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Computing Cycle Index Polynomial of a Permutation Group Is Known to Be #P -Complete

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Computing Cycle Index Polynomial of a Permutation Group Is Known to Be #P -Complete Computation of the cycle index polynomial of a Permutation Group CS497-report Rohit Gurjar Y5383 Supervisor: Prof Piyush P. Kurur Dept. Of Computer Science and Engineering, IIT Kanpur November 13, 2008 Abstract Computing cycle index polynomial of a permutation group is known to be #P -complete. In this report we give some introduction to the problem and describe some permutation groups for which the computation of cycle index polynomial is easy. 1 Introduction Definition 1. Cycle index polynomial of a group G of degree n is defined as 1 X c (g) c (g) P (z , z . z ) = z 1 z 2 . zcn(g) G 1 2 n |G| 1 2 n g∈G where ci(g) represents the number of cycles of length i in the disjoint cycle decomposition of g. 1.1 Application Let G be a permutation group acting on X = {0, 1, . n − 1} and Ω = {f : f is a function from X to C}, where C be a finite set of alphabet of cardinality k. We define a group action on the set Ω by g(φ)(x) = φ(g−1(x)) for each φ ∈ Ω The action of G partitions Ω into a number of orbits, being the equivalence classes of Ω under the equivalence relation that identifies α and β whenever there exists g ∈ G mapping α to β. Then we know the following theorem. 1 Theorem 1. Let C, X and G be defined as above. Then the number of orbits of Ω under the action of G is PG(k, k, . k). C can be a set of colors. Then the expression PG(k, k, . k) gives the num- ber of different color patterns of X. Definition 2. Let w : C → R be a map which assigns a weight from R (or any ring) to each element of C. Then the weight of φ ∈ Ω, with respect to the weight Q function w is given by, w(φ) = x∈X w(φ(x)). Now, for each g ∈ G and φ ∈ Ω, Y Y Y w (g(φ)) = w (g(φ)(x)) = w φ(g−1(x)) = w (φ(x)) = w(φ) x∈X x∈X x∈X So, the elements of Ω in the same orbit, under the action of G, will have same weight. So we can define the following. Definition 3. Suppose ∆ ⊂ Ω be an orbit under the action of G. Then the weight of ∆, denoted w(∆), is defined to be equal to w(φ) for any φ ∈ ∆. Theorem 2 (Polya’s Enumeration Theorem). With the above definitions and notations X I = w(∆) = PG(x1, x2 . xn), ∆ where the summation goes over distinct orbits of Ω under the action of G and P i xi = c∈C w(c) . We can see that cycle index polynomial of a permutation group G may be computed at a specified point, directly from definition(1) by explicit enumera- tion of the elements of G. In general, however, the order of G is exponentially larger than its input size, say a list of generators. Generally, degree n of the permutation group G is taken as the measure of input size. So, we want to compute the cycle index polynomial at a specified point in the time polynomial in degree n. It is known that the problem of computing cycle index polynomial for a group is #P -complete. The proof (given in [1]) involves a polynomial-time reduction from finding number of independent sets in a graph(a #P -complete problem) to computing cycle index polynomial of a permutation group. The proof suggests that that the problem of computing cycle index polynomial is #P -complete even for abelian groups. Here we describe some special permuta- tion groups for which the computation of cycle index polynomial is easy (can be computed in polynomial time). 