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Enumeration of Rooted Planar Triangulations with Respect to Diagonal Flips
Journal of Combinatorial Theory, Series A 88, 276296 (1999) Article ID jcta.1999.2996, available online at http:www.idealibrary.com on
Enumeration of Rooted Planar Triangulations with Respect to Diagonal Flips Zhicheng Gao 1 and Jianyu Wang School of Mathematics and Statistics, Carleton University, Ottawa, Canada K1S 5B6 Communicated by the Managing Editors Received January 13, 1999
We use the generating function approach to enumerate two families of rooted planar near-triangulations (2-connected, and 2-connected with no multiple edges) with respect to the number of flippable edges. It is shown that their generating functions are algebraic. Simple explicit expressions are obtained for the expected number of flippable edges in a random near-triangulation. Asymptotic estimates are obtained for the first two moments, which are then used to show that the numbers of flippable edges in a random near-triangulation and strict near-triangulation are sharply concentrated around 5n2 and 9n4, respectively. 1999 Academic Press
1. INTRODUCTION We consider planar near-triangulations in this paper. They are 2-connected graphs embedded in the plane such that all interior faces are triangles. Near-triangulations in this paper may have multiple edges, but not loops. Unless stated otherwise, all near-triangulations shall be planar. A neartriangulation is called strict if it does not contain any multiple edge and is called of type [n, m] if it has n interior vertices and m exterior vertices. A near-triangulation is called rooted if an edge on the exterior face and an end-vertex of the edge are distinguished. The distinguished edge and vertex are called the root edge and root vertex, respectively, of the near-triangulation. The exterior face is called the root face. A near-triangulation is called a triangulation if the root face is also a triangle. Two rooted near-triangulations are considered the same if there is a homeomorphism from the plane to itself which transforms one to the other and preserves the rooting. Much work has been done on enumerating rooted triangulations since Tutte's poineering works in the early 1960s [11]. In this paper, we enumerate rooted near-triangulations with respect to the number of flippable edges. 1
Research supported by NSERCC.
276 0097-316599 30.00 Copyright 1999 by Academic Press All rights of reproduction in any form reserved.
ROOTED PLANAR TRIANGULATIONS
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The operation of diagonal flips is an important tool in the study of triangulations (both planar and nonplanar) and it has been used extensively by both graph theorists and computer scientists. See [5, 10] for diagonal flips related to computer science, and [3, 6, 8] for graph theoretical aspects of diagonal flips. For enumeration purposes, we define the operation of diagonal flip as follows. Let e be an interior edge (i.e., not on the root face) in a near-triangulation with end vertices a and b. Let abc and abd be the two facial triangles incident with the edge ab. Note that e is a diagonal of the quadrangle acbd. The operation of removing the diagonal e and adding the other diagonal bd is called a diagonal flip. We say that the edge e is flippable if flipping it does not create a loop (i.e., b{d ). For e being flippable in a strict triangulation, we further require that flipping it does not create a multiple edge. In this paper, exterior edges (i.e., edges on the root face) are always unflippable. Throughout this paper, a random near-triangulation is chosen uniformly from a family of rooted near-triangulations. P, E, and Var are used to denote the probability, expectation, and variance of a random variable, respectively. For j 3, let f ijk be the number of rooted near-triangulations of type [i, j] with k flippable edges. Define the generating function f (x, y, z)= : f ijk x iy jz k =: f j (x, z) y j. i, j, k
j
Similarly define the generating function F (x, y, z)= : F ijk x iy jz k =: F j (x, z) y j, i, j, k
j
for rooted strict near-triangulations. We shall prove Theorem 1. f 3(x, z)= g 3(1&2xz 3g 3 ),
(1)
where g= g(x, z) is the power series satisfying 2x 2z 5(1&z) g 5 &2xz 3g 3 &2xz 2(1&z) g 2 +(1&zx+2z 2x&z 3x) g&1=0. (2) Theorem 2. Let * n, m be the number of flippable edges in a random rooted near-triangulation of type [n, m]. Then for any fixed m,
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44m&55nm&24m 2 +15mn 2 +14nm 2 &24+4m 3 &15n 2 +39n E(* n, m )= 2(m&1)(3n+2m&4) = E(* n, m(* n, m &1))=
(3)
5n +O(1), 2
(4)
25n 2 +O(n). 4
(5)
Theorem 3. G(2xzG 4 +(12xz 2(1&z)&1) G 2 +z(1+x&3xz 2 +2xz 3 ) G&3z(1&z)) F 3(x, z)= 3 , (6) xz (1+x&3xz 2 +2xz 3 ) G 3 +9xz 3(1&z) G 2 &3z 2(1&z) where G=G(x, z) is the power series satisfying x 2zG 7 +12x 2z 2(1&z) G 5 &xz(1+x&3xz 2 +2xz 3 ) G 4 +xz(36z 2x&78z 3x+4x 2z 7 + 2xz&6x 2z 3 +9x 2z 5 +4x 2z 4 &12x 2z 6 + 16z+40xz 4 +x 2z&15) G 3 + 12xz 2(1&z)(1+x&3xz 2 +2xz 3 ) G 2 + 3(1&z)(1&3xz 2 +3xz 3 ) G & 3z(1&z)(1+x&3xz 2 +2xz 3 )=0.
(7)
Theorem 4. Let 4 n, m be the number of flippable edges in a random rooted strict near-triangulation of type [n, m]. Then for any fixed m, 4m 3 +14m 2n&28m 2 +18mn 2 &75nm &27n 2 +63m+72n&45 E(4 n, m )= (2m&3)(4n+2m&5) = E(4 n, m(4 n, m &1))=
9n +O(1), 4 81n 2 +O(n). 16
(8) (9) (10)
ROOTED PLANAR TRIANGULATIONS
279
Corollary 1. Let m 3 be fixed, 0 n be any function satisfying 0 n Ä and 0 n =o(- n). Then P(|* n, m &5n2|
(11)
P( |4 n, m &9n4|
(12)
The rest of the paper is organized as follows: In the next section, we derive a functional equation for f (x, y, z). In Section 3 we use Brown's quadratic method to solve this functional equation and obtain algebraic equations for f (x, y, z) and f 3(x, z). The first two moments of * n, m are computed in Section 4. We derive corresponding results for strict neartriangulations in Section 5 and discuss sharp concentration results in Section 6.
2. A FUNCTIONAL EQUATION FOR f (x, y, z) In this section, the arguments of a function are often omitted for simplicity. We first prove Lemma 1. f (x, y, z) satisfies the functional equation f =y 3 +2yzf +y &1z 2f 2 +2xz 2f 2(zf +y 2 )+x(zf +y 2 ) +y &1xz 2( f &f 2 y 2 &2yzf 2(zf +y 2 )& y(zf +y 2 )).
(13)
Proof. We prove the lemma using a standard decomposition technique. Let T be a rooted near-triangulation with v 1 being the root vertex, v 1 v 2 being the root edge, and v 1 v 2 v 3 being the interior face incident with the root edge. We consider the following cases and subcases. Case A. v 3 Is on the Root Face (see Fig. 1). Removing the root edge decomposes T into an ordered pair of rooted near-triangulations: T 1 and T 2 . Note that T 1 andor T 2 might be a single edge, which we call degenerate. There are four subcases: Subcase A 1 .
T 1 and T 2 are both degenerate.
The contribution from this subcase is y 3. Subcase A 2 .
T 1 is degenerate, but T 2 is not.
Noting that v 3 v 2 is flippable in T but it is unflippable in T 2 , we obtain the contribution from this subcase to be yzf.
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FIG. 1.
Subcase A 3 .
A rooted near-triangulation with v 3 being an exterior vertex.
T 2 is degenerate, but T 1 is not.
The contribution is the same as in subcase A 2 , i.e. yzf. Subcase A 4 .
Neither T 1 nor T 2 is degenerate.
Note that in this subcase both edges v 1 v 3 and v 2 v 3 are flippable in T. Hence the contribution is y &1z 2f 2. Case B.
v 3 Is an Interior Vertex (see Fig. 2).
Removing the root edge v 1 v 2 , we obtain a rooted near-triangulation T $ which has at least three exterior vertices. We have the following subcases: Subcase B 1 . In T, edge v 1 v 3 is unflippable and edge v 2 v 3 is flippable (see Fig. 2(a)). In this subcase, v 2 v 3 must lie on the boundary of a digon. The contribution in this case is xz 2(zf +y 2 ) f 2 .
FIG. 2.
Rooted 2-c planar near triangulations with v 3 being an interior vertex.
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Subcase B 2 . In T, edge v 2 v 3 is unflippable and edge v 1 v 3 is flippable (see Fig. 2(b)). The contribution to f from this case is the same as in subcase B 1 . Subcase B 3 . Fig. 2(c)).
In T, edge v 1 v 3 and edge v 2 v 3 are both unflippable (see
Similar to subcase B 1 , we obtain the contribution from this subcase to be x(zf +y 2 ). Subcase B 4 .
Both edge v 1 v 3 and edge v 2 v 3 are flippable.
