Reconfiguration: From statistical physics to graph theory Nicolas - - PowerPoint PPT Presentation
Reconfiguration: From statistical physics to graph theory Nicolas Bousquet Joint works with Marthe Bonamy, Carl Feghali, Matthew Johnson, Guillem Perarnau. Journ ees Structures Discr` etes ENS Lyon 1/22 Spin systems Spin is one of two
Nicolas Bousquet
Joint works with Marthe Bonamy, Carl Feghali, Matthew Johnson, Guillem Perarnau.
Journ´ ees Structures Discr` etes ENS Lyon
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Spin is one of two types of angular mo- mentum in quantum mechanic. [...] In some ways, spin is like a vector quan- tity ; it has a definite magnitude, and it has a “direction”.
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Spin is one of two types of angular mo- mentum in quantum mechanic. [...] In some ways, spin is like a vector quan- tity ; it has a definite magnitude, and it has a “direction”.
Usually, spins take their value in {+, −}, but sometimes the range is larger...
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Spin is one of two types of angular mo- mentum in quantum mechanic. [...] In some ways, spin is like a vector quan- tity ; it has a definite magnitude, and it has a “direction”.
Usually, spins take their value in {+, −}, but sometimes the range is larger... A spin configuration is a function σ : S → {1, . . . , k}.
modelized via an interaction matrix.
with a graph :
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Spin is one of two types of angular mo- mentum in quantum mechanic. [...] In some ways, spin is like a vector quan- tity ; it has a definite magnitude, and it has a “direction”.
Usually, spins take their value in {+, −}, but sometimes the range is larger... A spin configuration is a function σ : S → {1, . . . , k}.
modelized via an interaction matrix.
with a graph :
Spin configuration ⇒ (non necessarily proper) graph coloring.
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2 4 T = 5, 1, 0.2, 0.05
H(σ) : number of monochromatic edges. = Edges with both endpoints of the same color. Gibbs measure at fixed temperature T : νT(σ) = e− H(σ)
T 3/22
2 4 T = 5, 1, 0.2, 0.05
H(σ) : number of monochromatic edges. = Edges with both endpoints of the same color. Gibbs measure at fixed temperature T : νT(σ) = e− H(σ)
T
Important points to notice :
colorings.
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2 4 T = 5, 1, 0.2, 0.05
H(σ) : number of monochromatic edges. = Edges with both endpoints of the same color. Gibbs measure at fixed temperature T : νT(σ) = e− H(σ)
T
Important points to notice :
colorings.
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2 4 T = 5, 1, 0.2, 0.05
H(σ) : number of monochromatic edges. = Edges with both endpoints of the same color. Gibbs measure at fixed temperature T : νT(σ) = e− H(σ)
T
Important points to notice :
colorings.
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2 4 T = 5, 1, 0.2, 0.05
H(σ) : number of monochromatic edges. = Edges with both endpoints of the same color. Gibbs measure at fixed temperature T : νT(σ) = e− H(σ)
T
Important points to notice :
colorings.
Limit of a k-state Potts model when T → 0. ⇔ Only proper colorings have positive measure. Definition (Glauber dynamics)
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In the statistical physics community, the following Monte Carlo Markov chain was proposed to sample a configuration :
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In the statistical physics community, the following Monte Carlo Markov chain was proposed to sample a configuration :
Questions :
random” ?
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Definition (k-Reconfiguration graph Ck(G) of G) All along the talk k denotes the number of colors.
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Definition (k-Reconfiguration graph Ck(G) of G) All along the talk k denotes the number of colors. Remark 1. Two colorings equivalent up to color permutation are distinct.
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Definition (k-Reconfiguration graph Ck(G) of G) All along the talk k denotes the number of colors. Remark 1. Two colorings equivalent up to color permutation are distinct.
Remark 2. All the k-colorings can be generated ⇔ The k-reconfiguration graph is connected.
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A Markov chain is irreducible if any solution can be reached from any other. ⇔ The reconfiguration graph is connected. A chain is aperiodic if there exists t0 such that Pr(Xt = a) is positive for every t > t0 and every state a.
