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Where Are We?

CS 3240 - Introduction

A Pushdown AutomatonMachine for Context-Free Languages

- Take an FA and add a stack
- A restricted form of unbounded memory
- Operations:
- Pop
- Push
- Ignore (λ)
- (Pop always precedes push)

CS 3240 - Pushdown Automata

PDA Operations

- Three items per transition edge
- <read>,<pop>,<push>
- any or all could be λ
- A PDA accepts when:
- An accepting state is reached, and
- The stack is empty
- Different from book!
- The machine crashes when no move is possible
- implicit jail

CS 3240 - Pushdown Automata

A PDA for anbn (n > 0)

A deterministic PDA (no choices; λ is okay sometimes). The stack alphabet (Γ = {X}) can differ from the input alphabet (Σ = {a, b}) .

Trace aabb…

CS 3240 - Pushdown Automata

Tracing the PDA(<state> | <input> | <stack>)

q0 aabb λ

q0 abb X

q0 bb XX

q1 b X

q1 λ λ

Traditional notation (see book):

(q0,aabb, λ) ⊢ (q0,abb,X) ⊢ (q0,bb,XX) ⊢ (q1,b,X) ⊢ (q1,λ, λ)

Also: (q0,aabb,λ) ⊢*(q1,λ, λ) (⇒ aabb ∈ L)

CS 3240 - Pushdown Automata

A PDA for anbn (n ≥ 0)

CS 3240 - Pushdown Automata

Transition Functions for PDAs

- Input: (state-1, input character, pop character)
- Output: (state-2, push character(s))
- Example: The edge (a,λ,a) on previous slide: δ(q0,a,λ) = (q0,a)
- Exercise:
- Rewrite wcwR in functional form

CS 3240 - Pushdown Automata

A PDA for All Palindromes

Is this deterministic?

Trace the string abaaaaaba.

Trace the string abaabaaba.

Trace the string abaaaaba.

CS 3240 - Pushdown Automata

A PDA for ambn, m ≤ n ≤ 2mNon-deterministic

The grammar:

S → aSb | aSbb | λ

Trace aaabbbb

CS 3240 - Pushdown Automata

A PDA for nb = na + 1

CS 3240 - Pushdown Automata

A PDA for Twicebnb = 2na

- Reasoning: There must be 2 b’s for every a. When we read an a, there are three possibilities:
- Its 2 b’s have already been read
- 1 b has been read and 1 remains
- Both b’s are yet to be read

Try abababbbb.

How would you do Twicea?

CS 3240 - Pushdown Automata

What Language Is This?

CS 3240 - Pushdown Automata

aibjck, i,j,k ≥ 0, i = jor i = k

CS 3240 - Pushdown Automata

anor anbn

CS 3240 - Pushdown Automata

Formal Definition of PDAs

- Q: set of states
- ∑: input alphabet
- Γ: stack alphabet
- δ: Q x (∑∪λ) x (Γ∪λ) →finite subsets of Q x Γ*
- q0: start state
- F ⊆ Q: final states
- (z ⋲ Γ: optional stack start symbol)

CS 3240 - Pushdown Automata

Using Stack Start Symbols

- Not strictly necessary
- Can serve as a “bottom marker” to allow detection of an empty stack
- A convenience
- Can make things easier to design/understand
- Sometimes handy when seeking an associated CFG
- Can eliminate some lambda pops
- Handy later

CS 3240 - Pushdown Automata

anbn with a Start Symbol

CS 3240 - Pushdown Automata

A PDA for EqualUsing a Stack Start Symbol (“Almost” Deterministic)

CS 3240 - Pushdown Automata

xxR (Even Palindrome)

CS 3240 - Pushdown Automata

anbm, n ≤ m≤ 2n

CS 3240 - Pushdown Automata

PDAs and Context-Free LanguagesSection 7.2

- How do we know that PDAs are the correct machine model for CFLs?
- 1) Must show that for any CFG, there is a PDA that accepts its language
- 2) Must show that for any PDA, there is a CFG that generates its language
- The first case is easier than the second!
- It can be handy to go back and forth

