INHERENT LIMITATIONS OF COMPUTER PROGAMS

CSci 4011 INHERENT LIMITATIONS OF COMPUTER PROGAMS

string pop push ε,ε → $ 0,ε→ 0 1,0→ε ε,$ → ε 1,0→ε The language of P is the set of strings it accepts.

CONTEXT-FREE GRAMMARS A → 0A1 A → B B → # 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.

GIVE A CFG FOR… L1 = { strings of balanced parens } L2 = { aibjak | i, j, k ≥ 0 and k = i + j }

VERY INTERESTINGCFGs… http://java.sun.com/docs/books/jls/third_edition/html/j3TOC.html http://docs.python.org/reference/grammar.html

COMPILER MODULES LEXER PARSER SEMANTIC ANALYZER TRANSLATOR/INTERPRETER

PARSE TREES A A A B 0 0 # 1 1 A  0A1  00A11  00B11  00#11

Suppose L is generated by a CFG G = (V, Σ, R, S) Construct P = (Q, Σ, Γ, , q, F) that recognizes L A Language is generated by a CFG  It is recognized by a PDA  Idea: P will derive w ∈L on its stack.

(1) Place the marker symbol $ and the start variable on the stack Suppose L is generated by a CFG G = (V, Σ, R, S) Construct P = (Q, Σ, Γ, , q, F) that recognizes L (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.

Suppose L is generated by a CFG G = (V, Σ, R, S) Construct P = (Q, Σ, Γ, , q, F) that recognizes L (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,A → w for rule A → w a,a → ε for terminal a ε,$ → ε

S → aTb T → Ta | ε ε,ε → $ ε,ε → T ε,S → b ε,ε → T ε,ε → S ε,T → a ε,$ → ε ε,ε → a ε,T → ε a,a → ε b,b → ε

A Language is generated by a CFG  It is recognized by a PDA 

A Language is generated by a CFG  It is recognized by a PDA  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

SIMPLIFY ε,ε → $ 0,ε→ 0 q1 q0 1,0→ε ε,ε → 0 ε,ε → ε ε,$ → ε q2 q3 1,0→ε ε,ε → 0 ε,ε → ε q4 ε,0 → ε 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,qQ } S = Aq0qaccept *starting from any stack S in p, P has stack S at q.

ε,ε → $ 0,ε→ 0 q1 q0 1,0→ε ε,ε → 0 ε,$ → ε q2 q3 1,0→ε ε,ε → 0 q4 ε,0 → ε q5 none What strings does Aq0q1 generate? {0n1n | n > 0} What strings does Aq1q2 generate? What strings does Aq1q3 generate? 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

1. The first repeat is at the end of x: stack height r s a b input string p q Apq→ aArsb

2. The stack repeats before the end of x: stack height input string p r q Apq→ AprArq

Formally: V = {Apq | p,qQ } S = Aq0qaccept 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 q1 q0 1,0→ε ε,ε → 0 ε,$ → ε q2 q3 1,0→ε ε,ε → 0 q4 ε,0 → ε q5 Aqq→ ε Apq→ AprArq Aq0q3 → εAq1q2ε Aq1q2 → 0Aq1q21 Aq1q2 → 0Aq1q11 none What strings does Aq0q1 generate? {0n1n | n > 0} What strings does Aq1q2 generate? What strings does Aq1q3 generate? none

Apq generates x  x can bring P from p to q without changing the stack  Proof (by induction on the number of steps in the derivation of x from Apq): Inductive Step: Base Case: The derivation has 1 step: Appε Assume true for derivations of length at most k and prove true for derivations of length k+1 Apq x in k+1 steps Apq → AprArq Apq → aArsb (r,t)  (p,a,ε) and (q, ε)  (s,b,t) state push state alphabet pop

Apq generates x  x can bring P from p to q without changing the stack  Proof (by induction on the number of steps in the computation of P on input x): Base Case: The computation has 0 steps We must show that App x But it must be that x = ε, so we are done

Inductive Step: Assume true for computations of length at most k and prove true for computations of length k+1 Suppose that P has a computation wherein x brings p to q in k+1 steps Two cases: 1. The stack repeats only at the end of this computation 2. The stack repeats somewhere in the middle of the computation

A Language is generated by a CFG  It is recognized by a PDA

INHERENT LIMITATIONS OF COMPUTER PROGAMS