Introduction to Symbolic Dynamics Part 4: Entropy Silvio Capobianco Institute of Cybernetics at TUT May 12, 2010 Revised: May 12, 2010 ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 1 / 32
Overview Constructions and algorithms on sofic shifts. Entropy of a shift subspace. Computing entropy via Perron-Frobenius theory. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 2 / 32
Sofic shifts Path labelings Let G = ( G , L ) be an A -labeled graph. The labeling of a path π = e 1 . . . e m on G is the sequence L ( π ) = L ( e 1 ) . . . L ( e m ) . The labeling of a bi-infinite path ξ ∈ E ( G ) Z is the sequence x = L ( ξ ) ∈ A Z s.t. x i = L ( ξ i ) for every i ∈ Z . We put � � x ∈ A Z | ∃ ξ ∈ X G | x = L ( ξ ) X G = Definition X ⊆ A Z is a sofic shift if X = X G for some A -labeled graph G . In this case, G is a presentation of X . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 3 / 32
Special kinds of presentations A labeled graph G = ( G , L ) is: right-resolving if initial state and label determine edge follower-separated if differen states have different follower sets Fischer’s theorem Two minimal right-resolving presentations of an irreducible sofic shifts are isomorphic. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 4 / 32
Unions Union of two graphs Let G 1 = ( G 1 , L 1 ) and G 2 = ( G 1 , L 2 ) be labeled graphs. Set V ( G ) = V ( G 1 ) ⊔ V ( G 2 ) . Set E ( G ) = E ( G 1 ) ⊔ E ( G 2 ) . Set L ( e ) = L i ( e ) if e ∈ E ( G i ) Then G = ( G , L ) = G 1 ∪ G 2 is the union of G 1 and G 2 . Union of two sofic shifts is sofic G 1 ∪ G 2 is a presentation of X G 1 ∪ X G 2 . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 5 / 32
Products Product of two graphs Let G 1 = ( G 1 , L 1 ) and G 2 = ( G 1 , L 2 ) be labeled graphs. Set V ( G ) = V ( G 1 ) × V ( G 2 ) . Set E ( G ) = E ( G 1 ) × E ( G 2 ) . Set L ( e ) = L ( e 1 , e 2 ) = ( L 1 ( e 1 ) , L 2 ( e 2 )) . Then G = ( G , L ) = G 1 × G 2 is the product of G 1 and G 2 . Product of two sofic shifts is sofic G 1 × G 2 is a presentation of X G 1 × X G 2 . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 6 / 32
Label products Label product of two graphs Let G 1 = ( G 1 , L 1 ) and G 2 = ( G 1 , L 2 ) be labeled graphs. Set V ( G ) = V ( G 1 ) × V ( G 2 ) . Set E ( G ) = { ( e 1 , e 2 ) ∈ E ( G 1 ) × E ( G 2 ) | L 1 ( e 1 ) = L 2 ( e 2 ) } . Set L ( e ) = L ( e 1 , e 2 ) = L 1 ( e 1 ) = L 2 ( e 2 ) . Then G = ( G , L ) = G 1 ∗ G 2 is the label product of G 1 and G 2 . Intersection of two sofic shifts is sofic G 1 ∗ G 2 is a presentation of X G 1 ∩ X G 2 . And it isn’t over here. . . If G 1 and G 2 are right-resolving, then G 1 ∗ G 2 is right-resolving. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 7 / 32
Equality of sofic shifts The problem Given G 1 and G 2 , determine whether X G 1 = X G 2 . The idea Express equality of sofic shifts through the constructions seen before. An useful lemma Let G = ( V , E ) be a graph with r states. Let S ⊆ V contain s states. For I ∈ V \ S let U I = { π path on G | i ( π ) = I , t ( π ) ∈ S } . If U I is nonempty then min {| π | | π ∈ U I } ≤ r − s . Thus, there is a path from I �∈ S to J ∈ S iff B I , J > 0, where B = � r − s i = 1 A i . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 8 / 32
Equality of sofic shifts is decidable The idea Given G 1 and G 2 , construct � G s.t. tfae : 1 There is a word in B ( X G i ) \ B ( X G j ) . 2 There is a path in � G from some state I to some set S i . The algorithm 1 Let G ′ i be G i plus a sink K i : If there is no edge from I labeled a , make an edge from I to K i labeled a ; Add all self-loops to K i . 2 Let � G i be the subset graph of G ′ i . Let K i = { K i } . 3 Let � G = � G 1 ∗ � G 2 . Set I = ( V 1 , V 2 ) . 4 Set S 1 = { ( J , K 2 ) | J � = K 1 } and S 2 = { ( K 1 , J ) | J � = K 2 } ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 9 / 32
Cost of the algorithm If G i has r i states. . . G has ( 2 r 1 + 1 − 1 ) · ( 2 r 2 + 1 − 1 ) . . . . then � Could one do better? In general, no. But maybe, in special cases. . . A hint from Fischer’s theorem Suppose G 1 and G 2 are irreducible and right-resolving. Let H i be the minimal right-resolving presentation of X G i . Then X G 1 = X G 2 . if and only if H 1 ∼ = H 2 . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 10 / 32
