1. 1 Parametrized complexity of Graph MOTIF By: Amr Koura
2. 2 Agenda Problem definition Preliminaries FPT algorithms
Parameters for which graph MOTIF is hard References
3. 3 Problem Definition
4. 4 Problem definition Input: A triple (G,c,M) where G=(V,E)
is graph, c:V-->C is coloration function of V on |C| colors, and
M is multiset of Colors of C. Output: A subset such that G(P) is
connected and c(P)=M. PV
5. 5 Example M={red,blue,red,yellow,black,orange,green}
7. 7 Problem Definition The problem is NP-hard. Application:
Biological,social and technical Networks.
8. 8 Problem definition
9. 9 Preliminaries
10. 10 Some Definitions For any vertex , the set of neighbors
of V is N(v) and for , , and . Vertex v dominate S if ,set R
dominate S if . Denote the multiplicity of x in multiset M, and .
vV SV N (S)= vS N (v) S SN (v) SN (R) mM x M= xM mM x N [v]=N
(v){v} N [S ]=N (S)S
11. 11 Some Definitions clique is a graph where each two
distinct vertices are connected. cluster is a graph set of disjoint
union of cliques. (I,K) is fixed parameter traceable FPT if it can
be solved in where f is computable function and c is constant. If C
is graph class , distance of graph G to C is number of vertices to
be removed from G to get C. f (K).I c
15. 15 Cluster editing Definition:Definition: number of edge
deletion or additionnumber of edge deletion or addition requited to
get cluster.requited to get cluster. Theorem: Graph MOTIF can be
solved in with cluster editing K. Proof: use parameter neighborhood
diversity. O (8 k )
16. 16 Neighborhood diversity Definition: Graph G has
neighborhood diversity K if its vertices can be partitioned into at
most K sets such that all vertices in the set have the same type.
Two vertices u,v have the same type if N(u) {v}=N(v){u}.
18. 18 Cluster editing Theorem: graph MOTIF can be solved in on
graph with neighborhood diversity K [1]. Compute neighborhood
diversity: - G is input graph, G' graph obtained after k edition on
G. - let X is set of vertices that are endpoints of the edit edges.
- Then - Let is L cliques of G'. - ,so number of neighborhood
diversity of G is bounded by . - applying the above theorem , Graph
MOTIF can be solved in O (2 k ) x2k C1,. .. ,C L i[L]vCi X , N
[v]=Ci x+ l2k+ k=3k O (23k )=O (8k )
19. 19 Distance to Clique
20. 20 Distance to clique Theorem: graph MOTIF can be solved in
where k is distance of input graph to clique. Proof: Algorithm: -
Find vertex cover S of size k in in time [2]. - S is also the
distance to clique in G. - let R be solution, trying all subset of
S, guess subset of S, which is in R. - ,then vertices of the clique
C with colors should complete the solution. - Problem: finding a
minimal (inclusion wise) set such that and G[ ] is connected. O (2
klogk ) G 1.2738k 2k S '=SR c(S ')M MS ' M '=M c(S ') R'C c(R')M '
R'S '
21. 21 Distance to clique Proof: Algorithm continue: - Let be
connected components of G[S'], - build graph G'=(V',E') from graph
G: +Keep the clique C as it is. + contract into single vertex and
draw and edge from to iff and + a minimal (inclusion wise) in a
dominating vertex in G', + try out all l-partitions of denoted by
st: dominate . note that . + number of partitions : Bell number.
