Pushdown Automata (PDA)

Question

I'm trying to write a pda pushdown automata that accept a^2n b^n, n>0 but I'm not sure if the last part is correct

(p0, a, z0) = (p0, az0)
(p0, a, a) = (p0, aa)
(p0, b, a) = (p1, λ)
(p1, λ, b) = (p2, λ)   <=
(p2, 0, b) = (p1, λ)  <=
(p2, λ, z0) = (p3, λ)  <=

Still not quite (not enough opportunities to push 1's). Did you draw it? That might help — harold, Mar 13 '15 at 11:53
@harold now? I was only taught with some similar examples in these format...I'm not sure how to draw it — userNew, Mar 13 '15 at 12:13
Very close, as far as I can see it just needs a rule from p2 back to p1 to read some more 0's — harold, Mar 13 '15 at 12:35
Looks good, personally I'd be consistent and use `a` and `b` for the input symbols but that's a minor thing — harold, Mar 13 '15 at 12:41

score 0 · Answer 1 · answered Nov 25 '16 at 11:56

As far as your answer is concerned, in your first two steps you are pushing a's in only one step. With the current design the machine would accept aaabb which is not in the form a^2nb^n. So it's better to divide it in two states separately. According to me the right answer might be something like:

(p0, a, z0) = (p0, az0)
(p0, a, a) = (p1, aa)
(p1, a, a) = (p0, aa)
(p1, b, a) = (p2, λ)

score 0 · Answer 2 · edited Jul 21 '17 at 17:52

0

Deterministic push down automata for a^2nb^n n>=0
Bypass altetnate a's and push rest of a's

edited Jul 21 '17 at 17:52

user1859022

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answered Jul 21 '17 at 08:00

Vishal

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This PDA will also work for a^(2n - 1)b^n. To correct this the first state and can bypass a's and the in the second one you can push a's. – Vardan Agarwal Nov 03 '19 at 06:15

score 0 · Answer 3 · edited Oct 22 '19 at 12:23

0

q0 is initial state and qf is final is state; ^ is for null:

Q(q0,a,z)=Q(q1,z)
Q(q0,a,a)=Q(q1,aa)
Q(q0,b,a)=Q(q0,^)
Q(q0,^,z)=Q(qf,z)
Q(q1,a,z)=Q(q2,az)
Q(q1,a,a)=Q(q2,a,a)
Q(q2,a,a)=Q(q0,az)

edited Oct 22 '19 at 12:23

Adrian Mole

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answered Oct 22 '19 at 12:01

vivek

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Pushdown Automata (PDA)

3 Answers3