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A Markov chain is irreducible if any solution can be reached from any other. ⇔ The reconfiguration graph is connected. A chain is aperiodic if there exists t0 such that Pr(Xt = a) is positive for every t > t0 and every state a. Every ergodic (aperiodic and irreducible) Markov chain converges to a unique stationnary distribution. Theorem
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A Markov chain is irreducible if any solution can be reached from any other. ⇔ The reconfiguration graph is connected. A chain is aperiodic if there exists t0 such that Pr(Xt = a) is positive for every t > t0 and every state a. Every ergodic (aperiodic and irreducible) Markov chain converges to a unique stationnary distribution. Theorem In our case :
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A Markov chain is irreducible if any solution can be reached from any other. ⇔ The reconfiguration graph is connected. A chain is aperiodic if there exists t0 such that Pr(Xt = a) is positive for every t > t0 and every state a. Every ergodic (aperiodic and irreducible) Markov chain converges to a unique stationnary distribution. Theorem In our case :
stationnary distribution is uniform.
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A Markov chain is irreducible if any solution can be reached from any other. ⇔ The reconfiguration graph is connected. A chain is aperiodic if there exists t0 such that Pr(Xt = a) is positive for every t > t0 and every state a. Every ergodic (aperiodic and irreducible) Markov chain converges to a unique stationnary distribution. Theorem In our case :
stationnary distribution is uniform. Question : How much time do we need to converge ?
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Mixing time = number of steps we need to be sure that we are “close” to the stationnary distribution. ⇔ Number of steps needed to guarantee that the solutions is sampled “almost” at random.
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Mixing time = number of steps we need to be sure that we are “close” to the stationnary distribution. ⇔ Number of steps needed to guarantee that the solutions is sampled “almost” at random. A chain is rapidly mixing if its mixing time is polynomial (and even better O(n log n).
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Mixing time = number of steps we need to be sure that we are “close” to the stationnary distribution. ⇔ Number of steps needed to guarantee that the solutions is sampled “almost” at random. A chain is rapidly mixing if its mixing time is polynomial (and even better O(n log n).
Mixing time and Reconfiguration graph ?
⇒ Mixing time ≥ 2 · D.
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Mixing time = number of steps we need to be sure that we are “close” to the stationnary distribution. ⇔ Number of steps needed to guarantee that the solutions is sampled “almost” at random. A chain is rapidly mixing if its mixing time is polynomial (and even better O(n log n).
Mixing time and Reconfiguration graph ?
⇒ Mixing time ≥ 2 · D.
reconfiguration graph.
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors.
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors. Known results :
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors. Known results :
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors. Known results :
6 ∆.
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors. Known results :
6 ∆.
polynomial if c = ( 11
6 − ǫ)∆.
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How many colors (in terms of the maximum degree ∆) do we need to ensure that the chain is rapidly mixing ? We denote by c(∆) the number of colors. Known results :
6 ∆.
polynomial if c = ( 11
6 − ǫ)∆.
If c ≥ ∆ + 2, the graph is c(∆)-mixing in time O(n log n). Conjecture
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Informal definition Two Markov chains (Xt, Yt) are coupled if :
transitions in Yt might depend on transitions of Xt).
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Informal definition Two Markov chains (Xt, Yt) are coupled if :
transitions in Yt might depend on transitions of Xt). If there exists a coupling defined only every Xt, Yt that only differ
E(d(Xt+1, Yt+1)) < (1 − 1 n) then the mixing time is O(n log n). Theorem d(X, Y )= Hamming distance = number of vertices on which they differ
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Let v be the vertex on which Xt and Yt differ. Coupling : If vertex u and color c are chosen in Xt, we make the same choice in Yt.
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Let v be the vertex on which Xt and Yt differ. Coupling : If vertex u and color c are chosen in Xt, we make the same choice in Yt. Analysis : Assume that k ≥ 3∆ + 1. P(d(Xt+1, Yt+1) = 0) > 1
n · 2∆+1 3∆+1.
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Let v be the vertex on which Xt and Yt differ. Coupling : If vertex u and color c are chosen in Xt, we make the same choice in Yt. Analysis : Assume that k ≥ 3∆ + 1. P(d(Xt+1, Yt+1) = 0) > 1
n · 2∆+1 3∆+1.
