This lecture covers Chapter 6 of HMU: Pushdown Automata

Transcription

1 This lecture covers Chapter 6 of HMU: ushdown Automata ushdown Automata (DA) Language accepted by a DA Equivalence of CFs and the languages accepted by DAs Deterministic DAs Additional Reading: Chapter 6 of HMU.

2 ushdown Automata Introduction to DAs DA = ɛ-nfa + Stack (LIFO) At each instant, the DA uses: (a) the input symbol, if read; (b) present state; and (c) symbol atop the stack to transition to a new state and alter the top of the stack. Once the string is read, the DA decides to accept/reject the input string. Note: The DA can only read a symbol once (i.e., the reading head is unidirectional). DA = -NFA + Stack Reading head Internal Stack Finite Control Accept/Reject 2 / 27

3 ushdown Automata DA: Formal Definition Definition A DA = (Q, Σ, Γ, δ, q 0, Z 0, F ) where Just like in DFAs: Q is the (finite) set of internal states; Σ is the finite alphabet of input tape symbols; q 0 Q is the (unique) start state; F is the set of final or accepting states of the DA. Γ is the finite alphabet of stack symbols; δ : Q (Σ {ɛ}) Γ 2 Q Γ (power set of Q Γ ) such that δ(q, a, γ) is always a finite set of pairs (q, γ ) Q Γ. Z 0 Γ is the sole symbol atop the stack at the start; and Input symbol (or ) Next State The number of possible transitions resent state (q; a; A) ={(q 0 i; i ):i =1;:::; } Stack symbol on top The string replacing A on top of the stack Convention: lower case symbols s, a, and b will denote input symbols; lower case symbols u, v, w will exclusively denote strings of input symbols; stack symbols are indicated by upper case letters (e.g., A, B, etc); strings of stack symbols are indicated by greek letters (e.g., α, β, etc); 3 / 27

4 ushdown Automata A DA Example Transition Diagram Notation Notation: The label a, A/γ on the edge from a state q to q indicates a possible transition from state q to state q by reading the symbol a when the top of the stack contains the symbol A. This stack symbol is then replaced by the string γ. (q 0 a; A= ; ) 2 (q;a; A), q q 0 (Note: q 0 can be q itself) DA that accepts L = {ww R : w {0.1} } 0;Z 0 =0Z 0 1;Z 0 =1Z 0 0; 0=00 1; 0=10 0; 1=01 1; 1=11 0; 0= 1; 1= ; Z 0 =Z 0 q 0 q 1 q 2 ; 0=0 ; Z 0 =Z 0 ; 1=1 4 / 27

5 Language Accepted by a DA Language Accepted by a DA Definitions The Configuration or Instantaneous Description (ID) of a DA is a triple (q, w, γ) Q Σ Γ where: (i) q is the state of the DA; (ii) w is the unread part of input string; and (iii) γ is the stack contents from top to bottom. An ID tracks the trajectory/operation of the DA as it reads the input string. One-step computation of a DA, denoted by, indicates configuration change due to one transition. Suppose (q, γ) δ(q, a, A). For w Σ, α Γ, (q, aw, Aα) (q, w, γα), [one-step computation] (multi-step) computation, denoted by, indicates configuration change due to zero or any finite number of consecutive DA transitions. ID ID if there are k IDs ID 1,..., ID k (for some k 2) such that: (i) ID 1 = ID and ID k = ID, and (ii) for each i = 1,..., k 1, either ID i = ID i+1 or ID i ID i+1. 5 / 27

6 Language Accepted by a DA Beware of IDs! Lemma Let DA = (Q, Σ, Γ, δ, q 0, Z 0, F ) be given. Let q, q Q, x, y, w Σ, and α, β, γ Σ. Then the following hold. (q, x, α) (q, y, β) (q, xw, α) (q, yw, β) (1) (q, x, α) (q, y, β) (q, x, αγ) (q, y, βγ) (2) roof Idea (1) What hasn t been read cannot affect configuration changes (2) DA transitions cannot occur on empty stack. So the (q, x, α) (q, y, β) must not access any location beneath the last symbol of x. Why is the reverse implication of (2) not true? 6 / 27