CS 3240 - Pushdown Automata

From Greibach Form to PDA

- The leading character is read
- The LHS variable is popped
- The RHS string is pushed
- Initial transition pushes S
- See next slide

CS 3240 - Pushdown Automata

CFG to PDA ExampleFrom Greibach Form

CS 3240 - Pushdown Automata

CFG to PDA ExampleFrom Greibach Form II

CS 3240 - Pushdown Automata

CFG => PDAGeneral Case

- Very trivial (Yay!)
- But the (2-state) PDA is non-deterministic
- State-2 is a final state
- 1) Push S on the stack; move to state-2
- 2) For every rule, have a transition that:
- pops the left-hand side and pushes the right-hand side
- 3) For every character, c, in Σ, have a rule that:
- consumes and pops c simultaneously

CS 3240 - Pushdown Automata

Without Greibach FormFor Lazy People

S → aSb | aSbb | λ

Trace aaabbbbb: (q0,aaabbbbb,λ) ⊢ (q1,aaabbbbb,S) ⊢ (q1,aaabbbbb,aSbb) ⊢ (q1,aabbbbb,Sbb) ⊢ (q1,aabbbbb,aSbbbb) ⊢ (q1,abbbbb,Sbbbb) ⊢ (q1,abbbbb,aSbbbbb) ⊢ (q1,bbbbb,Sbbbbb) ⊢

(q1,bbbbb,bbbbb) ⊢*(q1,λ,λ)

CS 3240 - Pushdown Automata

One More

S => (S) | SS | λ

Derive ()(()) from the grammar.

Then accept it by the machine.

Then find a one-state DPDA for this language.

CS 3240 - Pushdown Automata

PDA => CFG

- We need to relate PDA movement to a CFG production rule somehow
- Key: consuming a character in the PDA corresponds to generating a character in the CFG
- The stack contents must also be part of all this

CS 3240 - Pushdown Automata

Special Case

- PDAs of the form below have a natural translation to a CFG
- the reverse of CFG-to-PDA conversion
- (conveniently) call the stack start symbol S
- Translate each transition in state-2 into a rule:
- c,X,Y ⟺ X → cY (i.e., <pop> → <char> <push>)
- (X cannot be λ)

CS 3240 - Pushdown Automata

PDA-to-CFG ExampleEQUAL

S → aXS | bYS | λ

X → aXX | bYX | b

Y → bYY | aXY | a

Derive abbbaaba

CS 3240 - Pushdown Automata

Exercise

Go back and find grammars for slides 22,24-25.

Also, work backward from the (()) DPDA to a grammar.

CS 3240 - Pushdown Automata

PDA => CFGGeneral Case

- We must also track state changes
- Our variables must therefore contain state and stack information
- Very tricky!

CS 3240 - Pushdown Automata

PDA => CFGContinued

- Suppose X is on the stack and ‘a’ is read
- What can happen to X?
- It will be popped (and possibly put back)
- It may be replaced by one or more other stack symbols
- Which symbols will also later be popped, replaced, etc…
- The stack grows and shrinks and grows and shrinks …
- Eventually, as more input is consumed, the effect of having pushed X on the stack must be erased (or we’ll never reach an empty stack!)
- And the state may change many times!

CS 3240 - Pushdown Automata

Observing a PDA(Source: Aho et al)

… (qi, x1x2…xk, Y1Y2…Yk) ⊢*(qj, x2…xk, Y2…Yk) ⊢*(qn, xk, Yk) ⊢*(qf, λ, λ)

CS 3240 - Pushdown Automata

Converting from PDA to CFG

- Let the symbol <qAp> represent the sequence of movements in a PDA that:
- starts in state q
- eventually ends in state p
- eventually removes A and its after-effects from the stack
- The symbol <sλf> then represents accepting a valid string (if s is the start state and f is a final state)
- These symbols will be our variables
- Because they track the machine configuration that accepts strings
- Just as the grammar will generate those strings

