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On the convergence of Boolean automata networks without negative cycles Tarek Melliti and Damien Regnault e d IBISC - Universit Evry Val dEssonne, France Adrien Richard I3S - Universit e de Nice-Sophia Antipolis, France Sylvain

SLIDE 1

On the convergence of Boolean automata networks without negative cycles

Tarek Melliti and Damien Regnault

IBISC - Universit´ e d’´ Evry Val d’Essonne, France

Adrien Richard

I3S - Universit´ e de Nice-Sophia Antipolis, France

Sylvain Sen´ e

LIF - Universit´ e d’Aix-Marseille, France

Gießen, September 19, 2013

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 1/20

SLIDE 2

Boolean networks

= Finite and heterogeneous CAs on {0, 1}

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 2/20

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Boolean networks

= Finite and heterogeneous CAs on {0, 1} Classical models for Neural networks [McCulloch & Pitts 1943] Gene regulatory networks [Kauffman 1969, Tomas 1973]

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 2/20

SLIDE 4

Focus on interaction graphs

1 2 3 4 5 6 7 8 9 10 11 Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 3/20

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Focus on interaction graphs

1 2 3 4 5 6 7 8 9 10 11

Question

What can be said on the dynamics of a Boolean network according to its interaction graph ?

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 3/20

SLIDE 6

Focus on interaction graphs

1 2 3 4 5 6 7 8 9 10 11

[Arabidopsis Thaliana]

Question

What can be said on the dynamics of a Boolean network according to its interaction graph ? Application to gene networks: reliable information on the interaction graph only.

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 3/20

SLIDE 7

Definitions

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 4/20

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Setting

There are n components (cells) denoted from 1 to n The set of possible states (configurations) is {0, 1}n The local transition function of component i ∈ [n] is any map fi : {0, 1}n → {0, 1} The resulting global transition function is f : {0, 1}n → {0, 1}n, f(x) = (f1(x), . . . , fn(x))

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 5/20

SLIDE 9

Setting

There are n components (cells) denoted from 1 to n The set of possible states (configurations) is {0, 1}n The local transition function of component i ∈ [n] is any map fi : {0, 1}n → {0, 1} The resulting global transition function is f : {0, 1}n → {0, 1}n, f(x) = (f1(x), . . . , fn(x)) We consider the fully-asynchronous updating ֒ → very usual in the context of gene networks [Thomas 73]

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 5/20

SLIDE 10

Given a map v : N → [n], the fully-asynchronous dynamics is xt+1

v(t) = fv(t)(xt),

xt+1

i

= xt

i

∀i = v(t)

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 6/20

SLIDE 11

Given a map v : N → [n], the fully-asynchronous dynamics is xt+1

v(t) = fv(t)(xt),

xt+1

i

= xt

i

∀i = v(t) In practice, non information on v... → we regroup all the possible asynchronous dynamics under the form of a directed graph

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 6/20

SLIDE 12

Given a map v : N → [n], the fully-asynchronous dynamics is xt+1

v(t) = fv(t)(xt),

xt+1

i

= xt

i

∀i = v(t) In practice, non information on v... → we regroup all the possible asynchronous dynamics under the form of a directed graph

Definition

The asynchronous state graph of f, denoted by ASG(f), is the directed graph on {0, 1}n with the following set of arcs:

x → ¯

xi | x ∈ {0, 1}n, i ∈ [n], xi = fi(x)

Melliti, Regnault, Richard, Sen´

e Convergence of Boolean networks without negative cycles Automata 2013 6/20

SLIDE 13

Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

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Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

SLIDE 15

Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

SLIDE 16

Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

SLIDE 17

Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

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Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

The attractors of ASG(f) are its terminal strong components

Attractor of size one = fixed point
Attractor of size at least two = cyclic attractor

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

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Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 ASG(f)

000 001 010 011 100 101 110 111

The attractors of ASG(f) are its terminal strong components

Attractor of size one = fixed point
Attractor of size at least two = cyclic attractor

A path from a state x to a state y is a direct path if its length ℓ is equal to the Hamming distance between x and y (so ℓ ≤ n).

