Conjunctive networks Complexity of limit cycle problems with - PowerPoint PPT Presentation

Conjunctive networks Complexity of limit cycle problems with different schedules Julio Aracena, Florian Bridoux, Luis G´ omez, Lilian Salinas Florian BRIDOUX Conjunctive networks 2020 1/18

Boolean networks and interaction digraph Boolean networks: Global function: f : { 0 , 1 } n → { 0 , 1 } n . Local functions: f 1 , . . . , f n : { 0 , 1 } n → { 0 , 1 } . f ( x ) = ( f 1 ( x ) , f 2 ( x ) , . . . , f n ( x )) Local functions: Interaction digraph D f : 1 2 f 1 : x �→ 1. f 2 : x �→ x 2 . 3 f 3 : x �→ x 1 ∨ x 2 . N in (1) = ∅ , N in (2) = { x 2 } and N in (3) = { x 1 , x 2 } . Florian BRIDOUX Conjunctive networks 2020 2/18

Conjunctive networks A conjunctive networks f : { 0 , 1 } n → { 0 , 1 } n : � ∀ j ∈ [ n ] , f j : x �→ x i i ∈ N in ( j ) (if N in ( j ) = ∅ , f j ( x ) = 0). Local functions: Interaction digraph D f : 1 2 f 1 : x �→ 0. f 2 : x �→ x 2 . 3 f 3 : x �→ x 1 ∨ x 2 . Florian BRIDOUX Conjunctive networks 2020 3/18

Limit cycles x ∈ { 0 , 1 } n is in a limit cycle of f of length k if ∀ 1 ≤ q < k , f q ( x ) � = x , and ∀ f k ( x ) = x . Notations: φ k ( f ): number of limit cycles of length k of f : . Φ k ( f ): configurations in a limit cycle of length k of f : We have φ k ( f ) = | Φ k ( f ) | / k . Florian BRIDOUX Conjunctive networks 2020 4/18

Decision problems For any constant k , we define the following problems. Definition: k -Parallel Limit Cycle problem ( k -PLC) Given a conjunctive network f , does φ k ( f ) ≥ 1? Definition: k -Block-sequential Limit Cycle problem ( k -BLC) Given a conjunctive network f , does there exist a block-sequential schedule w such that φ k ( f w ) ≥ 1? Definition: k -Sequential Limit Cycle problem ( k -SLC) Given a conjunctive network f , does there exist a sequential schedule w such that φ k ( f w ) ≥ 1? Remark All this problems are trivial for k = 1. Florian BRIDOUX Conjunctive networks 2020 5/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. See: Disjunctive networks and update schedules, Eric Goles and Mathilde Noual, 2011. Florian BRIDOUX Conjunctive networks 2020 6/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 1 2 1 2 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 2 1 2 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 1 2 1 2 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 1 2 1 2 0 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 1 2 1 2 1 0 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 0 1 2 1 2 1 0 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 0 1 1 2 1 2 1 0 1 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Theorem For all k ≥ 2, The k -PLC problem can be resolved in polynomial time. When D f is strongly connected, it is equivalent to know if there exists a function c : [ n ] → [0 , k − 1] such that for all i , j ∈ [ n ], i ∈ N in ( j ) = ⇒ c ( j ) = c ( i ) + 1 mod k . Example: 2-PLC problem for the two following interaction digraphs. 0 1 0 1 1 2 1 2 1 0 1 4 3 4 3 Florian BRIDOUX Conjunctive networks 2020 7/18

k -PLC Proof of: c exists ⇒ φ k ( f ) ≥ 1. 2 1 4 5 1 3 0 2 3 1 x ( t ) : ∀ i ∈ [ n ] , c ( i ) = t ⇐ ⇒ x ( t ) = 1. i x (0) → x (1) f → x (2) f → x (3) f → x (0) f − − − − i i i i i f f f f 10000 → 01100 − − → 00010 − → 00001 − → 10000 Florian BRIDOUX Conjunctive networks 2020 8/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1 4 5 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 ?11? 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 0110 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 0110 0011 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 1001 0110 0011 3 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 1 4 5 1001 0110 0011 3 1100 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 0 1 4 5 1001 0110 0011 3 1 2 3 1100 0 Florian BRIDOUX Conjunctive networks 2020 9/18

k -PLC Proof of: φ k ( f ) ≥ 1 ⇒ c exists . Let x ∈ Φ k ( f ). Periodic trace: for all i ∈ [ n ], i ( x )) , . . . , ( f k − 1 p i ( x ) = ( x i , f i ( x ) , f 2 ( x ))) i In this example, consider that i ∈ [2] with the maximum 1 in its periodic trace is 2. 2 1100 0 1 4 5 1001 0110 0011 3 1 2 3 1100 0 Is to possible to have 0 − → 2 for example? = ⇒ No, because otherwise the period would not be minimum. Florian BRIDOUX Conjunctive networks 2020 9/18