pdmw 2011 rick brewster, ben moore, sean mcguinness, jon ... · rick brewster, ben moore, sean...

The complexity of circular recolouringsPDMW 2011

Rick Brewster, Ben Moore, Sean McGuinness, Jon Noel

Aug 9, 2015

Rick Brewster, Ben Moore, Sean McGuinness, Jon NoelThe complexity of circular recolourings Aug 9, 2015 1 / 34

Galena Trail, New Denver, BC

Photo credit: Our Mountain Bike Vacations: www.mtb-hol.com

Fox, chicken, seed

A collection of configurations : 24 − 6 = 10.

Starting configuration, ending configuration

Construct a graph with vertices as configurations and edges asallowed moves between.

Find the shortest path between starting and ending configs.

Another popular reconfig problem

About 43× 1019 configurations.

1995 Michael Reid finds a 20 move path [Cube20.org]

2010 a group using 35-CPU-years of donated by Google shows 20moves always suffice.

Reconfig Questions

Many puzzles are examples of reconfiguration problems (E. Demaine)

Questions one might ask:I Can we get from Config A to Config B?I Can we get to all configs?I Can we change one object?

We are interested in how the computational complexity grows withincreasing size.

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

Recolouring

Inital Colouring

Final Colouring

The graph L4

Example

χ = 2

col = 4

Colouring number

The colouring number is col(G ) = maxH⊆G{δ(H) + 1} and

χ(G ) ≤ χ`(G ) ≤ col(G )

Easily find a col(G )-colouring, but col(G ) can be far from χ(G ).

The graph L4

Example

χ = 2

col = 4

Colouring number

χ(G ) ≤ χ`(G ) ≤ col(G )

The graph L4

Example

χ = 2

col = 4

Colouring number

χ(G ) ≤ χ`(G ) ≤ col(G )

The graph L4

Example

χ = 2

col = 4

Colouring number

χ(G ) ≤ χ`(G ) ≤ col(G )

The graph L4

Example

χ = 2

col = 4

Colouring number

χ(G ) ≤ χ`(G ) ≤ col(G )

The graph L4

Example

χ = 2

col = 4

Colouring number

χ(G ) ≤ χ`(G ) ≤ col(G )

Reconfiguration for colourings

Given G can we generate all k-colourings recolouring one vertex at atime?

Given G and two colourings f , g , can we reconfig f to g?

Given G , a colouring, and a vertex v , can we change the colour of v?

The colour graph

A framework to capture mixing

Given a graph G and an nonnegative integer k.

Ck(G ) is the graph withI V the set of all k-colourings of G ,I f ∼ g if f and g differ on exactly one vertex

G is k-mixing iff Ck(G ) is connected.

f reconfigs to g if same component.

The colour graph

Theorem (Cereceda, van den Heuvel, and Johnson)

For any graph G , G is k-mixing for k ≥ col(G ) + 1.

Theorem (Choo and MacGillivray)

The graph Ck(G ) is hamiltonian for k ≥ col(G ) + 2.

The colour graph

Theorem (Cereceda, van den Heuvel, and Johnson)

For any graph G , f reconfigs to g if

Same fixed vertices

Same weighted cycles

Same weighted paths between fixed vertices

The graph C3(L4)

Non-surjective colourings from C2(L4)

Partition is refinement of non-surjective

New partition

The graph C3(L4)

New partition

The graph C3(L4)

New partition

Computational complexity k-colouring

k-Colouring (k fixed)

Instance: A graph G .

Question: Does G admit a k-colouring?

k = 1: trivial.

k = 2: YES iff G has no odd cycles. (Theorem from Graph Theory101)

k ≥ 3: NP-complete.

For k ≥ col(G ) we can answer YES.

k = 1: trivial.

Complexity k-recolouring

The k-Recolouring problem

Instance Graph plus two k-colourings: (G , f , g).

Question Does f recolour to g?

k-Recolouring complexity

k = 2 k = 3 k ≥ 4Trivial P – Weighted cycles PSPACE-complete

Bonsma, Cereceda, van den Heuvel, Johnson

Is there 3-colouring? NP-complete

Given a 3-colouring f can all others be obtained by recolouring fromf ? coNP-complete restricted to bipartite graphs

Given two: Can g be obtained from f ? Polynomial

Circular cliques

Definition

Circular clique Gp,q has V = Zp; uv ∈ E iff q ≤ |u − v | ≤ p − q.

