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Construct PDA for L = {0n1m2(n+m) | m,n >=1}
A push down automata (PDA) can be formally described as seven tuples
(Q, Σ,S, δ,q0,I,F)
Where,
- Q is finite number of states
- Σ is input alphabet
- S is stack symbol
- Δ is the transition function: QX(Σ∪{e})XSXQ
- q0 is the initial state (q0 belongs to Q)
- I is the initial state top symbol
- F is a set of accepting states(F belongs to Q)
Problem
Construct PDA for 0n1m2(n+m) where n,m>=1.
Solution
So, the strings which are generated by the given language are as follows −
L={0122,001222,000112222,….}
That is to add the number of 0's and 1's, and that will equal the number of 2's.
So for every 0's and 1's we will pop 2's from the stack.
Let’s count the number of 0's and 1's that total number is equal to the number of 2's.
It can be achieved by pushing 0's and 1's in STACK and then pop 0's and 1's whenever "2" comes.
Finally, at the end of the strings if nothing is left in the STACK then we can say that language is accepted in the PDA.
The PDA for the given problem is as follows −
Transition Function
Step 1: δ(q0, 0, Z) = (q0, 0Z)
Step 2: δ(q0, 0, 0) = (q0, 00)
Step 3: δ(q0, 1, 0) = (q0, 10)
Step 4: δ(q0, 1, 1) = (q0, 11)
Step 5: δ(q0, 2, 1) = (q1, ε)
Step 6: δ(q1, 2, 1) = (q1, ε)
Step 7: δ(q1, 2, 0) = (q1, ε)
Step 8: δ(q1, ε, Z) = (qf, Z)
Explanation
Step 1 − Let’s take the input string: "00112222” which satisfies the given condition.
Step 2 − Scan string from left to right.
Step 3 − For input '0' and STACK alphabet Z, then
Push the input '0' into STACK: (0,Z/0Z) and the state will be q0.
Step 4 − For input '0' and STACK alphabet 0', then
Push the input '0' into STACK: (0,0/00) and state will be q0.
Step 5 − For input '1' and STACK alphabet '0', then
Push the input '1' into STACK: (1,0/10) and state will be q0.
Step 6 − For input '1' and STACK alphabet '1', then
Push the input '1' into STACK: (1,1/11) and state will be q0.
Step 7 − For input '2' and STACK alphabet '1' and state is q0, then
Pop one '1' as: (2,1/ε) and the state will be q1.
Step 8 − For input ‘2' and STACK alphabet '1' and state is q1, then
Pop one '1': (2,1/ε) and the state remains q1.
Step 9 − For input '2' and STACK alphabet '0' and state is q1, then
Pop one ‘0': (2,0/ε) and the state will be q1.
Step 10 − For input '2’ and STACK alphabet '0' and state is q1, then
Pop one '0' as: (2,0/ε) and the state will remain q1.
Step 11 − We reached end of the string, for input ε and STACK alphabet Z, then
Go to final state(qf) : (ε, Z/Z)*.
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