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 7/20

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Definition

The interaction graph of f, denoted G(f), is the signed directed graph on {1, . . . , n} with the following arcs:

There is a positive arc j → i iff there is a state x such that

fi(x1, . . . , xj−1, 0, xj+1, . . . , xn) = 0 fi(x1, . . . , xj−1, 1, xj+1, . . . , xn) = 1

There is a negative arc j → i iff there is a state x such that

fi(x1, . . . , xj−1, 0, xj+1, . . . , xn) = 1 fi(x1, . . . , xj−1, 1, xj+1, . . . , xn) = 0

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 8/20

SLIDE 21

Definition

The interaction graph of f, denoted G(f), is the signed directed graph on {1, . . . , n} with the following arcs:

There is a positive arc j → i iff there is a state x such that

fi(x1, . . . , xj−1, 0, xj+1, . . . , xn) = 0 fi(x1, . . . , xj−1, 1, xj+1, . . . , xn) = 1

There is a negative arc j → i iff there is a state x such that

fi(x1, . . . , xj−1, 0, xj+1, . . . , xn) = 1 fi(x1, . . . , xj−1, 1, xj+1, . . . , xn) = 0 j → i ∈ G(f) ⇐ ⇒ fi(x) depends on xj

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 8/20

SLIDE 22

Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 Asynchronous State Graph ASG(f)

000 001 010 011 100 101 110 111

Interaction Graph

G(f)

1 2 3

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 9/20

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Example

x f(x) 000 100 001 110 010 100 011 110 100 010 101 110 110 010 111 111 Asynchronous State Graph ASG(f)

000 001 010 011 100 101 110 111

Interaction Graph

G(f)

1 2 3

Question

What can be said on ASG(f) according to G(f) ?

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 9/20

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Results

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 10/20

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Theorem [Robert 1980]

If G(f) has no cycles then

1. f has a unique fixed point
2. ASG(f) has no cycles

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 11/20

SLIDE 26

Theorem [Robert 1980]

If G(f) has no cycles then

1. f has a unique fixed point
2. ASG(f) has no cycles
3. ASG(f) has a direct path from every state to the fixed point

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 11/20

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Theorem [Robert 1980]

If G(f) has no cycles then

1. f has a unique fixed point
2. ASG(f) has no cycles
3. ASG(f) has a direct path from every state to the fixed point

⇒ complexity comes from cycles of the interaction graph Two kinds of cycles have to be considered:

Positive cycles: even number of negative arcs
Negative cycles: odd number of negative arcs

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 11/20

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Theorem on positive cycles [Aracena 2004]

If all the positive cycles of G(f) can be destroyed by removing k vertices, then ASG(f) has at most 2k attractors.

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 12/20

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Theorem on positive cycles [Aracena 2004]

If all the positive cycles of G(f) can be destroyed by removing k vertices, then ASG(f) has at most 2k attractors. Corollary If G(f) has no positive cycles then ASG(f) has a unique attractor

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 12/20

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Theorem on positive cycles [Aracena 2004]

If all the positive cycles of G(f) can be destroyed by removing k vertices, then ASG(f) has at most 2k attractors. Corollary If G(f) has no positive cycles then ASG(f) has a unique attractor

Theorem on negative cycles [Richard 2010]

If G(f) has no negative cycles then ASG(f) has a path from every state x to a fixed point

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 12/20

SLIDE 31

Theorem on positive cycles [Aracena 2004]

If all the positive cycles of G(f) can be destroyed by removing k vertices, then ASG(f) has at most 2k attractors. Corollary If G(f) has no positive cycles then ASG(f) has a unique attractor

Theorem on negative cycles [Richard 2010]

If G(f) has no negative cycles then ASG(f) has a path from every state x to a fixed point

