r+1 and r edges {u i, u i+1 } for 1 i r. (Note that C 0 consists of a single vertex.) It follows easily that I(C r ) = f r+3 for r 0. Hence {X I(C r ) : {u 1, u r+1 } X = } = f r+1, {X I(C r ) : u 1 X and u r+1 X} = f r, {X I(C r ) : {u 1, u r+1 } X = {u 1, u r+1 }} = f r 1. For r 0, let H r denote the r-stretch of H with respect to E. That is, let H r be the graph obtained from H by replacing each edge {v, w} E by a copy of the chain C r, making the identifications v = u 1, w = u r+1. Note that H 0 = G. Then, for r 0, I(H r ) = t=0 i=0 = f r+1 m = f r+1 m t n t,i f i r+1 f t i m t r 1 f r t t=0 i=0 t=0 i=0 ( ) t i ( ) m t fr 1 fr n t,i f r+1 f r+1 t n t,i (1 x r ) t i m t x r (12) 12

13 where x r = f r /f r+1. This can be rewritten as f r+1 m I(H r ) = c s x s r = g(x r ), (13) s=0 for some c 0,..., c m. The quotients x r are distinct for r 1, as proved in [10, Lemma A.1] (the result also follows from Lemma 1.1). Hence the equations (13), 1 r m + 1, can be expressed using an invertible Vandermonde matrix. If we knew the values I(H 1 ),..., I(H m+1 ), then the coefficients c 0,..., c m can be found in polynomial time by inverting this matrix. We can then evaluate g(x) for any x. In particular, we can evaluate g(0). But g(0) = g(x 0 ) = I(H 0 ) = I(G), the quantity we wish to compute. This completes the first reduction. REDUCTION 2 (#IND-SETS(3) #REGULAR-IND-SETS(3)). Let G = (V G, E G ) be a graph with maximum degree at most three, and let U d = {v V G : d(v) = d} for 0 d 3, where d(v) denote the degree of v. Let µ d = U d for 0 d 3. Obtain the graph H from G by deleting all vertices of degree zero. Then I(H) = 2 µ 0 I(G), so it suffices to calculate I(H). Let R = 2µ 1 + µ 2. For 0 r R, construct a 3-regular graph H r from H, as follows. Attach 3 d copies of the graph Q r to each vertex v of degree d < 3, using edges of the form {v, u 2r }. Clearly H r is a 3-regular graph, for 0 r R, and we can form these graphs in polynomial time. For 0 i µ 1, 0 j µ 2, define n i,j by Then Now µ 1 µ 2 n i,j = {X I(H) : X U 1 = i, X U 2 = j}. µ 1 µ 2 n i,j = I(H). (14) i=0 j=0 (0) I(H r ) = n i,j M 2i+j r M R (2i+j) R r = M r i=0 j=0 µ 1 i=0 µ 2 j=0 n i,j x r 2i+j = M r R g(x r ), where x r = M r (0) /M r. It follows from (9) and from Lemma 1.1 that the quotients x r are distinct for 0 r R. Suppose that the values I(H r ) were known, for 0 r R. By inverting a Vandermonde matrix, we can interpolate the polynomial g given in (15), and evaluate it at x r = 1 (in polynomial time). By (14), this value is I(H). This completes the polynomial-time reduction from the problem #IND-SETS(3) to the problem #REGULAR-IND-SETS(3), proving that the latter is #P -complete. (15) 13

14 3.3 Hypergraphs with maximum degree at most three where every edge has size three In Theorem 3.1 we proved that it is #P -complete to count independent sets in graphs with maximum degree at most three. This proof is adapted below to show that it is #P -complete to count independent sets in hypergraphs with maximum degree at most three, where every edge has size three. We will use an analogue of the chain graph C r, which has 2r + 1 vertices and r edges. If r = 0 then C r consists of a single vertex. For r 1, C r has r 1 vertices of degree two and r + 2 vertices of degree one: in particular, every edge contains at least one vertex of degree one, and every edge has size three. The chain hypergraph C r is depicted in Figure 5. Let N r = I(C r ) for r 0. Then N 0 = 2, u 1 w 1 u 2 w 2 u 3 w 3 u 4 u r- 1 w r- 1 u r w r u r+1 N 1 = 7 and Figure 5: The chain hypergraph C r N r+2 = 3N r+1 + 2N r (16) for r 0. Extend this sequence of numbers backwards, defining N 1, N 2 and N 3 by N 1 = 1/2, N 2 = 1/4, N 3 = 1/8. (17) Then these numbers also satisfy the relation (16). Let A r (0) A r (1) A r (2) Then it is not difficult to show that = {X I(C r ) : {u 1, u r+1 } X = }, = {X I(C r ) : u 1 X and u r+1 X}, = {X I(C r ) : {u 1, u r+1 } X = {u 1, u r+1 }}. A r (0) = 4N r 2, A r (1) = 2N r 2 + 4N r 3, A r (2) = 3N r 2 2N r 3 (18) for r 0. We now prove that the problem of counting independent sets in hypergraphs with maximum degree at most three, where every edge has size three, is #P -complete. For this proof we make explicit use of the correspondence between independent sets in hypergraphs and certain SAT instances, as described in Section 3.1. This allows us to take as our starting point the #P -complete counting problem #3SAT. Theorem 3.2 Let #IND-SETS(3, 3) denote the following counting problem. An instance is a hypergraph with maximum degree at most three where every edge has size three. The problem is to calculate the number of independent sets in that hypergraph. Then #IND-SETS(3, 3) is #P -complete. 14