2 2 Special Permutation Groups Notation 1. Let G be a permutation group with degree n. Consider the permu- tation g ∈ G, if g is the product of a cycle of length l1, a cycle of length l2 and so on till cycle of length lk, then we say g has a cycle struture P (g) = zl1 zl2 . zlk . Notation 2. Let G be a permutation group with degree n. Consider a set of permutations H ⊂ G, then we say X P (H) = P (g) g∈H 2.1 Cyclic Groups Suppose the cyclic group G with degree n acting on Ω = {1, 2, . n}, is generated by g having cycle structure P (g) = zl1 zl2 . zlk where 1 ≤ li ≤ n ∀i. It is easy to see that |G| = lcm{l1, l2 . lk} which can be exponential in n. In general we do not see any easy way to compute cycle index polynomial for cyclic groups. But it is easy in some special cases given below. Case 1 When k = 1 i.e., g = (1, 2, 3 . n) and |G| = n. So, the cycle index polynomial can be computed very easily. It is given by 1 1 X n/d P (G) = φ(d)z n n d d|n . Case 2 m |G| = p , where p is a prime, this case is also easy. Because |G| = lcm{l1, l2 . lk} = m m m p so, ∃li s.t. li = p . Hence |G| = p ≤ n. The cycle index polynomial is given by m 1 1 X i P (G) = φ(pm−i)P (gp ) pm pm i=1 . Case 3 If li 6= lj then gcd{li, lj} = 1 ∀i, j ∈ {1, 2 . k}, i.e., distinct lis are pairwise relatively prime. Let us assume that all lis are distinct. The results can be easily generalised to the other case. Let Si ⊂ Ω be the set of elements in the li-cycle of g. Consider the subgroup l1 H0 = hg i of group G. This is the subgroup of all permutations in G, which fixes all elements in S1. Let right cosets of H0 in G be {H0,H1,...Hl1−1} and 3 the corresponding representatives are {g0, g1 . gl1−1}. It is easy to see that Pl1−1 P (G) = i=0 P (Hi) 0 Let G1 and G1 be permutation groups acting on S1 and Ω\S1 respectively 0 0 0 and for every h ∈ G we can write h = h1 × h1 where, h1 ∈ G1 and h1 ∈ G1, 0 0 s.t. h1(i) = h(i) ∀i ∈ S1 and h1(i) = h(i) ∀i ∈ Ω\S1. Let g = g1 × g1, where 0 0 l1 0 l1 g1 ∈ G1 and g1 ∈ G1. Now, we can see that H0 = hg i ≈ hg1 i. Now, because 0 l1 0 0 ∀j ∈ {1, 2, . k} l1 and lj are relatively prime, hg1 i = G1. So, G1 ≈ H0. Now, i Hi = g H0 i 0 i Hi = (g1 × g1 )H0 i 0 i 0 Hi = (g1 × g1 )G1 0 i 0 0 i 0 0 g1 ∈ G1, so g1 G1 = G1. Hence, i 0 Hi = g1 × G1 So, l1−1 ! X i 0 P (G) = P (g1) × P (G1) i=0 0 P (G) = P (G1) × P (G1) Inductively, P (G) = P (G1)×P (G2) · · ·×P (Gk), where for every j ∈ {1, 2 . k}, Gj be permutation group acting on Sj, and for every h ∈ G we can write h = h1 × h2 · · · × hk where, hj ∈ Gj s.t. hj(i) = h(i) ∀i ∈ Sj. Now, |Gj| = lj ≤ n, so P (Gj) can be computed easily for all j ∈ {1, 2 . k}. Then the cycle index polynomial will be given by 1 1 P (G) = (P (G ) × P (G ) · · · × P (G )) |G| |G| 1 2 k 2.2 Symmetric Group Computing cycle index polynomial for symmetric group(whole permutation group) can be done easily because of its symmetry. Let G be the symmet- ric group acting on Ω = {a1, a2, . an}. Consider the following definition for group G ∀i ∈ {1, 2, . n}, X Pi(G) = P (g) g∈G an is in a i−cycle in g Pn We can see that P (G) = i=1 Pi(G). Consider the subgroup H0 of all permu- tations in G which fixes an. Consider H ≈ H0 as acting on Ω\{an}. For H, 4 Pi(H), would be defined like this, X Pi(H) = P (g) g∈H an−1 is in a i−cycle in g Note that Pn(H) = 0. Because in any g ∈ H maximum length of a cycle can be n − 1. We give the following recursive algorithm for computing cycle index polynomial for a symmetric group at the point (z1, z2 . zn). Algorithm 1 cycle-index(G) Require: A symmetric group G with degree n. Ensure: P (G) and Pi(G) ∀i ∈ {1, 2, . n} at the point (z1, z2 . zn) 1: if n == 1 then 2: return {z1, z1} 3: else 4: {P (H),P1(H),P2(H) ...Pn−1(H)} ← cycle-index(H) 5: P1(G) ← z1P (H) 6: for i ← 2 to n do zi 7: Pi(G) ← (n − 1) Pi−1(H) zi−1 8: end for 9: P (G) ← P1(G) + P2(G) ··· + Pn(G) 10: return {P (G),P1(G),P2(G) ...Pn(G)} 11: end if 1 Then the cycle index polynomial of G will be given by n! P (G). We can see that the algorithm takes O(n2) time. Theorem 3. With the notations given above, P1(G) = z1P (H) and Pi(G) = zi (n − 1) Pi−1(H) ∀i ∈ {2, 3, . n}. And hence, the algorithm is correct. zi−1 Proof. Consider the case when i = 1. Note that, {g ∈ G|an is in a 1 − cycle in g} = {g ∈ G|g fixes an} = H0.
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