2-connecting Outerplanar Graphs without Blowing Up the Pathwidth

0.45 MB·English

by  Jasine Babu

2-connecting Outerplanar Graphs without
Blowing Up the Pathwidth
Jasine Babu1, Manu Basavaraju2, L. Sunil Chandran1, and Deepak
Rajendraprasad1
1 Indian Institute of Science, Bangalore 560012, India.
2 Institute of Mathematical Sciences, Chennai 600113, India.
3
[email protected], [email protected], {sunil,
1
0 deepakr}@csa.iisc.ernet.in
2

n
a
Abstract. Given a connected outerplanar graph G with pathwidth p,
J
wegiveanalgorithmtoaddedgestoGtogetasupergraphofG,which
5
is 2-vertex-connected, outerplanar and of pathwidth O(p). This settles

an open problem raised by Biedl [2]: as a consequence of our algorithm,
]
M wegetaconstantfactorapproximationalgorithmtocomputeastraight
lineplanardrawingofanyouterplanargraph,withitsverticesplacedon
D
a two dimensional grid of minimum height.
.
s
c Keywords: Pathwidth, Outerplanar Graph, Bi-connected
[

2
1 Introduction
v
2
8
AgraphG(V,E)isouterplanar,ifithasaplanarembeddingwithallitsvertices
3
lyingontheouterface.Computingplanarstraightlinedrawingsofplanargraphs
6
. withverticesplacedonatwodimensionalgrid,isawellknownproblemingraph
2
drawing. The height of a grid is defined as the smaller of the two dimensions
1
of the grid. If G has a planar straight line drawing, with its vertices placed on
2
1 a two dimensional grid of height h, then we call it a planar drawing of G of
: height h. It is known that any planar graph on n vertices can be drawn on an
v
(n 1) (n 1) sized grid [10].
i
X − × −
We use pw(G) to denote the pathwidth of a graph G. Pathwidth is a struc-
r tural parameter of graphs, which is widely used in graph drawing and layout
a
problems [2,4,12]. The study of pathwidth, in the context of graph drawings,
was initiated by Dujmovic et al. [4]. It is known that any planar graph that has
a planar drawing of height h has pathwidth at most h [12]. However, there exist
planar graphs of constant pathwidth but requiring Ω(n) height in any planar
drawing [1]. In the special case of trees, Suderman [12] showed that any tree
T has a planar drawing of height at most 3pw(T) 1. Biedl [2] considered the
−
same problem for the bigger class of outerplanar graphs. For any bi-connected
outerplanargraphG,Biedl[2]obtainedanalgorithmtocomputeaplanardraw-
ing of G of height at most 4pw(G) 3. Since it is known that pathwidth is a
−
lower bound for the height of the drawing [12], the algorithm given by Biedl
2
[2] is a 4-factor approximation algorithm for the problem, for any bi-connected
outerplanar graph. The method in Biedl [2] is to add edges to the bi-connected
outerplanar graph G to make it a maximal outerplanar graph H and then draw
H onagridofheight4pw(G) 3.Thesamemethodwouldgiveaconstantfactor
−
approximationalgorithmforarbitraryouterplanargraphs,ifitispossibletoadd
edgestoanyarbitraryouterplanargraphGtoobtainabi-connectedouterplanar
graph G(cid:48) such that pw(G(cid:48)) = O(pw(G)). This was left as an open problem by
Biedl [2].
Notice that, if we relax the condition that G(cid:48) has to be outerplanar, this
problem becomes very easy. It is easy to verify that the supergraph G(cid:48) of G,
obtainedbymakinganytwoverticesofGuniversal,i.e.,makingthemadjacentto
eachotherandtoeveryothervertexinthegraph,haspathwidthatmostpw(G)+
2.Inthispaper,wegiveanalgorithmtoaugmentaconnectedouterplanargraph
G of pathwidth p by adding edges so that the resultant graph is a bi-connected
outerplanar graph of pathwidth O(p). The problem of augmenting outerplanar
graphs to make them bi-connected, while maintaining the outerplanarity and
optimizing some other properties like number of edges added [5,8] have been
investigated previously.