Our contribution

If G(f) has no negative cycles then ASG(f) has a direct path from every state x to a fixed point

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 12/20

SLIDE 32

Sketch of proof

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 13/20

SLIDE 33

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 14/20

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Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

It is sufficient to prove the theorem in the case where G(f) strongly connected (the general case follows by decomposition) So we suppose that G(f) is strong and has no negative cycles

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 14/20

SLIDE 35

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

It is sufficient to prove the theorem in the case where G(f) strongly connected (the general case follows by decomposition) So we suppose that G(f) is strong and has no negative cycles It is well known [Harary 1953] that G(f) has a set of vertices I such that an arc of G(f) is negative iff this arc leaves I or enters in I

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 14/20

SLIDE 36

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

It is sufficient to prove the theorem in the case where G(f) strongly connected (the general case follows by decomposition) So we suppose that G(f) is strong and has no negative cycles It is well known [Harary 1953] that G(f) has a set of vertices I such that an arc of G(f) is negative iff this arc leaves I or enters in I I

Melliti, Regnault, Richard, Sen´

e Convergence of Boolean networks without negative cycles Automata 2013 14/20

SLIDE 37

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

Let h be the network defined by h(x) = f(¯ xI)

I for all x ∈ {0, 1}n

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 15/20

SLIDE 38

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

Let h be the network defined by h(x) = f(¯ xI)

I for all x ∈ {0, 1}n

ASG(h) is isomorphic to ASG(f) and the isomorphism is x → ¯ xI The isomorphism preserves the Hamming distance

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 15/20

SLIDE 39

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

Let h be the network defined by h(x) = f(¯ xI)

I for all x ∈ {0, 1}n

ASG(h) is isomorphic to ASG(f) and the isomorphism is x → ¯ xI The isomorphism preserves the Hamming distance In addition, G(h) is obtained from G(f) by changing the sign of every arc that leaves I or enters in I Thus G(h) has only positive arcs

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 15/20

SLIDE 40

Theorem If G(f) has no negative cycles then ASG(f) has Theorem a direct path from any state x to a fixed point

Let h be the network defined by h(x) = f(¯ xI)

I for all x ∈ {0, 1}n

ASG(h) is isomorphic to ASG(f) and the isomorphism is x → ¯ xI The isomorphism preserves the Hamming distance In addition, G(h) is obtained from G(f) by changing the sign of every arc that leaves I or enters in I Thus G(h) has only positive arcs Conclusion: We can suppose that G(f) has only positive arcs This is equivalent to say that f is monotonous: ∀x, y ∈ {0, 1}n x ≤ y ⇒ f(x) ≤ f(y)

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 15/20

SLIDE 41

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 16/20

SLIDE 42

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point Lemma 1 f(0) = 0 and f(1) = 1

Suppose f(0) = 0, that is, fi(0) = 1 for some i Then since f is monotonous, fi(x) = 1 for all x ∈ {0, 1}n Thus fi = cst, so i has no in-neighbor in G(f) Thus G(f) is not strong, a contradiction We prove similarly f(1) = 1.

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 16/20

SLIDE 43

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point Lemma 2 The set of states reachable from x, denoted by R(x),

has a unique maximal element, reachable from x by a direct path

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 17/20

SLIDE 44

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point Lemma 2 The set of states reachable from x, denoted by R(x),

has a unique maximal element, reachable from x by a direct path Let P be an increasing path from x of maximal length. Let y the last state of P, so that f(y) ≤ y. We prove that z ≤ y, ∀z ∈ R(x)

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 17/20

SLIDE 45

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point Lemma 2 The set of states reachable from x, denoted by R(x),

has a unique maximal element, reachable from x by a direct path Let P be an increasing path from x of maximal length. Let y the last state of P, so that f(y) ≤ y. We prove that z ≤ y, ∀z ∈ R(x) If not there is a path x z → ¯ zi with z ≤ y and ¯ zi ≤ y. Thus ¯ zi

i = 1 and yi = 0, so z → ¯

zi increases component i. Thus fi(z) = 1 and since z ≤ y and fi is monotonous, fi(y) = 1.