The contribution from this case is y &1xz 2(( f &f 2 y 2 )&2y &1f 2 z(zf +y 2 )& y(zf +y 2 )), which is obtained by subtracting all the contributions of the other cases from the total f &f 2 y 2. Combining all these cases and subcases, we obtain Eq. (13). K
3. PROOF OF THEOREM 1 Now, we use Brown's Quadratic Method [2] to solve Eq. (13) and prove Theorem 1. To simplify the expression, let f (x, y, z)=zf +y 2. Then f 2(x, z)=[ y 2 ] f (x, y, z)=zf 2 +1, f 3(x, z)=[ y 3 ] f (x, y, z)=zf 3 , and (13) can be rewritten as f 2 +(xy+xz&z &1y&2xyz+xyz 2 +2xyzf 2(1&z)) f =xzy 2f 2 &z &1y 3. (14) Rearrange (14) as A(x, y, z) 2 =B(x, y, z) with A=f +xyz( f 2 &1)&xyz 2f 2 +(xy+xz+xyz 2 &z &1y)2, B=xzy 2f 2 &z &1y 3 +(xyz( f 2 &1)&xyz 2f 2 +(xy+xz+xyz 2 &z &1y)2) 2.
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Let h(x, z)= i 0 h ik x iz k be a power series solution to A(x, h, z)=0. Then B(x, h, z)=0,
and
B y
}
=0. y=h(x, z)
Applying the Maple function ``gsolve'' to the above two equations and setting h=xz 2g, we obtain Eq. (2) and f 2 = g(1&xz 3g 3 ).
(15)
Taking the coefficient of y 3 on both sides of (14), we obtain f 3 =(2xz 3 &2xz 2 ) f 22 +(2xz 2 &xz&xz 3 +1) f 2 &1. Using (15) and (2), we obtain f 3 =zg 3(1&2xz 3g 3 ), and hence f 3 = g 3(1&2xz 3g 3 ). Theorem 1 has been proven. The expression for f (x, y, z) is more complicated, and it will be needed for computing the moments of * n, m . Substituting (15) into (14) and solving for f , we obtain f =
1 ( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z) 2z & (( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z)) 2 + 4y 2z(xz 2g(1&xz 3g 3 )& y)) 12 ).
Note that we have chosen the negative square root in the above expression to ensure that f is a power series. From f =( f & y 2 )z, we obtain f=
1 (y&2y 2z&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z) 2z 2 & (( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z)) 2 + 4y 2z(xz 2g(1&xz 3g 3 )& y)) 12 ) .
(16)
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We can use (2) to get the first few terms in the power series expansion of g(x, z), (using iteration) and then use (1) to obtain the first few terms in f 3(x, z). The following is obtained using Maple, f 3(x, z)=1+(z 3 +3 z) x+(6 z 6 +6 z 4 +6 z 3 +6 z 2 ) x 2 +(28 z 9 +6 z 8 +54 z 7 +12 z 6 +36 z 5 +30 z 4 +10 z 3 ) x 3 +(173 z 12 +30 z 11 +354 z 10 +174 z 9 +300 z 8 +138 z 7 +182 z 6 +90 z 5 +15 z 4 ) x 4 +(1125 z 15 +258 z 14 +2685 z 13 +1152 z 12 +2814 z 11 +1668 z 10 +1494 z 9 +936 z 8 +693 z 7 +210 z 6 +21 z 5 ) x 5 +(7868 z 18 +1920 z 17 +21132 z 16 +9876 z 15 +24900 z 14 +14952 z 13 +18256 z 12 +10680 z 11 +7380 z 10 +4520 z 9 +2100 z 8 +420 z 7 +28 z 6 ) x 6 +(57524 z 21 +15348 z 20 +172032 z 19 +82152 z 18 +231672 z 17 +143244 z 16 +187444 z 15 +124656z 14 +102516z 13 +56716z 12 +34056z 11 +17064z 10 +5376z 9 +756z 8 +36z 7 ) x 7 +(436233 z 24 +124092 z 23 +1437996 z 22 +709140 z 21 +2155704 z 20 +1374252 z 19 +1998540 z 18 +1356804 z 17 +1244754 z 16 +802932 z 15 +515268 z 14 +268476 z 13 +138528 z 12 +53604 z 11 +12132 z 10 +1260 z 9 +45 z 8 ) x 8 + } } } Figure 3 shows all rooted triangulations with up to five vertices. The arrows indicate all the rootings in each triangulation. The flippable edges in each triangulation are indicated by darker edges.
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FIG. 3.
Rooted triangulations with number of vertices 5.
4. PROOF OF THEOREM 2 Now we can compute the first two moments of * n, m . Let q 1(x, y)=
f (x, y, z) z
}
,
q 2(x, y)=
z=1
2 f (x, y, z) z 2
}
. z=1
Then E(* n, m )=
[x ny m ] q 1(x, y) , [x ny m ] f (x, y, 1)
E(* n, m(* n, m &1))=
[x ny m ] q 2(x, y) , [x ny m ] f (x, y, 1)
where [x ny m ] q 1(x, y) denotes the coefficient of x ny m in the power series expansion of q 1(x, y) etc. Setting z=1 and g(x, 1)=1+t in (1) and (2), and solving for f 3(x, 1) and x in terms of t, we obtain f 3(x, 1)=(1&t)(1+t) 3, x=
t . 2(1+t) 3
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(17) (18)
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285
Substituting z=1, and x=t2(1+t) 3 into (16), we obtain f (x, y, 1)=
2y(1+t) 3 &t&( y(1+t) 2 &t) R , 4(1+t) 3
(19)
where (and in the following) R=- 1&4y(1+t) 2. Using (2), we get (10x 2z 4 &12x 2z 5 ) g 5 &6xz 2g 3 &(4xz&6xz 2 ) g 2 g &(x+3xz 2 &4xz) g = . 2 6 z 5(2x z &2x 2z 5 ) g 4 +6xz 3g 2 +2(&2xz 3 +2xz 2 ) g&1 +xz+xz 3 &2xz 2
(20)
Setting z=1 and using x=t(2(1+t) 3 ), we obtain g z
}
= z=1
(4+7t) t . 2(1&2t)
Differentiating (16) with respect to z, using the chain rule and (20) with a bit of algebra, we obtain q 1(x, y)=
y(4+6t+5t 2 &12y&36yt&36yt 2 +12yt 3 ) 4(1+t) 2 R &
y(4+6t+5t 2 &4y&8yt&4yt 2 ) . (1+t) 2
(21)
Similarly, we can differentiate (16) twice with respect to z and then set z=1 to obtain q 2(x, y)=
8(1+t) 2y 3 2(10t 3 +19t 2 +18t+7) y 2 & R2 (1+t) R 2 +
(10t 3 +6+19t 2 +17t) y 2(1+t) 3 R 2
&
30(1+t) 3y 3 y 2(55t 4 +112t 3 +86t 2 &26t&40) + R3 2(&1+2t)(1+t) R 3
&
(10t 3 +6+19t 2 +17t) y . 2(1+t) 3R 3
(22)
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Using (18) and Lagrange inversion formula, we obtain 1 [x n ](1+t) m = [t n&1 ](2 n(1+t) 3nm(1+t) m&1 ) n =
m n 3n+m&1 . 2 n&1 n
\
+
(23)
Using the binomial theorem and a bit of algebra, for m 2, we obtain from (19) that [ y m ] f (x, y, 1)= &
2m 1 t(1+t) 2m&3 4(2m&1) m
\ +
2m&2 1 (1+t) 2m&3. 4(2m&3) m&1
\
+
Using (23) with a bit of algebra, we obtain [x ny m ] f (x, y, 1)=
2 n+1(2m&3)! (3n+2m&4)! , (m&2)! (m&2)! n! (2n+2m&2)!
(24)
which agrees with the result obtained by Mullin [7]. Similarly we obtain from (21) that 2 n&1 [x ny m ] q 1(x, y)=
2m&2
\ m&1 + (3n+2m&5)! (44m&55nm&24m
2
n! (2n+2m&2)!
+15mn 2 +14nm 2 & 24+4m 3 &15n 2 +39n) . Therefore E(* n, m )=
44m&55nm&24m 2 +15mn 2 +14nm 2 &24+4m 3 &15n 2 +39n , 2(m&1)(3n+2m&4)
which is (3) in Theorem 2. When n=0, the above expression gives m&3, which agrees with the fact that all the interior edges are flippable in a neartriangulation with no interior vertices. The exact expression for [x ny m ] q 2(x, y) is a bit more complicated. (It does not seem to be hypergeometric in m, n.) So we only derive its asymptotic expression for each fixed m. The asymptotics of [x ny m ] f (x, y, 1) and [x ny m ] q 1(x, y) can be derived using their exact expressions and the Stirling's formula. In the following, we will derive asymptotics based on the analysis of their generating functions, since it explains more clearly why
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287
[x ny m ] f (x, y, 1), [x ny m ] q 1(x, y) and [x ny m ] q 2(x, y) form a geometric progression (asymptotically as n Ä ), which is the basis for the sharp concentration results. It follows from (18), (19), (21), and (22) that t(x), f (x, y, 1), q 1(x, y), and q 2(x, y) are all algebraic. Using (18), it is easy to show that t(x) is analytic in a region 2 1(=, ,)=[x: |x| 227+=, x{227, |Arg(x&227)| ,] for some 0<,0, and t(x) has the following asymptotic expansion around the singularity x=227, 1 -3 35 3 40 4 2 t(x)= & X+ X 2 & X + X +O(X 5 ) , 2 2 3 108 81
(25)
where X=- 1&256x27. Since t(227)=12, we have |t(x)| <12&$ in 2 1(=, ,) for any $ and sufficiently small =. Therefore f (x, y, 1), q 1(x, y) and q 2(x, y) are all analytic in 2 1(=, ,) for all small y (say, | y| 110). So we can apply Flajolet and Odlyzko's ``Transfer Theorem [4] to derive their asymptotics. Using (18), (19), (21), and (22), and doing the asymptotic expansions at the singularity x=227, we obtain f (x, y, 1)=
- 3 y2 (1&9y) &32X 3 +(a quadratic polynomial in x)+O(X 5 ), 3
q 1(x, y)= & q 2(x, y)=
5- 3 y 2 (1&9y) &32X+(a linear polynomial in x)+O(X 3 ), 4
25- 3 y 2 (1&9y) &32X &1 +(a constant in x)+O(X). 16
Therefore, for each fixed m and n Ä , we have [x ny m ] f (x, y, 1)=
&32 -3 (&9) m n &52(272) n (1+O(1n)), m 31(&32)
\
[x ny m ] q 1(x, y)= & [x ny m ] q 2(x, y)=
+
&32 5- 3 (&9) m n &32(272) n (1+O(1n)), 41(&12) m
\
+
&32 25- 3 (&9) m n &12(272) n (1+O(1n)). m 161(12)
\
+
Now (4) and (5) follow immediately by observing - ?=1(12)=(&12) 1(&12)=(&12)(&32) 1(&32).
