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# Construct Pushdown automata for L = {0m1(n+m)2n | m,n = 0} in C++

We are given with a language “L” and the task is to construct a pushdown automata for the given language which explains that the occurrences of 1’s will be the addition of occurrences of 0’s and 2’s and also, occurrence of 0 and 2 will be minimum one which can also makes the string NULL and it should be accepted by the automata.

## What is pushdown Automata?

A pushdown automata or pushdown automaton or PDA is a technique to implement a context−free grammar in a similar way we design Deterministic Finite Automaton or DFA for a regular grammar. A DFA can operate on finite data, but a PDA can operate on infinite data. We can understand pushdown automata as the combination of “finite state machine” and a “stack”.

A pushdown automaton has three components −

an input tape,

a control unit, and

a stack with infinite size.

A PDA can be formally described as a 7-tuple (Q, Σ, S, δ, q0, I, F) −

Q is the finite number of states

Σ is input alphabet

S is stack symbols

δ is the transition function: Q × (Σ υ {ε}) × S × Q × S*

q0 is the initial state (q0 Ε Q)

I is the initial stack top symbol (I Ε S)

F is a set of accepting states (F Ε Q)

**Let’s construct a pushdown Automata for the given language** −

The strings that are acceptable by this PDA are of the form −

0

^{m}1^{m}− 01, 0011, 000111 etc. No. of 0s is equal to no. of 1s. When n is 0 we will have no 2s. Keep pushing 0s and as soon as the first 1 is encountered then pop 0s. If we reach the end of the string and are left with no 0s then the string is accepted.1

^{n}2^{n}− 12, 1122, 111222 etc. No. of 1s is equal to no. of 2s. When m is 0 we will have no 0s. Keep pushing 1s and as soon as the first 2 is encountered then pop 1s. If we reach the end of the string and are left with no 1s then the string is accepted.0

^{n}1^{m+n}2^{m}− 0112, 001112, 001112 etc. No. of 1s is equal to sum of no. of 0s and 2s. Keep pushing 0s as soon as the first 1 is encountered then pop 0s until no 0s are left. Now again push 1s until the first 2 is encountered. Then for each 2 start popping 1s until we are left with no 1s. If we reach the end and no 1 is left then the string is accepted.NULL string is also accepted. 0

^{0}1^{0}2^{0}

## Let’s understand the machine

Transitions for state q0 −

( 0, I/0I ) − If top of stack is I and current input symbol is 0 then push 0 to top of stack and remain at q0. Stack becomes 0I...

( 0, 0/00 ) − If top of stack is 0 and current input symbol is also 0 then push 0 to top of stack and remain at q0. Stack becomes 00.... Keep pushing 0s until the next 1 or 2.

( 1, I/1I ) − If top of stack is I and current input symbol is 1 then push 1 to top of stack and move to q1. Stack becomes 1I...

( 1, 0/$ ) − If top of stack is 0 and current input symbol is 1 then pop 0 and move to q1.

( 1, 0/$ ) − If top of stack is 0 and current input symbol is 1 then pop 0 and move to q1.

Transitions for state q1 −

( 1, 1/11 ) − If top of stack is 1 and current input symbol is also 1 then push 1 to top of stack and remain at q1. Stack becomes 11.... Keep pushing 1s until the next 0 or 2.

( 1, 0/$ ) − If top of stack is 0 and current input symbol is 1 then pop 0 and remain at q1.

( $, I/I ) − If top of stack is I and there is no input then do nothing and move to qf.

( 2, 1/$ ) − If top of stack is 1 and current input symbol is 2 then pop 1 and move to q2.

Transitions for state q2 −

( 2, 1/$ ) − If top of stack is 1 and current input symbol is 2 then pop 1 and remain at q2.

( $, I/I ) − If top of stack is I and there is no input then do nothing and move to qf.

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