/5 Optimal Linear Arrangement of Interval Graphs Johanne Cohen Fedor Fomin Pinar Heggernes Dieter Kratsch Gregory Kucherov 4 LORIA/CNRS, Nancy, France. Department of Informatics, University of Bergen, Norway. LITA, Université de Metz, France. 4 LIFL/CNRS, Lille, France MFCS, August, 006
Outline /5 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
Outline /5 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
Definition /5 4 Definition A linear layout (or simply layout) of a given graph G = (V, E) is a linear ordering of its vertices. b a c d 4 4 graph G linear layout L linear layout L
Optimal linear arrangements /5 5 Definition The weight of L on G is W(G, L) = (u,v) E L(u) L(v). 4 4 linear layout L linear layout L W(G, L ) = + + + W(G, L ) = + + + Definition An optimal linear arrangement (OLA) of G is a layout with the minimum weight, i.e., argmin L W(G, L). We denote W(G) = min L W(G, L) and call it the minimum weight on G.
Previous work /5 6 Computing an OLA is NP-hard (for general graphs [Garey, Johnson 979], for bipartite graphs [Even, Shiloach 975]). But the problem is polynomial for trees [Goldberg, Klipker 976], for grids, for hypercubes [Diaz and al 00]. There exists an approximation algorithm with performance ratio O(log(n)) [Rao, Richa 998].
Interval graphs /5 7 Definition A graph G = (V, E) is an interval graph if there is an one-to-one correspondence between V and a set of intervals of the real line such that, for all u, v V, (u, v) E if and only if the intervals corresponding to u and v have a nonempty intersection. b b g a d g f a d f c c Fig.: An interval graph G Fig.: Its interval representation
Linear arrangements and interval graphs /5 8 The bandwidth (b(g, L) = max (u,v) E L(u) L(v) ) minimization problem is polynomial for interval graphs [Kleitman, Vohra990]. Interval graphs are used in bioinformatic (reconstruction of relative positions of DNA fragments, gene structure prediction, model of temporal relations in protein-protein interactions). An optimal linear arrangement of an interval graph models an optimal molecular pathway [Farach-Colton and al 004].
Our results /5 9 a proof of NP-hardness of the optimal linear arrangement (OLA) problem on interval graphs. a -approximation algorithm for interval graphs. a 8-approximation algorithm for cocomparability graphs.
Outline 0/5 0 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
OLA problem on complete graphs. Lemma Let K n be the complete graph on n vertices. Then W(K n ) = (n )n(n + ) 6. 8 7 5 4 6 4 7 8 5 6 8 7 5 4 6 /5
OLA problem on complete graphs. Lemma Let K n be the complete graph on n vertices. Then W(K n ) = (n )n(n + ) 6. 8 7 5 4 6 4 7 8 5 6 8 7 5 4 6 /5
OLA problem on star graphs. /5 Lemma Let S α be the star with a center vertex c and α leaves. Then,. a permutation L is an optimal linear arrangement if and only if L places c at the middle position. α α W(S α ) = ( + ). W(S α ) W(S α, L) W(S α ) for any layout L c
Proof of Lemma /5 Let S α be a star with a center vertex c and α leaves (α even). c
/5 Proof of Lemma Let S α be a star with a center vertex c and α leaves (α even). i i i + α + Case where L(c) = : W(S α, L) = α i= (i + ) = α i= i = α(α+)
Proof of Lemma /5 Let S α be a star with a center vertex c and α leaves (α even). k k k + k + α + Case where L(c) = : the worst case (W(S α, L) = α(α+) ). Case where L(c) = k : W(S α, L) = k i= i + α+ k i= i. Case where L(c) = α/ + : the better case (W(S α, L) = α ( α + )).
Simple graphs 4/5 4 Remark The stars and the complete graphs are interval graphs. c c 4 8 5 6 7
Partition of edge set. 5/5 5 G = c
Partition of edge set. 5/5 5 G = c can be split into graphs : G and G G = c G = c
Partition of edge set. 5/5 5 G = c W(G, L) = W(G, L) + W(G, L) G = 7 G = 7 4 5 6 8 9 0 4 5 6 8 9 0 W(G, L) W(G, L)
Partition of edge set. 5/5 5 G = 7 4 5 6 8 9 0 W(G) W(G ) + W(G ) G = 7 G = 7 4 5 6 8 9 0 4 5 6 8 9 0 W(G ) W(G )
Some results : 6/5 6 Lemma Let G = (V, E) be a graph, E = E E and E E =. Then W(G) W(G ) + W(G ), where G = (V, E ) and G = (V, E ).
Some results : 6/5 6 Lemma Let G = (V, E) be a graph, E = E E and E E =. Then W(G) W(G ) + W(G ), where G = (V, E ) and G = (V, E ). More generally, Corollary Let G = (V, E), V = V V n, and E = E E n, where E,, E n are pairwise disjoint. Then W(G) W(G ) +... + W(G n ), where G i = (V i, E i ), i n.
Outline 7/5 7 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
The NP-completeness result 8/5 8 Theorem The problem of deciding, for an interval graph G = (E, V ) and a constant K, whether W(G) K is NP-complete. This problem belongs to NP. The proof is by reduction from the -Partition problem : Instance : A finite set A of m integers {a,..., a m }, a bound B Z + such that m i= a i = mb. Question : Can A be partitioned into m disjoint sets A, A,..., A m such that, for all i m, a A i a = B?
