Construct Pushdown automata for L = {0ⁿ1^m2^m3ⁿ | m,n ≥ 0}

Last Updated : 20 Jul, 2018

Prerequisite – Pushdown automata, Pushdown automata acceptance by final state
Pushdown automata (PDA) plays a significant role in compiler design. Therefore there is a need to have a good hands on PDA. Our aim is to construct a PDA for L = {0ⁿ1^m2^m3ⁿ | m,n ≥ 0}

Examples –

Input : 00011112222333
Output : Accepted

Input : 0001122233
Output : Not Accepted

Approach used in this PDA –
There can be 4 cases while processing the given input string.

Case-1: m = 0 – In this cases the input string will be of the form {0ⁿ3ⁿ}. In this condition, keep on pushing 0’s in the stack until we encounter with 3. On receiving 3 check if top of stack is 0, then pop it (0) from the stack. Pop 0’s until all the 3’s of the input string are processed. If we reach to the end of input string and stack becomes empty, then reach to final state i.e. accept the input string else move to dead state.

Case-2: n = 0 – In this cases the input string will be of the form {1^m2^m}. In this condition, keep on pushing 1’s in the stack until we encounter with 2. On receiving 2 check if top of stack is 1, then pop it (1) from the stack. Keep on pop 1’s until all the 2’s of the input string are processed. If we reach to the end of input string and stack becomes empty, then reach to final state i.e. accept the input string else move to dead state.

Case-3: m, n>0 – In this cases the input string will be of the form {0ⁿ1^m2^m3ⁿ}. In this condition, keep on pushing 0’s and 1’s in the stack until we encounter with 2. On receiving 2 check if top of stack is 1, then pop it (1) from the stack. Keep on pop 1’s until all the 2’s of the input string are processed. Then on receiving 3 check if top of stack is 0, then pop it (0) from the stack. Pop 0’s until all the 3’s of the input string are processed. If we reach to the end of input string and stack becomes empty, then reach to final state i.e. accept the input string else move to dead state.