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Convex Recoloring of Trees

Reuven Bar-YehudaIdo Feldman

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Convex Coloring

A Coloring is Convex if each color induces a block.

Convex Coloring

Non-Convex Coloring

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Problem Definition

Input: Colored Tree (T,,W) Output: Convex Coloring ’ Cost: Weight of recolored

vertices

Cost=2

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is a cover if there is a recoloring which changes only vertices from .

X V

Coloring Cover

X

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Formal Definition Input: Weight vector W, Tree T, Coloring

. Output: Coloring Cover . Cost:

4

1

3

6

2

5

87 9

0,1n

xw x

0

1

1 0

0 00

0 0

Cost=2+9=11

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Motivation Phylogenetic trees

Each node represents a specie Each color represents an attribute “Real world” examples are convex

Linguistics

Strongly NP complete [Moran,Snir] Hardness holds for strings Exponential Algorithms / Polynomial Approximations

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r-approximation Given a weight vector w, Tree T, Coloring .

Minimize w∙xSubject to: x is a coloring cover

x is r-approximation if w∙x≤r∙w∙x* An algorithm is r-approximation if for any

w,T, it returns an r-approximation.

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Previous Results

[Snir,Moran]: Dynamic Programming algorithm

(EXP) 3 Approximation for trees 2 Approximation for strings

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Our Contribution

A 2+ε approximation algorithm for trees Time complexity: poly(n)exp(1/ ε)

Faster Dynamic Programming Though still exponential time.

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The Local Ratio Theoremx is an r-approximation with respect to w1

x is an r-approximation with respect to w2

x is an r-approximation with respect to w1+w2

Proof: Let x*, x1*, x2* opts w.r.t. w, w1, w2.

w1 · x r × w1x1*

w2 · x r × w2x2*

w · x r × (w1x1* + w2x2*)

r × ( w1 x* + w2 x* )

= r × ( w1+ w2 ) x*

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Example: 3-approximation

v separates 3 different colors. For every feasible x:

v

1 1

1

1 1

1

0

0

12 6w x

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3-aprx algorithmAlgorithm MinCR3(T,,W)

If there is no node separating 3 colorsReturn Optimal(T,,W)

Else,Let V’ be the set of the corresponding 6

verticesDefine

Let

Return MinCR3(T,,W-W1)

'minv V w v

1

'

0 '

v Vw v

v V

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Running Example

4

141 2 2

5

4

3

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Running Example (2)

0 1

4

1 1

4

3

4

2

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Running Example (3)

0 0

3

0 1

3

2

4

1

No More Weight Reductions. Need to calculate optimal cover.

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(t,d)-separator

v is a (t,d)-separator if in T-{v}, there are t colors such that each appear on at least d components.

{d1

1

1

2 2 2

t t t

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Example: (2,3)-seperator

3

2

22

1

11

4

v

2 colors ( , ) are separated to 3 components.

1 2

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Weight reduction On a (t,d)-separator, assign a weight of 1

to each one of the t∙d vertices.

1

1

1

1 1 1

1 1 1

For every feasible x:

1 1

tdr

t d

1

1

1

2 2 2

t t t 11 1t d w x td

At most one color left on

≥2 sides

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Weight Reduction #1

1 1

2 2

k k

1

1

1

1

1

1

2

2 1 1

kr

k

, 2t k d

22

1r

k

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Weight Reduction #2

1 2

1 2

1 2

1

1

1

1

1

1

K nodes 2,t d k

22

1r

k

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Weight Reduction #3

1 2

1 2

1 2

3 3 3

1 2

1

1

1

1

1

1

1

11

1 1

1

3

3 4

3 1 4 1r

3, 4t d

2r

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Weight Reduction #4

1 2

1 2

1 2

3 3 3

4 4 4

1

1

1

1

1

1

11 1

11 1

4, 3t d

2r

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Light Colorings

A coloring is lightk if it does not contain: (2,k)-separator (k,2)-separator (3,4)-separator (4,3)-separator