2 Background
A tree decomposition of a graph G(V,E) [9] is a pair (T,X), where T is a tree
and X = (X : t V(T)) is a family of subsets of V(G), with the following
t
∈
properties:
(cid:83)
1. (X :t V(T))=V(G).
t
∈
2. For every edge e of G there exists t V(T) such that e has both its end
∈
points in X .
t
3. Fort,t(cid:48),t(cid:48)(cid:48) V(T),ift(cid:48) isonthepathofT betweentandt(cid:48)(cid:48) then,X X
t t(cid:48)(cid:48)
∈ ∩ ⊆
X .
t(cid:48)
Thewidthofthetreedecompositionis max (X 1).EachX X isreferred
t t
t∈V(T) | |− ∈
to as a bag in the tree decomposition. The graph G has treewidth w if w is the
minimum such that G has a tree decomposition of width w.
A path decomposition (P,X) of a graph G is a tree decomposition of G
with the additional property that the tree P is a path. The width of the path
decomposition is max (X 1). The graph G has pathwidth w if w is the
t
t∈V(P) | |−
minimum such that G has a path decomposition of width w.
Without loss of generality we can assume that, in any path decomposition
( ,X) of G, the vertices of the path are labeled as 1,2,..., in the order
inPwhich they appear in . AccordinglPy, the bags in X also get indexed as
1,2,.... For each vertex vP V(G), define FirstIndexX(v) = min i Xi X
contains v , LastIndexX(v∈)=max i Xi X contains v and R{an|geX(∈v)=
} { | ∈ }
[FirstIndexX(v),LastIndexX(v)]. By the definition of a path decomposition,
if t RangeX(v), then v Xt. If v1 and v2 are two distinct vertices, define
∈ ∈
GapX(v1,v2) as follows:
3
– If RangeX(v1) RangeX(v2)= , then GapX(v1,v2)= .
∩ (cid:54) ∅ ∅
– IfLastIndexX(v1)<FirstIndexX(v2),GapX(v1,v2)is[LastIndexX(v1)+
1,FirstIndexX(v2)].
– IfLastIndexX(v2)<FirstIndexX(v1),GapX(v1,v2)is[LastIndexX(v2)+
1,FirstIndexX(v1)].
The motivation for this definition is the following. Suppose (P,X) is a path
decomposition of a graph G and v and v are two non-adjacent vertices of G.
1 2
If we add a new edge between v and v , a natural way to modify the path
1 2
decompositiontoreflectthisedgeadditionisthefollowing.IfGapX(v1,v2)= ,
there is an X X, which contains v and v together and hence, we nee∅d
t 1 2
∈
not modify the path decomposition. If LastIndexX(v1) < FirstIndexX(v2),
we insert v1 into all Xt X, such that t GapX(v1,v2). On the other hand,
if LastIndexX(v2) < Fi∈rstIndexX(v1), w∈e insert v2 to all Xt X, such that
∈
t GapX(v1,v2). It is clear from the definition of GapX(v1,v2), that this
∈
procedure gives a path decomposition of the new graph. Whenever we add an
edge (v ,v ), we stick to this procedure to update the path decomposition.
1 2
A block of a graph G is a maximal connected subgraph of G without a cut
vertex. Every block of a connected graph G is thus either a single edge which is
abridgeinG,oramaximalbi-connectedsubgraphofG.IfablockofGisnota
singleedge,wecallitasanon-trivialblockofG.Givenaconnectedouterplanar
graph G, we define a rooted tree T (hereafter referred to as the rooted block tree
of G) as follows. The vertices of T are the blocks of G and the root of T is an
arbitrary block of G which contains at least one non-cut vertex (it is easy to see
that such a block always exists). Two vertices B and B of T are adjacent if
i j
theblocksB andB shareacutvertexinG.ItiseasytoseethatT,asdefined
i j
above, is a tree. In our discussions, we restrict ourselves to a fixed rooted block
tree of G. If block B is a child block of block B in the rooted block tree of G,
i j
and they share a cut vertex x, we say that B is a child block of B at x.
