jonathan l. gross- embeddings of cubic halin graphs a surface-by-surface inventory

8/3/2019 Jonathan L. Gross- Embeddings of Cubic Halin Graphs A Surface-by-Surface Inventory

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Embeddings of Cubic Halin GraphsA Surface-by-Surface Inventory

Jonathan L. Gross

Columbia University, New York, NY 10027

Abstract

We derive an O(n2)-time algorithm for calculating the sequence ofnumbers g0(G), g1(G), g2(G), ... of distinct ways to embed a 3-regularHalin graph G on the respective orientable surfaces S0, S1, S2, ... .Key topological features are a quadrangular decomposition of planeHalin graphs and a new recombinant-strands reassembly process thatfits pieces together three-at-a-vertex. Key algorithmic features arereassembly along a post-order traversal, with just-in-time dynamic as-signment of roots for quadrangular pieces encountered along the tour.

1. Introduction

2. Some preexisting results

3. Quadrangulating a plane Halin graph

4. Reassembling from quadrangles

5. Partials and productions

6. Productions for Halin graphs

7. Sample calculation

8. Conclusions and open problems

* [email protected]; http://www.cs.columbia.edu/gross/.

1


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Embeddings of Cubic Halin Graphs: An Inventory 2

1 Introduction

We begin with a quick explanation of the title.

Halin graphs

A Halin graph is constructed as follows:

1. Let T be a plane tree

with at least 4 vertices no vertices of degree 2

2. Draw a cycle thru the leaves of the tree

in the order they occur in a preorder traversal.

Figure 1.1: A Halin graph for a 14-vertex tree with 8 leaves.


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Inventories of Embeddings

Def. The genus distribution for graph G is the sequence

dist(G) : g0(G), g1(G), g2(G),

EXAMPLE: K2 K3 a.k.a. CL3

Figure 1.2: g0 = 2, g1 = 38, g2 = 24

Notation Si denotes the orientable surface of genus i.

Notation gi(G) denotes # embeddings G Si.

Notation min(G) is the minimum genus of G.

Notation max(G) is the maximum genus of G.


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Counting Embeddings

Def. Let G be a labeled graph and S an oriented surface. Theembeddings : G S and : G S are equivalent iff thereexists an autohomeomorphism h : S S that

maps (G) isomorphically to (G);

preserves every label of G;

preserves the orientation of S.

2 1 3

0

2 13

0

Figure 1.3: Looking alike is not enough.


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Graph Amalgamations

In general, amalgamating two graphs means identifying asubgraph in one of them to an isomorphic subgraph in the other.

Vertex-amalgamation and edge-amalgamation are thetwo simplest kinds of amalgamation of two graphs.

=*V

Figure 1.4: Amalgamation of two graphs at a vertex.

=*E

Figure 1.5: Amalgamation of two graphs on an edge.


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Genus Distributions of Graph Amalgamations

=u v

*

Figure 1.6: Bar-amalgamation of two graphs.

dist(G H) = deg(u) deg(v) dist(G) dist(H)

here means convolution; [GrFu87]

g-dist of vertex amalg [GrKhPo10, KhPoGr10, Gr11].

g-dist of edge-amalg [PoKhGr10a, PoKhGr10b].


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2 Some Preexisting Results

About Minimum Genus

Thm 2.1 Calculating min(G) is NP-hard. [Th89]

min(G) is known for relatively few graphs:

complete graphs Kn [RiYo68]

complete bipartite graphs Km,n [Ri65]

hypercubes Qn [Ri55]

a bunch of less prominent families of graphs

K6 K3,4 Q3

Figure 2.1: Three families solved by Ringel.

Remark By way of contrast, there are combinatorial characterizations of

maximum genus by [Xu79] and [Ne81] and a polynomial-time algorithm by[FuGrMc88]. There are maximum genus determinations (e.g., see [Sk91])for families of graphs for which minimum genus calculation would seem tobe formidably hard. An up-to-date overview of maximum genus is given by[ChHu09].


