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Path-Based Supports for Hypergraphs
Ulrik Brandes, Sabine Cornelsen, Barbara Pampel, Arnaud Sallaberry
To cite this version:
Ulrik Brandes, Sabine Cornelsen, Barbara Pampel, Arnaud Sallaberry. Path-Based Supports for
Hypergraphs. Journal of Discrete Algorithms, Elsevier, 2012, 14, pp.248-261. �hal-00735894�
Path-Based Supports for Hypergraphs
1
Ulrik Brandesa, Sabine Cornelsena, Barbara Pampela, Arnaud Sallaberryb
2
aDepartment of Computer and Information Science, University of Konstanz, Box 67, 78457 Konstanz,
3
Germany
4
bCNRS UMR 5800 LaBRI, INRIA Bordeaux - Sud Ouest, Pikko, 351, cours de la Libration, 33405
5
Talence Cedex, France
6
Abstract
7
A path-based support of a hypergraphH is a graph with the same vertex set asHin which each hyperedge induces a Hamiltonian subgraph. While it isN P-hard to decide whether a path-based support has a monotone drawing, to determine a path-based support with the minimum number of edges, or to decide whether there is a planar path-based support, we show that a path-based tree support can be computed in polynomial time if it exists.
Keywords: graph algorithm, graph drawing, hypergraph, metro map layout
8
1. Introduction
9
A hypergraph is a pair H = (V, A) where V is a finite set and A is a (multi-)set of
10
non-empty subsets of V. The elements of V are called vertices and the elements of A
11
are calledhyperedges. Asupport (orhost graph) of a hypergraphH = (V, A) is a graph
12
G= (V, E) such that each hyperedge ofH induces a connected subgraph ofG, i.e., such
13
that the graphG[h] := (h,{e∈E, e⊆h}) is connected for everyh∈A. See Fig. 1(b) for
14
an example.
15
Applications for supports of hypergraphs are, e.g., in hypergraph coloring [2, 3],
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databases [4], or hypergraph drawing [5, 6, 7, 8]. E.g., see Fig. 1 for an application
17
of a support for designing Euler diagrams. AnEuler diagramof a hypergraphH = (V, A)
18
is a drawing of H in the plane in which the vertices are drawn as points and each hy-
19
peredge h∈A is drawn as a simple closed region containing the points representing the
20
vertices in h and not the points representing the vertices in V \h. There are various
21
well-formedness conditions for Euler diagrams, see e.g. [9, 8].
22
Recently, many papers have been devoted to the problem of deciding which classes
23
of hypergraphs admit what kind of supports. It can be tested in linear time whether a
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hypergraph has a support that is a tree [10], a path or a cycle [7]. It can be decided in
25
polynomial time whether a hypergraph has a tree support with bounded degrees [7] or a
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cactus support [11]. A minimum weighted tree support can be computed in polynomial
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time [12]. It isN P-complete to decide whether a hypergraph has a planar support [5],
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a compact support [5, 6] or a 2-outerplanar support [7]. A support with the minimum
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Email addresses: Ulrik.Brandes@uni-konstanz.de(Ulrik Brandes),
Sabine.Cornelsen@uni-konstanz.de(Sabine Cornelsen),Barbara.Pampel@uni-konstanz.de(Barbara Pampel),arnaud.sallaberry@labri.fr(Arnaud Sallaberry)
A preliminary version of this paper was presented at the 21st International Workshop on Combina- torial Algorithms (IWOCA 2010) and appears in the corresponding proceedings [1].
Corresponding author phone: +49 7531 88 4375, fax: +49 7531 88 3577.
{v1} {v2} {v3} {v4} {v5} {v6} {v7} V
h′
h
h1 h2 h3 h4 h5
(a) Hasse diagram
v1 v2 v3 v4 v5 v6
v7 (b) tree support
v1 v2 v3 v4 v5 v6
v7 (c) metro map like drawing
v1 v2 v3 v4 v5 v6
v7
(d) Euler diagram
Figure 1: Three representations of the hypergraph H = (V, A) with hyperedgesh1 ={v1, v2},h2 = {v2, v3},h3 ={v3, v4},h4 ={v4, v5},h5 ={v5, v6},h={v2, v3, v4, v5},h′ ={v2, v3, v4, v5, v7}, and V ={v1, . . . , v7}.
number of edges can be computed in polynomial time if the hypergraph is closed under
30
intersections [7]. If the set of hyperedges is closed under intersections and differences, it
31
can be decided in polynomial time whether the hypergraph has a planar or outerplanar
32
support [11].