i j
Itisknownthateverybi-connectedouterplanargraphhasauniquehamilto-
nian cycle [13]. Though the hamiltonian cycle of a bi-connected block of G can
be traversed either clockwise or anticlockwise, let us fix one of these orderings,
so that the successor and predecessor of each vertex in the hamiltonian cycle of
the blockis fixed. We call thisorder as theclockwise order. Considera non-root
block B of G such that B is a child block of its parent, at the cut vertex x.
i i
If B is a non-trivial block and y and y(cid:48) respectively be the predecessor and
i i i
successor of x in the hamiltonian cycle of B , we call y as the last vertex of B
i i i
and y(cid:48) as the first vertex of B . If B is a trivial block, the neighbor of x in B
i i i i
is regarded as both the first vertex and the last vertex of B . By the term path
i
we always mean a simple path, i.e., a path in which no vertex repeats.
3 An overview of our method
GivenaconnectedouterplanargraphG(V,E)ofpathwidthp,ouralgorithmwill
produce a bi-connected outerplanar graph G(cid:48)(cid:48)(V,E(cid:48)(cid:48)) of pathwidth O(p), where
E E(cid:48)(cid:48). Our algorithm proceeds in three stages.
⊆
4
(1) We use a modified version of the algorithm proposed by Govindan et al.
[6] to obtain a nice tree decomposition (defined in Section 4) of G. Using this
nicetree decompositionof G,weconstructa specialpathdecompositionof G of
width at most 4p+3.
(2) For each cut vertex x of G, we define an ordering among the child blocks
attached through x to their parent block. To define this ordering, we use the
specialpathdecompositionconstructedinthefirststage.Thisorderinghelpsus
inconstructinganouterplanarsupergraphG(cid:48)(V,E(cid:48))whosepathwidthisatmost
8p+7,andforeverycutvertexxinG(cid:48),G(cid:48) xhasexactlytwocomponents.The
\
properties of the special path decomposition of G obtained in the first stage is
cruciallyusedinourargumenttoboundthewidthofthepathdecompositionof
G(cid:48), produced in the second stage.
(3) We bi-connect G(cid:48) to construct G(cid:48)(cid:48)(V,E(cid:48)(cid:48)), using a straightforward algo-
rithm. As a by-product, this algorithm also gives us a surjective mapping from
thecutverticesofG(cid:48) totheedgesinE(cid:48)(cid:48) E(cid:48).Wegiveacountingargumentbased
\
onthismappingandsomebasicpropertiesofpathdecompositionstoshowthat
the width of the path decomposition of G(cid:48)(cid:48) produced in the third stage is at
most 16p+15.
4 Stage 1: Construct a nice path decomposition of G
In this section, we construct a nice tree decomposition of the outerplanar graph
G and then use it to construct a nice path decomposition of G. We begin by
giving the definition of a nice tree decomposition.
Given an outerplanar graph G, Govindan et al. [6, Section 2] gave a linear
time algorithm to construct a width 2 tree decomposition (T,Y) of G where
Y =(Y :t V(T)), with the following special properties:
t
∈
1. There is a bijective mapping b from V(G) to V(T) such that v Y .
b(v)
∈
(Hereafter,foranyv V(G),whilereferringtothevertexb(v)ofT,wejust
∈
call it as vertex v of T.)
2. If B is a child block of B at a cut vertex x, the induced subgraph T(cid:48) of T
i j
on the vertex set V(B x) is a spanning tree of B x and (T(cid:48),Y(cid:48)) where
i i
Y(cid:48) =(Y :t V(T(cid:48))) gi\ves a tree decomposition of B\.
t i
∈
Definition 1 (Nice tree decomposition of an outerplanar graph G). A
tree decomposition (T,Y) of G, where Y =(Y : t V(T)) having properties 1
t
∈
and 2 above, together with the following additional property, is called a nice tree
decomposition of G.
