TWO-WAY FINITE AUTOMATA & PEBBLE AUTOMATA. Written by Liat Peterfreund

Transcription

1 TWO-WAY FINITE AUTOMATA & PEBBLE AUTOMATA Written by Liat Peterfreund 1

2 TWO-WAY FINITE AUTOMATA A two way deterministic finite automata (2DFA) is a quintuple M Q,,, q0, F where: Q,, q, F are as before 0 : Q Q { L, R} o If ( q, a) ( p, L) then in state q, scanning the input symbol a, the 2DFA enters state p and moves its head left one square. o If ( q, a) ( p, R) then in state q, 2 scanning the input symbol a, the 2DFA enters state p and moves its head right one square.

3 COMMENTS: o o In describing the behavior of one-way finite automata, we extend to * Q. This notion is insufficient for 2DFA because the head may move left. Instead we introduce the notion of instantaneous description ( ID ) of 2DFA which describes the input string, current state and current position if the input head. o The relation on ID 's such that M I1 M I 2 if and only if M o 3 can go from I 1 to I 2 in one move. An ID of M is a sting in * * Q.

4 o The ID wqx whereas intended to represent : 1) wx is the input string 2) q is the current state * wx, and q Q is 3) the input head is scanning the first symbol of x (If x then the input head has moved off the right hand end of the input). 4

5 5 o We define the relation M understood, by or just if M is a a... a qa... a a a... a a pa... a 1) 1 2 i1 i n 1 2 i1 i i 1 n whenever ( q, a ) ( p, R) i a a... a a qa... a a a... a pa a... a 2) 1 2 i2 i1 i n 1 2 i2 i 1 i n whenever ( q, a ) ( p, L) and i 1(The i condition i 1 prevents any action in the event that the tape head would move off the left hand end of the tape). in (the tape o Note that no move is possible if 1 head has moved off the right hand end).

6 Let * be the reflexive and transitive closure of. * I1 I k whenever I1 I2... I k for some I2,..., k 1 define * L( M ) w q 0w wp, p F. That is w is I we accepted by M if, starting in state q 0 with w on the input tape and the head at the left end of w, M eventually enters a final state at the same time it falls off the right hand of the input tape. 6

7 Example: Consider a 2DFA M that behaves as follows: starting in state q 0, M repeats a cycle of moves wherein the tape head moves right until two ones have been encountered then left until encountering a zero at which point state q 0 is reentered and the cycle repeated. More precisely M has three states all of which are final. is given as follows: q q q 2 R L 0, 2,

8 Consider the input : 8 q q q q q q q q q q q 1 than LM ( )

9 Crossing sequences q 0 q 1 q 1 q 2 q 0 q 1 q 1 q 1 q 2 q 0 q 1 9 The list of states below each boundary between squares is termed a crossing sequence.

10 Observations: o o o 10 If a 2DFA accepts its input, no crossing sequence may have a repeated state with the head moving in the same direction. Otherwise the 2DFA being deterministic would be in a loop. The first time a boundary is crossed, the head must be moving right. Subsequent crossings must be in alternate directions: odd-numbered elements of the crossing sequence represents right moves even-numbered elements of the crossing sequence represents left moves

11 o o If the input is accepted then all crossing sequences are of odd length A crossing sequence q,..., 1 q k is said to be valid if : it is of odd length no two odd-numbered and no two evennumbered elements are identical o The number of valid crossing sequences is finite (a 2DFA with s states can have valid crossing sequence of length at most 2s-1). 11

12 Theorem: 2DFA and DFA (deterministic finite automata) have the same expressive power. 12

13 Proof: o o We construct a NFA (non-deterministic finite automata) that will simulate the 2DFA M and whose states are the valid crossing sequences of M. In order to construct the transition function we first examine the relationship between adjacent crossing sequences. 13

14 o Suppose we are given an isolated tape square holding the symbol a and are also given valid crossing sequences q1, q2,..., q k and p1, p2,..., p l at the left and the right boundaries of the square, respectively. Note that there may be no input strings that can be attached to the left and right of symbol a to actually produce these two crossing sequences 14

15 o We can test two sequences for local compatibility as follows: If the tape head moves left from the square holding a in state q i, restart the automaton on the square holding a in state qi 1 If the tape head moves right from the square holding a in state p i, restart the automaton on the square holding a in state pi 1 15

16 We take q1, q2,..., q k to appear at the left boundary of a and p1, p2,..., p l at the right boundary of a then, q1, q2,..., q k right-match p1, p2,..., p l on a if these sequences are consistent, assuming we initially reach a in state q 1 moving right q1, q2,..., q k left-match p1, p2,..., p l on a if these 16 sequences are consistent, assuming we initially reach a in state p 1 moving left

