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Raphael
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My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let $\Sigma = \{0,1\}$ and $B$ is the collection of all strings that contain at least one $1$ in the second half. To state it more precisely: $B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes $B$. Give a diagram to describe your PDA.

Let $\Sigma = \{0,1\}$ and $B$ is the collection of all strings that contain at least one $1$ in the second half. To state it more precisely: $B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes $B$. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a $1$, go to the accept state. For example if $u=1001010101$ and $v=000011$, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at $v$. Then when I read the first $1$, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let $\Sigma = \{0,1\}$ and $B$ is the collection of all strings that contain at least one $1$ in the second half. To state it more precisely: $B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes $B$. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a $1$, go to the accept state. For example if $u=1001010101$ and $v=000011$, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at $v$. Then when I read the first $1$, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let $\Sigma = \{0,1\}$ and $B$ is the collection of all strings that contain at least one $1$ in the second half. To state it more precisely: $B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes $B$. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a $1$, go to the accept state. For example if $u=1001010101$ and $v=000011$, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at $v$. Then when I read the first $1$, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

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David Richerby
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Push Down Automata Confusion PDA recognising all strings with Problema $1$ in the second half

LaTeX ahoy
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Luke Mathieson
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My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let sigma={0,1)$\Sigma = \{0,1\}$ and B$B$ is the collection of all strings that contain at least one 1$1$ in the second half. To state it more precisely: B={u,v|u is an element of sigma^, v is an element of sigma^ 1 sigma^* and |u|>=|v|$B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes B$B$. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a 1$1$, go to the accept state. For example if u=1001010101$u=1001010101$ and v=000011$v=000011$, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at v$v$. Then when I read the first 1$1$, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let sigma={0,1) and B is the collection of all strings that contain at least one 1 in the second half. To state it more precisely: B={u,v|u is an element of sigma^, v is an element of sigma^ 1 sigma^* and |u|>=|v|. Give a PDA that recognizes B. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a 1, go to the accept state. For example if u=1001010101 and v=000011, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at v. Then when I read the first 1, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

My professor gave us an old exam to look over for our final exam and I am having a hard time understanding the push down automata problem he gave. In the problem it says:

Let $\Sigma = \{0,1\}$ and $B$ is the collection of all strings that contain at least one $1$ in the second half. To state it more precisely: $B=\{uv\mid u \in \Sigma^{\ast}, v \in \Sigma^{\ast} 1 \Sigma^{\ast}, |u|\geq|v|\}$. Give a PDA that recognizes $B$. Give a diagram to describe your PDA.

My question is why do I need a PDA or really a stack for this because all I am looking at is the second half which I can just epsilon to the second half and then when I read a $1$, go to the accept state. For example if $u=1001010101$ and $v=000011$, wouldnt I just loop around for a bit for u and then epsilon over to say I am now looking at $v$. Then when I read the first $1$, I just accept. I wouldnt need to use the stack at all would I? I'm not sure if I understand it correctly or not and would appreciate any help.

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