7 Language Accepted by a DA Language Accepted by DAs Definition iven DA = (Q, Σ, Γ, δ, q 0, Z 0, F ), the language accepted by by final states is { } L() = w Σ : (q 0, w, Z 0) (q, ɛ, α) for some q F, α Γ. The language accepted by by empty stack is { } N() = w Σ : (q 0, w, Z 0) (q, ɛ, ɛ) for some q Q. Can L() and N() be different? ick a DFA A such that L(A). Convert it to a DA by pushing each symbol that is read onto the stack, increasing the stack size each time a symbol is read. For the derived DA, L() = L(A). However, N() =. Which of the two definitions accepts more languages? 7 / 27

8 Language Accepted by a DA Equivalence of the Two Notions of Language Acceptance Theorem iven DA, there exist DAs and such that L() = N( ) and N() = L( ). roof of Existence of ; X 0= DA DA 00 DA ; X 0= ; Z 0;Z 0X 0 ; X 0= ; X 0= Introduce a new start state and a new final state with the transitions as indicated. The start state first replaces the stack symbol Z 0 by Z 0X 0. If and only if w N() will the computation by end with the stack containing precisely X 0. The DA then transitions to the final state popping X 0. Hence, N() = L( ). 8 / 27

9 Language Accepted by a DA Equivalence of the two Notions of Language Acceptance roof of Existence of such that L( ) = N() DA DA 0 DA ; any= ; any= ; Z 0 ;Z 0 X 0 ; any= ; any= Introduce a new start state and a special state with the transitions as indicated. The start state first replaces the stack symbol Z 0 by Z 0X 0. If and only if w L() will the computation by end in a final state with the stack containing (at least) X 0. The DA then transitions to the special state and starts to pop stack symbols one at time until the stack is empty. Hence, L() = N( ). 9 / 27

10 CFs and DAs CFs and DAs Is every CFL accepted by some DA and vice versa? Theorem For every CF, there exists a DA such that N() = L(). roof Let = (V, T,, S) be given. Construct DA = ({q 0}, V, V T, δ, S, {q 0}) with δ defined by [Type 1] δ(q 0, a, a) = {(q 0, ɛ)}, whenever a Σ, [Type 2] δ(q 0, ɛ, A) = {(q 0, α) : A α is a production rule in }. This DA mimics all possible leftmost derivations. We use induction to show that L() = N() 10 / 27

11 CFs and DAs CFs and DAs roof of 1-1 Correspondence between DA Moves and Leftmost Derivations Suppose w T and S LM w i 2 T V i 2 (V [ T ) i 2 V Leftmost Derivation in rammar w 2 w 3 w 4 V 2 S)LM V 3..)LM )LM )LM )LM w k = w V w. x \ y:=suffix of y in x. S! 1 V 2! 2 V 3! 3 V 4! 4 { { { { Unread art of Input Tape w w w \ w 2 w \ w 2 w \ w 3 w \ w 3 w \ w 4 w \ w k 1 Stack Stack Symbols that have been popped S 1 V V V 4 4. w 2 w 3 w 3 w 4 k 1 k 1 w k 1 A \ B := The suffix of B in A w k [Start] [Type 2] [Type 1] w 2 [Type 2] [Type 1] [Type 2] [Type 1] [Type 2] [Type 1] Configurations in DA 11 / 27

12 CFs and DAs CFs and DAs Theorem For every DA, there exists a CF such that L() = N(). roof iven = (Q, Σ, Γ, δ, q 0, Z 0, F ), we define = (V, T,, S) as follows. T = Σ; V = {S} {[pxq] : p, q Q, X Γ}; Interpretation: Each variable [pxq] will generate a terminal string w iff w a move (in finite steps) from the state p to q popping X from the stack. contains only the following rules: S [q 0Z 0p] for all p Q. Suppose that (r, X 1 X l ) δ(q, a, X ). Then, for any states p 1,..., p l Q, [qxp l ] a[rx 1p 1][p 2X 2p 2] [p l 1 X l p l ]. Note that if (r, ɛ) δ(q, a, X ), then [qxr] a. We will show [qxp] q = q 0, X = Z 0. w (q, w, X ) (p, ɛ, ɛ). The proof is complete by choosing 12 / 27