17 We define right-matching and left matching pairs of crossing sequences recursively in (1) through (5): (1) The null sequence left- and right-matches the null sequence. (if we never reach the square holding a, then it is consistent that the boundaries on neither side should be crossed). 17

18 q (2) If,..., 3 qk right-matches p1, p2,..., pl and ( q, a) ( q, L) 1 2 q then 1, q2,..., qk right-matches p1, p2,..., p l (if the first crossing of the left boundary q is in state 1 and the head immediately moves left in 18 q state 2, then if we follow these two crossings by any consistent behavior starting from another crossing of the left boundary, we obtain a consistent pair of sequences with first crossing moving right i.e. a right matched pair). (3) If q,..., 2 q k left-matches p,..., 2 p l and ( q1, a) ( p1, R) then q1 q2,,..., q k right-matches p1, p2,..., p l.

19 (4) If q,..., 1 q k left-matches p,..., 3 l ( p, a) ( p, R) then q,..., 1 k 1 2 p1, p2,..., p l. (5) If q,..., 2 q k right matches p,..., 2 l ( p, a) ( q, L) then q,..., 1 k 1 1 p1, p2,..., p l. p and q left-matches p and q left-matches 19

20 The construction: Let M Q,,, q0, F M ' Q',, ', q ', F ' be a 2DFA. We contruct a NFA which accepts LM ( ) such that: 0 1) Q ' consists of all valid crossing sequences for M. 2) q ' is the crossing sequence consisting of 0 q 0 alone. 3) F ' is the set of all crossing sequences of length one 4) 20 consisting of a state in F. d d is a valid crossing sequence '( ca, ) that is right matched by c on input a

21 The intuitive idea is that M ' puts together pieces of computations of M as it scans the input string. This is done by guessing successive crossing sequences. If has guessed that c is the crossing sequence at a boundary, and a is the next input symbol then M ' can guess any valid crossing sequence that c right matches on input a. M ' If the guessed computation results in M moving off the right end of the input in an accepting state, then 21 accepts. M '

22 Claim: L( M) L( M ') Proof: L( M) L( M ') : Let w L( M ). Look at the crossing sequences generated by an accepting computation of M on w. Each crossing sequence right-matches the one at the next boundary, so 22 can guess the proper crossing sequence (among its other guesses) and accept i.e. w L( M '). M '

23 L( M ') L( M) : Let w L( M '). Consider the crossing sequence c,..., 0 c n of M corresponding the states of w a,..., 1 an. For each 0 i n i M ' as M ' scans c right matches ci 1 on We can construct an accepting computation of M on input w by determining when the head reverses 23 direction. a i

24 We prove by induction on i that M ' on reading a... 1 a i can enter state c q q conditions hold: i only if the following two 1,..., 1) M started in state q 0 on a... 1 a i will first move right from position i in state q 1 2) for j 2,4,..., if M is started at position i in state k q j, M will eventually move right from position i in 24 state q j 1 (this implies k must be odd).

25 Basis: 0 i then c q 0 0. (1) is satisfied since M begins its computation by "moving right from position 0" in state q 0. (2) holds vacuously. 25

26 Assume the hypothesis is true for 1 26 M ' on reading a... 1 i a can enter state i. Suppose that from state c q q c,..., i p1 p l i k. 1 1,..., o Since k and l are odd, and ci 1 right o matches c i on such that in state a i there must exist an odd j q j on input a i, M moves right. Let j 1 be the smallest such j. By definition of "right matches" it follows that ( q, a ) ( p, R). This proves (1). j1 1 1 We will prove (2) by induction: Basis: vacuously Inductive step: by the definition of "right-matches" (rule (3)) [ qj,..., qk] left-matches [ p2,..., p l ] 1 1

27 27 Now if ( p j, ai ) ( p j1, R) for all even j, then (2) follows immediately. In the case that for some smallest even j, 2 ( p j, a ) (, ) 2 i q L, then by definition of "left-matches" (rule (5)) q must be q j 1 and [ q,..., ] 1 j qk right matches [ p,..., p ] By induction hypothesis, (2) holds for [ qj,..., qk] [ pj,..., ] and 2 p 1 l 1 2 j l

28 With the induction hypothesis for all i established, the fact that c [ p] for some p F implies that M n accepts a... 1 a n. 28

29 Reminder - Example: Consider a 2DFA M that behaves as follows: starting in state q 0, M repeats a cycle of moves wherein the tape head moves right until two ones have been encountered then left until encountering a zero at which point state q 0 is reentered and the cycle repeated. More precisely M has three states all of which are final. is given as follows: q q q 2 R L 0, 2,

30 Consider the construction of NFA 2DFA M : M ' equivalent to the Since q 2 is only entered on a left move and q 1 and q3 are 30 only entered on right moves, all even numbered components of valid crossing sequences must be q 2. Since a valid crossing sequence must be of odd length, and no two odd numbered states can be the same, nor can two even-numbered states be the same, there are only four crossing sequences of interest: [ q0],[ q1 ],[ q0, q2, q1 ],[ q1, q2, q 0].