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5. STRICT NEAR-TRIANGULATIONS In this section we derive corresponding results for strict near-triangulations. The approach is similar to that used in previous sections for general near-triangulations. However, it seems more complicated to derive the functional equation here. To avoid considerable repeatitions, we will omit details which are essentially the same as in previous sections. The following lemma gives a functional equation for F. Lemma 2. Let u=1+x&xz 2 +2xz 2(1&z)(F 3 &1). Then ( y 2 +zF ) 2 zu+( y 2 +zF)(xz 2(1& yzF 3 ) u+xz 2y(1&z)(1+xz 2(F 4 &zF 23 ))) & y( y 2 +zF)+ y 3(1&xz 2(1&z)(1+xz 2(F 4 &zF 23 )))&xy 2z 2u=0 (26) Proof. Let T be a rooted simple planar near triangulation with v 1 being the root vertex, v 1 v 2 being the root edge and v 1 v 2 v 3 being the interior face incident with v 1 v 2 . As in the proof of Lemma 1, we consider two cases. Case A.
v 3 Is an Exterior Vertex
The contribution from this case is exactly the same as before, which is y 3 +2yzF+ y &1z 2F 2. Case B.
v 3 Is an Interior Vertex
This case is substantially more complicated than before. Let T $ be the near-triangulation obtained from removing the root edge of T. We note that T$ has at least four exterior vertices and there is no edge joining v 1 and v 2 . We also note that if an edge is unflippable in T $, then it is unflippable in T as well. We need to pay special attention to those edges (which we will call dart edges), which are flippable in T $, but are unflippable in T. It is clear that an edge ab in T $ is a dart edge if and only if flipping it creates an edge joining v 1 and v 2 . That means v 1 ab and v 2 ab form two triangular faces (see Fig. 4). We need to introduce a new generating function H(x, y, z, w) for T$ where x marks the number of interior vertices in T $, y marks the number of exterior vertices, z marks the number of flippable edges other than the dart edges, and the new indeterminate w marks the number of dart edges. We first express the contribution from Case B in terms of H(x, y, z, 1). Later we will deal with H. As in the proof of Lemma 1, we further consider the following subcases.
ROOTED PLANAR TRIANGULATIONS
FIG. 4.
289
Dart edge ab in rooted strict near-triangulation T $.
Subcase B 1 . In T, v 1 v 3 is unflippable and v 2 v 3 is flippable (see Figs. 5(a) and 5(b)). Let v 1 v 3 v 4 be a triangular face in T with v 4 {v 2 . Then v 2 v 4 must be an edge in T. There are two subsubcases: Subsubcase B 11 .
v 4 is an interior vertex (see Fig. 5(a)).
Noting that the four edges v 2 v 3 , v 3 v 4 , v 1 v 4 and v 2 v 4 are all flippable in T, we obtain the contribution in this situation to be y &1x 2z 4(F 3 &1) H(x, y, z, 1). Subsubcase B 12 .
v 4 is an exterior vertex (see Fig. 5(b)).
The contribution in this situation is xz 2(F 3 &1) y 3 +2xyz 3(F 3 &1) F+ y &1xz 4(F 3 &1) F 2, where the first term corresponds to the case where both v 1 v 4 and v 2 v 4 are exterior in T, the second term corresponds to the case where exactly one of v 1 v 4 and v 2 v 4 is exterior, and the last term corresponds to the case where neither edge v 1 v 4 nor edge v 2 v 4 is exterior. Hence the total contribution of Subcase B 1 is: y &1x 2z 4(F 3 &1) H(x, y, z, 1)+xz 2(F 3 &1) y 3 +2xyz 3(F 3 &1) F + y &1xz 4(F 3 &1) F 2 =y &1xz 2(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) Subcase B 2 . In T, v 2 v 3 is unflippable and v 1 v 3 is flippable (see Figs. 5(c) and 5(d)).
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Rooted strict near-triangulations with v 3 being an interior vertex.
FIG. 5.
The contribution from this subcase is the same as that from subcase B 1 . Subcase B 3 . 5(f)).
In T, v 1 v 3 and v 2 v 3 are both unflippable (see Figs. 5(e) and
The contribution from this subcase is y &1x 2z 2H(x, y, z, 1)+xy 3 +2xyzF+ y &1xz 2F
2
= y &1x(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ). Here the first term corresponds to the case where v 4 is an interior vertex, and the other terms correspond to the case where v 4 is an exterior vertex. Subcase B 4 .
v 1 v 3 and v 2 v 3 are both flippable.
The contribution of this subcase is y &1xz 2(H(x, y, z, 1)&(2z(F 3 &1)+1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 )), which is obtained by subtracting all the contributions of the other subcases from H(x, y, z, 1).
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Combining all these cases and subcases, we have F=y 3 +2yzF+ y &1z 2F 2 +2y &1xz 2(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) + y &1x 2z 2H(x, y, z, 1)+xy 3 +2xyzF+ y &1xz 2F 2 + y &1xz 2(H(x, y, z, 1) & 2z(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) &(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 )) .
(27)
Now we deal with H(x, y, z, w). We first note that H(x, y, z, z) is the generating function for T $ with z marking all the flippable edges in T $. Hence we have H(x, y, z, z)=(F(x, y, z)& y 3F 3(x, z))& yzF 3(x, z) F(x, y, z).
(28)
Here the first term corresponds to all the strict triangulations with at least four exterior vertices, including those which contain the edge v 1 v 2 , which are counted by the last term. Taking the coefficient of y 4 on both sides of Eq. (28), we have [ y 4 ] H(x, y, z, z)=F 4 &zF 23 .
(29)
Next we claim H(x, y, z, w)=H(x, y, z, 0)+xwz 2H(x, y, z, 0)(1+xz 2[ y 4 ] H(x, y, z, w)) + w(1+xz 2( y 2 +zF) 2 [ y 4 ] H(x, y, z, w).
(30)
Proof of Eq. (30). It is clear that H(x, y, z, 0) counts all T$ with no dart edges. Now suppose T $ contains at least one dart edge. Since any dart edge ab in T$ has the property that v 1 ab and v 2 ab are both triangular faces, there is a unique dart edge ab such that all dart edges of T $ are inside the quadrangle v 1 av 2 b. We call this edge ab the last dart edge of T $ (see Fig. 4). Note that if a=v 3 , then v 3 b is the only dart edge of T $. We consider the following two cases: Case A. Vertex b Is an Exterior Vertex (see Fig. 4(b)). If a=v 3 , the contribution is w( y 2 +zF ) 2; otherwise, the contribution is xwz 2( y 2 +zF ) 2 [ y 4 ] H(x, y, z, w). Hence the contribution in this case is w( y 2 +zF ) 2 (1+xz 2[ y 4 ] H(x, y, z, w)).
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GAO AND WANG
Case B.
Vertex b Is an Interior Vertex (see Fig. 4(a)).
If a=v 3 , the contribution is xwz 2H(x, y, z, 0); otherwise, the contribution is x 2wz 4H(x, y, z, 0)[ y 4 ] H(x, y, z, w). Hence the contribution from this case is xwz 2H(x, y, z, 0)+x 2wz 4[ y 4 ] H(x, y, z, w) H(x, y, z, 0). Combining Cases A and B, we obtain (30).