15 Proof. Let us extend the notation used for restricted SAT problems to distinguish instances where each clause contains the same number of variables. Define k SAT to be the subset of ksat containing all instances where each clause contains exactly k variables. Define k MON and lµ-k MON similarly. The counting problem #3SAT is #P -complete (see [7, p.442]). We give the following series of polynomial-time reductions #3SAT #3 SAT #3 MON #3µ 3 MON, with some modification due to the fact that the edges have size three. The existence of the given series of polynomial-time reductions implies that all the counting problems listed above are #P -hard. Since they are all also members of the class #P, we will be able to conclude that all the listed problems are #P-complete. REDUCTION 1 (#3SAT #3 SAT). Given a 3SAT instance, delete any clauses which contain only one variable and adjust all other clauses containing that literal (deleting those which become true, and deleting the literal from any clause where that literal takes the value false). Next, replace each clause v w of size two with the conjuction of two clauses (v w x) (v w x) of size three, where x is a new variable. If r new variables are introduced then this gives a 3SAT instance where every clause has three variables, and with exactly 2 r times as many satisfying assignments as the original. This reduction can be performed in polynomial time. REDUCTION 2 ( #3 SAT #3 MON). Given a 3 SAT instance f, form a hypergraph G as follows. Let the variables of f be u 1,..., u m. (The usual convention is to have m clauses and n variables; we reverse this convention so that subsequent notation is in keeping with that used in Theorem 3.1). For each variable u j let there be two vertices, one labelled v j and one labelled v j. Here v j corresponds to the literal u j and v j corresponds to the literal ū j. For each clause in f form an edge, as follows. Suppose that C = a 1 a 2 a 3 where the a i are literals and a i {u ji, ū ji } for 1 i 3. The edge corresponding to C is {b 1, b 2, b 3 } where { v ji if a i = u ji, b i = v ji if a i = ū ji.. Then every edge in G has size three. For 0 t m, 0 i t, define k t,i to be the number of independent sets X I(G) satisfying the following conditions: (i) {j : {v j, v j } X = } = i, (ii) {j : {v j, v j } X} = t i. Let X be an independent set in G, and suppose that, for all variables v j of f, we have {v j, v j } X = 1. Give u j the truth value TRUE if and only if v j X, for 1 j m. This is a satisfying assignment of f. This construction can be reversed, to give a one-one correspondence. Hence k 0,0 is the number of satisfying assignments of f. 15

