Why is it that the transition function for DPDA's only works for 1 alphabet symbol, and 1 stack symbol? Say f is the transition function, why does having
f(qi, 1, 0) -> (qj, 0) f(qi, epsilon, 0) -> (qj, 0)
I would understand that if I had
f(qi, 1, 0) -> (qj, 0) f(qi, 1, 0) -> (qz, 0)
it would cause nondeterminism, because looking at the same input symbol I can choose to move to either qj or qz, that's nondeterminism to me. However the former doesn't make sense at all... This is what Michael Sipser says in his book.