C1, C2,. .. ,Ck ' k '< k Ci ,i[k ' ] Ci vi vi vC uCi {u ,v}E Rd
R S ' Rd =r1 ,... ,rl {v1 ,... ,vk ' } {A1 ,... , Al } ri Ai Bk '
lk 'k
22. 22 Distance to clique Proof: Algorithm continue: - From G'
,build bipartite graph , H=( ,B) where and where there is an edge
between and all iff such that c(v)=x, v dominate and there is no ,v
dominate . - H has vertices and in can decided if H has perfect
matching in size l if exists [3]. H1H2 H1={uAi ,i[l]} H2= xM ' {ux
1 ,ux 2 ,... ,ux mM ' x } uAi {ux 1, ux 2,. .. ,ux mM ' x } vV ' Ai
i j Aj l+M 'k+M (k+ M )2.376
23. 23 Distance to clique Proof continue: - Build a match from
to where j is smallest integer not yet in the match. - Because ,j
will not exceed . U Ai uc(ri) j c(R' )M ' mM ' (c(ri))
24. 24 Distance to clique Proof: Algorithm continue: - from
perfect matching in H, graph MOTIF solution built: - there , namely
, and dominate . - set , G[Z] is connected and and then we extend z
by adding subset such that c(Z')=M'c(Z). il {uAi ,ux ji }B wi V '
c(wi)=x wi Ai Z=S ' il wi c(Z )M ' Z 'C Z
25. 25 Distance to clique Theorem: graph MOTIF can be solved in
where k is distance of input graph to clique. Proof: Algorithm
continue: - The algorithm needs: - to compute vertex cover - to
build G' from G. - to build H from G'. - to check perfect matching
in H. - to build perfect matching in H. - all together: - As ,so
the algorithm fulfill the above theorem. O (2 klogk ) (1.2738k )
p1(n) p2(n) (k+M) 2.376 p3(n) 1.2738k + 2k ( p1(n)+ Bk ( p2(n)+
(k+M) 2.376 + p3(n))) BK < ( 0.792k ln(k+ 1) ) k
26. 26 Vertex cover number
27. 27 Vertex cover number Theorem: Graph MOTIF can be solved
in on graph with vertex cover of size k. Proof: Algorithm: -same
previous algorithm but computing the vertex cover in G, up to
computing . -we guess in time , compute the order pair such that
(1) dominate , (2) has at least one neighbor in , (3) has no
neighbor in . O (2 2klog k ) Rd =r1 ,... ,rl O (k ! Bk ) < A1,
A2, .. , Al > riAi 1 j< i Aj ri i< jl Aj ri
28. 28 Vertex cover number Theorem: Graph MOTIF can be solved
in on graph with vertex cover of size k. Proof: continue: -in H={ }
there is an edge between and iff ,c(v)=x,v dominate , And v has one
neighbor in . -if H has perfect matching, then we can build MOTIF
solution. - and , then the complexity is O (2 2klog k ) H1H 2 , B
uAi {ux 1, ux 2,. .. ,ux mM ' (x) } vV ' Ai 1 j< i Aj k !k k Bkk
k O (kk kk )=O (22klogk )
29. 29 Parameters which Graph MOTIF is Hard
30. 30 Parameters which Graph MOTIF is Hard Graph MOTIF problem
still NP-Hard even if provided with some parameters. -Deletion set
Number.
31. 31 Deletion set Number Parameter
32. 32 Deletion set Number Parameter Definition: minimum number
of vertices to remove to make graph belong to restricted class.
Theorem: Graph MOTIF is NP-Hard for graph with distance 1 to
disjoint paths and colorful motif.
33. 33 Deletion set Number Parameter Proof: -Relation with X3C.
Given X={ },S={ }, find ,st. each element in X exists only once in
T. From I=(X,S), construct I'=(G=(V,E),c,M) ,M motif, by: one root
r , , two paths are built, first including , three elements of ,
.second includes . x1, x2,.... , x3q S1, S2,.... ,SS T S Si S ai 1
Si ai 2 ,bi 2 bi 1
34. 34 Deletion set Number Parameter C={1,2,...,2|S|+3q+1}.c(
)=c( )=i. c( )=c( )=|S|+i for . colors are assigned to nodes
vertices according to x,c(r)= .the construction is done in
polynomial time. ai 1 ai 2 ai 1 bi 1 bi 2 1iS 2S+ 1,... ,2S+ 3q 3q+
2S+ 1
35. 35 Deletion set Number Parameter C={1,2,...,2|S|+3q+1}.c(
)=c( )=i. c( )=c( )=|S|+i for . colors are assigned to nodes
vertices according to x,c(r)= .the construction is done in
polynomial time. ai 1 ai 2 ai 1 bi 1 bi 2 1iS 2S+ 1,... ,2S+ 3q 3q+
2S+ 1
36. 36 Deletion set Number Parameter : given is solution for I,
build solution P for I': -take root node. - take the full path from
to . - take the path . - all colors are taken only once. T S Si T
ai 1 bi 1 Si T ai 2 bi 2 X3C MOTIF
37. 37 Deletion set Number Parameter : Given solution P for I'
,build solution for I. - Root node is taken. - For each either or
is taken, same for and . - To add , should be added due to
connectivity constraint. MOTIF X3C 1iS ai 1 ai 2 bi 1 bi 2 bi 2 ai
2 T S
38. 38 Deletion set Number Parameter continue: -either three
nodes in are added or not. -T={ }.since P is solution , no color
repeated and each element of X is appear exactly once. Si S Siai 1
P MOTIF X3C
39. 39 Refrences [1] R. Ganian. Using neighborhood diversity to
solve hard problems. CoRR, abs/1201.3091,2012.. [2]M. Mucha and P.
Sankowski. Maximum Matchings via Gaussian Elimination. In 45th
Sym-posium on Foundations of Computer Science (FOCS 2004), pages
248255. IEEE ComputerSociety, 2004. [3] J. Chen, I. A. Kanj, and G.
Xia. Improved upper bounds for vertex cover. Theoretical Computer
Science, 411(4042):3736 3756, 2010.