13 CFs and DAs CFs and DAs roof of (q, w, X ) (p, ɛ, ɛ) [qxp] w. (Induction on # of steps of computation) Basis: Let w N(). Suppose there is a one-step computation (q, w, X ) (p, ɛ, ɛ). Then, w Σ {ɛ}. Since (p, ɛ) δ(q, w, X ), [qxp] w is a production rule. Induction: Let (q, w, X ) (p, ɛ, ɛ). Let a be read in the first step of the computation, and let w = ax. Then the following argument completes the proof. 1 Defn. (q;w;x) (r 1 ;x;y 1 ;:::;Y k ) ` (p; ; ) ) [qxp]! {z} a [r 1 Y 1 r 2 ][r 2 Y 2 r 3 ] [r k Y k p] ` {z } w=ax 2 A portion of x is read, and Y 1 is popped; more is read, Y 2 is popped,::: =x z } { (r 1 ; w 1 w 2 w ;Y 1 Y 2 Y k ) (r2 ;w 2 w k ;Y 2 Y k ) (r 3 ;w 3 w k ;Y 3 Y k ) (r k ;w k ;Y k ) ` ` ` 5 (p; ; ) ` (r 1 ;w 1 ;Y 1 ) (r 2 ; ; ) (r 2 ;w 2 ;Y 2 ) (r 3 ; ; ) ` + Induc. + Induc. [r 1 Y 2 r 2 ] ) w 1 [r 2 Y 3 r 3 ] ) w 2 + [r k Y k p] ) w k Induc. 13 / 27

14 CFs and DAs CFs and DAs roof of [qxp] w (q, w, X ) (p, ɛ, ɛ). (Induction on # of steps of derivation) Basis: Let [qxp] w in one step. Then, [qxp] w must be a production rule. Consequently, (p, ɛ) (q, w, X ) and (q, w, X ) (p, ɛ, ɛ). Induction: Let [qxp] w. 4 (r 0 ;Y 1 Y k ) 2 (q;a;x) () (q;a; X) (r 0 ; ;Y 1 Y k ) ` 3 Induc. (r 0 ;w 1 ;Y 1 ) (r 1 ; ; ) 5 1 [qxp] ) LM ` + a [r 0 Y 1 r 1 ][r 1 Y 2 r 2 ] [r k 1 Y k p] 2 + ) LM ) LM ) LM w 1 w 2 w k + Induc. (r 1 ;w 2 ;Y 2 ) (r 2 ; ; ) ` + ) LM w = aw 1 w k Induc. (r k 1 ;w k ;Y k ) (p; ; ) (q;aw 1 w 2 w k ;X) {z } (r 0 ;w 1 w k ;Y 1 Y k ) (r1 ;w 2 w k ;Y 2 Y k ) (p; ; ) ` ` ` ` w Lemma Lemma Lemma ` 14 / 27

15 Deterministic DAs Deterministic DAs (DDAs) DAs are (by definition) non-deterministic. Deterministic DAs are defined to have no choice in their transitions. Definition A DDA is a DA = (Q, Σ, Γ, δ, q 0, Z 0, F ) such that for each q Q and X Γ, δ(q, a, X ) 1 for any a Σ {ɛ}, i.e., a configuration cannot transition to more than one configuration. δ(q, a, X ) = 1 for some a Σ δ(q, ɛ, X ) =, i.e., both reading or not reading (a tape symbol) cannot lead to valid configurations. DDAs have a computation power that is strictly better than DFAs Example: L() = N() = {0 n 1 n : n 1} : 0;Z 0=0Z 0 0; 0=00 : q 1; 0= 0 q 1 q 2 ; Z 0=Z 0 DDAs have a computation power that is strictly worse that DAs. (We will discuss this later) 15 / 27

