Unlabelled Structures: Restricted Constructions

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1 Gao & Šana, Combinatorial Enumeration Notes 5 Unlabelled Structures: Restricted Constructions Let SEQ k (B denote the set of sequences of exactly k elements from B, and SEQ k (B denote the set of sequences of at least k elements from B. It is clear that the OGF of SEQ k (B is B(z k, and the OGF of SEQ k (B is B(z = k B(zk B(z. Integer compositions. Let B be the set of all positive integers. Then SEQ k (B is the class of compositions with exactly k parts. The OGF of SEQ k (B is ( z z k = n k ( n z n. k there are ( n k compositions of n with exactly k parts. Integer partitions. Let P o be the class of integer partitions with odd parts only. Then P o = MSET (B, where B is the set of odd integers. P o (z = m z 2m. We note P o (z = z = z2m 2m z2m z = 2m m m = ( z n ( + z n = ( + z n, z n n n n ( z2n n ( zn which is the same as the OGF for integer partitions with distinct parts. the number of partitions with odd parts is equal to the number of partitions with distinct parts. Unordered pairs. Let (B B := {(β, β : β B} denote the diagonal of B B. It is clear that the OGF of (B B is B n z 2n = B(z 2. n Let A = SET 2 (B be the class of unordered pairs of structures in B. Since we have B B = SET 2 (B + SET 2 (B + (B B, B(z 2 = 2A(z + B(z 2 and A(z = 2 ( B(z 2 B(z 2.

2 Gao & Šana, Combinatorial Enumeration Notes 5 2 To derive the OGF of general SET k, it is convenient to use bivariate generating functions. Detailed discussion of bivariate generating functions will be given later. The basic idea is to use a second variable u to keep track of the number of parts in a structure. For example, the OGF of SEQ k (B becomes u k B(z k, and the OGF of SEQ(B becomes u k B(z k = ub(z. k 0 Theorem. Let B be a combinatorial class without a neutral element, and with OGF B(z. The OGFs of SET k (B, MSET k (B, and CY C k (B are all polynomials in B(z,..., B(z k, namely: (a OGF of SET k (B: (b OGF of MSET k (B: (c OGF of CY C k (B: [u k ] exp [u k ] ( k [u k ] exp k = = u ( B(z ( k ϕ( = ln u B(z u B(z More over, the OGF of CY C(B is ϕ( ln B(z. Proof. (a The bivariate OGF for SET (B is ( ( ( + uz n Bn = exp ( B n ln( + uz n = exp n n n n ( B ( u z n ( u = exp ( B(z. (b Similarly, the bivariate OGF for MSET (B is ( ( uz n Bn = exp n u B(z. (c Let S = SEQ (B. A sequence in S is called primitive if it cannot be written as w := ww w for some w S and 2.

3 Gao & Šana, Combinatorial Enumeration Notes 5 3 Let PS be the class of primitive sequences, and PS k be the class of primitive sequences of length k. It is clear that each sequence in S can be written uniquely as w for some primitive sequence w and hence S = w PS {w, w2, w 3,...}. Then S(z, u = k u k B(z k = ub(z ub(z is the bivariate OGF of all sequences of structures in B. Note that u marks the length of a sequence. Define P S(z, u similarly as the bivariate OGF for primitive sequences, with u marking the length of a sequence. We have so S = { w, w 2, w 3,... }, k w PS k S(z, u = ( u k z w + u 2k z 2 w + + u k z w + k w PS k = ( z w ( u k w PS k k = P S(z, u Thus S n,k = n, k P S n,. k Let d = gcd(n, k and r = d. Define We have f(r = S rn d, rk d and f(d = r d g(r. g(r = P S rn d,. rk d It follows from Möbius inversion that g(d = d ( d µ(f, or equivalently, P S(z, u = n,k n, k P S n,k = n, k µ(s n,. k µ(s n, u k z n = µ( S k n,k uk z n n,k

4 Gao & Šana, Combinatorial Enumeration Notes 5 4 = µ(s(z, u = u B(z µ( u B(z. Let C(z, u be the bivariate OGF of all cyclic sequences, and P C(z, u be the bivariate OGF of primitive cyclic sequences. Since each cyclic sequence is a power of a primitive cyclic sequence, we have C(z, u = P C(z, u. Finally we note that each primitive cyclic sequence of length l corresponds to exactly l primitive sequences of length l. This gives and consequently P S(z, u = u P C(z, u, u P S(z, u P C(z, u = du = u µ( ln u B(z. Using the identity we obtain Therefore, k n µ(k k C(z, u = k = n = n k n C(z, = n = ϕ(n n, µ(k k ln u k B(z k µ(k k ln u n B(z n ϕ(n n ln u n B(z n ϕ(n n ln B(z n is the OGF of CY C(B, and is the OGF of CY C k (B. [u k ]C(z, u = [u k ] k = ϕ( ln u B(z

