Need a PDA for L={ a^n b^m c^m d^n n,m>=1 }
$begingroup$
I am trying to desing a PDA for automata lecture.Language is L={ a^n b^m c^m d^n n,m>=1 }
δ(q0, a, Z) = (q0, aZ)
δ(q0, a, a) = (q0, aa)
δ(q0, b, a) = (q1, ba)
δ(q1, b, b) = (q1, bb)
δ(q1, c, b) = (q2, ε)
δ(q2, c, b) = (q2, ε)
δ(q2,d , a) = (q3, ε)
δ(q3, d, a) = (q4, ε)
δ(q4, ε, Z) = (q4, Z)
States are q0,q1,q2,q3,q4.Z means empty stack.ε means push nothing to the stack.And q4 is final state.I designed it that way but i am not sure if it is correct.Can you check the answer and help me to fix it?
automata context-free-grammar
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
$endgroup$
add a comment |
$begingroup$
I am trying to desing a PDA for automata lecture.Language is L={ a^n b^m c^m d^n n,m>=1 }
δ(q0, a, Z) = (q0, aZ)
δ(q0, a, a) = (q0, aa)
δ(q0, b, a) = (q1, ba)
δ(q1, b, b) = (q1, bb)
δ(q1, c, b) = (q2, ε)
δ(q2, c, b) = (q2, ε)
δ(q2,d , a) = (q3, ε)
δ(q3, d, a) = (q4, ε)
δ(q4, ε, Z) = (q4, Z)
States are q0,q1,q2,q3,q4.Z means empty stack.ε means push nothing to the stack.And q4 is final state.I designed it that way but i am not sure if it is correct.Can you check the answer and help me to fix it?
automata context-free-grammar
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
$endgroup$
$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08
add a comment |
$begingroup$
I am trying to desing a PDA for automata lecture.Language is L={ a^n b^m c^m d^n n,m>=1 }
δ(q0, a, Z) = (q0, aZ)
δ(q0, a, a) = (q0, aa)
δ(q0, b, a) = (q1, ba)
δ(q1, b, b) = (q1, bb)
δ(q1, c, b) = (q2, ε)
δ(q2, c, b) = (q2, ε)
δ(q2,d , a) = (q3, ε)
δ(q3, d, a) = (q4, ε)
δ(q4, ε, Z) = (q4, Z)
States are q0,q1,q2,q3,q4.Z means empty stack.ε means push nothing to the stack.And q4 is final state.I designed it that way but i am not sure if it is correct.Can you check the answer and help me to fix it?
automata context-free-grammar
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
$endgroup$
I am trying to desing a PDA for automata lecture.Language is L={ a^n b^m c^m d^n n,m>=1 }
δ(q0, a, Z) = (q0, aZ)
δ(q0, a, a) = (q0, aa)
δ(q0, b, a) = (q1, ba)
δ(q1, b, b) = (q1, bb)
δ(q1, c, b) = (q2, ε)
δ(q2, c, b) = (q2, ε)
δ(q2,d , a) = (q3, ε)
δ(q3, d, a) = (q4, ε)
δ(q4, ε, Z) = (q4, Z)
States are q0,q1,q2,q3,q4.Z means empty stack.ε means push nothing to the stack.And q4 is final state.I designed it that way but i am not sure if it is correct.Can you check the answer and help me to fix it?
automata context-free-grammar
automata context-free-grammar
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
asked Dec 30 '18 at 15:58
Nilgün DağıdırNilgün Dağıdır
11
11
$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08
add a comment |
$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08
$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08
$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08
add a comment |
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$begingroup$
Seems correct if you ask me.
$endgroup$
– Wuestenfux
Dec 31 '18 at 12:08