K
Now we can express H(x, y, z, 1) in terms of F(x, y, z). Setting w=z in (30) and using (28) and (29), we have F& yzF 3 F& y 3F 3 =H(x, y, z, 0)+xz 3(1+z 2x(F 4 &zF 23 )) H(x, y, z, 0) + z(1+z 2x(F 4 &zF 23 ))( y 2 +zF ) 2. Hence H(x, y, z, 0)=
F& yzF 3 F&z(1+xz 2(F 4 &zF 23 ))( y 2 +zF) 2 . 1+xz 3(1+xz 2(F 4 &zF 23 ))
(31)
Taking the coefficient of y 4 on both sides of (31), we have [ y 4 ] H(x, y, z, 0)=
(1&xz 3 )(F 4 &zF 23 )&z . 1+xz 3(1+xz 2(F 4 &zF 23 ))
Taking the coefficient of y 4 on both sides of equation (30) and using (32), we have [ y 4 ] H(x, y, z, w)=
w&z+(1+xwz 2 &xz 3 )(F 4 &zF 23 ) . 1&xwz 2 +xz 3 &x 2z 4(w&z)(F 4 &zF 23 )
(32)
Setting w=1 in (30), and using (31) and (32), we obtain H(x, y, z, 1)=
F& yzF 3 F& y 3F 3 +(1&z)( y 2 +zF ) 2 (1+xz 2(F 4 &zF 23 )) . 1&xz 2(1&z)(1+xz 2(F 4 &zF 23 )) (33)
Substituting (33) into (27), and using a bit of algebra, we complete the proof of Lemma 2. K
ROOTED PLANAR TRIANGULATIONS
293
Theorem 3 can be proved using the same argument as that used in Section 3. The only difference is that there are three unknown functions F(x, y, z), F 3(x, z) and F 4(x, z) in (26), so we cannot use the quadratic method directly. Taking the coefficient of y 4 on both sides of (26), we obtain an equation relating F 3 and F 4 , from which it follows that 3x 2z 5(1&z) F 33 +(1+x&3xz 2 +2xz 3 )(xz 3F 23 &1) +(1&3xz 2 +3xz 3 ) F 3 . F4 = xz 2(3xz 2(1&z) F 3 +(1+x&3xz 2 +2xz 3 )) Substituting the above equation into (26), we obtain an equation involving only two unknown functions F and F 3 . This equation is still quadratic in F and is solved by the quadratic method to give Theorem 3. We omit the computational details here since they are quite tedious and are handled by the computer algebra system Maple. Using Maple and Theorem 3, we obtain the first few terms in the power series expansion of F 3(x, z), F 3(x, z)=1+x+3 x 2 z 2 +(z 9 +9 z 4 +3 z 3 ) x 3 +(3 z 12 +3 z 11 +7 z 9 +27 z 6 +27 z 5 +z 3 ) x 4 + (12 z 15 +9 z 14 +9 z 13 +33 z 12 +42 z 11 +21 z 9 +81 z 8 +162 z 7 +18 z 6 +12 z 5 ) x 5 + (59 z 18 +36 z 17 +27 z 16 +177 z 15 +216 z 14 +189 z 13 +174 z 12 +189 z 11 +243 z 10 + 845 z 9 +270 z 8 +90 z 7 +15 z 6 ) x 6 +(292 z 21 +198 z 20 +108 z 19 +920 z 18 +1035 z 17 + 1053 z 16 +1722 z 15 +1764 z 14 +1134 z 13 +1261 z 12 +4065 z 11 +2430 z 10 +710 z 9 + 270 z 8 +3 z 6 ) x 7 +(1488 z 24 +1038 z 23 +657 z 22 +5100 z 21 +5766 z 20 +4779 z 19 + 10524 z 18 +12609 z 17 +11772 z 16 +9744 z 15 +9411 z 14 +18459 z 13 +18018 z 12 + 6195 z 11 +2835 z 10 +210 z 9 +63 z 8 ) x 8 + } } } . Figure 6 shows all rooted strict planar triangulations with up to six vertices.
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GAO AND WANG
FIG. 6.
Rooted strict planar triangulations with up to six vertices.
Setting z=1 and G(x, 1)=1+t in (7), we obtain x=
t . (1+t) 4
(34)
Using Maple, we also find F(x, y, 1)=
(t& y(1+t) 3 )- 1&4y(1+t) 2 (1+t) 4 &
F z
}
= z=1
(t&3y(1+t) 3 +2y 2(1+t) 4 +2y(1+t) 2 ) , 2(1+t) 4
(35)
(&3+5(1+t)&3(1+t) 2 +(1+t) 3y) y (1+t) 3 (3&5(1+t)+3(1+t) 2 &6(1+t) 2y +9(1+t) 3y&6(1+t) 4y) y + , (1+t) 3 - 1&4y(1+t) 2
(36)
and 2F z 2
}
=((72yt 5 &578y 2t 4 +12yt 4 +26y 2t 2 &3yt&652y 2t 5 &630y 2t 6 z=1
&288y 2t 7 + 14y 2 +60yt 3 +40y 2t&8y 3 +160y 3t 7 +224y 3t 3 +672y 3t 5 +448y 3t 6 & 32y 3t&268y 2t 3 +560y 3t 4 +24y 3t 8 &3y) _ - 1&4y(1+t) 2 + 3y&20y 2 +30y 3 &648y 3t 9 +3yt+942y 2t 6
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ROOTED PLANAR TRIANGULATIONS
&2958y 3t 4 &5370y 3t 5 & 58y 2t&44y 2t 2 +382y 2t 3 +842y 2t 4 +964y 2t 5 &60yt 3 &6342y 3t 6 &12yt 4 & 72yt 5 +174y 3t 2 &678y 3t 3 &5346y 3t 7 +432y 2t 7 &2808y 3t 8 +138y 3t)((1+t) 5 (3t&1)(1&4y(1+t) 2 ) 32 ).
(37)
Note that (35) agrees with the result obtained by Brown [1] by using a new parameter u such that t=u(1&u), [x ny m ] F(x, y, 1)=
[x ny m ]
F z
2(2m&3)! (4n+2m&5)! , (m&3)! (m&1)! n! (3n+2m&3)!
}
2m&4
\ m&1 +(4n+2m&6)! (63m&75nm&28m = 2
z=1
2
n! (3n+2m&3)!
+18mn 2 +14nm 2 & 45+4m 3 &27n 2 +72n) . This proves (8) in Theorem 4. Using (34), it is easy to show that t(x) is analytic in 2 2(=, ,)=[x: |x| 27256+=, x{27256, |Arg(x&27256)| ,] for some 0<,0, and t(x) has the following asymptotic expansion around the singularity x=27256, 115- 6 3 353 4 10 1 2- 6 X+ X 2 & X + X +O(X 5 ). t(x)= & 3 9 27 972 1458
(38)
Since t(27256)=13, it follows from (35), (36), and (37) that F(x, y, 1), ( F z)| z=1 and ( 2F z 2 )| z=1 are analytic in 2(=, ,) for a small = and for all | y| <18, and they have the following asymptotic expansions F(x, y, 1)=c( y) X 3 +(a quadratic polynomial in x)+O(X 5 ), F z
} F z }
=& z=1
2
=
2
z=1
27 c( y) X+(a linear polynomial in x)+O(X 3 ), 8
243 c( y) X &1 +(a constant in x)+O(X), 64
where X=- 1&256x27 and c( y)=
-6 y(1&(27&288y)(9&64y) &32 ). 48
296
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Now the asymptotics in Theorem 4 follow immediately from Flajolet and Odlyzko's ``Transfer Theorem.''
6. SHARP CONCENTRATION RESULTS Proof of Corollary 1. By Theorem 2, we have Var(* n, m )=E(* n, m(* n, m &1))&(E(* n, m )) 2 +E(* n, m )=O(n). Now (11) follows immediately from Chebyshev's inequality. Similarly we obtain (12) using Theorem 4. K Corollary 1 shows that the number of flippable edges in a random rooted (strict) near-triangulation of type [n, m] is sharply concentrated around 5n2 (9n4). According to the results proved in [9], only an exponentially small fraction of (strict) near-triangulations have nontrivial symmetries. Hence Corollary 1 also holds for unrooted (strict) near-triangulations. We also note that, by (3) and (8), asymptotics (4) and (9) hold for all m>2 with m=o(- n). We believe that this is also true for the second moments.
REFERENCES 1. W. G. Brown, Enumeration of triangulations of the disk, Proc. London Math. Soc. (3) 14 (1964), 746768. 2. W. G. Brown, An algebraic technique for solving certain problems in the theory of graphs, in ``Theory of Graphs: Proc. Colloquium, Tihany, Hungary, 1966'' (P. Erdo s and G. Katona, Eds.), pp. 5760, Academic Press, New York, 1968. 3. R. Brunet, A. Nakamoto, and S. Negami, Diagonal flips of triangulations on closed surfaces preserving specified properties, J. Combin. Theory Ser. B 68 (1996), 295309. 4. A. M. Odlyzko, Asymptotic enumeration methods, in ``Handbook of Combinatorics'', (R. L. Graham, M. Grotschel, and L. Lovasz Eds.), Vol. II, Elsevier, Amsterdam, 1995. 5. F. Hurt, M. Noy, and J. Urrutia, Flipping edges in triangulations, in ``Proceeding 12th ACM Symposium on Computational Geometry, May 2426, 1996,'' pp. 214223. 6. H. Komuro, The diagonal flips of triangulations on the sphere, Yokohama Math. J. 44 (1997), 115122. 7. R. C. Mullin, On counting rooted triangular maps. Canad. J. Math. 17 (1965), 373382. 8. S. Negami, Diagonal flips in triangulations of surfaces, Discrete Math. 135 (1994), 225232. 9. R. C. Richmond and N. C. Wormald, Almost all maps are asymmetric, J. Combin. Theory Ser. B 63 (1995), 17. 10. D. Sleator, R. Tarjan, and W. Thurston, Rotation distance, triangulations, and hyperbolic geometry, J. Amer. Math. Soc. 1 (1988), 647681. 11. W. T. Tutte, The enumerative theory of planar maps, in ``A Survey of Combinatorial Theory'' (J. N. Srivastava, Ed.), pp. 437448, North-Holland, Amsterdam, 1973.
