Explicit enumeration of triangulations with multiple boundaries
aa r X i v : . [ m a t h . C O ] J u l Explicit enumeration of triangulations withmultiple boundaries
Maxim KrikunInstitut Elie CartanUniversite Henri PoincareNancy, [email protected] 3, 2018
Abstract
We enumerate rooted triangulations of a sphere with multiple holes bythe total number of edges and the length of each boundary component.The proof relies on a combinatorial identity due to W.T. Tutte.
A planar map is a class of equivalence of embedded graphs
G ֒ → S by thehomeomorphisms of S . We note by V ( G ), E ( G ) and F ( G ) the sets of vertices,edges and faces of the the map G , respectively.A map with holes , is a pair ( G, H ), H ⊂ F ( G ), such that no two faces h, h ′ ∈ H share a common vertex, and all vertices at the boundary of h i ∈ H are distinct (i.e. the boundary of h i is a cycle with no self-intersections). In thefollowing we refer to the faces h ∈ H as holes . A map is called a triangulation ,if every face of F ( G ) \ H has degree 3. If H = ∅ , such triangulation is called a complete triangulation . In the following we will consider rooted triangulations,that is triangulations with one distinguished directed edge, called the root. Inaddition to that, we assume that the holes of a triangulation are enumeratedby integers 0 , . . . , r and that the root is always located at the boundary of the0-th hole. In this paper we solve explicitly the recursive equations for generating func-tions planar triangulations with arbitrary number of holes, in terms of the totalnumber of edges and the length of each boundary component.1he class of triangulations we consider is the most wide possible — theunderlying graph may contain multiple edges and loops. Although this class issometimes thought of as ”pathological”, it turns out that the presence of loopsis a feature which greatly simplifies the calculations involved (e.g. comparedto [4]).Our main result is the following
Theorem 1
Let C r ( n, α ; α , . . . , α r ) be the number of rooted triangulationswith ( r + 1) hole, with α j edges on the boundary of the j -th hole and n edges intotal. Then we have, letting m = α + . . . + α r , C r ( n, α ; α , . . . , α r ) = 4 k (2 m + 3 k − k + 1 − r )!(2 m + k )!! α r Y j =0 (cid:18) α j α j (cid:19) , (1) if n = 2 m + 3 k , and C r ( n, α ; α , . . . , α r ) = 0 if n − m = 0 (mod ). The case r = 0 corresponds to the problem of enumeration of planar near-triangulations, solved by Tutte in [7] using the method of recursive decomposi-tion. The same method, applied to the problem of enumeration of triangulationson an orientable surface of genus g , leads in a natural way to enumeration oftriangulations (or maps) with multiple holes. We were unable to obtain anygeneral result in the non-planar case, but for completeness we provide the cor-responding recurrent relations in Section 2.3, as well as the generating functionsfor the triangulations of orientable a torus and double torus ( g = 1 and g = 2).The decomposition method used in our study and the equations involvedare not new. The similar ideas were applied by Bender and Canfield ([2]), andlater by Arqu´es and Gioretti ([1]), to the asymptotical enumeration of arbitraryrooted maps on surfaces.Similar equations appear under the name of loop , or Schwinger-Dyson equa-tions in some models of two-dimensional quantum gravity. Ambjørn et al. stud-ied the asymptotical number of triangulations (and some more general classes ofmaps) on the sphere and higher genera surfaces with multiple holes (see Chapter4 in [3]). We have found that the Proposition 1 in section 4 below looks verysimilar to the formula (4 .
95) in [3], which expresses the generating function ofplanar maps with multiple boundary components via the repeated applicationof the so-called loop insertion operator.A simplified version of loop insertion operation may be described as follows.Given a complete rooted triangulation, one can cut it along the root edge,and identify the obtained hole with two edges of an additional triangle. Thisoperation provides a bijection between the complete rooted triangulations with n edges, and triangulations with n + 2 edges and a single hole of length 1. Thus2aking C ( n + 2 , C ( n + 2 ,
1) = 2 · k − (3 k )!!( k + 1)!( k + 2)!! , n = 2 + 3 k, which gives, by duality, the number of almost trivalent maps with k vertices(sequence A002005 in [6]), computed by Mullin, Nemeth and Schellenberg in [5]. This paper is organized as follows. In section 2 we describe the recursive decom-position procedure for triangulations and derive equations on the correspondinggenerating functions, then solve explicitly these equations for r = 0 , , ,
3. InSection 3 we calculate explicitly the coefficients C r for r = 0 ,
2. This allows tosuggest the main formula of Theorem 1, which is then proved in section 4. Theproof closely follows that of [8].