16 Deterministic DAs Languages Accepted by DDAs The two notions of acceptance (empty stack and final state) are not equivalent in the case of DDAs. There are languages L such that L = L() for some DDA, but there exists no such that L = N( ). Theorem Every regular language L is the language accepted by the final states of some DDA. roof Simply view the DFA accepting L as a DDA (with the stack always containing Z 0). The regular language L = {0} cannot equal N() for any DDA. Suppose DDA accepts L by emptying its stack. Since 0 is accepted, eventually reaches a configuration (p, ɛ, ɛ) for some state p. Now, suppose that is fed with the input 00. Since is deterministic, reads a 0 and eventually has to get to (p, ɛ, ɛ). However, it hangs at this configuration and cannot read any further input symbols. Hence, cannot accept / 27

17 Deterministic DAs Languages Accepted by DDAs A language L is said to have the prefix property if no two distinct strings in the language are prefixes of one another. Theorem A language L = N() for some DDA iff L has the prefix property and L = L( ) for some DDA. roof Let L = N() for some DDA. Let w, ww be in L with w ɛ. Then (q 0, w, Z 0) (p, ɛ, ɛ) for some p Q. The DDA hangs at this state since the stack is empty. Hence, it cannot accept ww. The fact that L = L( ) for some DDA follows from Theorem since the construction yields a deterministic DA. ; X 0= DA DA 00 DA ; X 0= ; Z 0;Z 0X 0 ; X 0= ; X 0= 17 / 27

18 Deterministic DAs Languages Accepted by DDAs roof Let DDA be given. Let w L( ), (q 0, w, Z 0) (p, ɛ, γ) for some p F, and γ Γ. Since L( ) satisfies the prefix property, the DA cannot enter any final state before reading all of w. Then we can delete all transitions from final states; this X Γ does not alter L( ). Then, the construction of Theorem yields a deterministic DA such that N( ) = L( ) = L DA DA ; any= ; any= ; Z 0 ;Z 0 X 0 ; any= ; any= 18 / 27

19 Deterministic DAs DDAs and Unambiguous rammars Theorem If L = N() for some DDA, then L has an unambiguous CF. roof Let be the CF constructed in Theorem Suppose is ambiguous. Then, for some w L has 2 leftmost derivations. However, each derivation corresponds to a unique trajectory of configurations in that also accepts w by emptying stack. Since is deterministic, the trajectories, and hence, the derivations have to be identical. Hence, is unambiguous. 19 / 27

20 Deterministic DAs DDAs and unambiguous rammars Theorem If L = L() for some DDA, then L has an unambiguous CF. roof Let $ be a symbol not in the alphabet of L. Consider L = {w$ : w L}. Then, L has the prefix property. By Theorem 6.4.2, there must exist a DDA such that L = N( ). By Theorem 6.4.3, L has an unambiguous CF = (V, T,, S). Define CF = (V {$}, T \ {$}, {$ ɛ}, S). generates L. Suppose is ambiguous. Then, for some w L has 2 leftmost derivations. The last steps in the two leftmost derivations of w must use the production $ ɛ. Then, the portions of the two leftmost derivations without the last production step correspond to two leftmost derivations of w$. Hence, must be unambiguous, which is a contradiction. Hence, is also unambiguous. 20 / 27

21 Additional Slides Explanation for Slide 11 Suppose we want to show that if there is a derivation in generating w, then there is a trajectory in accepting w. To do that let S w. LM Then there must be a LM derivation as in the left column. In each step of the leftmost derivation, a part of the string w is uncovered, and the uncovered part is succeeded by a non-terminal. Let after i = 1,..., k 2 production uses: (1) the prefix w i+1 of w be uncovered (shown in purple); (2) the leftmost non-terminal be V i+1 (shown in orange); and (3) is the string to the right of the leftmost non-terminal α i+1 that contains both terminal and non-terminal symbols (shown in beige). After the k th production rule, we have derived w k = w. Now suppose S γ 1 = w 2V 2α 2, V 2 γ 2,..., V k 1 γ k 1 be the k 1 production rules used in the leftmost derivation. Now let us show that a trajectory exists for using the above information we have laid out. Since there is only one state for the DA, the right part of the slide presents only the portion of tape yet to be read, and the stack contents; additionally, it also gives the string of terminals that has been popped up until any point in time. Initially, the tape contains w, the stack contains S, and ɛ has been popped thus far. 21 / 27