33
In this paper we consider a restriction on the subgraphs of a support that are induced
34
by the hyperedges. A supportG of a hypergraphH = (V, A) is called path-based if the
35
subgraphG[h] contains aHamiltonian path for each hyperedgeh∈A, i.e.,G[h] contains a
36
path that contains each vertex ofh. This definition was motivated by by the aesthetics of
37
metro map layouts. I.e., the hyperedges could be visualized as lines along the Hamiltonian
38
path in the induced subgraph of the support like the metro lines in a metro map. See
39
Fig. 2 for examples of metro maps, Fig. 3 for an example of natural sciences drawn in the
40
metro map anthology, and Fig. 1(c) and 6(f) for a representation of some hyperedges in
41
such a metro map like drawing. For metro map layout algorithms see, e.g., [13, 14].
42
We briefly consider monotone, planar, and minimum path-based supports. Our main
43
result is a characterization of those hypergraphs that have a path-based tree support
44
and a polynomial time algorithm for constructing path-based tree supports if they exist.
45
E.g., Fig. 1 shows an example of a hypergraph H = (V, A) that has a tree support but
46
no path-based tree support. However, the tree support in Fig. 1(b) is a path-based tree
47
support for (V, A\ {V}).
48
The contribution of this paper is as follows. In Section 2 we give the necessary defi-
49
nitions. We then briefly discuss monotone path-based supports in Section 3 and mention
50
that finding a minimum path-based support or deciding whether there is a planar path-
51
based support, respectively, isN P-hard. We consider path-based tree supports in Sect. 4.
52
In Section 4.1 we review a method for computing tree supports using the Hasse diagram.
53
In Section 4.2 we show how to apply this method to test whether a hypergraph has a
54
path-based tree support and if so how to compute one in polynomial time. Finally, in
55
Section 4.3 we discuss the run time of our method.
56
(a) local trains of Zurich (b) metro of Amsterdam
Figure 2: Local train map of Zurich (www.zvv.ch) and the metro map of Amsterdam (www.amsterdam.info). In (b) the union of all lines forms a tree.
2. Preliminaries
57
In this section, we give the necessary definitions that were not already given in the
58
introduction. Throughout this paper let H = (V, A) be a hypergraph. We denote by
59
n=|V|the number of vertices, m=|A|the number of hyperedges, and N =P
h∈A|h|
60
the sum of the sizes of all hyperedges of a hypergraphH. Thesize of the hypergraph H is
61
thenN+n+m. A hypergraph is agraph if all hyperedges contain exactly two vertices.
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A hypergraphH = (V, A) isclosed under intersections ifh1∩h2∈A∪ {∅}forh1, h2∈A.
63
We say that two hyperedges h1, h2 overlap if h1∩h2 6= ∅, h1 6⊆ h2, and h2 6⊆ h1. A
64
hypergraphH= (V, A) isconnected if for any pair of verticesv, w∈V there is a sequence
65
of hyperedgesh1, . . . , hℓ∈Asuch thatv∈h1, w∈hℓ, andhi∩hi+16=∅, i= 1, . . . ℓ−1.
66
The Hasse diagram of a hypergraph H = (V, A) is the directed acyclic graph with
67
vertex setA∪ {{v};v∈V}and there is an edge (h1, h2) if and only ifh2(h1 and there
68
is no seth∈A withh2(h(h1. Fig. 1(a) shows an example of a Hasse diagram. Let
69
(v, w) be an edge of a directed acyclic graph. Then we say that w is achild ofv andv
70
a parent ofw. For a descendant dofv there is a directed path fromv to dwhile for an
71
ancestor aofvthere is a directed path fromatov. Asource does not have any parents,
72
a sink no children and an inner vertex has at least one parent and one child.
73
3. Path-Based Supports
74
In a metro map like drawing of a hypergraph vertices are drawn as disjoint simple
75
closed regions in the plane and each hyperedge h is drawn as a curve Ch with the end
76
points within the regions of different vertices ofh, visiting the region of every vertex of
77
hexactly once, not visiting the vertices not in h, and such that the pieces ofCh within
78
the region of a vertex or between two such regions are simple. Apath-based support of a
79
hypergraph H = (V, A) is a graph Gsuch thatG[h] contains a spanning path for every
80
hyperedgeh∈A.
81
On one hand, a metro map like drawing of a hypergraphH = (V, A) induces a path-
82
based support G = (V, E) of H: For a hyperedge h ∈ A let ph : v1, . . . , v|h| be the
83
Figure 3: A map of modern science (www.crispian.net).
sequence of vertices ofhin the order in which they are visited by the curve representing
84
h. Starting with an empty setE add for every hyperedgeh∈Awithph:v1, . . . , v|h|the
85
edges{vi−1, vi}, i= 2, . . . ,|h|toE. On the other hand, if we have a path-based support
86
GofH and we fix for every hyperedgeh∈Aa spanning pathphofG[h] then this induces
87
a metro map like drawing ofH.