31 Valid crossing sequences Right matches on 0 [ ] [ ] Right matches on 1 [ q ] q 0 q 0 1 q 1 [ q1 ],[ q1, q2, q 0] - [ ] [ q, q, q ] [ q, q, q ] [ q 1] Thus the automaton is: 0 0 [ 0] q 1 1 [ q ] [ q1, q2, q0]

32 32 Pebble Automata

33 Preface: We consider strings over infinite alphabet D. A D -string is a finite sequence d... 1 d n where n 0 and 1 i n : di D. dom( w) 1,2,..., w. For i dom( w) we write valw() i for d i. A language is a set of D -strings. 33

34 We delimit input strings by two special symbols, for the left and for the right of the string, neither of which are in D. Therefore the automata work on extended strings of the form w u where u * D. The positions of, are 0 and w 1 respectively 34 We write dom ( w) 0,..., w 1 of positions for the extended set We likewise extend the function val w by defining val (0) w and valw ( w 1)

35 Pebble automata are finite state machines equipped with a finite number of pebbles. The use of pebbles is restricted by a stack discipline (pebble i 1 can only be placed when pebble i is present on the string and pebble i can only be lifted when pebble i 1 is not present on the string). The highest numbered pebble present on the string acts as the head of the automaton. 35

36 36 A transition depends on: o The current state o The pebbles placed on the current position of the head o Equality type of the data values under the placed pebbles. The transition relation specifies: o Change of state o Movement of the head o Possibly whether the head pebble is removed or a new pebble is placed

37 Definition: A nondeterministic two-way k-pebble automaton A over D (infinite alphabet) is a tuple ( Q, q0, F, T ) where: o Q is a finite set of states o q0 Q is the initial state o F Q is a set of finite states 37

38 o T is a finite set of transitions of the form 38 where o is of the form ( i, s, P, V, q ) or ( i, P, V, q ) where i 1,..., k, s D, P, V 1,..., i 1, o is of the form ( qd, ) with q Q and stay,left,right,place-new-pebble, d lift-current-pebble

39 Given a string w, a configuration of A on w is of the form iq,, where i 1,..., k :{1,..., i} dom ( w)., q Q and We call pebble assignment and i the depth of configuration (and of pebble assignment). Initial configuration: : 1, q, where (1) Accepting configuration:, iq, where 39 F.

40 A transition ( i, s, P, V, q) jq,, if: 1) i j, p q applies to a configuration 2) V l i val ( ( l)) val ( ( i)) 3) P l i ( l) ( i) 4) val ( ( i)) s w w w 40 A transition ( i, P, V, q) applies to a configuration jq,, if 1-3 holds and no transition ( i, s, P, V, q) applies to.

41 We define the transition relation as follows: [ i, q, ] [ i', q', '] iff there is a transition ( pd, ) that applies to such that q' p and j i : '( j) ( j) and: 41 If d stay, then i' If d left, then i' If d right, then i' i and '( i) ( i) i and '( i) ( i) 1 i and '( i) ( i) 1 If d place-new-pebble, then i' i 1 and '( i) '( i 1) ( i) If d lift-current-pebble, then i' i 1

42 An automaton is deterministic if in each configuration at most one transition applies. If there are no left transitions then the automaton is oneway. The language of the automaton: A string w is accepted by A if * for some accepting configuration. 0 42

43 Example 1: Let L d1... dn n 0, i, j : i j di d j. We define a 2-pebble-automata A accepting this language. A ( Q, q, F, T) is defined as follows: 0 43 Q q q q q,,, acc 1 2 F { q acc } T consists of the following transitions: (1) (1,,, q1) ( q1,right) (2) (1,,, q1 ) ( q,place-new-pebble) (3) (2,{1},{1}, q ) ( q2,right) (4) (2,,, q2) ( q2,right) (5) (2,,{1}, q2) ( q acc,stay)