22 Additional Slides Explanation for Slide 11 (Continued) Now since S γ 1 is a valid production rule, by the definition of, there is a Type-22 transition that reads nothing from the input tape, reads S from the stack and pushes γ 1 := w 2V 2α 2 onto the stack. Thus, the following one-step computation is valid (q 0, w, S) (q 0, w, w 2V 2α 2). Note that w 1 is the prefix of w uncovered after the first step of the derivation, and hence matches the first few symbols of w. Then, it is clear that one can perform w Type-1 transitions that pop each of these symbols from the stack. Thus, after popping w 1 symbols, we see that: (q 0, w, S) (q 0, w, w 2V 2α 2) (q 0, w \ w 2, V 2α 2), where we let w \ w 2 to denote the suffix of w 2 in w. Now, note that V 2 γ 2 is a valid production rule; hence, there is a valid one-step computation from (q 0, w \ w 2, V 2α 2) that uses the corresponding Type-2 transition. The resultant configuration change will then be (q 0, w, S) (q 0, w, w 2V 2α 2) (q 0, w \ w 2, V 2α 2) (q 0, w \ w 2, (w 3 \ w 2)V 3α 3), where (w 3 \ w 2)V 3α 3 := γ 2α / 27

23 Additional Slides Explanation for Slide 11 (Continued) Again, we see that a portion of the top of the stack contains w \ w 2, which matches the initial segment of the input tape. Then there is a valid multi-step computation involving w 3 \ w 2 Type-1 transitions that pops w 3 \ w 2. The resultant configuration will then be q 0, w \ w 3, V 3α 3). Now, this proceeds until all of w is exhausted (read) from the input tape, and the configuration at the end will be (q 0, ɛ, ɛ). Since the stack is empty, the original string w will be accepted. The direction that a trajectory accepting w in implies a derivation of w in is simply arguing the above in the reverse direction using the facts that: a trajectory for accepting w in must consist only of Type-1 and Type-2 transitions, and each Type-2 transition corresponds to a unique production in. The argument is literally the same as above except that we now uncover the production rule from the corresponding Type-2 transition. 23 / 27

24 Additional Slides Explanation for Slide 13 Inductive proof for (q, w, X ) (p, ɛ, ɛ) [qxp] w based on length of computation. Basis: Let (q, w, X ) (p, ɛ, ɛ) be a one-step computation. Thus, w has to be an input symbol or ɛ. Then, by definition of one-step computation it must be true that (p, ɛ) (q, w, X ). Then, by the construction of, we have [qxr] w (see Slide 12 for the construction), and hence [qxr] Induction: (q, w, X ) (p, ɛ, ɛ) in say k > 1 steps. Let us assume that the in the first step of the computation, the symbol a is read from the input tape (or a = ɛ). Let w = ax. Let s break the k-step computation to a single step followed by a k 1-step computation as detained in 1 (encircled in black). Let r 1 be the state of the DA after the first step and let X be popped and Y 1 Y k be pushed onto the stack after the first step/transition/move. Now, the claim is that the k 1 step portion of the computation can be expanded into the sequence of computations as given in 2 (encircled in black). The reasoning is as follows. The ID (r 1, x, Y 1 Y k ) eventually changes to (p, ɛ, ɛ). There must be a finite number of moves after which the effective stack change is the popping of Y 1, i.e., after a finite number of steps Y 2 is at the top for the very first time. The steps until then could have popped Y 1, pushed a string, and then popped it eventually to reveal Y 2 at the top. w. 24 / 27