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 17/20

SLIDE 46

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point Lemma 2 The set of states reachable from x, denoted by R(x),

has a unique maximal element, reachable from x by a direct path Let P be an increasing path from x of maximal length. Let y the last state of P, so that f(y) ≤ y. We prove that z ≤ y, ∀z ∈ R(x) If not there is a path x z → ¯ zi with z ≤ y and ¯ zi ≤ y. Thus ¯ zi

i = 1 and yi = 0, so z → ¯

zi increases component i. Thus fi(z) = 1 and since z ≤ y and fi is monotonous, fi(y) = 1.

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 17/20

SLIDE 47

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 48

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 49

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 50

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x)

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 51

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 52

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 53

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 54

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 R(x) x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 55

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 56

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 57

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 fi(x) = 0 yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 58

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x ¯ xi y

Maximal element of R(x)

direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 fi(x) = 0 yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 59

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x ¯ xi z = f(z) y

Maximal element of R(x)

direct path direct path (induction)

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 fi(x) = 0 yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 60

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x ¯ xi z = f(z) y

Maximal element of R(x)

direct path direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 fi(x) = 0 yi = 1

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 61

Theorem If G(f) is strong and f is monotonous then ASG(f) Theorem has a direct path from any state x to a fixed point

We prove the theorem by induction on the number of ones in x. If x = 0 the theorem is true since f(0) = 0. Suppose that x > 0 x ¯ xi z = f(z) y

Maximal element of R(x)

direct path direct path

f(z) ≤ f(y) ≤ y, ∀z ∈ R(x) If f(y) = y nothing to prove Suppose fi(y) < yi for some i Then fi(z) = 0 for all z ∈ R(x) Component i cannot increase in R(x) xi = 1 fi(x) = 0 yi = 1

Algorithm in O(n2)

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 18/20

SLIDE 62

Further results & perspectives

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 19/20

SLIDE 63

Theorem

Suppose that G(f) has no negative cycles. The set of fixed points reachable from x has a unique maximal element x+ and a unique minimal element x−, which are reachable in at most 2n − 4 transitions (thigh bound).

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 20/20

SLIDE 64

Theorem

Suppose that G(f) has no negative cycles. The set of fixed points reachable from x has a unique maximal element x+ and a unique minimal element x−, which are reachable in at most 2n − 4 transitions (thigh bound). Are all the fixed points of R(x) reachable in at most 2n − 4 steps ? Can we obtain upper/lower bounds on the number of fixed points reachable from x according to G(f) ?

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 20/20

SLIDE 65

Theorem

Suppose that G(f) has no negative cycles. The set of fixed points reachable from x has a unique maximal element x+ and a unique minimal element x−, which are reachable in at most 2n − 4 transitions (thigh bound). Are all the fixed points of R(x) reachable in at most 2n − 4 steps ? Can we obtain upper/lower bounds on the number of fixed points reachable from x according to G(f) ? We also plain to understand the connexions with works on

Monotone maps on complete lattices [Tarski]
Monotone differential systems [Hirsch & Smith]

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 20/20

SLIDE 66

Theorem

Suppose that G(f) has no negative cycles. The set of fixed points reachable from x has a unique maximal element x+ and a unique minimal element x−, which are reachable in at most 2n − 4 transitions (thigh bound). Are all the fixed points of R(x) reachable in at most 2n − 4 steps ? Can we obtain upper/lower bounds on the number of fixed points reachable from x according to G(f) ? We also plain to understand the connexions with works on

Monotone maps on complete lattices [Tarski]
Monotone differential systems [Hirsch & Smith]

Thank you!

Melliti, Regnault, Richard, Sen´ e Convergence of Boolean networks without negative cycles Automata 2013 20/20