Probabilistic Cellular Automaton and Markov Chains Matthias Schulz University of Karlsruhe 1
Contents: - Introduction - PCAs with 2 states, M 1 -Neighbourhood - PCAs with k states, M 1 -Neighbourhood - PCAs with 2 states, M m -Neighbourhood - Conclusion 2
Probabilistic Cellular Automaton: - Space R, e.g. Z n (Rings) - Neighbourhood N, e.g. (-1, 0, 1) - Set of states S, e.g. { 0 , 1 } - Local transition rule δ , e.g. δ (012)(0) = 0 . 4 δ (012)(1) = 0 . 2 δ (012)(2) = 0 . 4 In each step, for every cell p in a configuration c = ( c 1 , ..., c n ) the probability to go into state i is δ ( c ( p + N ))( i ) . 3
Markov Chain: - Set of states S - Matrix P of transition probabilities: P ij contains the probability for state i changing to state j . Each Markov Chain induces a directed graph whose nodes are the states; there exists an edge beginning in i and ending in j , iff P ij > 0 . 4
Example: The matrix 0 0 . 5 0 . 5 0 0 1 0 1 0 induces the graph 5
A subset M in S is closed , iff P ij = 0 for i ∈ M, j / ∈ M. A closed subset M in S is irreducible , iff there is no nontrivial closed subset in M: Here the set { B, C } is closed and irreducible. 6
Let M v ⊆ N be the set of all lengths of paths beginning and ending in v ∈ S. v is aperiodic , iff gcd ( M v ) = 1 : There is the path ABCA of length 3 beginning and ending in A, and the path ACA of length 2 beginning and ending in A; since the gcd of 3 and 2 is 1, A is aperiodic. 7
A Markov Chain is irreducible , iff S is irreducible: 8
A Markov Chain is aperiodic , iff each state in S is aperiodic: 9
A Markov Chain is ergodic , iff it is irreducible and aperiodic: 10
Observation: PCAs can be interpreted as Markov Chains. Question: How hard is it determining whether a given PCA is ergodic? 11
- Introduction - PCAs with 2 states, M 1 -Neighbourhood -PCAs with k states, M 1 -Neighbourhood - PCAs with 2 states, M m -Neighbourhood - Conclusion 12
Definitions: A string ( c 1 , ..., c k ) ∈ { 0 , 1 , X } k is a structure , iff - none of the c i equals X or - k ≥ 3 and not all c i are X . A configuration d is an instance of a structure c , if it differs from c only where c i = X , e.g. 101 and 100 are instances of 10X. 13
A transition rule δ is homogenous , iff δ (000)(1) = 1 and δ (111)(0) = 1 . A transition rule δ is structure-invariant , iff - δ is not homogenous and - there exists a structure d = ( d 1 , ..., d k ) , such that each successor of an instance of d is an instance of d (possibly shifted), eg. for each δ satisfying δ (101)(1) = 1 , δ (010)(0) = 1 , the structure 10 is left invariant, since 10 has 01 as successor, which is 10 shifted to the right. 14
A transition rule δ is inhomogenous , iff - δ is neither homogenous nor structure-invariant, and - for odd n the set of inhomogenous configurations is closed. A transition rule is ergodic , iff each PCA with this transition rule and n ≥ 4 is ergodic. 15
Theorem 1: A transition rule is ergodic, iff it is neither homogenous, structure-invariant nor inhomogenous. This can be shown by analysis of 21 different cases. 16
- Introduction - PCAs with 2 states, M 1 -Neighbourhood - PCAs with k states, M 1 -Neighbourhood - PCAs with 2 states, M m -Neighbourhood - Conclusion 17
Theorem 2: The problem to decide whether or not a given PCA with k states is ergodic is PSPACE-complete in ( n + k ) . Theorem 3: The problem to determine for a given transition rule δ whether or not there exists n ∈ N so that the PCA defined by δ and n is ergodic is undecidable . 18
Proof of Theorem 2: The problem is in PSPACE: Irreducibility and aperiodicity can be checked nondeterminitically with algorithms needing polynomial space in ( n + k ) . With Savitch’s Theorem, it follows that there exists a deterministic algorithm needing poly- nomial space in ( n + k ) . 19
The problem is PSPACE-hard: There is a nondeterministic LBA so that it is PSPACE-hard in n to decide whether or not a given word of length n can be accepted by the LBA. Idea: For each n there is a PCA P constructed with O ( n ) states and a length in O ( n ) , such that P is ergodic ⇔ a given word w of length n is accepted by a fixed LBA A . 20
Construction: 0. The length of P shall be odd and ≥ n + 2 . 21