Sketch of proof : 9/5 9 Let f be the following reduction from PARTITION problem to OLA problem { G A f (A, B) = (G, k) with i= K a i S B with a center v k W(S B ) + A i= W(K a i ) A = {, 4,, } B = 6 G = c
Sketch of proof : 9/5 9 Let f be the following reduction from PARTITION problem to OLA problem { G A f (A, B) = (G, k) with i= K a i S B with a center v k W(S B ) + A i= W(K a i ) G is an interval graph. A = {, 4,, } B = 6 v A A
Sketch of proof : 9/5 9 Let f be the following reduction from PARTITION problem to OLA problem { G A f (A, B) = (G, k) with i= K a i S B with a center v k W(S B ) + A i= W(K a i ) A = {, 4,, } B = 6 A = {, 4} A = {, } 6 5 4 7=(B+) 0 9 8 A A
Sketch of proof : 9/5 9 Let f be the following reduction from PARTITION problem to OLA problem { G A f (A, B) = (G, k) with i= K a i S B with a center v k W(S B ) + A i= W(K a i ) A can be partitioned into disjoint sets A, A such that a A a = a A a = B there exists a linear layout L with W(G, L) k A polynomial-time reduction from PARTITION problem to OLA problem is based on the same idea.
Outline 0/5 0 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
Right endpoint orderings /5 Definition The layout of G consisting of vertices ordered by the right endpoints of their corresponding intervals is called the right endpoint ordering (reo) of G with respect to I. b b a d g f a d c c g f an interval graph G its interval representation
Right endpoint orderings /5 Definition The layout of G consisting of vertices ordered by the right endpoints of their corresponding intervals is called the right endpoint ordering (reo) of G with respect to I. b a c d g f 6 4 5 an interval graph G its interval representation
Right endpoint orderings /5 Definition The layout of G consisting of vertices ordered by the right endpoints of their corresponding intervals is called the right endpoint ordering (reo) of G with respect to I. 4 6 5 6 4 5 an interval graph G its interval representation
-Approximation algorithm /5 Remark In a right endpoint ordering reo, for every pair of adjacent vertices reo(u) < reo(w), each vertex between u and w is adjacent to w. 4 6 5 6 4 5
-Approximation algorithm /5 Remark In a right endpoint ordering reo, for every pair of adjacent vertices reo(u) < reo(w), each vertex between u and w is adjacent to w. 4 6 5 6 4 5 Theorem Let G = (V, E) be an interval graph, and let I be an interval model of G. Then, W(G, reo) W(G).
Proof of the Theorem : /5 E is split : E i = {(u, v) reo(u) = i reo(v) < i} E(G) 4 6 5 4 6 5 4 6 5 G and reo E 4 E 6
/5 Proof of the Theorem : E is split : E i = {(u, v) reo(u) = i reo(v) < i} E(G) G i = (V, E i ) is composed of a star of n i (= E i ) leaves 4 6 5 4 6 5 4 6 5 G and reo E 4 and graph G 4 E 6 and graph G 6
/5 Proof of the Theorem : E is split : E i = {(u, v) reo(u) = i reo(v) < i} E(G) G i = (V, E i ) is composed of a star of n i (= E i ) leaves 4 6 5 4 6 5 4 6 5 G and reo E 4 and graph G 4 E 6 and graph G 6 For G i = (V, E i ) :W(G i, reo) = v V (G i ) reo(u) reo(v)
/5 Proof of the Theorem : E is split : E i = {(u, v) reo(u) = i reo(v) < i} E(G) G i = (V, E i ) is composed of a star of n i (= E i ) leaves 4 6 5 4 6 5 4 6 5 G and reo E 4 and graph G 4 E 6 and graph G 6 For G i = (V, E i ) :W(G i, reo) = v V (G i ) reo(u) reo(v) can be the worst case for the star of n i leaves. W(S ni ) W(G i, reo) W(S ni )
/5 Proof of the Theorem : E is split : E i = {(u, v) reo(u) = i reo(v) < i} E(G) G i = (V, E i ) is composed of a star of n i (= E i ) leaves 4 6 5 4 6 5 4 6 5 G and reo E 4 and graph G 4 E 6 and graph G 6 For G i = (V, E i ) :W(G i, reo) = v V (G i ) reo(u) reo(v) can be the worst case for the star of n i leaves. W(S ni ) W(G i, reo) W(S ni ) For G :W(G) n i= W(G i) n i= W(S n i ) So W(G, reo) = n i= W(G i, reo) and W(G) W(G, reo)
Outline 4/5 4 Introduction Optimal linear arrangements Interval graphs Preliminaries OLA problem on simple graphs. Remarks The complexity of the OLA problem Approximation algorithm Conclusions
5/5 5 Conclusion Optimal linear arrangement of interval graphs is NP-hard. There exists a fast -approximation algorithm based on any interval model of the input graph. In the paper, we extend this result to cocomparability graphs (NP-completeness, 8-approximation polynomial-time algorithm) The complexity of several other linear layout problems, like Cutwidth is not resolved for the class of interval graphs.