16 For each (r, p) with 1 r pm + 1, 1 p m + 1, we form a hypergraph H (r,p), as follows. Recall the chain hypergraph C r with r edges, depicted in Figure 5. Add p copies of the chain C r between each pair of vertices {v j, v j } corresponding to a variable v j of f, making the identifications v j = u 1 and v j = u r+1 in each copy of the chain. Then t I(H (r,p) ) = k t,i A (0) r pi A (2) r p(t i) A (1) r p(m t). (19) t=0 i=0 Let x r = N r 3 / A r (1) for 1 r mp + 1. Dividing (19) by A r (1) mp, we obtain I(H (r,p) ) A r (1) mp = t=0 t k t,i 2 p(i t) (2 8x r ) pi (3 16x r ) p(t i) (20) i=0 for 1 r mp + 1. Using Lemma 1.1, we see that the quotients x r are distinct for 1 r mp + 1. Suppose that we knew the values I(H (r,p) ) for 1 r pm + 1. Then we could obtain the value of the right hand side of (20) at x r = 3/16, in polynomial time, by inverting a Vandermonde matrix. This value is equal to k t,t 2 pt. t=0 If we can do this for 1 p m + 1 then we obtain another Vandermonde matrix, as the coefficients 2 p are distinct. Thus the values k t,t can be found in polynomial time by inverting this final matrix. In particular we have k 0,0, the number of satisfying assignments of the original 3 SAT instance f. This completes the second reduction. REDUCTION 3 (#3 MON #3µ 3 MON). This reduction is based on Reduction 1 of Theorem 3.1, with some modification due to the fact that all edges have size three. Let G = (V G, E G ) be a hypergraph with maximum degree at most, where every edge has size three and 3. Form a hypergraph H = (V, E) from G as follows. Consider each vertex v V G with degree d > 3 in turn. Replace v by a copy of the chain hypergraph C d 3, attaching each edge containing v to a unique vertex u i of the chain. This can be done in such a way that each vertex u i has degree three and each vertex w i has degree one. Let each edge in the chain hypergraph be placed in the set E, while all other edges become members of E. The hypergraph H can be constructed from G in polynomial time. Let m = E. For 0 t m, 0 i t let n t,i = S t,i, where S t,i is defined in (10). Then I(G) = n m,i, (21) i=0 since all edges are well-behaved. For r 0, let H r denote the r-stretch of H with respect to E. That is, obtain H r from H by replacing each edge {v 1, w, v 2 } E by a copy of the chain C r, making the identifications v 1 = u 1, v 2 = u r+1, ensuring that w has degree 16

17 one. Note that H 0 = G. For r 0, using (18) we obtain I(H r ) = t=0 i=0 t n t,i A (0) r i A (2) r t i A (1) r m t = (4N r 2 ) m m t=0 i=0 t n t,i 4 i m (3 2x r ) t i (2 + 4x r ) m t, (22) where x r = N r 3 /N r 2 (using the values defined in (17) where necessary). The quotients x r are distinct for r 1, for example by appealing to Lemma 1.1. Suppose that the values I(H r ) were known, for 1 r m + 1. Then the value of the right hand side of (22) could be interpolated at x r = 1/2, in polynomial time, by inverting a Vandermonde matrix. Now x r = 1/2 corresponds to r = 0, using the values N 2 and N 3 as defined in (17). Thus we obtain the value I(H 0 ) = I(G) = n m,i. i=0 This completes the third reduction, and this series of reductions proves that #IND- SETS(3, 3) is #P -complete. 4 Acknowledgements I would like to thank Martin Dyer for many helpful discussions about this work, and Mark Jerrum for his assistance in adapting the methods from colourings to independent sets. I am grateful to both referees for their helpful comments. References [1] R. Bubley and M. Dyer, Graph orientations with no sink and an approximation for a hard case of #SAT, in 8th Annual Symposium on Discrete Algorithms, ACM SIAM, New York Philadelphia, 1996 pp [2] R. Bubley, M. Dyer, C. Greenhill, and M. Jerrum, On approximately counting colourings of small degree graphs, SIAM Journal on Computing, 29 (1999), pp [3] M. Dyer and C. Greenhill, A more rapidly mixing Markov chain for graph colourings, Random Structures and Algorithms, 13 (1998), pp [4] K. Edwards, The complexity of colouring problems on dense graphs, Theoretical Computer Science, 43 (1986), pp [5] F. Jaeger, D. L. Vertigan, and D. Welsh, On the computational complexity of the Jones and Tutte polynomials, Mathematical Proceedings of the Cambridge Philosophical Society, 108 (1990), pp [6] M. Luby and E. Vigoda, Approximately counting up to four, in Twenty-Ninth Annual Symposium on Theory of Computing, ACM, New York, 1997 pp [7] C. H. Papadimitriou, Computational Complexity, Addison Wesley, Reading, MA, (1994). [8] D. Roth, On the hardness of approximate reasoning, Artificial Intelligence, 82 (1996), pp [9] S. Vadhan, The complexity of counting, (1995), Undergraduate thesis, Harvard University. 17

18 [10] S. P. Vadhan, The complexity of counting in sparse, regular, and planar graphs, Preprint (Available from salil/), (May 1997). [11] L. G. Valiant, The complexity of enumeration and reliability problems, SIAM Journal on Computing, 8 (1979), pp [12] J. H. van Lint and R. M. Wilson, A course in combinatorics, Cambridge University Press, Cambridge, (1992). 18