88
In order to have a readable metro map like drawing of a hypergraph it is typically
89
desirable to draw any curve representing a hyperedge without self intersection or even
90
monotone.
91
3.1. Monotone Path-Based Supports
92
Adrawing of a graph is a mapping of each vertex to a distinct point in the plane and
93
of each edge to a simple curve between the image of its adjacent vertices not containing
94
the image of any other vertex. In a straight-line drawing of a graph each edge is drawn
95
as a line segment. Given a drawing of G, a path pof Gis monotone with respect to a
96
straight lineℓ– called theaxis of monotonicity – if every line perpendicular toℓintersects
97
the drawing of pin at most one point. Note that a pathpin a straight-line drawing is
98
monotone with respect to the axisℓif and only if the orthogonal projections of the vertices
99
ofponℓ appear alongℓin the order induced byp.
100
Let G= (V, E) be a path-based support of a hypergraphH = (V, A). A drawing of
101
Gis monotone with respect to H if for each hyperedgeh∈A there is a spanning path
102
phofG[h] and a straight line ℓhsuch thatphis monotone with respect to the axisℓh. G
103
is amonotone path-based support ofH ifGhas a monotone drawing with respect toH.
104
Remark 1. If G has a monotone drawing with respect to a hypergraph H then G has
105
a straight-line drawing that is monotone with respect to H with the same axes of mono-
106
tonicity.
107
Proof. Let a drawing D of G that is monotone with respect to H = (V, A) be given
108
and let ph, h∈A be a spanning path ofG[h] that is monotone with respect to the axis
109
ℓh. If for each edge{v, w} ofGthe line segment betweenv and wdoes not contain any
110
vertex ofG other than v or wthen the straight-line drawing of Gin which the vertices
111
are mapped to the same points as inDis monotone with respect toH.
112
Consider now for two verticesv, win a hyperedgehthe distances disth(v, w) between
113
the orthogonal projections of v and w to ℓh. Let ∆ be the minimum of all distances
114
disth(v, w) over allh∈Aandv, w∈hwithv6=w. Let 0< ε≤∆/3.
115
Consider now the vertices of V in an arbitrary order v1, . . . , vn, n = |V|. For k =
116
1, . . . , n, we can now placevk on the circle with radiusεaround the position ofvk in D
117
but not on the intersection with the line through the already fixed drawings ofvi andvj,
118
1 ≤i < j < k. The corresponding straight-line drawing is monotone with respect toH
119
with the axes of monotonicityℓh, h∈A.
120
Remark 2. Not every path based support of a hypergraph is monotone.
121
Proof. Consider the following hypergraph. LetI ={(i, j, k, ℓ); 1≤i < j ≤5,1≤k <
122
ℓ≤5, i < k,{i, j} ∩ {k, ℓ}=∅}be an index set representing unordered pairs of disjoint
123
edges of the complete graph K5. Let VI = {vi; i = 1, . . . ,5} ∪ {vi,j,k,ℓ,x; (i, j, k, ℓ) ∈
124
I, x = 1, . . . ,3}, let hijkℓ = {vi, vi,j,k,ℓ,1, vj, vi,j,k,ℓ,2, vk, vi,j,k,ℓ,3, vℓ}, (i, j, k, ℓ) ∈ I, let
125
AI ={hijkℓ; (i, j, k, ℓ)∈I}, and letHI = (VI, AI). LetEcontain the edges{vi, vi,j,k,ℓ,1},
126
{vi,j,k,ℓ,1, vj}, {vj, vi,j,k,ℓ,2},{vi,j,k,ℓ,2, vk}, {vk, vi,j,k,ℓ,3}, {vi,j,k,ℓ,3, vℓ} for (i, j, k, ℓ)∈I.
127
The resulting path-based support G = (V, E) of HI is shown in Fig. 4(a). Note that
128
G[hijkℓ] is a path for any hyperedge hijkℓ ∈ A visiting the vertices vi, vj, vk, vℓ in this
129
order. Consider now any drawing of G. Since a K5 is not planar, there are two straight
130
line segmentsvivj, vk, vℓ,(i, j, k.ℓ)∈I that intersect. Hence, the pathG[hijkℓ] cannot be
131
drawn monotonously.
132
Remark 3. Every hypergraph has a monotone path-based support.