3. If y and y(cid:48) are respectively the last and first vertices of a non-root, non-
i i
trivial block B , then the bag Y Y contains both y and y(cid:48).
i yi ∈ i i
In the discussion that follows, we will show that any outerplanar graph G has
a nice tree decomposition (T,Y) of width at most 3. Initialize (T,Y) to be the
tree decomposition of G, constructed using the method proposed by Govindan
5
et al. [6], satisfying properties 1 and 2 of nice tree decompositions. We need to
modify (T,Y) in such a way that, it satisfies property 3 as well.
For every non-root, non-trivial block B of G, do the following. If y and y(cid:48)
i i i
are respectively the last and first vertices of B , then, for each t V(B x), we
i i
∈ \
insert y(cid:48) to Y , if it is not already present in Y and we call y(cid:48) as a propagated
i t t i
vertex.
Claim. After the modification, (T,Y) remains as a tree decomposition of G.
Proof. Clearly,weonlyneedtoverifythatthethirdpropertyinthedefinitionof
atreedecompositionholds,forallthepropagatedvertices.Lety(cid:48)beapropagated
i
vertex,whichgotinsertedtothebagscorrespondingtoverticesofB x,during
i
\
the modification. Let V(cid:48) = t y(cid:48) Y , after the modification and Let V =
yi(cid:48) { | i ∈ t } yi(cid:48)
t y(cid:48) Y , before the modification . Then, clearly, V(cid:48) =V V(B x).
{ | i ∈ t } yi(cid:48) yi(cid:48) ∪ i\
Clearly, the induced subgraph of T on the vertex set V is connected, since
y(cid:48)
i
wehadatreedecompositionofGbeforethemodification.Byproperty2ofnice
decompositions, the induced subgraph of T on the vertex set V(B x) is also
i
\
connected. Moreover, by property 1 of nice decompositions, y(cid:48) V and hence,
y(cid:48) (B x) V . This implies that the induced subgraph oif∈T oyni(cid:48) the vertex
seit∈V(cid:48) iis\conn∩ectyei(cid:48)d.
yi(cid:48) (cid:116)(cid:117)
Claim. After the modification, (T,Y) becomes a nice tree decomposition of G
of width at most 3.
Proof. Itiseasytoverifythatallthethreepropertiesrequiredbynicedecompo-
sitionsaresatisfied,afterthemodification.Moreover,foranyblockB ,attached
i
toitsparentthroughthecutvertexx,atmostone(propagated)vertexisgetting
newly inserted into the bags corresponding to vertices of B x. Since the size
i
of any bag in Y was at most two initially and it got increase\d by at most one,
the new decomposition has width at most three.
(cid:116)(cid:117)
From the claims above, we can conclude the following.
Lemma 1. Every outerplanar graph G has a nice tree decomposition (T,Y) of
width 3, constructable in polynomial time.
Definition 2 (Nice path decomposition of an outerplanar graph). Let
( ,X) be a path decomposition of an outerplanar graph G. If, for every non-
roPot, non-trivial block B , there is a bag X X simultaneously containing both
i t
the first and last vertices of B , then ( ,X∈) is called a nice path decomposition
i
P
of G.
Lemma 2. Let G be an outerplanar graph of pathwidth p. A nice path decom-
position ( ,X) of G, of width at most 4p+3, is constructable in polynomial
P
time.
Proof. Let ( ,Y) be a nice tree decomposition of G of width 3, obtained using
Lemma 1. OTbtain an optimal path decomposition ( ,X ) of the tree in
T T
P T
6
polynomial time, using a standard algorithm (See [11]). Since is a spanning
T
tree of G, the pathwidth of is at most that of G. Therefore, the width of the
path decomposition ( ,XT) is at most p; i.e. there are at most p+1 vertices
T T
of in each bag X P X .
T Ti ∈ T (cid:91)
Let = and for each X X , we define X = Y . Clearly,
P PT Ti ∈ T i vT
vT∈XTi
( ,X), with X = (X ,...,X ), is a path decomposition of G (See [6]).