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Minimum Genus of Graph Amalgamations

Thm 2.2 (Additivity of minimum genus)

min(G V H) = min(G) + min(H)

Proof [BHKY62]

By way of contrast, consider the edge-amalgamation K5EK5.

Figure 2.2: Amalgamation of two copies of K5 on an edge.

Obviously, the minimum genus is at least 1 and at most 2.

But is it really 2?


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NO! min(K5 E K5) = 1.

Y 0 2 X 1 Y

20Y1X2

K5

-> S1

Y 0 2 X 1 Y

20Y1X2

K5

-> S1

Y 0 2 X 1 Y

20Y1X2

K5 *E

K5

-> S1

Figure 2.3: K5 E K5 S1


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About Genus Distributions

Calculating dist(G) requires calculating min(G). The graphswhose genus distributions are known are planar or one edge awayfrom planarity.

Closed-end ladders [FuGrSt89] (derived 1984).

Figure 2.4: The closed-end ladder L4.

Circular ladders and Mobius ladders [McG87] (derived 1986).

Figure 2.5: Circular ladder CL4. and Mobius ladder M L4.

Ringel ladders [Te00].

Figure 2.6: Ringel ladder RL4.


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Some g-dist deriviations use [Ja87] (group representations).

Bouquets [GrRoTu89].

Figure 2.7: Bouquets B1, B2, and B3.

Dipoles [Ri90], [KwLe93].

Figure 2.8: Dipoles D1, D2, and D3.

Fans [ChMaZo11].

Figure 2.9: Fans F3 and F5.


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Cubic outerplanar graphs [Gr11].

Figure 2.10: A cubic outerplanar graph.

4-regular outerplanar graphs [PoKhGr11].

Figure 2.11: A 4-regular outerplanar graph.

Remark A survey of genus distributions, including average genus, is givenby [Gr09].


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3 Quadrangulating a Plane Halin Graph

We decompose a plane Halin graph H S0 into atomic partsthat will facilitate a genus distribution calculation.

We regard the vertices and the edges of the Halin graph itselfas black.

Step 1. In each edge on the exterior cycle, insert a red midpoint.

Figure 3.1: Halin graph with red midpoints on exterior cycle.

Since H is a Halin graph, there is exactly one exterior edgeon each polygonal face of the plane embedding.


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Step 2. Join each red vertex v to all of the non-leaf vertices on

the boundary of the face in whose boundary v lies.

Figure 3.2: Halin graph with all of its red edges.

Prop 3.1 The red and black edges together triangulate the re-gion inside the exterior cycle of the Halin graph.

Prop 3.2 Every black tree edge lies on two of the triangles formedby Steps 1 and 2.


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Step 3a. For each black tree edge, we pair the two incident

triangles into a quadrangle.Step 3b. We assign (unseen) colors blue, green, and brown theto tree edges so as to form a proper edge 3-coloring.

Step 3c. We visibly color each quadrangle with the unseencolor of the tree edge that bisects it.

Figure 3.3: Quadrangulation of a plane Halin graph.

Remark Any tree of maximum degree 3 is edge-3-colorable(via greedy algorithm).


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Step 4. Separate the quadrangulated map into quadrangles.

Figure 3.4: Separated quadrangles of a plane Halin graph.

Remark This quadrangulation works for any Halin graph, re-gardless of the degrees of its vertices.

Whats Next??

4 a puzzle: reassembling quadrangles into the Halin graph.

5 how this leads to a genus distribution.

6 the new set of productions for Halin graphs.

7 application to our running example of a Halin graph.

8 conclusions and open problems


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4 Reassembling from Quadrangles

u

w

v x

z

y

Quadrangulation puzzle for a plane cubic Halin graph H S0:

1. Each quadrangle Q is regarded as an initial fragment.

2. An RR-path on a fragment boundary is a

2-path with two red edges,

from a red vertex

through a black vertex

to another red vertex.