133
Proof. Order the vertices of H = (V, A) with respect to an arbitrary ordering<. The
134
support G< = (V, E<) of H with respect to the ordering < is constructed as follows.
135
For each hyperedge {v1, . . . , vk} ∈ A with v1 < · · · < vk the edge set E< contains the
136
edges {vi−1, vi}, i = 1, . . . , k. Assume now that in a drawing of G< the x-value of a
137
vertex v is smaller than the x-value of the vertex w if v < w and that the edges are
138
drawn monotonously in x-direction. Then for each hyperedgeh={v1, . . . , vk} ∈Awith
139
v1 <· · ·< vk the pathph:v1, . . . , vk is drawn monotonously with respect to the x-axis.
140
See Fig 4(c) for an example.
141
Note that the problem of deciding whether a given support is a support with respect
142
to an ordering and if so, finding such an ordering, is closely related to the betweenness
143
problem [15].
144
Theorem 1. Given a supportG of a hypergraphH
145
1. it isN P-hard to decide whether Gis a monotone path-based support ofH, and
146
2. it is N P-complete to decide whether there exists an ordering < of the vertex set
147
such that Gis the support ofH with respect to<,
148
even ifGhas the minimum number of edges among all supports ofH.
149
v1 v2
v3
v4
v5
7
6 7 5 5
(a) maximum path-based support
v1 v2
v3
v4
v5
(b) minimum path-based support
v1 v2 v3 v4 v5
(c) minimum path-based support w.r.t. an ordering
Figure 4: Three different supports for the hypergraphHI introduced in Section 3.1. The small black vertices are the verticesvσ,x, σ∈I, x= 1,2,3. The thick red path indicates the hyperedgeh1324.
Proof.
150
1. Consider an instance of thestrictly monotone trajectory drawing problem consisting
151
of a set of paths P on a set of vertices Vt. It is N P-hard to decide whether the
152
vertices can be mapped to points in the plane such that each path is monotone with
153
respect to some axis (one for each path) [16].
154
Consider the hypergraphH= (V, A) withV containingVtand for each pathp∈P
155
and each edgee∈pa vertexvep. The setAcontains for each pathp∈Pa hyperedge
156
hp=S
{v,w}∈p{v, v{v,w}p, w}as well as the hyperedges{v, v{v,w}p}and{v{v,w}p, w}
157
for each edge{v, w} ∈p. The graphG= (V, E) withE=S
p∈P
S
e∈p{{v, vep}; v∈
158
e} is a path-based support ofH and has the minimum number of edges among all
159
supports ofH. Gis monotone if and only ifP is drawable with each path monotone
160
with respect to some axis.
161
2. Consider an instance of the betweenness problem consisting of a set of verticesVb
162
and a set of constraints C. Each constraint c ∈C consists of a sequence of three
163
vertices. It is N P-complete to decide whether the vertices can be totally ordered
164
such that for each constraintc= (u, v, w) the vertexvis between the verticesuand
165
w [15].
166
Consider the hypergraphH = (V, A) withV containingVb and for each constraint
167
c ∈ C vertices vc2 and vc4. The set A contains for each c = (vc1, vc3, vc5) ∈ C a
168
hyperedge hc = {vc1, . . . , vc5} and hyperedges hci = {vci, vc(i+1)} for 1 ≤ i ≤ 4.
169
The graph G= (V, E) withE =S
c∈C{hci; 1≤i≤4} is a path-based support of
170
H and has the minimum number of edges among all supports ofH.
171
There is an ordering < of V such that G is the support of H with respect to <
172
if and only if for each constraint c = (vc1, vc3, vc5)∈C the five vertices in hv are
173
either ordered vc1 < vc2 < vc3 < vc4 < vc5 or vc5 < vc4 < vc3 < vc2 < vc1. Since
174
the vertices vc2 andvc4 do not appear in a hyperedgehc′ for any constraintc6=c′
175
it follows that there is an ordering < of V such that Gis the support of H with
176
respect to<if and only ifVbcan be totally ordered while satisfying all betweenness
177
constraints inC.
178
3.2. Minimum Path-Based Supports
179
Assuming that each hyperedge contains at least one vertex, each hypergraph H =
180
(V, A) has a monotone path-based support G= (V, E) with at mostN−m edges. Just
181
take the support G< with respect to an arbitrary ordering <of the vertex set V. It is,
182
however, N P-hard to find an ordering that minimizes the number of edges among all
183
path-based supports ofH with respect to an ordering of the vertex set [17].