P 1 |V(PT)|
The width of this path decomposition is at most 4(p+1) 1 = 4p+3, since
Y 4, for each bag Y Y and X p+1, for each−bag X X .
| vT|≤ vT ∈ | Ti|≤ Ti ∈ T
Let B be a non-root, non-trivial block in G and y and y(cid:48) respectively be
i i i
the first and last vertices of B . Since y is a vertex of the tree , there is
i i
some bag X X , containing y . The bag Y Y contains bothTy and y(cid:48),
since ( ,Y)Tjis∈a niTce tree decompiosition of G.yIit∈follows from the defiinition oif
X , thTat X X contains both y and y(cid:48). Therefore, ( ,X) is a nice path
j j ∈ i i P
decomposition of G.
(cid:116)(cid:117)
5 Edge addition without spoiling the outerplanarity
Inthissection,weprovesometechnicallemmaswhichwillbelaterusedtoprove
that the intermediate graph G(cid:48) obtained in Stage 2 and the bi-connected graph
G(cid:48)(cid:48) obtained in Stage 3 are outerplanar. These lemmas are easy to understand
intuitively, by visualizing the outerplanar drawing. However, for completeness,
we prove them combinatorially.
Lemma 3. Let G(V,E) be a connected outerplanar graph. Let u and v be two
distinctnon-adjacentverticesinGandletP =(u=x ,x ,x ,...,x ,x =v)
0 1 2 k k+1
where k 1 be a path in G such that:
≥
– P shares at most one edge with any block of G and
– for 0 i k, if the block containing the edge (x ,x ) is non-trivial, then
i i+1
≤ ≤
x is the successor of x in the hamiltonian cycle of that block.
i+1 i
Then the graph G(cid:48)(V,E(cid:48)), where E(cid:48) =E (u,v) , is outerplanar.
∪{ }
Proof. ThereadermayrefertoFig.1,togetageometricideaofstatementofthe
lemma. It is well known that a graph G is outerplanar if and only if it contains
nosubgraphwhichisasubdivisionofK orK [3].ConsiderapathP between
4 2,3
u and v as stated in the lemma.
Property 1. In every path in G from u to v, vertices x ,...,x should appear
1 k
and for 0 i k, x should appear before x in any such path.
i i+1
≤ ≤
Proof. For any 1 i k, the two consecutive edges e = (x ,x ) and e =
i i−1 i i+1
≤ ≤
(x ,x ) of the path P belong to two different blocks of G, by assumption.
i i+1
Therefore, each internal vertex x , 1 i k, is a cut vertex in G. As a result,
i
≤ ≤
in every path in G between u and v, vertices x ,...,x should appear and for
1 k
0 i k, x should appear before x in any such path.
i i+1
≤ ≤ (cid:116)(cid:117)
7
a
u
b
k
v
j
l c
g d
i
e
f
h
u0 v0
Fig.1. The path between vertices u and v and the path between vertices u(cid:48) and v(cid:48)
(showninthickedges)satisfytheconditionsstatedinLemma3.Byaddinganyoneof
the dotted edges (u,v) or (u(cid:48),v(cid:48)), the graph remains outerplanar.
Property 2. For any 0 i k, there are at most two internally vertex disjoint
≤ ≤
paths in G between x and x .
i i+1
Proof. Any(simple)pathfrom(x ,x )liesfullyinsidetheblockB ,containing
i i+1 i
the edge (x ,x ). If B is trivial, the only path from x to x is the direct
i i+1 i i i+1
edge between them. If this is not the case, B is bi-connected. It is a property
i
of bi-connected outerplanar graphs, that between consecutive vertices in the
hamiltonian cycle of the graph, there are exactly two internally vertex disjoint
paths. By the assumption of Lemma 3, if B is non-trivial, then x is the
i i+1
successor of x in the hamiltonian cycle of B . Hence, the property follows.
i i
(cid:116)(cid:117)
We will show that if G(cid:48) is not outerplanar, then G also was not outerplanar,
which is a contradiction. Assume that G(cid:48) is not outerplanar. This implies that
thereisasubgraphH(cid:48) ofG(cid:48) thatisasubdivisionofK orK .SinceGdoesnot
4 2,3
have a subgraph which is a subdivision of K or K , H(cid:48) cannot be a subgraph
4 2,3
of G. Hence, the new edge (u,v) should be an edge in H(cid:48) and all other edges of
H(cid:48) are edges of G.