3. Initially, all RR-paths are said to be live.


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4. A legal move is initiated by choosing a vertex v such that

v is previously unchosen;

at least one fragment at v is a quadrangle;

all three RR-paths through v are live RR-paths.

Then the three fragments that meet at v are merged into asingle (larger) fragment.

If there is more than one live RR-path on the boundaryof the merged fragment, then all but one of the live RR-

paths are deemed to be dead.

5. You LOSE if you run out of legal moves before the map isfully reassembled.

This happens whenever there occurs an unmerged vertex wsuch that

there is a dead RR-path through w;

none of the fragments meeting at w is a quadrangle.

6. You WIN the game by reassembling the plane map.


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Attempt 1. Start with a merger at v.

Analysis. There are three live RR-paths on the boundary ofthe merged fragment.

u

w

vx

z

y

live RR-path

Figure 4.1: Attempt #1.

Result. LOSE, because RR-paths through two of the unmergedvertices u,w,x become dead.


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Attempt 2. First choose u and then choose v.

Analysis. There are two live RR-paths on the boundary of themerged fragment.

u

w

vx

z

y

live RR-path


Result. LOSE, because the RR-path through one of the un-merged vertices w, x becomes dead.


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Attempt 3. Start with u , w , y, z.

u

w

v x

z

y


Result. LOSE, since after there is a merger at v or x, there willbe no quadrangle at the remaining unmerged vertex.


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Solution: post-order traversal

post-order traversal for a plane tree: trace the boundary ofthe only region and call out the name of a vertex whenever yousee it for the last time.

u y

v x

w z

Figure 4.4: Post-order traversal.

z y x u v w


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u

w

v x

z

y

1. As a root for the inscribed tree of the Halin graph, chooseany leaf-vertex. (Must be a leaf to win.)

2. Choose vertices in the order in which they occur on a post-order traversal of the tree.

SOLUTION: for a root at the lower left corner

SOLUTION: z y x u v w


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z y x u v w

u

w

v x

z

y u

w

v x

z

y

u

w

vx

z

y

u

w

vx

z

y

u

w

v x

z

y u

w

v x

z

y

Figure 4.5: Solving the puzzle with post-order traversal.


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5 Partials and Productions

Partial Genus Distributions

In a single-rooted graph (G, r) with deg(r) = 2, the root r lieseither on two distinct fb-walks (type di) or twice on the same fb-walk (type si). We partition gi(G, r) into single-root partials:

di(G, r) = the number of embeddings of type-di, and

si(G, r) = the number of embeddings of type-si.

type-d0 type-s1

rr

Figure 5.1: The types of single-root partials for a 2-valent root.

Example 5.1

d0(B2, r) = 4 d1(B2, r) = 0s0(B2, r) = 0 s1(B2, r) = 2

g0(B2, r) = 4 g1(B2, r) = 2

Example 5.2

d0(K4, r) = 2 d1(K4, r) = 8s0(K4, r) = 0 s1(K4, r) = 6

g0(K4, r) = 2 g1(K4, r) = 14


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Double-Rooted Graphs

When two single-rooted graphs are pasted together acrosstheir single roots, there is no surviving root. However, if twodouble-rooted graphs are pasted together with one root of onegraph matched to one root of the other graph, then the mergedgraph has two roots.

(G,b,c) H(d, e) (X,b,e)

This permits iterated amalgamations.

Figure 5.2: An iterated edge-amalgamation.

Example 5.3 By following iterated amalgamation with a self-amalgamation, we can obtain graphs like circular ladders andMobius ladders.


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Whereas a single-edge-rooted graph (G, c) has only two par-

tials, namelydi(G, c) and si(G, c)

a double-edge-rooted graph (G,c,d) has the four partials

ddi(G,c,d), dsi(G,c,d), sdi(G,c,d), and ssi(G,c,d)

Example 5.4 (double-root partial g-dist for K4)

Figure 5.3: K4 with two roots at midpoints of two edges.

i ddi dsi sdi ssi gi0 2 0 0 0 21 4 4 0 2 14

Table 5.1: Partial genus distribution for (K4, u , v).

