# acceptance of a^n b^m c^n ; n>1, m≥0

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is it possible with deterministic PDA?

yes,it is possible with DPDA.

push all 'a' s into the stack.

whenevr 'b' comes change to next state and skip all 'b' s.

and 'c' encounters when 'a' is on top of the stack pop it from stack.finally whenever it sees $\epsilon$ it accept the string.

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but if theres no 'b', as m≥0
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should (c,a/E) be on the same transition as skip b to next state say q1 and then again define (c,a/E) on q1 for the case if the previous transition was due to b
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if no 'b' 's directly go to the next state that is where 'c' is defined then perform remaing opeartions.

c,a/$\epsilon$