Case 1. H(cid:48) is a subdivision of K .
4
Let k ,k ,k and k denote the four vertices of H(cid:48), that correspond to the
1 2 3 4
verticesofK .WecallthemasbranchverticesofH(cid:48).Fori,j 1,2,3,4 ,i=j,
4
∈{ } (cid:54)
let P denote the path in H(cid:48) from the branch vertex k to the branch vertex
i,j i
k , such that each intermediate vertex of the path is a degree two vertex in H(cid:48).
j
Without loss of generality, assume that the edge (u,v) is part of the path P
1,2
of H(cid:48).
8
Claim. All of the vertices x ,...,x appear in P . The order < in which the
1 k 1,2
vertices appear in P should be one of the following three: (without loss of
1,2
generality, assuming u<v): (1) u<v <x <x < <x (2) x <x <
k k−1 1 k k−1
···
< x < u < v (3) x < x < < x < u < v < x < x for some
1 j j−1 1 k j+1
··· ··· ···
j 1,2,...,k 1 .
∈{ − }
Proof. Suppose x , 1 i k, does not belong to the path P . Then, there is
i 1,2
≤ ≤
a path in H(cid:48) (u,v) between vertices u and v, avoiding the vertex x , since H(cid:48)
i
\
is a subdivision of K . Since H(cid:48) (u,v) is a subgraph of G, this implies that
4
\
there is a path in G, between u and v, which avoids x . This is a contradiction
i
to Property 1. Therefore, x P . Notice that there is a path in H(cid:48) between u
i 1,2
∈
andv whichgoesthroughthevertexk .TosatisfyProperty1,x shouldappear
3 i
before x , for 0 i k, in this path. Hence, one of the orderings mentioned
i+1
≤ ≤
in the claim should happen in P .
1,2
(cid:116)(cid:117)
Let us denote the first vertex in the ordering by z and the last vertex in the
1
orderingbyz .(Inthefirstcase,z =uandz =x .Inthesecondcase,z =x
2 1 2 1 1 k
and z = v. In the third case, z = x and z = x .) In all the three cases of
2 1 j 2 j+1
theordering,thereisadirectedgeinG,betweenz andz .Noticethatinanyof
1 2
thesethreepossibleorderings,wedonothavez =uandz =v simultaneously.
1 2
Since (z ,z ) = (u,v), by deleting the intermediate vertices between z and z
1 2 1 2
(cid:54)
from the path P and including the direct edge between z and z , we get a
1,2 1 2
path P(cid:48) between k and k in G. All vertices in P(cid:48) are from the vertex set of
1,2 1 2 1,2
P . Therefore, by replacing the path P in H(cid:48) by P(cid:48) , we get a subgraph H
1,2 1,2 1,2
of G which is a subdivision of K . This means that G is not outerplanar, which
4
is a contradiction. Therefore, H(cid:48) cannot be a subdivision of K .
4
Case 2. H(cid:48) is a subdivision of K .
2,3
As earlier, let k ,k ,k ,k and k denote the branch vertices of H(cid:48), that
1 2 3 4 5
correspond to the vertices of K . Let k ,k ,k be the degree 2 branch vertices
2,3 1 3 5
in H(cid:48) and k ,k be the degree 3 branch vertices of H(cid:48). For i 1,3,5 and
2 4
∈ { }
j 2,4 , let P denote the path in H(cid:48) from vertex k to vertex k , such
i,j i j
∈ { }
that each intermediate vertex of the path is a degree two vertex in H(cid:48). Also,
for i 1,3,5 and j 2,4 let P denote the path from j to i in which the
j,i
∈ { } ∈ { }
vertices in P appear in the reverse order compared to P . Without loss of
j,i i,j
generality, assume that the edge (u,v) is part of the path P of H(cid:48). Let P
1,2 4,1,2
denote the path in H(cid:48) between vertices k and k , obtained by concatenating
4 2
the paths P and P .