44 Intuitive explanation: (1-2) A stays in state q 1 while moving to the right or places a pebble (3) After the placement of the pebble A moves one position to the right (4-5) A continues moving to the right and when it reads a symbol equal to the symbol under the first pebble, it moves to the final state 44

45 Example 2: * Let L u$ v u, v D $ D [ u] [ v] whereas [ u],[ v ] are the sets of all the letters of uv, respectively. We define a 2-pebble-automata A accepting this language that operates as follows: o The first pebble passes through all of the positions of 45 o o the word u (until it encounters the sign $ ) and for each such position it checks whether there is a position in v with the same letter. The above check is done by placing the second pebble and iterating through all of the positions in v - if there is such a position it lifts the pebble and continues with the next position in u, if there is not it rejects. When the first pebble reaches the $ it moves to an accepting state.

46 Formally, A ( Q, q0, F, T) whereas Q q,,,,, 0 q q q q q, u uv v rej acc the following transitions: F q acc and T contains ( a1)(1,,, q ) ( q,right) 0 ( b1)(1,$,,, q ) ( q,stay) u ( c1)(1,,, q ) ( q,place-new-pebble) u u uv acc 46 ( a2)(2,{1},{1}, q ) ( q,right) ( b2)(2,$,,, q ) ( q,right) ( c2)(2,,, q ) ( q,right) uv ( d2)(2,,{1}, q ) ( q,right) uv uv uv ( e2)(2,,{1}, q ) ( q,lift-current-pebble) ( f 2)(2,,,, q ) ( q,stay) ( g2)(2,,, q ) ( q,right) v v v uv 0 v uv uv v rej

47 Intuitive explanation: ( a 1) is used to skip a sign ( b1),( c 1) are used to iterate all over u 's elements ( a2) ( d2) skip the end of u and the $ ( e2) ( f2) iterates through v and compares the sign under the current pebble with the sign under pebble 1 47

48 Example 3: We define a two-way nondeterministic pebble automaton with three pebbles that accepts all words w where w 2 and there exists some position i in dom( w ) such that the set of symbols occurring at positions smaller than i is disjoint from the set of symbols occurring at position larger than or equal to i. For example the automaton accepts abb and 48 rejects abab

49 49 The automaton will work as follows: o The first pebble is used nondeterministically to guess i o The second pebble is used to step through position to the left of i one at a time, from o o i 1 to For each such position of the second pebble, the third pebble is used to check that all symbols at positions i are distinct from the symbol under the second pebble. The input is excepted if the second pebble can reach the leftmost symbol

50 Formal definition: A Q, q, F, T Q q q 0,..., 7, F q5 transitions: where o { } and T is the set of 50

51 ( a1) (1,,,, q ) ( q,right) } skip 0 0 (b1) (1,,, q ) ( q,right) } i (c1) (1,,,, q ) ( q,stay) } reached the end without guessing i 1 2 q1 q1 q1 q3 (d1) (1,,, ) (,right) (e1) (1,,, ) (,place-new-pebble) guess i ( a2) (2,{1},{1}, q ) ( q,left) } move to position i ( b2) (2,,, q ) ( q,left) } move one position left (after lifting pebble 3) 3 4 ( c2) (2,,,, q ) ( q,stay) } finished reading positions i ( d2) (2,,, q ) ( q,place-new-pebble) } "save" position 4 6 ( a3) (3,{2},{2}, q6) ( q6,right) (b3) (3,,,q 6) ( q6,right) move right to position i 1 ( c3) (3,,{2}, q6) ( q6,right) ( d3) (3,{1},{1}, q6) ( q7,right) 51 (e3) (3,,{1},q 7) ( q7,right) ( f 3) (3,,,, q7) ( q3,lift-pebble) scan positions > i ( g3) (3,,, q7) ( q7,right)

52 For example, an accepting run on abb will be: 52

53 53 In the above definition, new pebbles are placed at the position if the most recent pebble. An alternative would be to place new pebbles at the beginning of the string. While the choice makes no difference in the two-way case, it is significant in the one way case. We refer to the model as defined above as weak pebble automata and to the latter as strong pebble automata. Strong pebble automata are formally defined by setting '( i 1) 0 in the place-new-pebble case of the definition of the transition relation.

