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I have a problem in constructing a Context-free Grammar for the Language $$L = \{a^mb^n : m≠n,m>0,n>0\} .$$ Though I can able to construct a Pushdown Automata.

pda

I can construct a CFG, but it accepts both $m \neq n$ and $m=n$. My CFG is:

$S \to aS/Sb/aSb/aab/abb $.

I cannot able to consturct the CFG to accept for $m \neq n$ alone.

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up vote 3 down vote accepted

How about

S ::= aSb | A | B
A ::= a | Aa
B ::= b | Bb
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This will accept if my input is 'a' but the condition is both m,n>0 – EAGER_STUDENT Sep 18 '11 at 16:02
    
@EAGER_STUDENT it is trivial to modify Henning's answer to make both m,n > 0. Just add T ::= aSb as the first production rule (and always start at T, this will force us to have at least one a and b. – Artem Kaznatcheev Sep 18 '11 at 16:06

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