4,1 1,2
Claim. All of the vertices x ,...,x appear in P . The order < in which the
1 k 4,1,2
vertices appear in P should be one of the following three (without loss of
4,1,2
generality, assuming u<v): (1) u<v <x <x < <x (2) x <x <
k k−1 1 k k−1
···
< x < u < v (3) x < x < < x < u < v < x < x for some
1 j j−1 1 k j+1
··· ··· ···
j 1,2,...,k 1 .
∈{ − }
This can be proved in a similar way as in Case 1. The remaining part of the
proof is also similar. Let us denote the first vertex in the ordering by z and
1
9
the last vertex in the ordering by z . Repeating similar arguments as in Case 1,
2
we can prove that by deleting the intermediate vertices between z and z from
1 2
the path P and including the direct edge between z and z , we get a path
4,1,2 1 2
P(cid:48) between k and k in G. All vertices in P(cid:48) are from the vertex set of
4,1,2 4 2 4,1,2
P .Therefore,byreplacingthepathP inH(cid:48) byP(cid:48) ,wegetasubgraph
4,1,2 4,1,2 4,1,2
H of G. If P(cid:48) has at least one intermediate vertex, the subgraph H of G,
4,1,2
obtained by replacing the path P in H(cid:48) by P(cid:48) , is a subdivision of K ,
4,1,2 4,1,2 2,3
where an intermediate vertex of P(cid:48) takes the role of the branch vertex k .
4,1,2 1
This contradicts the assumption that G is outerplanar.
Therefore, assume that P(cid:48) has no intermediate vertices, i. e., z =k and
4,1,2 1 4
z = k . In the first case of ordering mentioned in the claim above, we have
2 2
z =u=x and z =x . In the second case, z =x and z =v =x . In the
1 0 2 1 1 k 2 k+1
third case, z = x and z = x . In each of these cases, by Property 2, there
1 j 2 j+1
canbeatmosttwovertexdisjointpathsinGbetweenz andz .But,inallthese
1 2
cases, there is a direct edge between z and z in G. Since H(cid:48) is a subdivision of
1 2
K , other than this direct edge, in H(cid:48) (u,v) there are two other paths from
2,3
\
z = k to z = k which are internally vertex disjoint and containing at least
1 4 2 2
one intermediate vertex. This will mean that there are at least three internally
vertex disjoint paths from z to z in G, which is a contradiction. Therefore, H(cid:48)
1 2
cannot be a subdivision of K .
2,3
Since, G(cid:48) does not contain a subgraph H(cid:48) which is a subdivision of K or
4
K , G(cid:48) is outerplanar.
2,3
(cid:116)(cid:117)
Let us now analyze how the addition of an edge (u,v), as mentioned in Lemma
3, affect the block structure and the hamiltonian cycle of each block. Assume
that for each 0 i k, the edge (x ,x ) belongs to the block B .
i i+1 i
≤ ≤
Lemma 4. – Other than the blocks B to B of G merging together to form
0 k
a new block B(cid:48) of G(cid:48), blocks in G and G(cid:48) are the same.
– Vertices in blocks B to B , except x , 0 i k+1, retains their successor
0 k i
≤ ≤
andpredecessorinthehamiltoniancycleofB(cid:48) sameasitwasinitsrespective
block’s hamiltonian cycle in G.
– Each x , 0 i k, retains its hamiltonian cycle predecessor in B(cid:48) same as
i
≤ ≤
itwasintheblockB ofGandeachx ,1 i k+1,retainsitshamiltonian
i i
≤ ≤
cycle successor in B(cid:48) same as in the block B of G.
i−1
Proof. When the edge (u,v) is added, it creates a cycle containing the vertices
u = x ,x ,...,x = v. Hence, the blocks B to B of G merge together to
0 1 k+1 0 k
form a single block B(cid:48) in G(cid:48). It is obvious that other blocks are unaffected by
this edge addition.