1. There is a state X which can change into each state and can spread over the PCA; all other states shall consist of several registers: � � � � ... X ... 22
1. There is a state X which can change into each state and can spread over the PCA; all other states shall consist of several registers: � � � � ... XXX ... 23
1. There is a state X which can change into each state and can spread over the PCA; all other states shall consist of several registers: � � � � ... XXXXX ... It follows that P is ergodic iff from every possible configuration there is a configuration containing X reachable. 24
2. Exactly one cell is marked with START, or there is a configuration reachable containing X (this can be achieved for any odd length): � START �� �� � � � ... So if there are configurations from which no configurations containing X can be reached, these contain exactly one START-cell. 25
3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes the word START w END into the first register and returns to the START-cell: � START �� �� � � � ... INIT 1 26
3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes the word START w END into the first register and returns to the START-cell: � START �� �� � � � w 1 ... INIT 2 27
3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes the word START w END into the first register and returns to the START-cell: � START �� w 1 �� � � � w 2 ... INIT 3 28
3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes the word START w END into the first register and returns to the START-cell: � START �� w 1 �� w 2 � � � END ... INIT n +2 29
4. When the INIT-signal reaches again the START-cell, a HEAD-signal is started in the START-cell simulating the head of the LBA A : corresponds to �� w 1 �� w 2 � w n � START � � END � � ... s 0 30
4. When the INIT -signal reaches again the START-cell, a HEAD-signal is started in the START-cell simulating the head of the LBA A : corresponds to � START �� v 1 �� v 2 � v n � � END � � ... s So from each configuration from which no configuration containing an X is reachable, a configuration containing the word START w END and a HEAD-signal in the START-cell can be reached. 31
5. If the HEAD-signal goes into an accepting state, the cell containing the signal changes to X: � START �� v 1 �� v 2 � v n � � END � � ... f + 32
5. If the HEAD-signal goes into an accepting state, the cell containing the signal changes to X: � START �� v 1 � v n � � END � � X ... So if there is a configuration from which no configuration containing an X is reachable, the HEAD-signal cannot go into an accepting state. If there is none, then the only way to reach a configuration containing X from a configuration containing exactly one START-cell and no signals, is by the HEAD-signal going into an accepting state. 33
It follows that there can only be configurations from which no configuration containing X can be reached, iff the HEAD-signal cannot go into an accepting state ⇔ P is ergodic iff w can be accepted by the LBA A , which proves Theorem 2. 34
Proof of Theorem 3: With a similar construction a Turing machine T with input w can be simulated, so that a PCA with odd length m is ergodic iff T stops for input w on a tape of length m − 2 ; therefore it is undecidable whether there exists m such that the PCA with length m and given transition rule is ergodic, which proves Theorem 3. 35
- Introduction - PCAs with 2 states, M 1 -Neighbourhood - PCAs with k states, M 1 -Neighbourhood - PCAs with 2 states, M m -Neighbourhood - Conclusion 36
Simulation of a PCA with k states by a PCA P’ with 3 states: P has states { 0 , ..., k − 1 } , P’ has states { 0 , 1 , 2 } i �→ 20 i 1 { 0 , 1 } k − 2 − i if i < k − 1 , k − 1 �→ 20 k − 1 ; e.g. 20011 and 20010 are both coding the state 2 in P. Section of the configuration each cell has to know: ... 200101 200010 211101... 37
If in P a cell changes with probability p j to j , then the corresponding block shall change with the same probability to a block coding the state j . Let ˜ p i be the probability for the i th cell in the block to go into state 1; if the ˜ p i solve the following equations (and solutions always exist), then P’ shows the wanted behaviour: � j i =1 (1 − ˜ p i ) · ˜ p j +1 = p j for j < k − 1 � k − 1 i =1 (1 − ˜ p i ) = p k − 1 Then P’ simulates P. 38
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