Let G = (AN , AT , S, P) be a context-free grammar in Chomsky
normal...
Let G = (AN , AT , S, P) be a context-free grammar in Chomsky
normal form. Prove that if there exists a word w ∈ L(G) generated
by a derivation that uses more than |P| + |AT | steps, then L(G) is
infinite.
Consider the context-free grammar G = ( {S}, {a, b}, S, P) where
P = { S -> aaSb | aab }.
Construct a NPDA M that such that L(M) = L(G).
I would like the transition graph for this NPDA please.
Theory of Computation Problem
Let G be an arbitrary CFG (Context-free Grammer), and
let DG be the string-pushing
PDA(pushdown automata) for it. Let w be
some string of length n in L(G).
Suppose you know that the leftmost derivation of
w in G consists of m
substitutions. How many transitions would be in the corresponding
computation of DG for input string
w? Justify your answer
carefully.
Consider the following context-free grammar
G:
E ® T +
E ® * T i
E ® f i
E ® * f +
T ® +
Questions:
(5 points) Compute the Canonical LR(1) Closure
set for state I0 for grammar G.
(10 points) Compute (draw) the DFA that
recognizes the Canonical LR(1) sets of items for grammar G.
(5 points) Construct the corresponding
Canonical LR(1) parsing table.
(10 points) Compute (draw) the DFA for
LALR(1).
(5 points) Construct LALR(1)...
For a grammar G with the productions where G = ( {S, A, B},
{a, b}, S, P ) with productions
S --> AB |
bbbB,
A --> b |
Ab, B -->
a..
1.Show that the grammar G is ambiguous.
2.Give language L that is generated by G, L = L(G), in a formal
expression (including a regular expression).
3.Can you construct an unambiguous grammar that is equivalent to
G? Otherwise, show that G is inherently ambiguous.
Given a grammar G, G = (Ν, Σ, Π, S), where
Ν = { ... }
Σ = { ... }
Π = { ... }
S is ... For a string ... write
(a) A leftmost derivation
(b) A rightmost derivation
Let G be a group and K ⊂ G be a normal subgroup. Let H ⊂ G be a
subgroup of G such that K ⊂ H Suppose that H is also a normal
subgroup of G. (a) Show that H/K ⊂ G/K is a normal subgroup. (b)
Show that G/H is isomorphic to (G/K)/(H/K).