184
Further, note that a path-based support that minimizes the number of edges among
185
all path-based support of a hypergraph H with respect to some ordering of the vertex
186
set might not be a path-based support ofH with the minimum number of edges over all
187
path-based supports of H. E.g., consider the hypergraphHI from the previous section
188
(Fig. 4) or the hypergraph H with hyperedges {1,2,4}, {1,3,4}, and {2,3,4} for an
189
easier example: the unique minimum path-based support of H is a star centered at 4
190
which cannot be created from any ordering of the vertex set. The problem of finding a
191
minimum path-based support remains, however,N P-hard.
192
Theorem 2. It isN P-hard to minimize the number of edges among all path-based sup-
193
ports (or among all monotone path-based supports) of a hypergraph – even if the hyper-
194
graph is closed under intersections.
195
Proof. Reduction from Hamiltonian path. Let G = (V, E) be a graph. Let H =
196
(V, E∪ {V} ∪ {{v}; v∈V}) andK=|E|. Note that any support of H containsGas a
197
subgraph. Hence,H has a path-based support with at mostKedges if and only ifGis a
198
path-based support ofH which is true if and only ifGcontains a Hamiltonian path.
199
3.3. Planar Path-Based Supports
200
A graph isplanar if it has a drawing in which no pair of edges intersect but in common
201
end points. For the application of Euler diagram like drawings, planar supports are of
202
special interest. However, like for general planar supports, the problem of testing whether
203
there is a path-based planar support is hard.
204
Theorem 3. It is N P-complete to decide whether a hypergraph – even if it is closed
205
under intersections – has a path-based planar support.
206
Proof. The support that Johnson and Pollak [5] constructed to prove that it is N P-
207
complete to decide whether there is a planar support, was already path-based.
208
4. Path-Based Tree Supports
209
In this section we show how to decide in polynomial time whether a given hypergraph
210
has a path-based tree support. If such a support exists, it is at the same time a path-based
211
support of minimum size, a monotone path-based support [18], and a planar path-based
212
support. Moreover the intersection of any subset of hyperedges induces again a path in a
213
path-based tree support. So far it is known how to decide in linear time whether there is
214
a path-based tree support ifV ∈A[7].
215
{v1} {v2} {v3} {v4} {v5} {v6} {v7} V
h31
h22
h11 h12 h14 h15 h21
(a) Hasse diagramD
{v1} {v2} {v3} {v4} {v5} {v6} {v7} V
h31
h22
h11 h13 h14 h15 h21
h12
(b) ClosureD
{v1} {v2} {v3} {v4} {v5} {v6} {v7} V
h31
hs
h11 h12 h13 h14 h15
(c) AugmentedD′
Figure 5: Illustration of the augmented Hasse Diagram for the hypergraphH= (V, A) indicated in 5(a).
A=A∪ {h13,{v2}, . . . ,{v5}}. The two hyperedgesh21 andh22are both implied but no summery edges.
They are not present in the augmented Hasse diagram. The summary hyperedgehs is added toA′.
4.1. Constructing a Tree Support from the Hasse Diagram
216
A support with the minimum number of edges and, hence, a tree support if one exists
217
can easily be constructed from the Hasse diagram if the hypergraph is closed under inter-
218
sections [7]. Note, however, that the number of intersections of any subset of hyperedges
219
could be exponential in the size of the hypergraph.
220
To construct a tree support of an arbitrary hypergraph H = (V, A), it suffices to
221
consider theaugmented Hasse diagram – a representation of “necessary” intersections of
222
hyperedges. The definition is as follows. First consider the smallest setAof subsets ofV
223
that containsAand that is closed under intersections. Consider the Hasse diagramD of
224
H = (V, A). Note that any tree support ofH is also a tree support ofH: The intersection
225
of two subtrees is again a subtree.
226
Leth1, . . . , hk be the children of a hyperedgehin D. The hyperedgeh∈Aisimplied
227
if the hypergraph (h1∪· · ·∪hk,{h1, . . . , hk}) is connected andnon-implied otherwise. Let
228
{h1, . . . , hk} be a maximal subset of the children of a non-implied hyperedge inA such
229
that (h1∪ · · · ∪hk,{h1, . . . , hk}) is connected. Thenh1∪ · · · ∪hk is asummary hyperedge.
230
Note that a summary hyperedge might not be in A. Let A′ be the set of subsets of V
231
containing the summary hyperedges, the hyperedges inA that are not implied, and the
232
sources ofD. For an example consider Fig. 5(c). In this example, the hyperedgehsis a
233
summary hyperedge, h31 andh11, . . . , h15 are non-implied, andV is a source.
