INHERENT LIMITATIONS OF COMPUTER PROGAMS CSci 4011.

INHERENT LIMITATIONS OF COMPUTER PROGAMS

CSci 4011

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

string pop push

The language of P is the set of strings it accepts.

A → 0A1A → BB → #

CONTEXT-FREE GRAMMARS

A 0A1 00A11 00B11 00#11

A derives 00#11 in 4 steps.

The language of G is the set of strings derived by S.

VERY INTERESTING CFGs…

http://java.sun.com/docs/books/jls/third_edition/html/j3TOC.html

http://docs.python.org/reference/grammar.html





PARSE TREES

A 0A1 00A11 00B11 00#11

A

B

0 1#

A

A

0 1

Definition. T is a parse tree for the derivation S ⇒* w under grammar G = (V,Σ,R,S) if

1. The root of T has label S

2. The leaves of T have labels wi Σ ∈

4. For each node with label v V and children ∈with labels ri (V Σ), (v→r) R.∈ ∪ ∈

3. The non-leaf nodes have labels v V∈

<EXPR> → <EXPR> + <EXPR>

<EXPR> → <EXPR> x <EXPR>

<EXPR> → ( <EXPR> )

<EXPR> → a

Build a parse tree for a + a x a

<EXPR>

a x+ a

<EXPR><EXPR>

a

<EXPR> <EXPR>

<EXPR>

a+ xa

<EXPR><EXPR>

a

<EXPR><EXPR>

COMPILER MODULES

LEXER

PARSER

SEMANTIC ANALYZER

TRANSLATOR/INTERPRETER

A Language is generated by a CFG

It is recognized by a PDA

Suppose L is generated by a CFG G = (V, Σ, R, S)

Construct P = (Q, Σ, Γ, , q, F) that recognizes L

Idea: P will derive w ∈ L on its stack.



(1) Place the marker symbol $ and the start variable on the stack

(2) Repeat forever:

(a) If v is in the stack, and (v → s) R, ∈push s on the stack(b) If stack is a string, goto (3)

(3) Loop until stack is empty: (a) if top of stack matches input, pop. (b) on (ε,$), accept.



(qstart) Push S$ and go to qloop

(qloop) Repeat the following steps forever:

(a) On (ε,v) where (v → s) R, push s and ∈go to qloop

(b) On (,), pop and go to qloop

(c) On (ε,$) go to qaccept

(else) get stuck!

ε,ε → S$

ε,$ → ε

ε,A → w for rule A → w a,a → ε for terminal a

S → aTb

T → Ta | ε

ε,ε → $

ε,$ → ε

ε,ε → Sε,S

→ b

ε,ε → T

ε,T → a

ε,ε → a

ε,ε → T

ε,T → εa,a → εb,b → ε



Given PDA P = (Q, Σ, Γ, , q, F)

Construct a CFG G = (V, Σ, R, S) such that L(G)=L(P)

First, simplify P so that:

(1) It has a single accept state, qaccept

(2) It empties the stack before accepting

(3) Each transition either pushes a symbol or pops a symbol, but not both

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

SIMPLIFY

q0 q1

q2q3

ε,ε → ε

ε,ε → ε

ε,0 → ε

ε,ε → 0

ε,ε → 0

q4

q5

Idea: for each pair of states p and q in P, the grammar will have a variable Apq that generates all strings that that can take P from p to q without changing the stack*

V = {Apq | p,qQ }

S = Aq0qaccept

*starting from any stack S in p, P has stack S at q.

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

q0 q1

q2q3

ε,0 → ε

ε,ε → 0

ε,ε → 0

q4

q5

What strings does Aq0q1 generate?



none

{0n1n | n > 0}

none

Apq generates all strings that take P from p to q without changing the stack

Let x be such a string• P’s first move on x must be a push

• P’s last move on x must be a pop

Consider the stack while reading x. Either:

1. The first repeat comes at the end of x

2. The stack repeats before the end of x

stackheight

inputstring p q

Apq → aArsb

r s

1. The first repeat is at the end of x:

ba

stackheight

inputstring p r q

Apq → AprArq

2. The stack repeats before the end of x:

V = {Apq | p,qQ }

S = Aq0qaccept

Formally:

For each p,q,r,s Q, t Γ and a,b Σε

If (r,t) (p,a,ε) and (q, ε) (s,b,t)

Then add the rule Apq → aArsb

For each p,q,r Q,

add the rule Apq → AprArq

For each p Q,

add the rule App → ε

ε,ε → $0,ε → 0

1,0 → ε

1,0 → εε,$ → ε

q0 q1

q2q3

ε,0 → ε

ε,ε → 0

ε,ε → 0

q4

q5




Aq0q3 → εAq1q2ε

Aqq → ε

Apq → AprArq

Aq1q2 → 0Aq1q21

Aq1q2 → 0Aq1q11

none

{0n1n | n > 0}

none

INHERENT LIMITATIONS OF COMPUTER PROGAMS CSci 4011.

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