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2+2/(k-1)-aprx algorithmAlgorithm MinCR(T,,W)

If is a lightk coloringReturn SolveLight(T,,W)

Find (t,d)-separator ( )Let V’ be the set of it’s t∙d verticesDefine

Let

Return MinCR(T,,W-W1)

, 2, , , 2 , 3,4 , 4,3t d k k

'minv V w v

1

'

0 '

v Vw v

v V

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Algorithm Correctness

Claim: MinCR returns an -apx Proof: By induction on recursion

depth: Base: (Light Coloring) X is optimal. Step:

X is an r-aprx w.r.t. W-W1 (Induction hyp.).

Every solution is an r-aprx w.r.t. W1. Thus, X is an r-aprx w.r.t. to W (LR Theorem).

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1r

k

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What’s Next?

We show how to handle lightk colorings…

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Separated colors

Sep(v,c) – Number of connected components containing c in T-{v}.

For a vertex v, let si=sep(v,i). Assume wlog si≥si+1.

3

3

22

1

11

4

v

Sep(v,1)=3

Sep(v,3)=2

S=(3,2,2,1)

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In a Light Coloring,for any v: s1≤deg(v) s2≤k-1 s3≤3 s4≤2 sk≤1

Lemma 1

Trivial. There are at most deg(v) components.

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Otherwise, s1≥s2≥k.

Not a light coloring. Contradiction.

For every v: s1≤deg(v) s2≤k-1 s3≤3 s4≤2 sk≤1

Lemma 1 (cont.)

1 21 2

1 2

K nodes

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Otherwise, s1≥s2≥s3≥4.

Not a light coloring. Contradiction.

For every v: s1≤deg(v) s2≤k-1 s3≤3 s4≤2 sk≤1

Lemma 1 (cont.)

1 21 21 2

3 3 3

1 2

3

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Otherwise, s1≥s2≥s3≥s4≥3.

Not a light coloring. Contradiction.

For every v: s1≤deg(v) s2≤k-1 s3≤3 s4≤2 sk≤1

Lemma 1 (cont.)

1 21 21 2

3 3 3

4 4 4

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Otherwise, s1≥s2≥…≥sk≥2.

Not a light coloring. Contradiction.

For every v: s1≤deg(v) s2≤k-1 s3≤3 s4≤2 sk≤1

Lemma 1 (cont.)

1 1

2 2

k k

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Conclusion We would like to assign colors to

connected components. We will assign each color only to

components containing it.

1 2

3 ...

Deg

(v)

1 2

3 …

k-1

1 2

3

1 2

1 2..

1 2

1k 25 4 3…

#

2 deg

i

k

Part s

O k v

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Definitions

Color(v) – The color of v. NewColor(v) – The new color of v. OutDeg(v) – The outgoing degree

of v. vi – The i’th son of v (0=father). Subtree(v) – The subtree rooted at

v.

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Observation 1

v

vjvi

NewColor v i jNC SubTreeN vC SubTree v

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Conclusion 2

v

v2v1

NewColor(v)= ,

NewColor(v2)≠

c

c

No c No c No c

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Dynamic Programming - Idea

v

Recoloring subtree(v)

Sum cost of subtrees

Add cost of root

Recoloring subtree(v)

Sum cost of subtrees

Add cost of root

Reserved Colors

Allowed Colors

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Dynamic Programming

OPT(v) - the optimal recoloring of Subtree(v), while are reserved colors.

OPT,c(v) - the optimal recoloring of Subtree(v), while are reserved colors, and NewColor(v)=c.

Observation:

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Cost of recoloring a subtree

- cost of recoloring subtree(vi) while i are reserved colors and NewColor(v)=c.

v

vi

Case 1:

Case 2:

v

vi

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Dynamic Programming – Cont.

Lemma: It is enough to scan

partitions.

2 degkO k v

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Summary

Total time complexity:

Setting we get: Approximation Ratio:

Time Complexity:

2 22kO k n n

log 2k n

2O n

82 2 1

logo

n

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Thank you!