234
The augmented Hasse diagram of H is the Hasse diagram D′ of H′ = (V, A′). If
235
H has a tree support, then the augmented Hasse diagram has O(n+m) vertices and
236
can be constructed inO(n3m) time [7] (without explicitly constructing the closure under
237
intersection A). Further note that ifH has a tree support and h∈ A′ is non-implied,
238
then all children ofhinD′ are disjoint: Otherwise there would be a summary hyperedge
239
betweenhand intersecting children.
240
If a tree supportG= (V, E) ofH exists, it can be constructed as follows [7]. Starting
241
with an empty graph G, we proceed from the sinks to the sources of D′. If h ∈ A′
242
is not implied, choose an arbitrary ordering h1, . . . , hk of the children of h in D′. We
243
assume that at this stage,G[hi], i= 1, . . . , k are already connected subgraphs of G. For
244
j= 2, . . . , k, choose verticesvj∈Sj−1
i=1hi,wj ∈hj and add edges{vj, wj} toE.
245
If we want to construct a path-based tree support, then G[hj], j= 1, . . . , k are paths
246
and as verticesvj+1andwj for the edges connectingG[hj] to the other paths, we choose
247
the end vertices of G[hj]. The only choices that remain are the ordering of the children
248
ofhand the choice of which end vertex ofG[hj] iswj and which one isvj+1. The implied
249
hyperedges give restrictions on how these choices might be done.
250
4.2. Choosing the Connections: A Characterization
251
When we want to apply the general method introduced in Sect. 4.1 to construct a
252
path-based tree supportG, we need to make sure that we do not create vertices of degree
253
greater than 2 inG[h] when processing non-implied hyperedges contained in an implied
254
hyperedgeh.
255
Definiton 1 (Conflicting Hyperedges). Two overlapping hyperedgesh′, h′′∈A′ have
256
a conflictif there is some hyperedge inA′ that contains bothh′ andh′′. Two overlapping
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hyperedgesh′, h′′∈A′ have a conflict with respect toh∈A′ ifh′ has a conflict with h′′,
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h′∩h′′⊆handhis a child ofh′ orh′′. In that case we say thath′ andh′′are conflicting
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hyperedges of h. Let A′h be the set of conflicting hyperedges of h. Let Ach be the set of
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children hi ofhsuch that h∈A′hi.
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E.g., consider the hypergraph in Fig. 5(c). Thenh31andh11are both contained in the
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hyperedge V and they both contain {v2}. Hence, they have a conflict. Further, is the
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intersectionh31∩h11={v2}contained in the child hsofh31. Hence,h31 has a conflict with
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h11 with respect tohs. Similarly doesh31 have a conflict with h15 with respect tohs and
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we have on one handA′hs ={h11, h12, h31}. On the other hand doeshshave a conflict with
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h11with respect toh12 and withh15with respect toh14 and we haveAchs ={h12, h14}. Note
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that there might be hyperedges that have a conflict but not with respect to any of their
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children. As an example see the hyperedgesh41 andh42in Fig. 6(a). In the lemmas in this
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section, we will prove that it suffices if the algorithm considers only conflicts with respect
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to some child.
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Assume now thatH has a path-based tree supportGand leth′, h′′∈A′be such that
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h′ andh′′ have a conflict with respect to a childhofh′′. Since h′∪h′′ is contained in a
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hyperedge it follows thatG[h′∪h′′] is the subgraph of a path. Since in additionh′andh′′
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intersect and G[h′] and G[h′′] are paths, it follows thatG[h′∪h′′] is also a path. Hence,
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we have the following situation.
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h′
z }| {
h
| {z }
h′′
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Note especially that among the two end vertices of G[h′′] exactly one is contained inh′
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and that this end vertex is also an end vertex of G[h]. This yields the following three
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types of restrictions on the connections of the paths.
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1. G[h′\h] andG[h′′\h] must be paths that are attached to different end vertices of
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G[h].
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2. Assume further that h′′′ does also have a conflict withh′′ with respect toh. Then
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both, G[h′\h] andG[h′′′\h], must be appended to the common end vertex ofG[h]
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andG[h′′].
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3. Assume further that h2, h1 ∈ Ach, h2 6= h1. Let hi ∈ A′h have a conflict with h
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with respect to hi, i = 1,2, respectively. Then G[hi\h] has to be appended to
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the common end vertex of G[h] and G[hi]. Hence,G[h1\h] andG[h2\h] must be
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appended to different and vertices of G[h].
