Assignment 1 Question 7 on PDAs

Te PDAs after page 302 leave out the REJECT states, can we do this too for assignment questions or must we include every possible REJECT state?

I put in all the reject states. (Just because they wanted a DPDA). Don’t know if it's right thou.

It seems that only on a non-deterministic PDA is the input allowed to crash, but on a deterministic PDA it must be accepted or rejected by ending up in said state. So rather keep them reject states in the diagram...

Yes..

I believe you can leave out the reject state, cause if you did not reach the accept state, then obviously you have crashed somewhere, and the word not accepted.

Nope you have to put in a reject state, or at least one and "everything" thats rejected can go to the one state. That's what the marker commented on my Assignment1. (I had multiple reject states)

just got my assignment back in the post, seems they wanted us to balance the (ab)* with the (ba)*. am i the only one who missed this?

i was also told that my reject states were redundant... that's only if you think of them as being seperate and like cluttered diagrams...

lycium Wrote:
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> just got my assignment back in the post, seems
> they wanted us to balance the (ab)* with the
> (ba)*. am i the only one who missed this?
>

Probably not. No mistakes are unique

> i was also told that my reject states were
> redundant... that's only if you think of them as
> being seperate and like cluttered diagrams...

Look on the bright side, at least you got some
valuable feedback. My assignment came back with
"See my comments at the end" written on the first
page of the assignment cover. Turning to the
last page (q7?) all I see was "good!" and an additional
comment about writing a coherent/readable symbol.

I'm still not too sure it was all correct

i don't recall seeing anything in the question about needing to balance the ab and ba terms. it's not that i find pushing (i'm a pimp) and popping (used to be a raver) difficult :|

lycium Wrote:
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> i don't recall seeing anything in the question
> about needing to balance the ab and ba terms.

From the tut101, a1, q7:
Draw a deterministic PDA that will accept the language
{(ab)^nbb(ba)^n | n >= 0}

From that it's pretty clear that you need to balance
the (ab) and (ba) strings, because n will be the same
on both sides.

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indeed, and i usually take pride in noticing more subtle stuff like that :| kinda dived right into it...