Circular colouring

A (p, q)-colouring of G is an assignment of p-colours s.t. adjacentvertices in G receive adjacent colours in Gp,q.

χc(G ) = min{p/q : G → Gp,q}; χ(G ) = dχc(G )e

Circular cliques

Definition

Circular clique Gp,q has V = Zp; uv ∈ E iff q ≤ |u − v | ≤ p − q.

Circular colouring

A (p, q)-colouring of G is an assignment of p-colours s.t. adjacentvertices in G receive adjacent colours in Gp,q.

χc(G ) = min{p/q : G → Gp,q}; χ(G ) = dχc(G )e

Computational complexity

(p, q)-Colouring (p, q fixed)

Question: Does G admit a (p, q)-colouring?

p/q < 2: trivial.

p/q = 2: YES iff G has no odd cycles. (Theorem from Graph Theory101)

p/q > 2: NP-complete.

p/q < 2: trivial.

Our starting point

Questions

Can we find dichotomy for (p, q)-Recolouring? Can we find hardcolouring problems that have tractable recolouring problems?

Recolouring complexity

k = 2 k = 3 k ≥ 4p/q = 2 2 < p/q < 4 p/q ≥ 4Trivial P – Weighted cycles PSPACE-complete

Our starting point

Questions

Theorem (BMMN)

The (p, q)-recolouring problem is:

polynomial time solvable if p/q < 4; and

PSPACE-complete if p/q ≥ 4.

Our starting point

Questions

Recolouring C5 when k = 3

Observation

Vertex recolourable if a colour missing from neighbourhood.

Initially v might be not be recolourable, but becomes recolourablewhen a vertex in its neighbourhood is recoloured.

Coded through directed edges.

Observation

No vertex can be recoloured.

We call such vertices fixed.

Observation

f does not recolour to g .

Necessary condition Ff = Fg . (fixed vertices)

To find fixed vertices in poly time, recursively delete sources and sinks(and isolates).

Observation

Finding fixed vertices when k = 3

Fixed vertices. Strongly connected components plus directed pathsbetween.

(7, 2)-Recolouring

Suppose f (v) = 0 and v has neighbours coloured {3, 4, 5}.

Pictorially,

The allowed colours for v are {0, 1}.

(7, 2)-Recolouring

Suppose f (v) = 0 and v has neighbours coloured {3, 4, 5}.Pictorially,

(7, 2)-Recolouring

Allowed colours for (p, q).

Allowed cols

When p/q < 4 the allowed colours form an interval.

v cannot be recolour iff neighbours with colours f (v)− q andf (v) + q.

We can find fixed vertices in polynomial time.I f is a (p, q)-colouringI Direct arc uv if f (v) = f (u) + q (nothing otherwise).I Recursively delete sources, sinks, and isolates.

Allowed colours for (p, q).

Allowed cols

When p/q < 4 the allowed colours form an interval.

v cannot be recolour iff neighbours with colours f (v)− q andf (v) + q.

We can find fixed vertices in polynomial time.I f is a (p, q)-colouringI Direct arc uv if f (v) = f (u) + q (nothing otherwise).I Recursively delete sources, sinks, and isolates.

Weighting cycles

(3− 2) + (3− 2)+

1 + 1 + (3− 1)

(3− 1) + (3− 1)+

2 + 2 + (3− 2)

Weighting

Assign wt(uv) = f (v)− f (u) mod q (wrt orientation).

wt(C , f ) = 6 and wt(C , g) = 9.

Recolouring does not change the weight.

Weighting cycles

(3− 2) + (3− 2)+

1 + 1 + (3− 1)

(3− 1) + (3− 1)+

2 + 2 + (3− 2)

Weighting

wt(C , f ) = 6 and wt(C , g) = 9.

Weighting cycles

(3− 2) + (3− 2)+

1 + 1 + (3− 1)

(3− 1) + (3− 1)+

2 + 2 + (3− 2)

Weighting

wt(C , f ) = 6 and wt(C , g) = 9.

A matroid view

Vertex colouring + orientation give group labelling of the edges:ϕ(evu) = f (u)− f (v). (Zp)

Reconfig step is to relabel a cut ∂(X ) by −α on ∂+(X ) and by +αon ∂−(X ).

Equivalent to changing the colour of all vertices in X by +α.

Two edge-labellings can be reconfig iff all cycles have the sameweight.