Constructing the minimal right-resolving presentation The idea Start from an irreducible right-resolving presentation. Its merged graph is the minimal right-resolving presentation. Deciding equality of follower sets 1 Let G ′ be G with a sink K , as before. G = G ′ ∗ G ′ , I = V × V , S = ( V × { K } ) ∪ ( { K } × V ) . 2 Set � 3 Let I and J be two distinct nodes in G . tfae . ◮ F G ( I ) � = F G ( J ) . ◮ There is a path from ( I , J ) to S in � G . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 11 / 32
Determining finiteness of type of sofic shifts Theorem A Let G be a right-resolving labeled graph. Suppose that every w ∈ B N ( X G ) is synchronizing for G . Then X G is an N -step sft . Theorem B Let X be an irreducible sofic shift. And let G = ( G , L ) be its minimal right-resolving presentation. Suppose that X is an N -step sft . Then: ◮ Every w ∈ B N ( X G ) is synchronizing for G . ◮ L ∞ is a conjugacy. ◮ If G has r states then X is ( r 2 − r ) -step. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 12 / 32
Proof of Theorems A and B Proof of Theorem A Suppose uw , wv ∈ B ( X G ) with w ≥ N —then w is synchronizing. If uw = L ( ρπ ) and wu = L ( τσ ) , then t ( π ) = t ( τ ) . Then uwv = L ( ρπσ ) ∈ B ( X L ) . Proof of Theorem B Suppose | w | = N and w = L ( π ) = L ( τ ) with t ( π ) � = t ( τ ) . . . ◮ Let v ∈ F G ( t ( π )) \ F G ( t ( τ )) , u synchronizing word focusing on i ( τ ) . ◮ Then uw , wv ∈ B ( X ) but uwv �∈ B ( X ) , against X being N -step. If x = L ∞ ( y ) = L ∞ ( z ) , then y [ i , ∞ ) = z [ i , ∞ ) because L ( y [ i − N , i − 1 ] ) = L ( z [ i − N , i − 1 ] ) is synchronizing and G is right-resolving. The graph G ∗ G minus diagonal vertices checks precisely non-synchronizing words and has r 2 − r states. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 13 / 32
Entropy Definition The entropy of a nonempty shift X is 1 1 h ( X ) = lim n log | B n ( X ) | = inf n log | B n ( X ) | n →∞ n ≥ 1 The limit above exists and the equality holds because for every m , n ≥ 1 | B m + n ( X ) | ≤ | B m ( X ) | · | B n ( X ) | If X = ∅ we put h ( X ) = − ∞ . Quick examples If X is a full shift on an alphabet of r elements then h ( X ) = log r . If G is a graph on k nodes with r outgoing edges per node then h ( X G ) = log r . ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 14 / 32
� � The entropy of the golden mean shift The idea for the computation Consider the even shift as the vertex shift of � 1 0 For n ≥ 2 there is a one-to-one correspondence between B n ( � X G ) and B n − 1 ( X G ) . We can compute the size of this through the adjacency matrix � 1 � 1 A = 1 0 because | B m ( X G ) | = ( A m ) 0 , 0 + ( A m ) 0 , 1 + ( A m ) 1 , 0 + ( A m ) 1 , 1 ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 15 / 32
The entropy of the golden mean shift (cont.) Eigenvalues and eigenvectors The characteristic polynomial of A is χ A ( t ) = t 2 − t − 1 which has solutions √ √ λ = 1 + 5 ; µ = 1 − 5 2 2 λ is known as the golden mean. Corresponding eigenvectors of A are � λ � � µ � v λ = ; v µ = 1 1 ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 16 / 32
The entropy of the golden mean shift (end) Diagonalizing � λ � µ Let P = . Then 1 1 � λ m � � λ m + 1 − µ m + 1 � λ m − µ m 0 1 A m = P P − 1 = √ λ m − µ m λ m − 1 − µ m − 1 µ m 0 5 But λ m + 2 = λ m + 1 + λ m because λ 2 = λ + 1 , and similar with µ . 5 ( λ n + 2 − µ n + 2 ) , from which Hence | B n ( � 1 X G ) | = √ � 1 � � n + 2 �� � µ 1 h ( � λ n + 2 √ X G ) = lim n log 1 − 5 λ n →∞ = log λ ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 17 / 32
� � The entropy of the even shift The idea for the computation Consider the even shift as presented by 0 � B A 1 0 Each word with a 1 has one presentation. 0 n has two presentations. Then, | B n ( X G ) | = | B n ( X G ) | − 1 . Then clearly h ( even shift ) = h ( golden mean shift ) = log λ. ioc-logo Silvio Capobianco (Institute of Cybernetics at TUT) May 12, 2010 18 / 32
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