# How do you obtain transition relation of a PDA?

I know how to figure out the start state, accepting state, input alphabet, and all that stuff. But how do you develop the transition relation of a PDA? For an FSM, (q0,a),q1) means if you start at q0 and get an a, you transition to q1. But what does (S,a,e),(S,a) mean? (S is start state and e is epsilon)

Here's a picture if that helps. I want to understand the circled part. I will be extremely appreciative of any help.

• I'm not sure what you're asking. At one part, you ask what $(S,a,\epsilon),(S,a)$ means -- is that what you want to know? If so, what research/self-study have you done, and where exactly did you get stuck? Have you read your textbook's chapter on PDA's? Have you read Wikipedia's article on PDAs? It even has a sentence that describes what this means, starting with the phrase "It has the intended meaning that...." Alternatively, if that isn't what you meant to ask, I think you should edit the question to make things clearer. – D.W. Nov 21 '13 at 17:02

As you can see, $\Delta$ is a set of pairs that represent a transition function. First element of the pair is a triple $(q, a, X)$, where $q$ is a state, $a$ is an input symbol (possibly empty string), and $X$ is the top stack symbol.
The other element of the pair describes the action of PDA when its configuration fits the above triple. This other element is again a pair $(p, \gamma)$, where $p$ is a new state and $\gamma$ is a sequence of symbols replacing $X$. If $\gamma = \epsilon$, then $X$ is just popped.
It might help to view $((q, a, X), (p, \gamma))$ as
$$(q, a, X) \rightarrow (p, \gamma)$$