For simplicity, if B is a trivial block containing the edge (x ,x ), we say
i i i+1
that x and x are neighbors of each other in the hamiltonian cycle of B . For
i i+1 i
eachB ,0 i k,letx ,x ,z ,z ,...,z ,x bethehamiltoniancycleofB
i ≤ ≤ i i+1 i1 i2 iti i i
inG.For0 i k,letusdenotethepathx ,z ,z ,...,z byP .Then,the
≤ ≤ i+1 i1 i2 iti i
hamiltoniancycleofB(cid:48) isu=x P P ...P u,where denotesthecon-
0 k k−1 0
◦ ◦ ◦ ◦ ◦
catenation of the paths. (For example, in Fig. 1, u,v,a,b,c,d,e,f,g,h,i,j,k,l,u
isthehamiltoniancycleofthenewblockformed,whentheedge(u,v)isadded.)
10
From this, we can conclude that the second and third parts of the lemma holds.
(cid:116)(cid:117)
6 Stage 2: Construction of G(cid:48) and its path decomposition
ForeachcutvertexxofG,wedefineanorderingamongthechildblocksattached
through x to their parent block, using the nice path decomposition ( ,X) of
P
G obtained using Lemma 2. This ordering is then used in defining a supergraph
G(cid:48)(V,E(cid:48)) of G such that for every cut vertex x in G(cid:48), G(cid:48) x has exactly two
\
components. Using repeated applications of Lemma 3, we then show that G(cid:48) is
outerplanar. We extend the path decomposition ( ,X) of G to a path decom-
position( (cid:48),X(cid:48))ofG(cid:48),asdescribedinSection2.BPyacountingargumentusing
the propePrties of the nice path decomposition ( ,X), we show that the width
ofthepathdecomposition( (cid:48),X(cid:48))ofG(cid:48) isatmPost2p(cid:48)+1,wherep(cid:48) isthewidth
of ( ,X). P
P
6.1 Defining an ordering of child blocks
If ( ,X) is a nice path decomposition of G, then, for each non-root block B of
G,aPtleastonebaginX simultaneouslycontainsboththefirstandlastvertices
of B.
Definition 3 (Sequence number of a non-root block). Let ( ,X) be the
P
nice path decomposition of G obtained using Lemma 2. For each non-root block
B of G, we define the sequence number of B as min i X X contains both
i
{ | ∈
the first and last vertices of B .
}
For each cut vertex x, there is a unique block Bx such that Bx and its child
blocks are intersecting at x. For each cut vertex x, we define an ordering among
the child blocks attached at x, as follows. If B ,...,B are the child blocks
1 k
attached at x, we order them in the increasing order of their sequence numbers
in ( ,X). If B and B are two child blocks with the same sequence number,
i j
P
their relative ordering is arbitrary.
From the ordering defined, we can make some observations about the ap-
pearance of the first and last vertices of a block B in the path decomposition.
i
Theseobservationsarecruciallyusedforboundingthewidthofthepathdecom-
position ( (cid:48),X(cid:48)) of G(cid:48). Let B ,...,B are the child blocks attached at a cut
1 k
P
vertexx,occurringinthatorderaccordingtotheorderingwedefinedabove.For
1 i k, let y and y(cid:48) respectively be the last and first vertices of B .
≤ ≤ i i i
Property 3. Forany1≤i≤k−1,ifGapX(yi(cid:48),yi+1)(cid:54)=∅,thenGapX(yi(cid:48),yi+1)=
[LastIndexX(yi(cid:48))+1,FirstIndexX(yi+1)]andx∈Xtforallt∈GapX(yi(cid:48),yi+1).
Proof. If GapX(yi(cid:48),yi+1) (cid:54)= ∅, either LastIndexX(yi(cid:48)) < FirstIndexX(yi+1)
or LastIndexX(yi+1) < FirstIndexX(yi(cid:48)). The latter case will imply that, se-
quence number of B < sequence number of B , which is a contradiction.
i+1 i