Pushdown Automata

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# Pushdown Automata - PowerPoint PPT Presentation

CS 3240 – Chapter 7. Pushdown Automata. Where Are We?. A Pushdown Automaton Machine 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). PDA Operations.

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CS 3240 – Chapter 7

### Pushdown Automata

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
• 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

(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

A PDA for anb2n (n ≥ 0)

How would you do a2nbn?

CS 3240 - Pushdown Automata

a2nbn

Each b must pop 2 X’s

CS 3240 - Pushdown Automata

A PDA for wcwR

Is this deterministic?

Trace the string abaacaaba.

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 Equal

Trace ababbaab

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:
• 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 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

Now You Do One

S → aSbSbS | bSaSbS | bSbSaS | λ

And trace bbabaabbb

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

SimplifySubstitute <fXf> → b
• S → a<sXf>
• <sXf> → a<sXf>b | b

CS 3240 - Pushdown Automata

SimplifyRename
• S → aX
• X → aXb | b
• Done! (finally :-)

CS 3240 - Pushdown Automata

Exercise
• Find a CFG for nb = na+ 1 using the generic conversion process

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

Examples
• Revisit previous PDAs and check for determinacy

CS 3240 - Pushdown Automata

A Hierarchy of Languages

Languages accepted bynondeterministic PDA

Languages accepted by DFA

Languages accepted bydeterministic PDA

CS 3240 - Pushdown Automata