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    <p>you can decide where to go from a current state based on the value on top of the stack, you use symbols on the stack to keep notes about the state of the parsing.</p> <p>here is how i think this would work: <strong>this</strong> are symbols read from the input, <strong>THIS</strong> are symbols on a stack. the symbol <strong>X</strong> on the botoom of the stack means that n &lt;= m do not confuse <strong>X</strong> with <strong>Z</strong>, which is the initial symbol of the stack and helps determine when the stack is empty.</p> <p>there are probably some problems with this solution, but it the overall approach should be correct.</p> <p>... and good luck with your test :-)</p> <hr> <p>first you read all the <strong>a</strong> symbols from the beginning of the string and add <strong>A</strong> to a stack for each of them, or push <strong>X</strong> if there was no <strong>a</strong></p> <p>then you read all the <strong>b</strong> symbols:</p> <ul> <li>if the stack is empty (<strong>Z</strong> is on top), <strong>B</strong> is on top or <strong>X</strong> is on top, then you push another <strong>B</strong> to the stack.</li> <li>if stack has <strong>A</strong> on top, then you remove it.</li> </ul> <p>the last step is to read the final <strong>a</strong> symbols.</p> <ul> <li>if there is <strong>B</strong> on the stack, remove it.</li> <li>if there is <strong>X</strong> on the stack, then keep it there</li> <li>if the stack is empty (<strong>Z</strong> on top), then this must be the end of the string</li> </ul> <hr> <p>another edit:</p> <p>sorry if the above isn't clear ... i'll try to formalize it.</p> <p>accepting states are (4) and (5), starting state is (1). and it's nondeterministic</p> <p>and the transition rules:</p> <p>state (1) : read the first batch of <strong>a</strong> symbols</p> <ul> <li>(1) <strong>a</strong>; <strong>Z</strong> / <strong>AZ</strong> -> (1)</li> <li>(1) <strong>a</strong>; <strong>A</strong> / <strong>AA</strong> -> (1)</li> <li>(1) epsilon; <strong>A</strong> / <strong>A</strong> -> (2)</li> <li>(1) epsilon; <strong>Z</strong> / <strong>Z</strong> -> (2)</li> </ul> <p>state (2) : read <strong>b</strong> symbols</p> <ul> <li>(2) <strong>b</strong>; <strong>Z</strong> / <strong>BZ</strong> -> (2)</li> <li>(2) <strong>b</strong>; <strong>X</strong> / <strong>BX</strong> -> (2)</li> <li>(2) <strong>b</strong>; <strong>B</strong> / <strong>BB</strong> -> (2)</li> <li>(2) <strong>b</strong>; <strong>A</strong> / epsilon -> (2)</li> <li>(2) epsilon; <strong>B</strong> / <strong>B</strong> -> (3)</li> <li>(2) epsilon; <strong>X</strong> / <strong>X</strong> -> (3)</li> <li>(2) epsilon; <strong>Z</strong> / <strong>Z</strong> -> (3)</li> </ul> <p>state (3) : read the last <strong>a</strong>s</p> <ul> <li>(3) <strong>a</strong>; <strong>B</strong> / nothing -> (3)</li> <li>(3) epsilon; <strong>X</strong> / <strong>X</strong> -> (4)</li> <li>(3) epsilon; <strong>Z</strong> / <strong>Z</strong> -> (5)</li> </ul> <p>state (4) : the trailing <strong>a</strong>s if m > n</p> <ul> <li>(4) <strong>a</strong>; <strong>X</strong> / <strong>X</strong> -> (4)</li> </ul> <p>state (5) is for accepting the exact string when m &lt; n</p> <p>(and just to be absolutely clear -- when there is no way of a state and the reading cursor is not at the end of the word, then the word is not accepted)</p> <p>This could maybe be made a bit simpler by using adidional states instead of the stack symbol <strong>X</strong>, but i guess you can do it yourself :-)</p>
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    1. POHow do I design the transition functions for this pushdown automaton?
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    1. POHow do I design the transition functions for this pushdown automaton?
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    1. COYou are using X as other people use Z right? I mean, the symbol being at the bottom of the stack when starting.
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    2. COno, that's another symbol :-) X is added to the stack only if n <= m. I kinda forgot about Z and assumed that you can tell that the stack is empty by some other means. Thanks for pointing it out, I'll fix that in the answer.
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    3. CO"f the stack is empty (Z is on top), B is on top or X is on top, then you push another B to the stack." If there's X on top do I push BX or just B?
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