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h1
z }| {
h2
z }| {
h1 h2
| {z }
h
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E.g., consider the hypergraphH= (V, A) in Fig. 5(c). Then on one hand,h31has a conflict
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withh11andh15with respect tohs. Hence, by the first type of restrictionsG[h11\hs] and
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G[h15\hs] must be appended to the same end vertex ofG[hs], i.e. the end vertex ofG[hs]
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to whichG[h31\hs] is not appended. On the other hand,h11 andhshave a conflict with
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respect to h12, while h15 and hs have a conflict with respect toh14. Hence, by the third
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type of restrictions it follows thatG[h11\hs] andG[h15\hs] must be appended to different
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end vertices ofG[h]. Hence, there is no path-based tree support forH.
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This motivates the following definition of conflict graphs.
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Definiton 2 (Conflict Graph). The conflict graphCh, h∈A′ is a graph on the vertex
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setA′h∪Ach. The conflict graphCh contains the following three types of edges.
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1. {h′, h′′}, h′, h′′∈A′h ifh′ andh′′ have a conflict with respect toh.
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2. {h′, h1}, h′ ∈A′h, h1∈Ach if h′ ∈A′h1 and h′ and hhave a conflict with respect to
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h1.
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3. {h1, h2}, h1, h2∈Ach, h16=h2.
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E.g., consider the hypergraph H= (V, A) in Fig. 5(c).
Then the conflict graph Chs contains the edges {h31, h15} and{h31, h11} of type one, the edges {h12, h11} and{h14, h15} of type 2 and the edge{h12, h14} of type 3. (See the figure on the right.) Hence,Chs contains a cycle of odd length, reflecting that there is no suitable assignment of the end vertices ofG[hs] toh11, h15 andh31.
h31 Ch
s: h11
h12 h14 h15 type 1 type 1
type 2 type 2
type 3
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Theorem 4. A hypergraphH = (V, A)has a path-based tree support if and only if
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1. H has a tree support,
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2. no hyperedge contains three pairwise overlapping hyperedges h1, h2, h3 ∈ A′ with
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h1∩h2=h2∩h3=h1∩h3, and
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3. all conflict graphsCh, h∈A′,|h|>1 are bipartite.
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From the observations before the definition of the conflict graph it is clear that the
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conditions of Theorem 4 are necessary for a path-based tree support. In the remainder
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of this section, we prove that the conditions are also sufficient.
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In the following assume that the conditions of Theorem 4 are fulfilled. We show
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in Algorithm 1 how to construct a path-based tree support G of H. We consider the
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vertices of the augmented Hasse diagram D′ from the sinks to the sources in a reversed
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topological order, i.e., we consider a hyperedge only if all its children in D′ have already
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been considered. During the algorithm, a conflicting hyperedge h′ of a hyperedge h is
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labeled with the end vertex v of G[h] if the path G[h′\h] will be appended tov. We
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will call this label sideh(h′). Concerning the choice of the ordering of the children in
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Line 8 of Algorithm 1: the sets Ach, h ∈A′ contain at most two hyperedges – otherwise
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the subgraph ofCh induced by Achcontains a triangle and, hence, is not bipartite.
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Algorithm 1 constructs a tree support GofH [7]. Before we show thatGis a path-
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based tree support, we illustrate the algorithm with an example. Consider the hypergraph
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H in Fig. 6. We show how the algorithm proceedsh51 and all its descendants inD′. For
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Algorithm 1: Path-based tree support
Input : augmented Hasse diagram D′ of a hypergraphH= (V, A)
fulfilling the conditions of Theorem 4;
conflict graphsCh on vertex setsA′h∪Ach,hnon-source vertex ofD′ Output: path-based tree supportG= (V, E) ofH
Data : labels sideh(h′)
indicating the end vertex ofG[h] to whichh′\hshould be appended begin
E← ∅;
foreachvertex hof D′ in a reversed topological order ofD′ do if h={v} for somev∈V then
foreachvertex h′ ofCh do sideh(h′)←v;
else
8 Leth1, . . . , hk be the children ofhsuch thath2, . . . , hk−1∈/ Ach; if his non-implied then
Letwi, vi+1, i= 1, . . . , k be the end vertices ofG[hi] such that
• sideh1(h) =v2 ifh∈A′h1 and
• sidehk(h) =wk ifh∈A′hk;
Add the edges{vi, wi}, i= 2, . . . , ktoE;
else
Letw16=vk+1be the end vertices of G[h] such that
• vk+1∈/ h1and
• w1∈/ hk;
if h1∈Ach thensideh(h1)←vk+1; if hk∈Achthensideh(hk)←w1;
Label the remaining vertices ofCh withvk+1 orw1
such that no two adjacent vertices have the same label;
end
h1
1 h12 h13 h14 h1
5 h16 h17 h18 h19 h110 h1
11 h112 h1
13
h2
1 h2
2 h2
3 h2
4 h2
5
h42 h41
v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 v13 h2 h26 7
h51 h5
2 h5
3
h114 v14 h31 h32 h33 h34
(a) augmented Hasse diagramD′of a hypergraphH
side=v5 side=v6
h31 h2
2
h1
5
(b) conflict graph of hyperedgeh23
side=v7 side=v9
h17 h1
9
h2
5 h41
(c) conflict graph of hyperedgeh24
h4
2
h2
3
side=v5 side=v7
h1
7
h2
2
(d) conflict graph of hyperedgeh31
v4 v5 v6 v7 v8 v9 v10
v11
v12 v13
v1
v2
v3 v14
(e) a path-based tree support ofH
v9
v3 v4 v5 v6 v7 v8 v10 v14 v13
v12 v11 v1
v2
(f) metro map like drawing of the sources ofD′ Figure 6: Illustration of Algorithm 1.