Enumeration of Rooted Planar Triangulations with Respect to Diagonal Flips Zhicheng Gao 1 and Jianyu Wang School of Mathematics and Statistics, Carleton University, Ottawa, Canada K1S 5B6 Communicated by the Managing Editors Received January 13, 1999
We use the generating function approach to enumerate two families of rooted planar near-triangulations (2-connected, and 2-connected with no multiple edges) with respect to the number of flippable edges. It is shown that their generating functions are algebraic. Simple explicit expressions are obtained for the expected number of flippable edges in a random near-triangulation. Asymptotic estimates are obtained for the first two moments, which are then used to show that the numbers of flippable edges in a random near-triangulation and strict near-triangulation are sharply concentrated around 5n2 and 9n4, respectively. 1999 Academic Press
1. INTRODUCTION We consider planar near-triangulations in this paper. They are 2-connected graphs embedded in the plane such that all interior faces are triangles. Near-triangulations in this paper may have multiple edges, but not loops. Unless stated otherwise, all near-triangulations shall be planar. A neartriangulation is called strict if it does not contain any multiple edge and is called of type [n, m] if it has n interior vertices and m exterior vertices. A near-triangulation is called rooted if an edge on the exterior face and an end-vertex of the edge are distinguished. The distinguished edge and vertex are called the root edge and root vertex, respectively, of the near-triangulation. The exterior face is called the root face. A near-triangulation is called a triangulation if the root face is also a triangle. Two rooted near-triangulations are considered the same if there is a homeomorphism from the plane to itself which transforms one to the other and preserves the rooting. Much work has been done on enumerating rooted triangulations since Tutte's poineering works in the early 1960s [11]. In this paper, we enumerate rooted near-triangulations with respect to the number of flippable edges. 1
Research supported by NSERCC.
276 0097-316599 30.00 Copyright 1999 by Academic Press All rights of reproduction in any form reserved.
ROOTED PLANAR TRIANGULATIONS
277
The operation of diagonal flips is an important tool in the study of triangulations (both planar and nonplanar) and it has been used extensively by both graph theorists and computer scientists. See [5, 10] for diagonal flips related to computer science, and [3, 6, 8] for graph theoretical aspects of diagonal flips. For enumeration purposes, we define the operation of diagonal flip as follows. Let e be an interior edge (i.e., not on the root face) in a near-triangulation with end vertices a and b. Let abc and abd be the two facial triangles incident with the edge ab. Note that e is a diagonal of the quadrangle acbd. The operation of removing the diagonal e and adding the other diagonal bd is called a diagonal flip. We say that the edge e is flippable if flipping it does not create a loop (i.e., b{d ). For e being flippable in a strict triangulation, we further require that flipping it does not create a multiple edge. In this paper, exterior edges (i.e., edges on the root face) are always unflippable. Throughout this paper, a random near-triangulation is chosen uniformly from a family of rooted near-triangulations. P, E, and Var are used to denote the probability, expectation, and variance of a random variable, respectively. For j 3, let f ijk be the number of rooted near-triangulations of type [i, j] with k flippable edges. Define the generating function f (x, y, z)= : f ijk x iy jz k =: f j (x, z) y j. i, j, k
j
Similarly define the generating function F (x, y, z)= : F ijk x iy jz k =: F j (x, z) y j, i, j, k
j
for rooted strict near-triangulations. We shall prove Theorem 1. f 3(x, z)= g 3(1&2xz 3g 3 ),
(1)
where g= g(x, z) is the power series satisfying 2x 2z 5(1&z) g 5 &2xz 3g 3 &2xz 2(1&z) g 2 +(1&zx+2z 2x&z 3x) g&1=0. (2) Theorem 2. Let * n, m be the number of flippable edges in a random rooted near-triangulation of type [n, m]. Then for any fixed m,
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44m&55nm&24m 2 +15mn 2 +14nm 2 &24+4m 3 &15n 2 +39n E(* n, m )= 2(m&1)(3n+2m&4) = E(* n, m(* n, m &1))=
(3)
5n +O(1), 2
(4)
25n 2 +O(n). 4
(5)
Theorem 3. G(2xzG 4 +(12xz 2(1&z)&1) G 2 +z(1+x&3xz 2 +2xz 3 ) G&3z(1&z)) F 3(x, z)= 3 , (6) xz (1+x&3xz 2 +2xz 3 ) G 3 +9xz 3(1&z) G 2 &3z 2(1&z) where G=G(x, z) is the power series satisfying x 2zG 7 +12x 2z 2(1&z) G 5 &xz(1+x&3xz 2 +2xz 3 ) G 4 +xz(36z 2x&78z 3x+4x 2z 7 + 2xz&6x 2z 3 +9x 2z 5 +4x 2z 4 &12x 2z 6 + 16z+40xz 4 +x 2z&15) G 3 + 12xz 2(1&z)(1+x&3xz 2 +2xz 3 ) G 2 + 3(1&z)(1&3xz 2 +3xz 3 ) G & 3z(1&z)(1+x&3xz 2 +2xz 3 )=0.
(7)
Theorem 4. Let 4 n, m be the number of flippable edges in a random rooted strict near-triangulation of type [n, m]. Then for any fixed m, 4m 3 +14m 2n&28m 2 +18mn 2 &75nm &27n 2 +63m+72n&45 E(4 n, m )= (2m&3)(4n+2m&5) = E(4 n, m(4 n, m &1))=
9n +O(1), 4 81n 2 +O(n). 16
(8) (9) (10)
ROOTED PLANAR TRIANGULATIONS
279
Corollary 1. Let m 3 be fixed, 0 n be any function satisfying 0 n Ä and 0 n =o(- n). Then P(|* n, m &5n2|
(11)
P( |4 n, m &9n4|
(12)
The rest of the paper is organized as follows: In the next section, we derive a functional equation for f (x, y, z). In Section 3 we use Brown's quadratic method to solve this functional equation and obtain algebraic equations for f (x, y, z) and f 3(x, z). The first two moments of * n, m are computed in Section 4. We derive corresponding results for strict neartriangulations in Section 5 and discuss sharp concentration results in Section 6.
2. A FUNCTIONAL EQUATION FOR f (x, y, z) In this section, the arguments of a function are often omitted for simplicity. We first prove Lemma 1. f (x, y, z) satisfies the functional equation f =y 3 +2yzf +y &1z 2f 2 +2xz 2f 2(zf +y 2 )+x(zf +y 2 ) +y &1xz 2( f &f 2 y 2 &2yzf 2(zf +y 2 )& y(zf +y 2 )).
(13)
Proof. We prove the lemma using a standard decomposition technique. Let T be a rooted near-triangulation with v 1 being the root vertex, v 1 v 2 being the root edge, and v 1 v 2 v 3 being the interior face incident with the root edge. We consider the following cases and subcases. Case A. v 3 Is on the Root Face (see Fig. 1). Removing the root edge decomposes T into an ordered pair of rooted near-triangulations: T 1 and T 2 . Note that T 1 andor T 2 might be a single edge, which we call degenerate. There are four subcases: Subcase A 1 .
T 1 and T 2 are both degenerate.
The contribution from this subcase is y 3. Subcase A 2 .
T 1 is degenerate, but T 2 is not.
Noting that v 3 v 2 is flippable in T but it is unflippable in T 2 , we obtain the contribution from this subcase to be yzf.
280
GAO AND WANG
FIG. 1.
Subcase A 3 .
A rooted near-triangulation with v 3 being an exterior vertex.
T 2 is degenerate, but T 1 is not.
The contribution is the same as in subcase A 2 , i.e. yzf. Subcase A 4 .
Neither T 1 nor T 2 is degenerate.
Note that in this subcase both edges v 1 v 3 and v 2 v 3 are flippable in T. Hence the contribution is y &1z 2f 2. Case B.
v 3 Is an Interior Vertex (see Fig. 2).
Removing the root edge v 1 v 2 , we obtain a rooted near-triangulation T $ which has at least three exterior vertices. We have the following subcases: Subcase B 1 . In T, edge v 1 v 3 is unflippable and edge v 2 v 3 is flippable (see Fig. 2(a)). In this subcase, v 2 v 3 must lie on the boundary of a digon. The contribution in this case is xz 2(zf +y 2 ) f 2 .
FIG. 2.
Rooted 2-c planar near triangulations with v 3 being an interior vertex.
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ROOTED PLANAR TRIANGULATIONS
281
Subcase B 2 . In T, edge v 2 v 3 is unflippable and edge v 1 v 3 is flippable (see Fig. 2(b)). The contribution to f from this case is the same as in subcase B 1 . Subcase B 3 . Fig. 2(c)).
In T, edge v 1 v 3 and edge v 2 v 3 are both unflippable (see
Similar to subcase B 1 , we obtain the contribution from this subcase to be x(zf +y 2 ). Subcase B 4 .
Both edge v 1 v 3 and edge v 2 v 3 are flippable.
The contribution from this case is y &1xz 2(( f &f 2 y 2 )&2y &1f 2 z(zf +y 2 )& y(zf +y 2 )), which is obtained by subtracting all the contributions of the other cases from the total f &f 2 y 2. Combining all these cases and subcases, we obtain Eq. (13). K
3. PROOF OF THEOREM 1 Now, we use Brown's Quadratic Method [2] to solve Eq. (13) and prove Theorem 1. To simplify the expression, let f (x, y, z)=zf +y 2. Then f 2(x, z)=[ y 2 ] f (x, y, z)=zf 2 +1, f 3(x, z)=[ y 3 ] f (x, y, z)=zf 3 , and (13) can be rewritten as f 2 +(xy+xz&z &1y&2xyz+xyz 2 +2xyzf 2(1&z)) f =xzy 2f 2 &z &1y 3. (14) Rearrange (14) as A(x, y, z) 2 =B(x, y, z) with A=f +xyz( f 2 &1)&xyz 2f 2 +(xy+xz+xyz 2 &z &1y)2, B=xzy 2f 2 &z &1y 3 +(xyz( f 2 &1)&xyz 2f 2 +(xy+xz+xyz 2 &z &1y)2) 2.