Remark The number of partials increases exponentiallyaccording to the number of roots. It follows that the num-ber of roots employed by a method of genus distribution calcu-lation must be limited.


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There are three sub-partials within the dd-type:

dd the two roots sharing two fb-walks;

dd the two roots sharing one fb-walk;

dd0 no shared fb-walks for the two roots.

type dd0

type dd'

type dd"

Figure 5.4: The three types of dd-subpartials.

Remark There are 10 double-root sub-partials for two edge-roots that both have 2-valent endpoints.

Remark The number of partials increases exponentially(also) with the degrees of vertex-roots and with degreesof the endpoints of the edge-roots, corresponding to a largernumber of possible configurations of fb-walks at the root.


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Deriving Productions with Recombinant Strands

Example 5.5

ddi (G,b,c) ddj (H,d,e) 4dd

i+j(X,b,e) + 2ss

2i+j+1(X,b,e)

b

e

b

e

b

e

b

e

b

eb

b

c

d

e

edd"i

dd"j

dd'i+j

dd'i+j dd'i+j dd'i+j

ss2i+j ss2i+j

Figure 5.5: Derivation of a production for vertex-amalgamation.


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6 Productions for Halin Graphs

We now introduce a new form of graph amalgamation, in whichseveral graphs are merged in a cycle around a vertex.

For cubic Halin graphs, we merge three graphs at a time,exactly as for the puzzle, so that one of them is a quadrangleQ = K4 e.

A

Q

X

B

vv

vv

r

s

z

y

s'

t'

t

r'

Figure 6.1: A three-way pie-merge (A , B , Q) X.

Prop 6.1 In a pie-merge (A , B , Q) X, each rotation system for X is consistent with exactly two rotation systems for A and

exactly two for B.

Remark In this oral presentation, we use the picture to definea pie-merge.

Remark A series of fundamental papers ([GrKhPo10], [Gr10a], [KhPoGr10],[PoKhGr10a], [PoKhGr10b], and [Gr10b]) derives the productions that cor-respond to various ways of synthesizing graphs from graphs whose partialgenus distributions are known.


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Six Relevant Partials

For Halin graphs, we split gi into six double-rooted partials:

ss1 ss2sd'ds'dd"dd'

u uuuuu

v v vvvv

Figure 6.0: The 6 double-rooted partials for 3-way pie-merge.

Here is what they count:

dd Each of the roots u and v lies on two distinct fb-walks. Oneand only one of these fb-walks traverses both roots.

dd Each of the roots u and v lies on two distinct fb-walks. Bothof these fb-walks traverse both roots.

ds Root u lies on two distinct fb-walks. One of these fb-walkstraverses root v twice.

sd Root v lies on two distinct fb-walks. One of these fb-walkstraverses root u twice.

ss1 A single fb-walks traverses roots u and v twice. The occur-rences of each root are consecutive.

ss2 A single fb-walks traverses roots u and v twice. The occur-rences of the two roots alternate.


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Computational Complexity

Thm 6.2 (a) The values of the six partials for the amalga-mands A and B completely determine those for the result

X of a pie-merge (A , B , Q) X.

(b) For |VA| = k and |VB| = m, there is an O(km)-time al-gorithm for calculating the partial genus distribution of the

resulting graph X of a pie-merge (A , B , Q) X.

Cor 6.3The post-order traversal using the 36 productions cor-

responding to the six partials yields an O(n2) algorithm for thegenus distribution of a cubic Halin graph with n vertices.

Practical Calculation with Spreadsheet

Express each of the six partial genus distributions of X as asum of convolutions of partials of A and B, as prescribed by theproductions. (See 7.)


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Thirty-Six Productions

1. ddi ddj dd

i+j + 2dd

i+j+1 + ss

2i+j+1.

2. ddi ddj 2dd

i+j + 2ss

2i+j+1.