A matroid view

Connecting vertex cols and edge labels

1-1 correspondence between (p, q)-colourings and Zp-edge-labellingswhere

I edges have weight q, q + 1, . . . , p − qI cycles have total weight 0 (mod p)

Recolouring a vertex is a special case of relabelling an edge cut.

Colour graph is a spanning subgraph of edge-labelling graph.

The Algorithm

Recolouring algorithm

Input: (G , f , g)Output: Recolouring sequence or a no-certificate.

1 Construct edge-labellings of G : ϕf and ϕg .

2 Find fixed vertices. Verify Ff = Fg or answer NO with certificate.

3 Find ∂(X ) to relabel ϕf closer to ϕg or discover a cycle C withwt(C , ϕf ) 6= wt(C , ϕg ). Answer NO.

4 Realize edge relabel as a sequence of vertex recols.1 Is G [X ] directed cycle free? If yes, topologically sort and increase the

colour of each vertex by 1.2 Else if G [X ] is directed cycle free, use X .3 Else we have discovered a path wt(P, ϕf ) 6= wt(P, ϕg ). Answer NO.

5 Repeat 3-4 until ϕf = ϕg .

6 If there are no fixed vertices shift all vertices by 1 until f (v) = g(v)for some vertex v .

The characterization

Theorem

Let G be a graph and f , g be two (p, q)-colourings with p/q < 4. Then frecolours to g iff

1 Ff = Fg ,

2 For all cycles C , W (C , f ) = W (C , g), and

3 For all paths P whose end points are fixed, W (P, f ) = W (P, g).

A beautiful application

Theorem (Wrochna)

If G has no cycle of length 0 mod 3, then G is 3-colourable.

Proof.

Let G be a minimal counter-example with edge uv . Then G − uv has a3-colouring f . Clearly f (u) = f (v).Let g(x) = f (x) + 1. Neither g nor f have fixed vertices. The weight ofall cycles and paths are equal. Therefore f recolours to g .At some point u is increased by 1 (before v is changed). Gives a3-colouring of G , a contraction.

First proved by: Chudnovsky, Plumettaz, Scott and Seymour.

Theorem (Wrochna)

Proof.

Theorem (Wrochna)

Proof.

Monochromatic neighbourhood (Wrochna)

Concluding remarks on Algorithm

CHJ fixed vertices, weighted cycles and paths. k = 3.

BMMN extend ideas to (p, q)-colourings. p/q < 4.

Wrochna extends to H-colourings. Monochromatic neighbourhoodproperty.

Topology and homotopy the right idea, but in HOM(G ,H).

PSPACE-completeness

Theorem (Bonsma, Cereceda 2009)

For every k ≥ 4, the decision problem k-Colour Path is PSPACE-complete.Moreover, it remains PSPACE-complete for the following restrictedinstances:

1 bipartite graphs and any fixed k ≥ 4;

2 planar graphs and any fixed 4 ≤ k ≤ 6; and

3 bipartite planar graphs and k = 4.

Reduction is from a token sliding problem.

PSPACE-completeness for (p, q)

Theorem (BMMN)

If p/q ≥ 4, then (p, q)-Recolouring is PSPACE-complete.

Proof.

By example with (9, 2).

(G , f , g) instance of 4-Recolouring.

(G ′, f ′, g ′), instance of (9, 2)-Recolouring, is constructed by putting a(0, 2)-forbidding path on each edge uv .

Confirm recolouring G if and only if recolouring G ′.

(0, 2)-forbidding paths

G is instance of 4-Recolouring.

For each uv in G add u − {8, 0, 1, 2, 3} − {2, 3, 4, 5} − {4, 5, 6, 7} −{6, 7, 8, 0} − {8, 0, 1, 2, 3} − v to form G ′ instance of(9, 2)-Recolouring.

If u receives colour 0, then v cannot receive colour 2.

Each 02-free (9, 2) colouring yields a 4-colouring and vice versa.

Other hardness results

3-mixing is coNP-complete for bipartite graphs. Certificate is afolding of G down to C6. Similar partial results for p/q < 4.

Quesiton: k-mixing is PSPACE-complete for k ≥ 4?

Indicator construction std tool in homomorphism reductions. Proofthat reconfig homs to odd wheels is PSPACE-complete.

Thank you!

pdmw 2011 rick brewster, ben moore, sean mcguinness, jon ... · rick brewster, ben moore, sean...

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