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GAO AND WANG
Let h(x, z)= i 0 h ik x iz k be a power series solution to A(x, h, z)=0. Then B(x, h, z)=0,
and
B y
}
=0. y=h(x, z)
Applying the Maple function ``gsolve'' to the above two equations and setting h=xz 2g, we obtain Eq. (2) and f 2 = g(1&xz 3g 3 ).
(15)
Taking the coefficient of y 3 on both sides of (14), we obtain f 3 =(2xz 3 &2xz 2 ) f 22 +(2xz 2 &xz&xz 3 +1) f 2 &1. Using (15) and (2), we obtain f 3 =zg 3(1&2xz 3g 3 ), and hence f 3 = g 3(1&2xz 3g 3 ). Theorem 1 has been proven. The expression for f (x, y, z) is more complicated, and it will be needed for computing the moments of * n, m . Substituting (15) into (14) and solving for f , we obtain f =
1 ( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z) 2z & (( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z)) 2 + 4y 2z(xz 2g(1&xz 3g 3 )& y)) 12 ).
Note that we have chosen the negative square root in the above expression to ensure that f is a power series. From f =( f & y 2 )z, we obtain f=
1 (y&2y 2z&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z) 2z 2 & (( y&xyz&xz 2 &xyz 3 +2xyz 2 &2xyz 2g(1&xz 3g 3 )(1&z)) 2 + 4y 2z(xz 2g(1&xz 3g 3 )& y)) 12 ) .
(16)
ROOTED PLANAR TRIANGULATIONS
283
We can use (2) to get the first few terms in the power series expansion of g(x, z), (using iteration) and then use (1) to obtain the first few terms in f 3(x, z). The following is obtained using Maple, f 3(x, z)=1+(z 3 +3 z) x+(6 z 6 +6 z 4 +6 z 3 +6 z 2 ) x 2 +(28 z 9 +6 z 8 +54 z 7 +12 z 6 +36 z 5 +30 z 4 +10 z 3 ) x 3 +(173 z 12 +30 z 11 +354 z 10 +174 z 9 +300 z 8 +138 z 7 +182 z 6 +90 z 5 +15 z 4 ) x 4 +(1125 z 15 +258 z 14 +2685 z 13 +1152 z 12 +2814 z 11 +1668 z 10 +1494 z 9 +936 z 8 +693 z 7 +210 z 6 +21 z 5 ) x 5 +(7868 z 18 +1920 z 17 +21132 z 16 +9876 z 15 +24900 z 14 +14952 z 13 +18256 z 12 +10680 z 11 +7380 z 10 +4520 z 9 +2100 z 8 +420 z 7 +28 z 6 ) x 6 +(57524 z 21 +15348 z 20 +172032 z 19 +82152 z 18 +231672 z 17 +143244 z 16 +187444 z 15 +124656z 14 +102516z 13 +56716z 12 +34056z 11 +17064z 10 +5376z 9 +756z 8 +36z 7 ) x 7 +(436233 z 24 +124092 z 23 +1437996 z 22 +709140 z 21 +2155704 z 20 +1374252 z 19 +1998540 z 18 +1356804 z 17 +1244754 z 16 +802932 z 15 +515268 z 14 +268476 z 13 +138528 z 12 +53604 z 11 +12132 z 10 +1260 z 9 +45 z 8 ) x 8 + } } } Figure 3 shows all rooted triangulations with up to five vertices. The arrows indicate all the rootings in each triangulation. The flippable edges in each triangulation are indicated by darker edges.
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GAO AND WANG
FIG. 3.
Rooted triangulations with number of vertices 5.
4. PROOF OF THEOREM 2 Now we can compute the first two moments of * n, m . Let q 1(x, y)=
f (x, y, z) z
}
,
q 2(x, y)=
z=1
2 f (x, y, z) z 2
}
. z=1
Then E(* n, m )=
[x ny m ] q 1(x, y) , [x ny m ] f (x, y, 1)
E(* n, m(* n, m &1))=
[x ny m ] q 2(x, y) , [x ny m ] f (x, y, 1)
where [x ny m ] q 1(x, y) denotes the coefficient of x ny m in the power series expansion of q 1(x, y) etc. Setting z=1 and g(x, 1)=1+t in (1) and (2), and solving for f 3(x, 1) and x in terms of t, we obtain f 3(x, 1)=(1&t)(1+t) 3, x=
t . 2(1+t) 3
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(17) (18)
ROOTED PLANAR TRIANGULATIONS
285
Substituting z=1, and x=t2(1+t) 3 into (16), we obtain f (x, y, 1)=
2y(1+t) 3 &t&( y(1+t) 2 &t) R , 4(1+t) 3
(19)
where (and in the following) R=- 1&4y(1+t) 2. Using (2), we get (10x 2z 4 &12x 2z 5 ) g 5 &6xz 2g 3 &(4xz&6xz 2 ) g 2 g &(x+3xz 2 &4xz) g = . 2 6 z 5(2x z &2x 2z 5 ) g 4 +6xz 3g 2 +2(&2xz 3 +2xz 2 ) g&1 +xz+xz 3 &2xz 2
(20)
Setting z=1 and using x=t(2(1+t) 3 ), we obtain g z
}
= z=1
(4+7t) t . 2(1&2t)
Differentiating (16) with respect to z, using the chain rule and (20) with a bit of algebra, we obtain q 1(x, y)=
y(4+6t+5t 2 &12y&36yt&36yt 2 +12yt 3 ) 4(1+t) 2 R &
y(4+6t+5t 2 &4y&8yt&4yt 2 ) . (1+t) 2
(21)
Similarly, we can differentiate (16) twice with respect to z and then set z=1 to obtain q 2(x, y)=
8(1+t) 2y 3 2(10t 3 +19t 2 +18t+7) y 2 & R2 (1+t) R 2 +
(10t 3 +6+19t 2 +17t) y 2(1+t) 3 R 2
&
30(1+t) 3y 3 y 2(55t 4 +112t 3 +86t 2 &26t&40) + R3 2(&1+2t)(1+t) R 3
&
(10t 3 +6+19t 2 +17t) y . 2(1+t) 3R 3
(22)
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Using (18) and Lagrange inversion formula, we obtain 1 [x n ](1+t) m = [t n&1 ](2 n(1+t) 3nm(1+t) m&1 ) n =
m n 3n+m&1 . 2 n&1 n
\
+
(23)
Using the binomial theorem and a bit of algebra, for m 2, we obtain from (19) that [ y m ] f (x, y, 1)= &
2m 1 t(1+t) 2m&3 4(2m&1) m
\ +
2m&2 1 (1+t) 2m&3. 4(2m&3) m&1
\
+
Using (23) with a bit of algebra, we obtain [x ny m ] f (x, y, 1)=
2 n+1(2m&3)! (3n+2m&4)! , (m&2)! (m&2)! n! (2n+2m&2)!
(24)
which agrees with the result obtained by Mullin [7]. Similarly we obtain from (21) that 2 n&1 [x ny m ] q 1(x, y)=
2m&2
\ m&1 + (3n+2m&5)! (44m&55nm&24m
2
n! (2n+2m&2)!
+15mn 2 +14nm 2 & 24+4m 3 &15n 2 +39n) . Therefore E(* n, m )=
44m&55nm&24m 2 +15mn 2 +14nm 2 &24+4m 3 &15n 2 +39n , 2(m&1)(3n+2m&4)
which is (3) in Theorem 2. When n=0, the above expression gives m&3, which agrees with the fact that all the interior edges are flippable in a neartriangulation with no interior vertices. The exact expression for [x ny m ] q 2(x, y) is a bit more complicated. (It does not seem to be hypergeometric in m, n.) So we only derive its asymptotic expression for each fixed m. The asymptotics of [x ny m ] f (x, y, 1) and [x ny m ] q 1(x, y) can be derived using their exact expressions and the Stirling's formula. In the following, we will derive asymptotics based on the analysis of their generating functions, since it explains more clearly why
ROOTED PLANAR TRIANGULATIONS
287
[x ny m ] f (x, y, 1), [x ny m ] q 1(x, y) and [x ny m ] q 2(x, y) form a geometric progression (asymptotically as n Ä ), which is the basis for the sharp concentration results. It follows from (18), (19), (21), and (22) that t(x), f (x, y, 1), q 1(x, y), and q 2(x, y) are all algebraic. Using (18), it is easy to show that t(x) is analytic in a region 2 1(=, ,)=[x: |x| 227+=, x{227, |Arg(x&227)| ,] for some 0<,0, and t(x) has the following asymptotic expansion around the singularity x=227, 1 -3 35 3 40 4 2 t(x)= & X+ X 2 & X + X +O(X 5 ) , 2 2 3 108 81
(25)
where X=- 1&256x27. Since t(227)=12, we have |t(x)| <12&$ in 2 1(=, ,) for any $ and sufficiently small =. Therefore f (x, y, 1), q 1(x, y) and q 2(x, y) are all analytic in 2 1(=, ,) for all small y (say, | y| 110). So we can apply Flajolet and Odlyzko's ``Transfer Theorem [4] to derive their asymptotics. Using (18), (19), (21), and (22), and doing the asymptotic expansions at the singularity x=227, we obtain f (x, y, 1)=
- 3 y2 (1&9y) &32X 3 +(a quadratic polynomial in x)+O(X 5 ), 3
q 1(x, y)= & q 2(x, y)=
5- 3 y 2 (1&9y) &32X+(a linear polynomial in x)+O(X 3 ), 4
25- 3 y 2 (1&9y) &32X &1 +(a constant in x)+O(X). 16
Therefore, for each fixed m and n Ä , we have [x ny m ] f (x, y, 1)=
&32 -3 (&9) m n &52(272) n (1+O(1n)), m 31(&32)
\
[x ny m ] q 1(x, y)= & [x ny m ] q 2(x, y)=
+
&32 5- 3 (&9) m n &32(272) n (1+O(1n)), 41(&12) m
\
+
&32 25- 3 (&9) m n &12(272) n (1+O(1n)). m 161(12)
\
+
Now (4) and (5) follow immediately by observing - ?=1(12)=(&12) 1(&12)=(&12)(&32) 1(&32).