25 Additional Slides Explanation for Slide 13 (Continued) Let w 1 be the portion of the input tape read and r 2 be the state pf the DA when this intermediate ID where Y 2 is at the top of the stack (i.e., the stack contains Y 2, Y k ) is attained. Thus, (r, x, Y 1 Y k ) (r 2, x \ w 1, Y 2, Y k ) (p, ɛ, ɛ), where again we let w \ w 1 to be the suffix of w 1 in w. By a similar argument, after reading another segment, say w 2, of the input tape and reaching (some) state r 3, the top of the stack of the DA contains Y 3 for the very first time. Thus, (r, x, Y 1 Y k ) (r 2, x \ w 1, Y 2, Y k ) (r 3, x \ (w 1w 2), Y 3, Y k ) (p, ɛ, ɛ). roceeding inductively, we see that 2 (encircled in black) holds. Note that x is then equal to the concatenation of the w i s, i.e., x = w 1 w k. Now focus on the computation within the blue block in 2. In no intermediate ID of the computation is Y 2 at the top of the stack (since (r 2, x \ w 1, Y 2, Y k ) is the very first time Y 2 is at the top of the stack). Thus, the stack contents Y 2 Y k are never visited in this first set of moves, and hence, we see that (r 1, x, Y 1 Y k ) (r 2, x \ w 1, Y 2, Y k ) (r 1, w 1, Y 1) (r 2, ɛ, ɛ). (3) 25 / 27

26 Additional Slides Explanation for Slide 13 (Continued) Similarly, we see that the in portion of the computation in orange, no intermediate ID of the computation has Y 3 at the top of the stack (since (r 3, x \ (w 1w 2), Y 3, Y k ) is the very first time Y 3 is at the top of the stack). Hence, (r 2, x \ w 2 w k, Y 2, Y k ) (r 3, w 2 w k, Y 3 Y k ) (r 2, w 2, Y 2) (r 3, ɛ, ɛ). (4) We can proceed inductively to argue that (r i, w i, Y i ) (r i+1, ɛ, ɛ) for i = 1,..., k 1. Now each of these derivations (r i, w i, Y i ) (r i+1, ɛ, ɛ) for i = 1,..., k 1 contain k 1 or less steps, because the number of steps they contain is at least one-less than the number of steps in the computation in 1 (encircled in black). Consequently, by the induction hypothesis, we have [r i Y i r i+1 ] By the very same argument [r k Y k p] w k. w i, i = 1,..., k 1. Now focus on the yellow box at the top, the first one-step computation guarantees that there exists a production rule [qxp] a[r 1Y 1r 2][r 2Y 2r 3] [r k 1 Y k 1 r k ][r k Y k p]. (5) Now combining the above production with the known derivations in 4 (encircled in black), we see that [qxp] aw 1 w k = ax = w. 26 / 27

27 Additional Slides (q, w, X ) (r 0, w 1 w k, Y 1 Y k ) (r 1, w 2 w k, Y 2 Y k ) (r k, ɛ, ɛ) = (p, ɛ, ɛ). 27 (6) / 27 Explanation for Slide 14 Inductive proof for (q, w, X ) (p, ɛ, ɛ) [qxp] derivation. Basis: [qxp] w based on length of leftmost w be a one-step derivation. This can be possible only if LM (p, ɛ, ɛ). (p, ɛ) (q, w, X ), which then means (q, w, X ) Induction: Let [qxp] w in k > 1 steps. As in the previous direction, let us split the leftmost derivation into the first step and then rest. The first step must involve the application of some production rule, say, [qxp] a[r 0Y 1r 1][r 1Y 2r 2] [r k 1 Y k p]. By 1 (encircled in 1) each non-terminal [r i 1 Y i r i ] i = 1,..., k must derive (via a leftmost derivation) a segment of w, say w i in k 1 steps or less. [w i is the yield of the parse subtree in the parse tree of [qxp] with yield w, and the depth of the subtree is at most 1 less than the depth of the parse tree of [qxp].). Hence, [r i 1 Y i r i ] LM w i for i = 1,..., k in k 1 steps or less (I ve set r k = p here). By induction hypothesis, then (r i 1, w i, Y i ) (r i, ɛ, ɛ). Then by Lemma 6.2.1, (r i 1, w i w k, Y i Y k ) (r i, w i+1 w k, Y i+1 Y k ). Thus,