P(d(Xt+1, Yt+1) = 2) ≤ ∆
n · 2 3∆+1
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Let v be the vertex on which Xt and Yt differ. Coupling : If vertex u and color c are chosen in Xt, we make the same choice in Yt. Analysis : Assume that k ≥ 3∆ + 1. P(d(Xt+1, Yt+1) = 0) > 1
n · 2∆+1 3∆+1.
P(d(Xt+1, Yt+1) = 2) ≤ ∆
n · 2 3∆+1
⇒ E(d(Xt+1, Yt+1) = 1 −
1 (3∆+1)n.
⇒ The chain is rapidly mixing if k > 3∆.
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v = Unique vertex on which Xt and Yt differ. Coupling :
⇒ Choose v and c′ color of v in Yt.
⇒ Choose v and c color of v in Xt in Yt.
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v = Unique vertex on which Xt and Yt differ. Coupling :
⇒ Choose v and c′ color of v in Yt.
⇒ Choose v and c color of v in Xt in Yt.
Analysis : Assume that k ≥ 2∆ + 1. P(d(Xt+1, Yt+1) = 0) > 1
n · ∆+1 2∆+1.
P(d(Xt+1, Yt+1) = 2) ≤ ∆
n · 1 2∆+1.
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v = Unique vertex on which Xt and Yt differ. Coupling :
⇒ Choose v and c′ color of v in Yt.
⇒ Choose v and c color of v in Xt in Yt.
Analysis : Assume that k ≥ 2∆ + 1. P(d(Xt+1, Yt+1) = 0) > 1
n · ∆+1 2∆+1.
P(d(Xt+1, Yt+1) = 2) ≤ ∆
n · 1 2∆+1.
⇒ E(d(Xt+1, Yt+1) = 1 −
1 (2∆+1)n.
⇒ The chain is rapidly mixing if k > 2∆.
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If the reconfiguration is connected then there exists a polynomial delay algorithm that enumerate all the solutions. Theorem An algorithm is polynomial delay if it enumerates all the solutions and the delay between two solutions is polynomial in n.
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If the reconfiguration is connected then there exists a polynomial delay algorithm that enumerate all the solutions. Theorem An algorithm is polynomial delay if it enumerates all the solutions and the delay between two solutions is polynomial in n. Sketch of the proof :
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If the reconfiguration is connected then there exists a polynomial delay algorithm that enumerate all the solutions. Theorem An algorithm is polynomial delay if it enumerates all the solutions and the delay between two solutions is polynomial in n. Sketch of the proof :
Remark : The algorithm might need an exponential space !
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Is the reconfiguration graph connected ?
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Is the reconfiguration graph connected ?
Given two vertices of the reconfiguration graph, are they in the same connected component ?
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Is the reconfiguration graph connected ?
Given two vertices of the reconfiguration graph, are they in the same connected component ?
What is the diameter of the reconfiguration graph ?
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Is the reconfiguration graph connected ?
Given two vertices of the reconfiguration graph, are they in the same connected component ?
What is the diameter of the reconfiguration graph ?
algorithmic point of view) ? Can we find a path between two vertices of the reconfiguration graph in polynomial time ?
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) A graph is k-degenerate if there exists an order v1, . . . , vn such that for every i, vi has at most k neighbors after it in the order.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) A graph is k-degenerate if there exists an order v1, . . . , vn such that for every i, vi has at most k neighbors after it in the order.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) A graph is k-degenerate if there exists an order v1, . . . , vn such that for every i, vi has at most k neighbors after it in the order.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) A graph is k-degenerate if there exists an order v1, . . . , vn such that for every i, vi has at most k neighbors after it in the order.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) A graph is k-degenerate if there exists an order v1, . . . , vn such that for every i, vi has at most k neighbors after it in the order. The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07)
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07)
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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The (k + 2)-recoloring diameter of any k-degenerate graph is at most n · (k + 1)n. Theorem (Cereceda ’07) Any (k + 2)-coloring can be transformed into any other by reco- loring at most (k + 1)n times each vertex. Lemma
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When do we need to recolor the leftmost vertex ?
(k + 1)n−1 (k + 1)n−1 (k + 1)n−1
⇒ k · (k + 1)n−1 recolorings.
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When do we need to recolor the leftmost vertex ?
