GATE | GATE CS 1997 | Question 55
Consider the grammar
S→ bSe S→ PQR P→ bPc P→ ε Q→ cQd Q→ ε R→ dRe R→ ε
where S,P,Q,R are non-terminal symbols with S being the start symbol; b,c,d,e are terminal symbols and ‘ε’ is the empty string. This grammar generates strings of the form bi, cj, dk, em for some i, j, k, m ≥ 0.
- (a). What is the condition on the values of i, j, k, m ?
- (b). Find the smallest string that has two parse trees.
Explanation: (a). Condition on the values of i, j, k, m
i+k = j+m
where i, j, k, m >= 0
(b). Smallest string that has two parse trees = bcde
Production used to generate the smallest string is :-
S- > bSe S- > bSe S- > bSe S -> PQR P-> null Q ->cQd Q -> null R- > null
Finally you will get the string like ” bbbcdeee”
which means that i=3, j=1, k=1, m=3 and hence the answer i+k=j+m .
Now production used to generate the smallest string is :-
S-> bSe S -> bPQRe S -> bεQRe S -> bcQdRe S -> bcεdRe S -> bcdεe S -> bcde
Hence smallest string is bcde.
Therefore we can see same number of b, c, d, e is generated.
Power of b, c, d, e are respectively i, j, k, m.
Hence relation between i, j, k, m is :-
i+k = j+m
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