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GAO AND WANG
5. STRICT NEAR-TRIANGULATIONS In this section we derive corresponding results for strict near-triangulations. The approach is similar to that used in previous sections for general near-triangulations. However, it seems more complicated to derive the functional equation here. To avoid considerable repeatitions, we will omit details which are essentially the same as in previous sections. The following lemma gives a functional equation for F. Lemma 2. Let u=1+x&xz 2 +2xz 2(1&z)(F 3 &1). Then ( y 2 +zF ) 2 zu+( y 2 +zF)(xz 2(1& yzF 3 ) u+xz 2y(1&z)(1+xz 2(F 4 &zF 23 ))) & y( y 2 +zF)+ y 3(1&xz 2(1&z)(1+xz 2(F 4 &zF 23 )))&xy 2z 2u=0 (26) Proof. Let T be a rooted simple planar near triangulation with v 1 being the root vertex, v 1 v 2 being the root edge and v 1 v 2 v 3 being the interior face incident with v 1 v 2 . As in the proof of Lemma 1, we consider two cases. Case A.
v 3 Is an Exterior Vertex
The contribution from this case is exactly the same as before, which is y 3 +2yzF+ y &1z 2F 2. Case B.
v 3 Is an Interior Vertex
This case is substantially more complicated than before. Let T $ be the near-triangulation obtained from removing the root edge of T. We note that T$ has at least four exterior vertices and there is no edge joining v 1 and v 2 . We also note that if an edge is unflippable in T $, then it is unflippable in T as well. We need to pay special attention to those edges (which we will call dart edges), which are flippable in T $, but are unflippable in T. It is clear that an edge ab in T $ is a dart edge if and only if flipping it creates an edge joining v 1 and v 2 . That means v 1 ab and v 2 ab form two triangular faces (see Fig. 4). We need to introduce a new generating function H(x, y, z, w) for T$ where x marks the number of interior vertices in T $, y marks the number of exterior vertices, z marks the number of flippable edges other than the dart edges, and the new indeterminate w marks the number of dart edges. We first express the contribution from Case B in terms of H(x, y, z, 1). Later we will deal with H. As in the proof of Lemma 1, we further consider the following subcases.
ROOTED PLANAR TRIANGULATIONS
FIG. 4.
289
Dart edge ab in rooted strict near-triangulation T $.
Subcase B 1 . In T, v 1 v 3 is unflippable and v 2 v 3 is flippable (see Figs. 5(a) and 5(b)). Let v 1 v 3 v 4 be a triangular face in T with v 4 {v 2 . Then v 2 v 4 must be an edge in T. There are two subsubcases: Subsubcase B 11 .
v 4 is an interior vertex (see Fig. 5(a)).
Noting that the four edges v 2 v 3 , v 3 v 4 , v 1 v 4 and v 2 v 4 are all flippable in T, we obtain the contribution in this situation to be y &1x 2z 4(F 3 &1) H(x, y, z, 1). Subsubcase B 12 .
v 4 is an exterior vertex (see Fig. 5(b)).
The contribution in this situation is xz 2(F 3 &1) y 3 +2xyz 3(F 3 &1) F+ y &1xz 4(F 3 &1) F 2, where the first term corresponds to the case where both v 1 v 4 and v 2 v 4 are exterior in T, the second term corresponds to the case where exactly one of v 1 v 4 and v 2 v 4 is exterior, and the last term corresponds to the case where neither edge v 1 v 4 nor edge v 2 v 4 is exterior. Hence the total contribution of Subcase B 1 is: y &1x 2z 4(F 3 &1) H(x, y, z, 1)+xz 2(F 3 &1) y 3 +2xyz 3(F 3 &1) F + y &1xz 4(F 3 &1) F 2 =y &1xz 2(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) Subcase B 2 . In T, v 2 v 3 is unflippable and v 1 v 3 is flippable (see Figs. 5(c) and 5(d)).
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Rooted strict near-triangulations with v 3 being an interior vertex.
FIG. 5.
The contribution from this subcase is the same as that from subcase B 1 . Subcase B 3 . 5(f)).
In T, v 1 v 3 and v 2 v 3 are both unflippable (see Figs. 5(e) and
The contribution from this subcase is y &1x 2z 2H(x, y, z, 1)+xy 3 +2xyzF+ y &1xz 2F
2
= y &1x(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ). Here the first term corresponds to the case where v 4 is an interior vertex, and the other terms correspond to the case where v 4 is an exterior vertex. Subcase B 4 .
v 1 v 3 and v 2 v 3 are both flippable.
The contribution of this subcase is y &1xz 2(H(x, y, z, 1)&(2z(F 3 &1)+1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 )), which is obtained by subtracting all the contributions of the other subcases from H(x, y, z, 1).
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ROOTED PLANAR TRIANGULATIONS
291
Combining all these cases and subcases, we have F=y 3 +2yzF+ y &1z 2F 2 +2y &1xz 2(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) + y &1x 2z 2H(x, y, z, 1)+xy 3 +2xyzF+ y &1xz 2F 2 + y &1xz 2(H(x, y, z, 1) & 2z(F 3 &1)(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 ) &(xz 2H(x, y, z, 1)+( y 2 +zF ) 2 )) .
(27)
Now we deal with H(x, y, z, w). We first note that H(x, y, z, z) is the generating function for T $ with z marking all the flippable edges in T $. Hence we have H(x, y, z, z)=(F(x, y, z)& y 3F 3(x, z))& yzF 3(x, z) F(x, y, z).
(28)
Here the first term corresponds to all the strict triangulations with at least four exterior vertices, including those which contain the edge v 1 v 2 , which are counted by the last term. Taking the coefficient of y 4 on both sides of Eq. (28), we have [ y 4 ] H(x, y, z, z)=F 4 &zF 23 .
(29)
Next we claim H(x, y, z, w)=H(x, y, z, 0)+xwz 2H(x, y, z, 0)(1+xz 2[ y 4 ] H(x, y, z, w)) + w(1+xz 2( y 2 +zF) 2 [ y 4 ] H(x, y, z, w).
(30)
Proof of Eq. (30). It is clear that H(x, y, z, 0) counts all T$ with no dart edges. Now suppose T $ contains at least one dart edge. Since any dart edge ab in T$ has the property that v 1 ab and v 2 ab are both triangular faces, there is a unique dart edge ab such that all dart edges of T $ are inside the quadrangle v 1 av 2 b. We call this edge ab the last dart edge of T $ (see Fig. 4). Note that if a=v 3 , then v 3 b is the only dart edge of T $. We consider the following two cases: Case A. Vertex b Is an Exterior Vertex (see Fig. 4(b)). If a=v 3 , the contribution is w( y 2 +zF ) 2; otherwise, the contribution is xwz 2( y 2 +zF ) 2 [ y 4 ] H(x, y, z, w). Hence the contribution in this case is w( y 2 +zF ) 2 (1+xz 2[ y 4 ] H(x, y, z, w)).
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GAO AND WANG
Case B.
Vertex b Is an Interior Vertex (see Fig. 4(a)).
If a=v 3 , the contribution is xwz 2H(x, y, z, 0); otherwise, the contribution is x 2wz 4H(x, y, z, 0)[ y 4 ] H(x, y, z, w). Hence the contribution from this case is xwz 2H(x, y, z, 0)+x 2wz 4[ y 4 ] H(x, y, z, w) H(x, y, z, 0). Combining Cases A and B, we obtain (30).