Let C k ( n, m ; α , . . . , α k ) be the number of rooted planar triangulations with( k + 1) holes H = ( h , h , . . . , h k ), such that there are m edges at the boundaryof h , α j edges on the boundary of h j , j = 1 , . . . , k and n edges total.First, let us remind the recursive decomposition method. Given a rootedplanar triangulation G with one hole (that is, a triangulation of a disk), andassuming that there is at least one triangle, one deletes the triangle t thatcontains the root. In function of the position of a vertex v , opposite to the rootedge in t , there are two possibilities:(A) if v is an internal vertex of the triangulation, one obtains a new triangu-lation with one face less and one more edge on the boundary.(B) if v lies on the boundary of G , one cuts the resulting map in two parts, with( n , n ) edges and the boundaries of length ( m , m ), such that n + n = n − m + m = m +1, ( n, m ) being the number of edges and boundarylength of the original configuration.As the final object one obtains a planar map, consisting of a single edge, whichwe treat as a triangulation with 0 faces, 1 edge and one hole with boundarylength 2.Now if G is a triangulation with multiple ( k + 1, say) holes, there exists athird possibility for v , namely(C) if v is located at the boundary of the hole h j , then after erasing the rootedge one cuts the resulting map along v , obtaining a map with one holeless, and with boundary of h having length m ′ = m + α j + 1, α j beingthe length of the boundary of h j in the original triangulation.3ow let U k ( x, y, z , . . . , z k ) be the multivariate generating function U k ( x, y, z , . . . , z k ) = X N ≥ X m ≥ X α j ≥ C ( n, m ; α , . . . , α k ) x n y m z α · · · z α k k . Translating the above decomposition procedure into the language of generatingfunction, we get the following
Lemma 2.1
The following equations hold U ( x, y ) = xy + xy (cid:16) U ( x, y ) − yL ( x ) (cid:17) + xy U ( x, y ) (2) U k ( x, y ; z ) = xy (cid:16) U k ( x, y ; z ) − yL k ( x ; z ) (cid:17) + xy X ω ⊂ I k U | ω | ( x, y ; z ω ) U k −| ω | ( x, y ; z I k \ ω )+ k X j =1 h xy − z j (cid:16) z j y U k − ( x, y ; ˆ z j ) − yz j U k − ( x, z j ; ˆ z j ) (cid:17) + xL k − ( x ; ˆ z j ) i (3) where L k ( x ; z ) = [ y ] U k ( x, y ; z ) ,I k = { , , . . . , k } is the index set, the sum is over all subsets ω of I k (includingempty set and I k itself ), z stands for z , . . . , z k , z ω is the list of variables z j with j ∈ ω , and ˆ z j stands for z , . . . , z k without z j . Proof.
The equation (2) is a classical relation for the generating function ofnear-triangulations: the term xy accounts for the special single-edged map,the term, linear in U , corresponds to the case (A) above, and the quadraticterm to the case (B).In (3), the first term on the right-hand side is derived exactly the same wayas in (2); the summation over ω corresponds to the possible ways to distributethe k enumerated holes between the two parts of a triangulation in case (B).To see how the summation over j in (3) arises, consider first the case k = 1,i.e. a triangulation with two holes. When the rule (C) above applies, removingof the root edge merges the two holes, of lengths α and α , into a single hole4f length ( α + α + 1). This gives the following contribution to U ( x, y, z ): X n ≥ X α ≥ ,α ≥ C ( n − , α + α + 1) x n y α z α = x X n ≥ X m ≥ C ( n − , m ) x n − ( yz m − + y z m − . . . + y m − z )= x X n ≥ X m ≥ C ( n − , m ) x n − zy m − − yz m − y − z = xy − z h zy (cid:16) U ( x, y ) − U ( x, − y [ t ] U ( x, t ) − y [ t ] U ( x, y ) (cid:17) − yz (cid:16) U ( x, z ) − U ( x, − z [ t ] U ( x, t ) − z [ t ] U ( x, y ) (cid:17)i = xy − z (cid:16) zy U ( x, y ) − yz U ( x, z ) (cid:17) + x [ t ] U ( x, t ) . A general case k ≥ h j with thehole h , all other holes remain intact. The equations (2), (3) may be solved exactly. First, (2) is solved using thequadratic method, giving U ( x, y ) = h − y h p − h y − x − y x , (4)where h = h ( x ) is a positive power series in x , satisfying the relation8 h x − h + x = 0 , (5)namely h ( x ) = ∞ X k =0 k (3 k − k !( k + 1)!! x k +1 = x ∞ X k =0 k (3 k − k !( k + 1)!! (2 x ) k (6)(cf. sequence A078531 in [6]).Next, one may solve (3) with respect to L k ( x ; z ) and group the terms con-taining U k ( x, y ; z ), obtaining xL k ( x, t ; z ) = 1 y (cid:16) x − y + 2 xU ( x, y ) (cid:17) U k ( x, y ; z ) + W k ( x, y ; z ) , (7)where W k ( x, y ; z ) is the sum of terms in (3), not containing U k , W k ( x, y ; z ) = xy X ω ⊂ Ik < | ω | The following relations hold: T g,k ( x, y ; z ) = xy (cid:16) T g,k ( x, y ; z ) − y [ t ] T g,k ( x, t ; z ) (cid:17) + xy g X i =1 X ω ⊂ I k T i, | ω | ( x, y ; z ω ) T g − i,k −| ω | ( x, y ; z I k \ ω )+ k X j =1 h xy − z j (cid:16) z j y T g,k − ( x, y ; ˆ z j ) − yz j T g,k − ( x, z j ; ˆ z j ) (cid:17) + x [ t ] T g,k − ( x, t ; ˆ z j ) i + x ∂∂t T g − ,k +1 ( x, y ; z , . . . , z k , t ) (cid:12)(cid:12)(cid:12) t = y . (15) Proof. When the case (D) applies, after removing the root edge we get a tri-angulation with an additional hole, and with a distinguished vertex on theboundary of this hole (the image of v ). This gives the last term in (15), andthe rest is similar to (3).The equation (15) may be solved analogously to (3). In particular, we findgenerating function for triangulations of genus 1 and 2 with one hole T , ( x, y ) = (1 − h y ) h y (1 − h ) (1 − h y ) / (16) T , ( x, y ) = P , ( h, y )(1 − h ) (1 − h y ) / , (17)where P , ( h, y ) = 3 h y (35 + 184 h + 48 h ) × (1024 h y + 1024 h y − y h + 1)+ 128 h y (545 + 1488 h − h + 2560 h )+ 64 h y ( − − h + 256 h + 324 h )7 Extracting exact coefficients Letting h = x √ ζ and t = x in (5) we get ζ = 8 t (1 + ζ ) / , (18)so the Lagrange’s inversion theorem applies, and we have, assuming n = m + 3 k ,[ x n ] h m = [ x n − m ]( h/x ) m = [ t k ](1 + ζ ) m/ = 1 k [ λ k − ] n m λ ) m/ − (1 + λ ) k/ o = mk ! 4 k ( m + 3 k − m + k )!! . (19)In particular this gives the formula (6) for h ( x ).For U we have the following series expansion in yU ( x, y ) = h − x + 2 h x hx y + ∞ X m =0 m + 1 (cid:18) mm (cid:19)(cid:16) − m + 2 m + 2 h (cid:17) h m +1 y m +2 . Letting n = 2 m + 3 k , k ≥ − x n y m ] U ( x, y ) = m (cid:18) mm (cid:19) k (2 m + 3 k − k + 1)!(2 m + k )!! , (20)and [ x n y m ] U ( x, y ) = 0 if n + m = 0 ( mod U ( sym )2 has the product form, so the expansion is particularlyeasy to calculate. First we’ll need the coefficients[ x n ] n h − h (4 h ) m − o = 12 ∞ X j =0 m + j [ x n ] h m +3 j +3 = 18 2 m +2 k (2 m + 3 k − k X j =1 m + 3 j ( k − j )!(2 m + 2 j + k )!!