1

Finding the smallest H-subgraph in real weighted graphs and

related problems

Raphael Yuster

University of Haifa

Joint work with:Virginia Vassilevska, Ryan Williams

CMU

2

The min H-subgraph problem

Input: A vertex-weighted graph G=(V,E), where |E|=m and |V|=n.

Output: An (induced) H-subgraph with minimum total weight, if exists.

2/3

1/23/4

1/3

1/2 H = K3

real vertex weights

Similarly define the edge-weighted version.

3

The min H-subgraph problem

• We work in the addition-comparison model.

• All results apply to directed graphs as well.

• The naïve algorithm solves themin H-subgraph problem in O(nh) time.

• The goal: reduce the exponent (problem is conjectured to be not fixed parameter tractable already in the unweighted case).

• Case h=3 (triangles) studied in [VW – STOC06] who gave the first sub-cubic algorithm.

4

Results – the vertex weighted case

For fixed H with h vertices, a minimum weight (induced) H-subgraph can be found in O(nt(ω,h)) time.

ω < 2.376 [Coppersmith-Winograd ’90] is the exponent of fast square matrix multiplication(possibly ω = 2+on(1) someday).

The exponent t(ω,h) in our theorem is obtained by solving a small integer program.

5

t(ω,h) : definition

21

22

2

111

1

,min),(

22

:)(min

2)3(,

2max)(

4

24:max

ssht

hbbh

bss

bhh

bhbhbs

bbhs

bhbNbb

t(ω,h) → 3h/(6-ω)≤ 0.828h

6

ht(ω,h)unweightedω=2.376ω=2.376

unweighted

32+1)/4-ω(ω2.6162.376

4ω+1ω+13.3763.376

5ω+2ω+24.3764.376

64+2)/4-ω(2ω5.2324.752

74+3)/4-ω(2ω+15.8485.752

82ω+22ω+26.7526.752

92ω+33ω7.7527.128

106+4)/4-ω(3ω+18.4638.128

113ω+26+5)/4-ω) (ω=2(

3ω+29.1288.5) ω=2(

9.1288) ω=2(

Improves (ω+3)/2

VW-STOC06

7

Proof – the vertex weighted case

• Assume H=Kh

• G=(V,E) V={1,2,…,n} w: V → R

• For a positive integer t let St be the set of allt-subsets of V sorted by their total weight.|St| < nt and can be sorted in O(nt log n) time.

• Suppose h=a+b+c where a,b,c are positive integers.

• Create two 0-1 matricesM1=M1(G,a,b) M2=M2(G,b,c)

8

Proof – the vertex weighted case

The matrix M1

Sa

Sb (sorted)

X

Y1: X UY = Ka+b

0: otherwise

9

Proof – the vertex weighted case

The matrix M2

Sb

(sorted)

Sc

Y

Z1: Y UZ = Kb+c

0: otherwise

10

Proof – the vertex weighted case

• Compute M3=M1M2

• Suppose (i) M3(X,Z) > 0 (ii) X U Z induces a Ka+c

YSb so that M1(X,Y)=1 M2(Y,Z)=1hence X U Y U Z is a Kh

• Y is called a witness for the pair (X,Z).If Y is the minimal witness in the sorted Sb thenw(X)+w(Y)+w(Z) is the weight of the smallest Kh containing X U Z.

11

Proof – the vertex weighted case

• Let W be the minimal witness matrix of the product of two 0-1 matrices C=AB.

W[i,j]=0 : C[i,j]=0k : A[i,k]=B[k,j]=1, A[i.k’]B[k’,j]=0 k’<k

•Computing W for dimensions n1,n2,n3 denotedf(n1,n2,n3)

•Need to prove:

mina+b+c=h f(na,nb,nc) = O(nt(ω,h))

12

Computing minimal witnesses

• Computing a matrix of witnesses can be done in essentially the same time needed to perform the product ([Seidel ’95], [Alon+Naor ’96]).

• Problem: Gives no clue on the minimum witness.

• [Kowaluk and Lingas ’05] f(n,n,n) = O(n2+1/(4-ω))They considered a different problem:

computing all pairs lca in a dag

• This already gives the case t(ω,3)=2+1/(4-ω)

• We need to extend their result to other dimensions and optimize upon the choices of a,b,c.

13

na

… nb

nb

nμ nμ

nc

nμ

nμ

...A1 Ap

B1

Bp

…...

a+b+c=h 0< μ≤b

μ split

Ci = AiBi i=1,…p

C=AB= ∑Ci

14

• Let r be the smallest index for which Cr[i,j]>0

• If r does not exist then W[i,j]=0.

• Otherwise let k be the smallest index forwhich Ar[i,k]=Br[k,j]=1.

• W[i,j] = (r-1)nμ+k.