(k + 1)n−1 (k + 1)n−1 (k + 1)n−1
⇒ k · (k + 1)n−1 recolorings.
The total number of recolorings is at most (k + 1)n.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda)
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
For other values of k ?
d + 1
d-degenerate Chordal graphs
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The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
For other values of k ?
d + 1
d-degenerate Chordal graphs
Not Conn. Not Conn. 17/22
The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
For other values of k ?
d + 1
d-degenerate Chordal graphs
Not Conn. Not Conn.
O(n2) d + 2 17/22
The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
For other values of k ?
d + 1
d-degenerate Chordal graphs
Not Conn. Not Conn.
O(n2) O(dn) O(dn) 2d + 2 d + 2 17/22
The (k + 2)-recoloring diameter of any k-degenerate graph is O(n2). Conjecture (Cereceda) Known results :
[B., Perarnau] k = 8 works.
For other values of k ?
d + 1
d-degenerate Chordal graphs
Not Conn. Not Conn.
O(n2) O(dn) O(dn)
2d + 2 d + 2 17/22
The 3-recoloring diameter of the path Pn is Ω(n2). Theorem (Bonamy et al.)
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The 3-recoloring diameter of the path Pn is Ω(n2). Theorem (Bonamy et al.) Sketch of the proof
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The 3-recoloring diameter of the path Pn is Ω(n2). Theorem (Bonamy et al.) Sketch of the proof
Claim : A recoloring performs the following :
a a + 1 a a a − 1 a
⇒ The surface is only modified by “one” at each step.
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The 3-recoloring diameter of the path Pn is Ω(n2). Theorem (Bonamy et al.) Sketch of the proof
Claim : A recoloring performs the following :
a a + 1 a a a − 1 a
⇒ The surface is only modified by “one” at each step. Claim : Ω(n2) steps are needed to transform 123....123 into 132....132.
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Impossible to sample - count (∆ + 1)-colorings ?
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Impossible to sample - count (∆ + 1)-colorings ? A coloring is frozen if all the colors appear in the (closed) neighborhood of all the vertices.
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Impossible to sample - count (∆ + 1)-colorings ? A coloring is frozen if all the colors appear in the (closed) neighborhood of all the vertices. Theorems :
exists between any pair of non-frozen (∆ + 1)-colorings (when ∆ ≥ 3).
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Impossible to sample - count (∆ + 1)-colorings ? A coloring is frozen if all the colors appear in the (closed) neighborhood of all the vertices. Theorems :
exists between any pair of non-frozen (∆ + 1)-colorings (when ∆ ≥ 3).
(∆ + 1)-colorings is exponentially smaller than the number of (∆ + 1)-colorings (when ∆ is small enough).
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At the beginning of the talk (a long time ago...). “At each step, change the color of a single vertex”
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At the beginning of the talk (a long time ago...). “At each step, change the color of a single vertex” Question : Why a single vertex ?
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At the beginning of the talk (a long time ago...). “At each step, change the color of a single vertex” Question : Why a single vertex ?
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At the beginning of the talk (a long time ago...). “At each step, change the color of a single vertex” Question : Why a single vertex ?
Question 2 : Can we imagine another rule ?
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At the beginning of the talk (a long time ago...). “At each step, change the color of a single vertex” Question : Why a single vertex ?
Question 2 : Can we imagine another rule ?
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We can generate all the (∆ + 1)-colorings using Kempe chains. Theorem (Mohar)
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We can generate all the (∆ + 1)-colorings using Kempe chains. Theorem (Mohar) We can generate all the ∆-colorings of any graph using Kempe chains. Conjecture (Mohar)
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We can generate all the (∆ + 1)-colorings using Kempe chains. Theorem (Mohar) We can generate all the ∆-colorings of any graph except the 3-prism using Kempe chains. Theorem (Bonamy, B., Feghali, Johnson 2017+)
2 3 1 2 1 3 1 2 3 2 1 3
Counter-example proposed by Jan van den Heuvel (2013)
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Relation with other fields :
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Relation with other fields :
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Relation with other fields :
recoloring ?
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Relation with other fields :
recoloring ?
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Relation with other fields :
recoloring ?
More questions :
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Relation with other fields :
recoloring ?
More questions :
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