5 Gao & Šana, Combinatorial Enumeration Notes 5 5 Pointing and Substitution The following constructions correspond to the differentiation and composition, respectively, of OGFs. Pointing Let {ɛ, ɛ 2,...} be a fixed set of distinct neutral structures. The pointing of a class B is defined as ΘB = n 0 B n {ɛ,..., ɛ n }. Let A = ΘB. Then each structure β B n corresponds to n structures in A n, obtained from β by pointing at a different atom. (Note that we assume that the atoms in β are distinguishable, though unlabelled. A n = nb n, and hence the OGF of ΘB is z B(z. Substitution (composition The substitution of C into B, denoted by B[C] or B C, is defined as B[C] = B C = SEQ k (C. k 0 β B That is, each structure in B[C] is obtained by replacing each atom in β B with a structure in C. (Again, we are assuming that the atoms in β are distinguishable, though unlabelled. the OGF of B[C] is B k C(z k = B(C(z. k 0 Example. Let B be the class of rooted plane trees where the size of a tree is equal to the number of edges. Then a pointed rooted plane tree is obtained by selecting ( pointing at an edge of the tree. Since a tree with n edges has n + vertices, the number of pointed rooted plane trees with n edges is equal to nb n = nt n+ = nc n = n ( 2n. n + n Example. The class P of permutations can be constructed recursively as follows. Each permutation of [n] can be constructed from a permutation of [n ] by pointing at one of the n gaps between the n elements, and then inserting the element n at the gap. Symbolically we have P = {ɛ} + Θ(Z P. Note that the factor Z increases the size of a permutation by. Consequently P (z = + z (zp (z. It follows that P n = np n, which we know to be true.

6 Gao & Šana, Combinatorial Enumeration Notes 5 6 Example. Smirnov words: words such that adacent letters are distinct. We note that an unconstrained word can be turned into a Smirnov word by compressing the sequence of consecutive identical letters into a single letter. We also note that this operation is reversible. Let S be the class of Smirnov words over [r] and W be the class of all words over [r]. Then W = S SEQ (Z, where Z denotes the set containing a single fixed atom. That is, any word is obtained by a Smirnov word by replacing each letter by a sequence of identical letters. ( z rz = W (z = S z and S(z = r z +z = + z (r z. S n = r(r n. This is consistent with the direct counting argument: there are r choices for the first position, after that there are exactly r choices for each position so that it is different from the previous choice. Example. Let B be the class of rooted plane binary trees, where the size of a tree is the number of its vertices. Then B = Z + Z B B, and consequently B(z = z + zb(z 2. B(z = 4z 2. 2z Now let M be the class of rooted plane unary-binary trees (so each vertex has at most 2 children. Then M = Z + Z M + Z M M, and consequently M(z = z + zm(z + zm(z 2. M(z = z ( z 2 4z 2. 2z On the other hand, we may obtain the class M by replacing each vertex of a tree in B by a path (above that vertex. That is, M = B SEQ (Z. ( z M(z = B = z ( 4 2 z z z z 2 = z ( z 2 4z 2. 2z

7 Gao & Šana, Combinatorial Enumeration Notes Example: Series-Parallel Networks The class SP of series-parallel networks is defined recursively as follows. The size of each network is the number of vertices. (i SP = S P, where S P = Z, containing the single-vertex network. (ii A network is in S if and only if it is either in Z or is obtained from (smaller networks in P using a series construction. (iii A network is in P if and only if it is either in Z or is obtained from (smaller networks in S using a parallel construction. We consider two versions, plane and non-plane. Plane series-parallel networks: in the parallel construction, vertical order matters. So the parallel construction corresponds to the sequence construction. Then we have S = Z + SEQ 2 (P, P = Z + SEQ 2 (S, and S + P = SP + Z. This gives S(z = z + P (z2 P (z and P (z = z + S(z2 S(z. S(z = P (z and S(z is determined by the equation or S(z = z + S(z2 S(z, 2S(z 2 ( + zs(z + z = 0. Solving this quadratic equation, we obtain (taking the negative square root S(z = ( + z 6z + z 4 2 = z + z 2 + 3z 3 + z z 5 +.

8 Gao & Šana, Combinatorial Enumeration Notes 5 8 S... P... Non-plane series-parallel networks: in the parallel construction, the vertical order does not matter (see the figure above. So the parallel construction corresponds to the multiset construction. Now we have S = Z + SEQ 2 (P, P = Z + MSET 2 (S, and S + P = SP + Z. S(z = z + P (z2 P (z and P (z = z + n ( z n Sn S(z. It is difficult to obtain explicit expressions for S(z and P (z in this case. But we may still use the above equations to compute the coefficients S n and P n recursively, starting with S = P =, then computing S 2, P 2, S 3,....