54 Our next result shows that the variants of strong pebble automata, one-way and two-way, collapse. Theorem: Two-way weak 2-pebble automata and two-way strong 2- pebble automata have the same expressive power. 54

55 Proof: Let A Q, q, F, T be a two-way strong 2-pebble o automaton. We want to construct a two-way weak 2-pebble automaton that will simulate A 's behavior. Idea of construction: For every state in q Q we will add a new state q ' and transition that such that the automaton head will move left until it encounters the left delimiter and then move to the appropriate state. 55

56 Let A Q, q, F, T be a two-way weak 2-pebble o automaton. We construct a two-way 2-pebble strong automaton that will simulate A 's behavior. The only difference between the automata is in placing a new pebble. Idea of construction: For every state in q new state Q we will add a q ' and transition that such that the automaton head will move right until it encounters the first pebble and then move to the appropriate state. 56

57 Theorem: Two-way strong 2-pebble automata and one-way strong 2- pebble automata have the same expressive power. Proof: A one-way strong 2-pebble automaton is a particular case of two-way strong automaton. 57

58 Claim: For each two-way non-deterministic pebble automata A there is a strong one-way deterministic pebble automata that accepts the same language. Proof: Assumptions: (1) A lifts pebble only in the right delimiter (2) A accepts its input only in configurations [1, q, θ], i.e., with only one pebble. 58

59 By virtually adding two steps we view an accepting computation as consisting of: (1) a first step in which the first pebble is placed at the left delimiter, i.e., position 0. (2) a computation in which always at least one pebble is present. (3) a final step in which the only remaining pebble is removed. 59

60 Writing [0, p, ] for a (virtual) configuration without pebble, to determine whether A accepts, one has to find out whether * [0, q, ] [0, q, ] for some final state q. 0 The latter can be done by recursively solving subproblems of * the form [ i, q, ] [ i, q', ] i where the subscript > i indicates that only subcomputations are considered in which, at every step, more than i pebbles are present. 60

61 More formally, we show the following claim by induction on i (starting from i = k). Claim: For each i {0,...,k} and each finite set of states R, there is a strong one-way deterministic pebble automata B i (with k pebbles) such that, whenever B i starts from a configuration[i, p, ], where p R, the next configuration of depth i of B i is [i,(p, S), ], * S {(q,q ) QQ [i,q, ] i [i, q, ]}. The set of states of B i includes R and 2 R R R 61. where

62 The theorem indeed follows from the claim: Let i = 0 and let R {p 0} (the intended initial state of B 0 ). We obtain an automaton which ends up in a state (p 0, S) where S {(q,q ) QQ [0,q, ] [0,q, ]} * 0 The set of final states of B 0 consists of all states 0 where S contains a pair (q 0, q) with q F. (p, S) 62

63 For i k the proof of the claim is trivial, as there are no configurations of depth k. Hence, B k can compute (p, S) by a stay-transition. Therefore, let i k and suppose the claim holds for all j i 63

64 64 Let the input string w of length n be fixed. Hence B i works on the string w with positions from dom ( w). For ln 1, let l denote the ( i 1) -pebble assignment that coincides with in the first i pebbles l and for which ( i1) l. S ( l ) is the set of pairs ( q, q ) of states such that there is a computation starting at [ i 1, q, l ] and reaching [ i 1, q', l ] which only includes configurations [ j, q, ] that fulfill j i 1 or ( ji 1 and ( i1) l ). Intuitively, this says that pebble i 1 is not allowed to move to the right of position l.

65 We write S ( l ) for the set of pairs ( q, q) of states for which [ i1, q, l ] can be reached from 1 [ i 1, q, ] by a subcomputation satisfying the same property. 65 The set S ( l ) can be computed as follows: Let R ( l ) be the set of pairs ( q, q ) for which one of the following holds: (a) There exist p1, p 2 such that 1 [ 1,, l l i q ] [ i 1, p, ], (b) 1 l 1 ( p1, p2) S ( ) and 1 [ 1, 2, l l i p ] [ i 1, q, ] l * l [ i 1, q, ] i 1 [ i 1, q, ]. (c) [ 1,, l l i q ] [ i 1, q, ]

66 S ( l ) is the transitive closure of R ( l ). They are computed simultaneously. The information needed for (a) can be computed in one left-to-right pass of pebble i 1. By induction we can assume a subautomaton Bi 1 that 66 computes, for each position l, the part of R ( l ) contributed by condition (b). (c) and the computation of the transitive closure do not require any pebble movements. During the same pass, the automaton can compute, for each position l, the set S ( l ). The computation of ( l m S ) makes use of the sets S ( ), m l.

67 From n 1 S ( ), n 1 S ( ) and the transition relation of A one can deduce, during a lift-pebble step, the set S as in the claim. Note that n 1is the position of the right delimiter and recall that A lifts its pebbles only at that position. This completes the proof of the claim and of the theorem. 67