the hyperedgesh13, h14, h16, andh18 the conflict graphs are empty. For the other leaves we
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have
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sideh15(h22) = sideh15(h23) = sideh15(h31) = sideh15(h42) = v5, sideh17(h24) = sideh17(h31) = v7, and sideh19(h24) = sideh19(h41) = sideh19(h25) = sideh19(h26) = sideh19(h27) = v9.
When operating h22 and h23, respectively, we add edges {v4, v5} and {v5, v6}, respec-
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tively, toG. While the conflict graph ofh22does only containh15with sideh22(h15) =v4, the
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assignment of side inCh23 is illustrated in Fig. 6(b). h24 has a conflict with respect to the
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childrenh17 andh19. Hence, we add edges{v7, v8}and {v8, v9} toG. The conflict graph
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ofh24is shown in Fig. 6(c). When operatingh31we can chooseh1=h23andh2=h17, since
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sideh23(h31) =v6 and sideh17(h31) =v7. We add the edge {v6, v7} toG. The conflict graph
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Ch31 is shown in Fig. 6(d). The hyperedgeh41 is implied and we set sideh41(h24) =v4. We
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can finally connect v3to v4 orv9 when operatingh51.
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To prove the correctness of Algorithm 1, it remains to show that all hyperedges ofH
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induce a path in G. Since we included all inclusion maximal hyperedges of H in A′, it
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suffices to show this property for all hyperedges inA′. We start with a technical lemma.
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Lemma 5. Let h′ andh′′ be two overlapping hyperedges and leth′ be not implied. Then
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there is a hyperedgeh∈A′ withh′∩h′′⊆h(h′.
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Proof. Lethc∈Abe maximal with h′∩h′′⊆hc (h′. The hyperedgehc is a child of
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the non-implied hyperedgeh′ inD. Consider the summary hyperedgehwithhc ⊆h(h′.
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By definition ofA′ it follows that h∈A′.
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For an edge{v, w}ofGlethvwbe the intersection of all hyperedges ofA′ that contain
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v and w. Note thathvw is not implied since v andw are contained in different children
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ofhvw inD and{v, w}is an edge of the tree supportGofH. Hence,hvw∈A′.
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Lemma 6. Let Conditions 1-3 of Theorem 4 be fulfilled and letG= (V, E)be the graph
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computed in Algorithm 1. Let h′, h′′ ∈A′ have a conflict with respect to a child hof h′
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and letG[h′]andG[h′′] be paths. Then
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1. sideg(h′′) = sideh(h′′)for allg∈A′ with h′∩h′′⊆g⊆h,
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2. sideh(h′′)∈h′′,
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3. sideh(h′′)is an end vertex of G[h′],
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4. G[h′\h′′]is a path, and
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5. sideh(h′′)is adjacent in Gto a vertex of h′′\h′.
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Proof. We prove the lemma by induction on the sum of the steps in which h′ and h′′
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were considered in Algorithm 1. If h′ andh′′ had been considered in the first two steps,
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then at least one of them is a leaf ofD′ and, hence,h′ andh′′ have no conflict. So there
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is nothing to show. Let now h′ andh′′ be considered in later steps. Leth′′∈A′ have a
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conflict withh′ with respect to a childhofh′ and letG[h′] andG[h′′] be paths.
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1. + 2. if h′∩h′′∈A′: There is nothing to show ifh=h′∩h′′. So leth1be the child
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of h with h1 ⊇ h′∩h′′. Then h, h′′ have a conflict with respect to h1. Hence,
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Ch contains the path h′, h′′, h1. By the inductive hypothesis on Property 3, it
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