CS 461 – Sept. 28 Section 2.2 – Pushdown Automata { 0n 1n } Palindromes Equal Next: Converting CFG  PDA

{ 0n 1n } While reading 0s, push them. If you read 1, change state. Then, as you read 1s, pop 0s off the stack. Watch out for bad input! Unspecified transition  crash (reject) Now, let’s write this in the form of a table.

POP & Go to state “reading 1” PDA δ for { 0n 1n } State Reading 0 Reading 1 Tos (don’t care) Empty Input 1 Action Push 0 POP & Go to state “reading 1” Crash Pop Notes: Action depends on input symbol AND what’s on top of stack. Action includes manipulating stack AND/OR changing state.

Example #2 { w # wR } (easier form of palindrome) Let’s design a PDA that accepts its input by empty stack, as before. Think about: state, top-of-stack, input, action

PDA δ for { w # wR } Before # After # (don’t care) 1 # State Tos Input 1 Input # Action Push 0 Push 1 Go to “after #” Pop Crash

Palindrome PDA? Before # After # (don’t care) 1 # State Tos Input 1 Input # Action Push 0 Push 1 Go to “after #” Pop Crash Changes needed: Non-deterministically go to “after #” when you push 0 or 1. Also, non-deterministically don’t push, in case we are dealing with odd-length palindrome!

Other examples Think about these “equal” language More 1s than 0s. How many states do we need? … More 1s than 0s. Twice as many 1s as 0s. Hint: think of the 0s as counting double.

Equal PDA State (Just one state) Tos ε 1 Input Action Push 0 Push 1 1 Input Action Push 0 Push 1 Pop