Mixing
PDA: Symbol in first half
How do I construct a PDA where:
$L = {w in {0, 1} ∗ : |w| text{ is even and } w text{ contains at least one 1-Symbol in the first Half}}$
To me it seems impossible to know when I reached the midpoint.
computer-science formal-languages automata
asked Nov 21 '18 at 17:19
ZerataX
164
add a comment |
How do I construct a PDA where:
$L = {w in {0, 1} ∗ : |w| text{ is even and } w text{ contains at least one 1-Symbol in the first Half}}$
To me it seems impossible to know when I reached the midpoint.
computer-science formal-languages automata
asked Nov 21 '18 at 17:19
ZerataX
164
Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
1
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14
add a comment |
How do I construct a PDA where:
$L = {w in {0, 1} ∗ : |w| text{ is even and } w text{ contains at least one 1-Symbol in the first Half}}$
To me it seems impossible to know when I reached the midpoint.
computer-science formal-languages automata
asked Nov 21 '18 at 17:19
ZerataX
164
How do I construct a PDA where:
$L = {w in {0, 1} ∗ : |w| text{ is even and } w text{ contains at least one 1-Symbol in the first Half}}$
To me it seems impossible to know when I reached the midpoint.
computer-science formal-languages automata
computer-science formal-languages automata
asked Nov 21 '18 at 17:19
ZerataX
164
asked Nov 21 '18 at 17:19
ZerataX
164
edited Nov 22 '18 at 2:45
asked Nov 21 '18 at 17:19
ZerataX
164
asked Nov 21 '18 at 17:19
ZerataX
164
asked Nov 21 '18 at 17:19
ZerataX
164
164
Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
1
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14
add a comment |
Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
1
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14
Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
1
1
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14
add a comment |
1 Answer
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So this is my solution:
I just stack 1s until I find the first 1-Symbol, then I pop them to see if the first 1 was in the first or second half and then I just make sure that the word is even.
answered Nov 23 '18 at 0:01
ZerataX
164
add a comment |
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So this is my solution:
I just stack 1s until I find the first 1-Symbol, then I pop them to see if the first 1 was in the first or second half and then I just make sure that the word is even.
answered Nov 23 '18 at 0:01
ZerataX
164
add a comment |
So this is my solution:
I just stack 1s until I find the first 1-Symbol, then I pop them to see if the first 1 was in the first or second half and then I just make sure that the word is even.
answered Nov 23 '18 at 0:01
ZerataX
164
add a comment |
So this is my solution:
I just stack 1s until I find the first 1-Symbol, then I pop them to see if the first 1 was in the first or second half and then I just make sure that the word is even.
answered Nov 23 '18 at 0:01
ZerataX
164
So this is my solution:
I just stack 1s until I find the first 1-Symbol, then I pop them to see if the first 1 was in the first or second half and then I just make sure that the word is even.
answered Nov 23 '18 at 0:01
ZerataX
164
edited Nov 23 '18 at 19:27
answered Nov 23 '18 at 0:01
ZerataX
164
answered Nov 23 '18 at 0:01
ZerataX
164
answered Nov 23 '18 at 0:01
ZerataX
164
164
add a comment |
add a comment |
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Parallelly, try also the pumping lemma.
– Berci
Nov 21 '18 at 17:34
The assignment doesn't say that the PDA needs to be deterministic.
– rici
Nov 21 '18 at 19:11
...although a DPDA does exist, too.
– rici
Nov 21 '18 at 21:13
1
Maybe it helps to reformulate the set definition. Equivalently we could say: $|w|$ is even, $w$ contains at least one 1-symbol, and the first 1-symbol that occurs is in the first half of $w$.
– Peter Leupold
Nov 22 '18 at 12:27
figured it out yesterday by myself, but @PeterLeupold reformulation really helps a lot. thanks!
– ZerataX
Nov 23 '18 at 0:14