CS 3240 - Pushdown Automata

Converting from PDA to CFGcontinued

- Consider the transition δ(q,a,X) = (p,Y)
- This means that a is consumed, X is popped, we move directly to state p, and subsequent processing must eventually erase Y and its follow-on effects
- A corresponding grammar rule is:
- <qX?> → a<pY?> (?’s represent the same state)
- We don’t know where we’ll eventually end up when the stack is back to where it was before the X was first pushed
- All we know is we immediately go throughp
- So we entertain all possibilities (see 3 and 4 on next slide)

CS 3240 - Pushdown Automata

From Transitions to Grammar Rules4 types of rules generated

- 1) S → <sλf> for all final states, f
- 2) <qλq> → λ for all states, q
- These will serve as terminators
- 3) For transitions δ(q,a,X) = (p,Y):
- <qXr> → a<pYr> for all states, r
- 4) For transitions δ(q,a,X) = (p,Y1Y2):
- <qXr> → a<pY1s><sY2r> for all states, r, s
- And so on, for longer pushed strings

CS 3240 - Pushdown Automata

Preparing a Grammar for Conversion

- The process requires:
- No multiple pops
- So spread them out over multiple states if needed
- Also, for every lambda-pop c,λ,X:
- Add an equivalent rule for each letter of Γ (the stack alphabet) as follows:
- (c,X,XX)
- (c,Y,XY)
- etc.
- And keep the original lambda pop!!!

CS 3240 - Pushdown Automata

Convert PDA for anbn (n > 0)

- Call the start state, s, and the final state, f.
- Add the following transition on s: (a,X,XX)
- 1) S → <sλf>
- 2) <sλs> → λ; <fλf> → λ
- Now go through each transition…

CS 3240 - Pushdown Automata

Converting Transitions to Rules

- δ(s,a,λ) = (s,X):
- <sλs> → a<sXs>
- <sλf> → a<sXf>
- δ(s,a,X) = (s,XX):
- <sXs> → a<sXs><sXs>
- <sXf> → a<sXs><sXf>
- <sXs> → a<sXf><fXs>
- <sXf> → a<sXf><fXf>

CS 3240 - Pushdown Automata

Converting Transitions to RulesContinued

- δ(s,b,X) = (f,λ):
- <sXs> → b<fλs>
- <sXf> → b<fλf>
- δ(f,b,X) = (f,λ):
- <fXs> → b<fλs>
- <fXf> → b<fλf>

CS 3240 - Pushdown Automata

What’s Left?

- S → <sλf>
- <sλs> → a<sXs> | λ
- <fλf> → λ
- <sλf> → a<sXf>
- <sXs> → a<sXs><sXs>
- <sXf> → a<sXs><sXf> | a<sXf><fXf> | b<fλf>
- <fXf> → b<fλf>

CS 3240 - Pushdown Automata

SimplifySubstitute lambda for <fλf>

- S → <sλf>
- <sλs> → a<sXs> | λ
- <sλf> → a<sXf>
- <sXs> → a<sXs><sXs>
- <sXf> → a<sXs><sXf> | a<sXf><fXf> | b
- <fXf> → b

CS 3240 - Pushdown Automata

Simplify<sλs> is unreachable; <sXs> doesn’t terminate

- S → <sλf>
- <sλf> → a<sXf>
- <sXf> → a<sXf><fXf> | b
- <fXf> → b

CS 3240 - Pushdown Automata

SimplifyEliminate unit production S → <sλf> → …

- S → a<sXf>
- <sXf> → a<sXf><fXf> | b
- <fXf> → b

CS 3240 - Pushdown Automata

Deterministic PDAs

- A PDA is deterministic if there are no choices:
- Depends on a state’s <character, pop symbol> pair:
- 1) δ(q,a,X) has only one choice (a could be λ), and
- 2) If a is not λ, then there is no δ(q,λ,X) edge, and
- 3) if X is not λ, then there is no δ(q,a,λ) edge.
- NOTE: Some PDAs are inherently non-deterministic (no deterministic equivalent, e.g., wwR)

CS 3240 - Pushdown Automata

A Hierarchy of Languages

Languages accepted bynondeterministic PDA

Languages accepted by DFA

Languages accepted bydeterministic PDA

CS 3240 - Pushdown Automata

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