Computing W[i,j]

15

Computing W

Computing a single witness: O(p+nμ)

Computing all witnesses: O(na+c(p +nμ))

Computing a single Ci: O(na-μnc-μnω μ)

Computing all Ci: O(p(na-μnc-μnω μ))

Overall running time

O(na+c(p+nμ) + p(na+c+(ω-2)μ)) =

O(nh-(3- ω) μ+nh-b+μ)

16

Computing W

•For h=3 take a=1 b=1 c=1 μ=1/(4-ω)

•For h=4 take a=1 b=2 c=1 μ=1

•For h=5 take a=2 b=2 c=1 μ=1

•For h=7 take a=2 b=3 c=2 μ=3/(4-ω)

•For h=9 take a=3 b=4 c=2 μ=2

17

Additional features

• Number of comparisons is pretty small:h=3 n log n + mh=4 n2 log n + mh=5 n2 log n + mnh=6 n2 log n + m2

• Can solve the decision problem:Given a real interval I R is there an (induced)H-subgraph whose weight is within I ?

Use binary search within the matrices C1, …,Cp

Running time increases only by log n factor.

18

Additional features

• Can find smallest K2,k in O(n2+1/(4-ω)) < O(n2.616) time.

• Result is meaningful also when the weights are large integers:

β(G,H) the H edge-covering numbermaximum number of edges incident with H-subgraph

β(G,K3)=82

3

4

34

1

1

19

Sparse graphsA triangle can be found in O(m2ω/(ω+1)) < O(m1.41) time

[Alon+Y+Zwick ’97]

• Let Δ be a parameter.

• Vertices in Y having degree > Δ. |Y| < 2m/Δ

• Triangles with vertex in V-Y examined in O(mΔ) time.

• Smallest triangle inside Y found in O(|Y|2+1/(4-ω)) time.

• Overall: O(mΔ+(m/Δ)2+1/(4-ω))

• Optimizing on Δ: O(m(18-4ω)/(13-3ω)) < O(m1.45)

Finding the smallest weighted triangle

20

Edge weighted graphs

•H has k vertices.

•Color V(G) with k colors, randomly.

•Colorful H sometimes easier to find.

•Probability of being colorful: k!/kk > e-k

•Can be derandomized.

The color coding method [Alon+Y+Zwick ’95]

21

Edge weighted graphs

• A and B two compatible matrices in R U ∞

• The distance product D=AB defined by

D[i,j] = mink A[i,k]+B[k,j]

• If A and B are square matrices of order n, then D can be computed in O(n3) time.

• [Chan ’05] Only O(n3/log n) time needed!

• Consequence: Min H-subgraph in O(nk/log n) time.

Distance products

22

Edge weighted graphs

A minimum weight k-cycle can be found, w.h.p., in 2O(k) n3/log n time.For k=o(log log n) this is sub-cubic.

• For each pair u,v find the shortest colorful path of length k-1 connecting them in 2O(k) n3/log n time.

• Assume k is a power of 2 (we use recursion).

• Let C1 be a set of k/2 colors. C2 = C-C1.

• Vi – the vertices colored with a color from Ci.

• Gi – the subgraph induced by Vi.

23

Edge weighted graphs• Recursively find, for each pair in Vi, a colorful

path in Gi of length k/2-1. Record the results in matrices A1 and A2.

• B – the matrix with rows indexed by V1 and columns indexed by V2. B[u,v]=w(u,v).

• The distance product DC1,C2=A1 B A2 gives for each pair of vertices, the shortest colorful path of length k-1 where the first k/2 vertices are colored from C1 and the last k/2 vertices are colored from C2.

24

Edge weighted graphs

• By considering all choices for (C1,C2) (less than 2k choices) we obtain an nn matrix D where D[u,v] is the shortest colorful path of length k-1 between u and v.

• The number of distance products using this approach satisfies t(k) ≤ 2kt(k/2).Thus, t(k) = 2O(k).

• Each product costs O(n3/log n).

25

Chromatic subgraphs

• G=(V,E) an edge-colored graph (number of colors is arbitrary).

We consider

• The monochromatic H-subgraph problem

• The rainbow H-subgraph problem

• Both problems are not easier than the (uncolored) H-subgraph existence problem.

• Are they also not harder?

26

Rainbow subgraphs• We use color reduction.

• Suppose H has t edges.

• f : C → {1,2,…,t}. We now have only t colors.

• A rainbow H now is also originally rainbow.

• f random – an original rainbow H will remain rainbow after reduction with probability t!/tt.

• Do this 2O(t) times and it will happen w.h.p. at least once.

• Can be derandomized. Cost is only O(log n) factor.

27

Rainbow subgraphs• When we have only t colors, same algorithm for

the uncolored version can be used to detect a monochromatic H.

Example K6

28

Monochromatic subgraphsFor any connected fixed graph H with 3k+j vertices (j=0,1,2), a monochromatic H can be detected in O(nωk+j), except of H=K3.

Example K6

•Running time identical to (uncolored) detection.

•Triangles can be found only in O(n(3+ω)/2) < o(n2.688).