The 25th International Symposium on Algorithms and Computation (ISAAC 2014)

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Polynomial-Time Algorithm for

Sliding Tokens on Trees

December 15 - 17, 2014

Erik D. DemaineMIT, USA

Martin L. DemaineMIT, USA

Eli Fox-EpsteinBrown University, USA

Duc A. HoangJAIST, Japan

Ryuhei UeharaJAIST, Japan

Takeshi YamadaJAIST, Japan

Yota OtachiJAIST, Japan

Hirotaka OnoKyushu University, Japan

Takehiro ItoTohoku University, Japan

Reconfiguration Problems

The problem arises when we want to find a step-by-step transformation between two feasible solutions of a problem such that all intermediate results are also feasible.

Many kind of reconfiguration problems have been studied recently, including the independent set reconfiguration problem (ISRECONF).

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van den Heuvel, J.: The complexity of change. Surveys in Combinatorics 2013,London Mathematical Society Lecture Notes Series 409 (2013)

Independent set

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Let be a graph with vertices, edges. A set of vertices is independent if for any two vertices , .

An independent set

Not an independent set

Independent set reconfiguration problem (ISRECONF)

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Two independent sets with . A token (coin) is placed at each vertex of .

Reconfiguration ruleso Token Sliding (TS)

• A token can be moved to one of its neighbors.o Token Jumping (TJ)

• A token can jump from one vertex to another vertex.o Token Addition and Removal (TAR)

• Add or remove some tokens, but there is always at least tokens.

GIVEN

PROBLEMCan we reconfigure to using one of the given rules such that all intermediate sets of tokens are independent?

ISRECONF under TS rule

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Let be two independent sets of a graph .

𝐈 𝐈 ′

YES (why?)

|𝐈|=¿𝐈 ′∨¿

Question: Can we reconfigure to using TS rule such that all intermediate sets of tokens are independent?

ISRECONF under TS rule

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Let be two independent sets of a graph .𝐈 𝐈 ′

Question: Can we reconfigure to using TS rule such that all intermediate sets of tokens are independent?

NO (why?)

|𝐈|=¿𝐈 ′∨¿

ISRECONF under TS rule

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general

perfect

chordal

tree

interval

distance-hereditary (D.H)

Ptolemaic

block

bipartite

trivially perfectproper interval

bipartite D.H.

planar

PSPACE-completePolynomialOpen

A B B is a subclass of A

Our Result

P, NP or PSPACE?Hearn and Demaine (2005)

Kaminski et. al. (2012)

Ito et. al. (2011)

Demaine et. al. (2014)

Why study the problem for trees?

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o Trees (connected + no cycle) are simple. Almost every problem can be solved in polynomial time for trees. This also holds for the ISRECONF problem.

o Kaminski et. al. (2012) gave a linear time algorithm for even-hole-free graphs (included trees) under TJ and TAR rules.• The answer is always YES.• Tokens never make detours.

Not for TS rule.

Key concept – Rigid tokens

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-rigid tokens -movable tokens

Claim: • All rigid tokens can be determined in time.• If there are no -rigid and -rigid tokens, can be

reconfigured to .

Intuitively, a token is rigid if there is no way to slide .

Non-trivial

Reconfigure non-rigid tokens

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Let be two independent sets of a graph .𝐈 𝐈 ′ |𝐈|=¿𝐈 ′∨¿

?o Order of sliding.o Detour.

Idea: o Find a safe degeree-1 vertex .o Move a red and a black token to .o Remove and its unique neighbor.o Repeat the steps.

Polynomial-Time Algorithm

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Step 1: Find all -rigid and -rigid tokens. If the two sets are different, return NO. Otherwise go to Step 2.Step 2: Delete all rigid tokens and their neighbors. Compare the number of tokens in each component of the obtained forest. If they are the same, return YES. Otherwise, return NO.𝐈 𝐈 ′

= 5

NO YES

Open Problems

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general

perfect

chordal

tree

interval

distance-hereditary (D.H)

Ptolemaic

block

bipartite

trivially perfectproper interval

bipartite D.H.

planar

PSPACE-completePolynomialOpen

A B B is a subclass of A

P, NP or PSPACE?Hearn and Demaine (2005)

Kaminski et. al. (2012)

Ito et. al. (2011)

Demaine et. al. (2014)