3. ddi dsj 2dd

i+j + 2ss

2i+j+1.

4. ddi sdj 2sd

i+j + 2ss

1i+j+1.

5. ddi ss1j 4sd

i+j.

6. ddi ss

2j 2ds

i+j + 2sd

i+j.

7. ddi ddj 2dd

i+j + 2ss

2i+j+1.

8. ddi ddj 4dd

i+j.

9. ddi dsj 4ds

i+j.

10. ddi sdj 4sd

i+j.

11. ddi ss1j 4ss

1i+j.

12. ddi ss2j 2dd

i+j1 + 2ss

2i+j.

13. dsi ddj 2ds

i+j + 2ss

1i+j+1.

14. dsi ddj 4ds

i+j.

15. dsi dsj 4ds

i+j.

16. dsi sdj 4ss

1i+j.

17. dsi ss1j 4ss1i+j.

18. dsi ss2j 2ds

i+j1 + 2ss

1i+j.


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19. sdi ddj 2dd

i+j + 2ss

2i+j+1.

20. sdi ddj 4sd

i+j.

21. sdi dsj 2dd

i+j1 + 2ss

2i+j.

22. sdi sdj 4sd

i+j.

23. sdi ss1j 4sd

i+j1.

24. sdi ss2j 2dd

i+j1 + 2ss

2i+j.

25. ss1

i dd

j 4ds

i+j.26. ss1i dd

j 4ss

1i+j.

27. ss1i dsj 4ds

i+j1.

28. ss1i sdj 4ss

1i+j.

29. ss1i ss1j 4ss

1i+j1.

30. ss1i ss2j 4ds

i+j1.

31. ss2i ddj 2ds

i+j + 2sd

i+j.

32. ss2i ddj 2dd

i+j1 + 2ss

2i+j.

33. ss2i dsj 2dd

i+j1 + 2ss

2i+j.

34. ss2i sdj 2sd

i+j1 + 2ss

1i+j.

35. ss2i ss1j 4sd

i+j1.

36. ss2i ss

2j 2dd

i+j1 + dd

i+j2 + ss

2i+j1.


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Derivations of Pie-Merge Productions

Figure 6.1: Prod #1: ddi ddj dd

i+j + 2dd

i+j+1 + ss

2i+j+1.

Figure 6.2: Prod #18: dsi ss2j 2ds

i+j1 + 2ss

1i+j.


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7 Sample Calculation

Merger at z

Graph A (K4 e):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 0 0 0 0 2 2

Graph B (K4 e):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 0 0 0 0 2 2

Use Productions 1, 6, 31, and 36.

Merged Graph K4:

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 4 4 4 0 2 14

Merger at y

Just like the merger at z.

Merged Graph: K4

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 4 4 4 0 2 14


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Merger at x

Graph A (result from merger at z):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 4 4 4 0 2 14

Graph B (result from merger at y):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 4 4 4 0 2 14

Use 25 productions (all those without the partial ss1).

Merged Graph:

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 40 4 12 12 0 2 70

2 0 16 48 48 32 40 184

Merger at u

Just like the merger at z.

Merged Graph: K4

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 2

1 0 4 4 4 0 2 14


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Merger at v

Graph A (result from merger at x):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 40 4 12 12 0 2 702 0 16 48 48 32 40 184

Graph B (result from merger at u):

i dd

i dd

i ds

i sd

i ss1

i ss2

i gi0 2 0 0 0 0 0 21 0 4 4 4 0 2 14

Use 30 productions.