= 18 2 m +2 k (2 m + 3 k − k − m + k )!! . where n = 2 m + 3 k . Then we obtain[ x n y α z α z α ] U ( sym )2 ( x, y, z , z )= Y i =0 (2 α i − α i − ( α i − · [ t n ] n h − h (4 h ) m − o = 2 m +2 k (2 m + 3 k − k − m + k )!! (2 α − α − α − α − α − α − , = α α α (cid:18) α α (cid:19)(cid:18) α α (cid:19)(cid:18) α α (cid:19) · k (2 m + 3 k − k − m + k )!! (21)8here m = α + α + α , n = 2 m + 3 k ; the coefficient is C ( n, . . . ) is null if n − m = 0 ( mod x n z α z α . . . z α k ] U ( sym ) k ( x, z ; z ) = 4 k (2 m + 3 k − k + 1 − r )!(2 m + k )!! r Y j =0 α j (cid:18) α j α j (cid:19) (22)where m = α + . . . + α r and n = 2 m + 3 k .Clearly, this formula is equivalent to (1), and it further agrees with the aboveexpressions for U ( sym )1 and U ( sym )3 (as can be seen by calculating explicitly fewfirst terms in the power series expansions of these functions). The above expression (22) resembles a formula obtained by Tutte in [8], for thenumber of slicings with k external faces of degrees 2 n , . . . , n k γ ( n , n , . . . , n k ) = ( n − n − k + 2)! k Y i =1 (2 n i )! n i !( n i − X ω ⊂ I D | ω |− k { λ · f ω } · D | ¯ ω |− l { µ · f ¯ ω } = X ω ⊂ I | ω | Let ˆ u r ( t, y ; z , . . . , z r ) = t − r U r ( t , t − y ; t − z , . . . , t − z r ) . (28) Then for all r ≥ u r ( t, y ; z ) = (cid:16) ∂∂t (cid:17) r − ζ ( t ) Z y (1 − sy ) / r Y j =1 (cid:16) p − sz j − (cid:17) ds. (29) Proof. First, applying the transformation (28) to U , U we findˆ u ( t, y ) = 12 (cid:16) t − y √ ζ (cid:17)p − ζ ) y − t − y , ˆ u ( t, y, z ) = y p − ζ ) z y − z ) p − ζ ) y − p − ζ ) y − z y − z ) . It can be verified by explicit integration that ˆ u satisfies (29).Next, for all r ≥ u k ( t, y ; z ) = (cid:16) ∂∂t (cid:17) r − n y ζ ′ ( t )(1 − ζ ) y ) / r Y j =1 (cid:16) p − ζ ) z j − (cid:17)o . (30)From (18) we have ζ ′ = 16(1 + ζ ) / − ζ , so ˆ u ( t, y ; z , z ) = 8 t (1 + ζ ) / − ζ/ y (1 − ζ ) y ) / × (cid:16) p − ζ ) z − (cid:17)(cid:16) p − ζ ) z − (cid:17) satisfies (29) as well.Now suppose that (29) holds for r = 0 , , . . . , k − k ≥ 3, and letus show that it holds as well for r = k .The equation (3) leads to the following equation on ˆ u k :ˆ u k ( t, y ; z ) = ty (cid:16) ˆ u k ( t, y ; z ) − y ˆ l k ( x ; z ) (cid:17) + 1 y X ω ⊂ I k ˆ u | ω | ( t, y ; z ω )ˆ u k −| ω | ( t, y ; z I k \ ω )+ k X j =1 h y − z j (cid:16) z j y ˆ u k − ( t, y ; ˆ z j ) − yz j ˆ u k − ( t, z j ; ˆ z j ) (cid:17) + ˆ l k − ( t ; ˆ z j ) i , (31)11ith ˆ l k ( t ; z ) = [ y ]ˆ u k ( t, y ; z ) . Rewrite (31) as (cid:16) y − t − u ( t, y ) (cid:17) ˆ u k ( t, y, z )= X ω ⊂ Ik < | ω | TheoreticalComputer Science , 234:255–272, 2000.[2] E.A. Bender and E.R. Canfield. The asymptotic number of rooted maps ona surface. Journal of Combinatorial Theory Series A , 43(2):244–257, 1986.[3] B.J. Durhuus J. Ambjørn and T. Jonsson. Quantum geometry: A StatisticalField Theory Approach . Cambridge Univ. Press, 1997.[4] M. Krikun. Boundaries of random triangulation of a disk. Discrete Mathe-matics and Applications , 14(3):301–315, 2003.[5] E. Nemeth R.C. Mullin and P.J. Schellenberg. The enumeration of almostcubic maps. Proceedings of the Louisiana Conference on Combinatorics,Graph Theory and Computer Science , 1:281–295, 1970.[6] N.J.A. Sloane. The On-Line Encyclopedia of Integer Sequences. .[7] W.T. Tutte. A census of planar triangulations. Canad. J. Math , 14(1):21–38,1962.[8] W.T. Tutte. A census of slicings.