K
Now we can express H(x, y, z, 1) in terms of F(x, y, z). Setting w=z in (30) and using (28) and (29), we have F& yzF 3 F& y 3F 3 =H(x, y, z, 0)+xz 3(1+z 2x(F 4 &zF 23 )) H(x, y, z, 0) + z(1+z 2x(F 4 &zF 23 ))( y 2 +zF ) 2. Hence H(x, y, z, 0)=
F& yzF 3 F&z(1+xz 2(F 4 &zF 23 ))( y 2 +zF) 2 . 1+xz 3(1+xz 2(F 4 &zF 23 ))
(31)
Taking the coefficient of y 4 on both sides of (31), we have [ y 4 ] H(x, y, z, 0)=
(1&xz 3 )(F 4 &zF 23 )&z . 1+xz 3(1+xz 2(F 4 &zF 23 ))
Taking the coefficient of y 4 on both sides of equation (30) and using (32), we have [ y 4 ] H(x, y, z, w)=
w&z+(1+xwz 2 &xz 3 )(F 4 &zF 23 ) . 1&xwz 2 +xz 3 &x 2z 4(w&z)(F 4 &zF 23 )
(32)
Setting w=1 in (30), and using (31) and (32), we obtain H(x, y, z, 1)=
F& yzF 3 F& y 3F 3 +(1&z)( y 2 +zF ) 2 (1+xz 2(F 4 &zF 23 )) . 1&xz 2(1&z)(1+xz 2(F 4 &zF 23 )) (33)
Substituting (33) into (27), and using a bit of algebra, we complete the proof of Lemma 2. K
ROOTED PLANAR TRIANGULATIONS
293
Theorem 3 can be proved using the same argument as that used in Section 3. The only difference is that there are three unknown functions F(x, y, z), F 3(x, z) and F 4(x, z) in (26), so we cannot use the quadratic method directly. Taking the coefficient of y 4 on both sides of (26), we obtain an equation relating F 3 and F 4 , from which it follows that 3x 2z 5(1&z) F 33 +(1+x&3xz 2 +2xz 3 )(xz 3F 23 &1) +(1&3xz 2 +3xz 3 ) F 3 . F4 = xz 2(3xz 2(1&z) F 3 +(1+x&3xz 2 +2xz 3 )) Substituting the above equation into (26), we obtain an equation involving only two unknown functions F and F 3 . This equation is still quadratic in F and is solved by the quadratic method to give Theorem 3. We omit the computational details here since they are quite tedious and are handled by the computer algebra system Maple. Using Maple and Theorem 3, we obtain the first few terms in the power series expansion of F 3(x, z), F 3(x, z)=1+x+3 x 2 z 2 +(z 9 +9 z 4 +3 z 3 ) x 3 +(3 z 12 +3 z 11 +7 z 9 +27 z 6 +27 z 5 +z 3 ) x 4 + (12 z 15 +9 z 14 +9 z 13 +33 z 12 +42 z 11 +21 z 9 +81 z 8 +162 z 7 +18 z 6 +12 z 5 ) x 5 + (59 z 18 +36 z 17 +27 z 16 +177 z 15 +216 z 14 +189 z 13 +174 z 12 +189 z 11 +243 z 10 + 845 z 9 +270 z 8 +90 z 7 +15 z 6 ) x 6 +(292 z 21 +198 z 20 +108 z 19 +920 z 18 +1035 z 17 + 1053 z 16 +1722 z 15 +1764 z 14 +1134 z 13 +1261 z 12 +4065 z 11 +2430 z 10 +710 z 9 + 270 z 8 +3 z 6 ) x 7 +(1488 z 24 +1038 z 23 +657 z 22 +5100 z 21 +5766 z 20 +4779 z 19 + 10524 z 18 +12609 z 17 +11772 z 16 +9744 z 15 +9411 z 14 +18459 z 13 +18018 z 12 + 6195 z 11 +2835 z 10 +210 z 9 +63 z 8 ) x 8 + } } } . Figure 6 shows all rooted strict planar triangulations with up to six vertices.
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FIG. 6.
Rooted strict planar triangulations with up to six vertices.
Setting z=1 and G(x, 1)=1+t in (7), we obtain x=
t . (1+t) 4
(34)
Using Maple, we also find F(x, y, 1)=
(t& y(1+t) 3 )- 1&4y(1+t) 2 (1+t) 4 &
F z
}
= z=1
(t&3y(1+t) 3 +2y 2(1+t) 4 +2y(1+t) 2 ) , 2(1+t) 4
(35)
(&3+5(1+t)&3(1+t) 2 +(1+t) 3y) y (1+t) 3 (3&5(1+t)+3(1+t) 2 &6(1+t) 2y +9(1+t) 3y&6(1+t) 4y) y + , (1+t) 3 - 1&4y(1+t) 2
(36)
and 2F z 2
}
=((72yt 5 &578y 2t 4 +12yt 4 +26y 2t 2 &3yt&652y 2t 5 &630y 2t 6 z=1
&288y 2t 7 + 14y 2 +60yt 3 +40y 2t&8y 3 +160y 3t 7 +224y 3t 3 +672y 3t 5 +448y 3t 6 & 32y 3t&268y 2t 3 +560y 3t 4 +24y 3t 8 &3y) _ - 1&4y(1+t) 2 + 3y&20y 2 +30y 3 &648y 3t 9 +3yt+942y 2t 6
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ROOTED PLANAR TRIANGULATIONS
&2958y 3t 4 &5370y 3t 5 & 58y 2t&44y 2t 2 +382y 2t 3 +842y 2t 4 +964y 2t 5 &60yt 3 &6342y 3t 6 &12yt 4 & 72yt 5 +174y 3t 2 &678y 3t 3 &5346y 3t 7 +432y 2t 7 &2808y 3t 8 +138y 3t)((1+t) 5 (3t&1)(1&4y(1+t) 2 ) 32 ).
(37)
Note that (35) agrees with the result obtained by Brown [1] by using a new parameter u such that t=u(1&u), [x ny m ] F(x, y, 1)=
[x ny m ]
F z
2(2m&3)! (4n+2m&5)! , (m&3)! (m&1)! n! (3n+2m&3)!
}
2m&4
\ m&1 +(4n+2m&6)! (63m&75nm&28m = 2
z=1
2
n! (3n+2m&3)!
+18mn 2 +14nm 2 & 45+4m 3 &27n 2 +72n) . This proves (8) in Theorem 4. Using (34), it is easy to show that t(x) is analytic in 2 2(=, ,)=[x: |x| 27256+=, x{27256, |Arg(x&27256)| ,] for some 0<,0, and t(x) has the following asymptotic expansion around the singularity x=27256, 115- 6 3 353 4 10 1 2- 6 X+ X 2 & X + X +O(X 5 ). t(x)= & 3 9 27 972 1458
(38)
Since t(27256)=13, it follows from (35), (36), and (37) that F(x, y, 1), ( F z)| z=1 and ( 2F z 2 )| z=1 are analytic in 2(=, ,) for a small = and for all | y| <18, and they have the following asymptotic expansions F(x, y, 1)=c( y) X 3 +(a quadratic polynomial in x)+O(X 5 ), F z
} F z }
=& z=1
2
=
2
z=1
27 c( y) X+(a linear polynomial in x)+O(X 3 ), 8
243 c( y) X &1 +(a constant in x)+O(X), 64
where X=- 1&256x27 and c( y)=
-6 y(1&(27&288y)(9&64y) &32 ). 48
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GAO AND WANG
Now the asymptotics in Theorem 4 follow immediately from Flajolet and Odlyzko's ``Transfer Theorem.''
6. SHARP CONCENTRATION RESULTS Proof of Corollary 1. By Theorem 2, we have Var(* n, m )=E(* n, m(* n, m &1))&(E(* n, m )) 2 +E(* n, m )=O(n). Now (11) follows immediately from Chebyshev's inequality. Similarly we obtain (12) using Theorem 4. K Corollary 1 shows that the number of flippable edges in a random rooted (strict) near-triangulation of type [n, m] is sharply concentrated around 5n2 (9n4). According to the results proved in [9], only an exponentially small fraction of (strict) near-triangulations have nontrivial symmetries. Hence Corollary 1 also holds for unrooted (strict) near-triangulations. We also note that, by (3) and (8), asymptotics (4) and (9) hold for all m>2 with m=o(- n). We believe that this is also true for the second moments.
REFERENCES 1. W. G. Brown, Enumeration of triangulations of the disk, Proc. London Math. Soc. (3) 14 (1964), 746768. 2. W. G. Brown, An algebraic technique for solving certain problems in the theory of graphs, in ``Theory of Graphs: Proc. Colloquium, Tihany, Hungary, 1966'' (P. Erdo s and G. Katona, Eds.), pp. 5760, Academic Press, New York, 1968. 3. R. Brunet, A. Nakamoto, and S. Negami, Diagonal flips of triangulations on closed surfaces preserving specified properties, J. Combin. Theory Ser. B 68 (1996), 295309. 4. A. M. Odlyzko, Asymptotic enumeration methods, in ``Handbook of Combinatorics'', (R. L. Graham, M. Grotschel, and L. Lovasz Eds.), Vol. II, Elsevier, Amsterdam, 1995. 5. F. Hurt, M. Noy, and J. Urrutia, Flipping edges in triangulations, in ``Proceeding 12th ACM Symposium on Computational Geometry, May 2426, 1996,'' pp. 214223. 6. H. Komuro, The diagonal flips of triangulations on the sphere, Yokohama Math. J. 44 (1997), 115122. 7. R. C. Mullin, On counting rooted triangular maps. Canad. J. Math. 17 (1965), 373382. 8. S. Negami, Diagonal flips in triangulations of surfaces, Discrete Math. 135 (1994), 225232. 9. R. C. Richmond and N. C. Wormald, Almost all maps are asymmetric, J. Combin. Theory Ser. B 63 (1995), 17. 10. D. Sleator, R. Tarjan, and W. Thurston, Rotation distance, triangulations, and hyperbolic geometry, J. Amer. Math. Soc. 1 (1988), 647681. 11. W. T. Tutte, The enumerative theory of planar maps, in ``A Survey of Combinatorial Theory'' (J. N. Srivastava, Ed.), pp. 437448, North-Holland, Amsterdam, 1973.