Merged Graph:

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 112 4 28 12 0 2 1582 544 96 544 352 80 112 17283 0 64 448 448 704 544 2208


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Merger at w

Graph A (result from merger at v):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 112 4 28 12 0 2 1582 544 96 544 352 80 112 17283 0 64 448 448 704 544 2208

Graph B (K4 e):

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 21 0 0 0 0 0 2 2

Merged Graph: final result

i ddi ddi ds

i sd

i ss

1i ss

2i gi

0 2 0 0 0 0 0 2

1 144 4 60 4 0 2 2142 1440 224 1632 224 56 144 37203 1024 1088 4800 1088 1088 1440 105284 0 0 0 0 896 1024 1920


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8 Conclusions and Open Problems

New features of this solution:

quadrangulation

pie-merge (invented for application to Halin graphs)

post-order traversal (also used in [Gr11] and [PoKhGr11])

dynamic assignment of roots to graphs

calculation for graphs of tree-width 3

Some open genus-distribution calculation problems:

1. wheels

2. planar cubic Hamiltonian graphs

3. truncated polyhedra


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Some High-Level Theoretical Questions

1. shape of genus distributions

unimodality of genus distributions

unimodality of the partials

approximation by Stirling cycle numbers

2. average genus (without calculating the entire g.d.)

The method presented here works for Halin graphs of boundedvalence. Halin graphs are known [Wi87] to be of treewidth 3.

Conjecture 8.1 There is a quadratic-time algorithm for thegenus distribution of any cubic graph of treewidth 3.


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9 Appendix

Derivations of the Productions

Figure 9.1: ddi ddj dd

i+j + 2dd

i+j+1 + ss

2i+j+1.

Figure 9.2: ddi ddj 2dd

i+j + 2ss

2i+j+1.

Figure 9.3: ddi dsj 2dd

i+j + 2ss

2i+j+1.


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Figure 9.4: ddi sdj 2sd

i+j + 2ss

1i+j+1.

Figure 9.5: ddi ss1j 4sd

i+j.

Figure 9.6: ddi ss2j 2ds

i+j + 2sd

i+j.


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i+j + 2ss

2i+j+1.


i+j.

Figure 9.9: ddi dsj 4ds

i+j.


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Figure 9.10: ddi sdj 4sd

i+j.

Figure 9.11: ddi ss1j 4ss

1i+j.

Figure 9.12: ddi ss2j 2dd

i+j1 + 2ss

2i+j.


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Figure 9.13: dsi ddj 2ds

i+j + 2ss

1i+j+1.

Figure 9.14: dsi ddj 4ds

i+j.

Figure 9.15: dsi dsj 4ds

i+j.


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Figure 9.16: dsi sdj 4ss

1i+j.

Figure 9.17: dsi ss1j 4ss

1i+j.

Figure 9.18: dsi ss2j 2ds

i+j1 + 2ss

1i+j.


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Figure 9.19: sdi ddj 2dd

i+j + 2ss

2i+j+1.

Figure 9.20: sdi ddj 4sd

i+j.

Figure 9.21: sdi dsj 2dd

i+j1 + 2ss

2i+j.


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Figure 9.22: sdi sdj 4sd

i+j.

Figure 9.23: sdi ss1j 4sd

i+j1.

Figure 9.24: sdi ss2j 2dd

i+j1 + 2ss

2i+j.


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Figure 9.25: ss1i ddj 4ds

i+j.

Figure 9.26: ss1i ddj 4ss

1i+j.

Figure 9.27: ss1i dsj 4ds

i+j1.


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Figure 9.28: ss1i sdj 4ss

1i+j.

Figure 9.29: ss1i ss1j 4ss

1i+j1.

Figure 9.30: ss1i ss2j 4ds

i+j1.


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Figure 9.31: ss2i ddj 2ds

i+j + 2sd

i+j.

Figure 9.32: ss2i ddj 2dd

i+j1 + 2ss

2i+j.

Figure 9.33: ss2i dsj 2dd

i+j1 + 2ss

2i+j.


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Figure 9.34: ss2i sdj 2sd

i+j1 + 2ss

1i+j.

Figure 9.35: ss2i ss1j 4sd

i+j1.

Figure 9.36: ss2i ss2j 2dd

i+j1 + dd

i+j2 + ss

2i+j1.

jonathan l. gross- embeddings of cubic halin graphs a surface-by-surface inventory

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