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Construct DPDA for anbncm where, n,m≥1 in TOC
DPDA is the short form for the deterministic push down automata (DPDA).
Problem
Construct DPDA for anbncm where m,n>=1
Solution
So, the strings which are generated by the given language are −
L={abc,aabbc,aaabbbcc,….}
That is we have to count equal number of a’s, b’s and different number of c’s
Let’s count the number of a's which is equal to the number of b's.
This can be achieved by pushing a's in STACK and then we will pop a's whenever "b" comes.
Then for c nothing will happen.
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
The transition functions are as follows −
Step 1 − δ(q0, a, Z) = (q0, aZ)
Step 2 − δ(q0, a, a) = (q0, aa)
Step 3 − δ(q0, b, a) = (q1, ε)
Step 4 − δ(q1, b, a) = (q1, ε)
Step 5 − δ(q1, c, Z) = (qf, Z)
Step 6 − δ(qf, c, Z) = (qf, Z)
Explanation
Step 1 − Consider an input string: "aabbcc" which satisfies the given condition.
Step 2 − Scan string from left to right.
Step 3 − First input is 'a' and the rule:
for input 'a' and STACK alphabet Z, then
push the input 'a' into STACK: (a,Z/aZ) and state will be q0
Step 4 − Second input is 'a' and the rule:
for input 'a' and STACK alphabet 'a', then
push the input 'a' into STACK : (a,a/aa) and state will be q0
Step 5 − Third input is 'b' and the rule:
for input 'b' and STACK alphabet 'a', then
pop STACK with one 'a' : (b,a/&epsiloon;) and state will be now q1
Step 6 − Fourth input is 'b' and the rule −
For input 'b' and STACK alphabet 'a' and state is q1, then
pop STACK with one 'a' : (b,a/&epsiloon;) and state will be q1
Step 7 − Fifth input is 'c' and top of STACK is Z the rule:
for input 'c' and STACK alphabet 'Z' and state is q1, then
do nothing: (c,Z/Z) and state will be qf
Step 8 − Sixth input is 'c' and top of STACK is Z the rule:
for input 'c' and STACK alphabet 'Z' and state is qf, then
do nothing: (c,Z/Z) and state will be qf
Step 9 − We reached end of the string, and the rule:
If we are standing at final state, qf then string is accepted in the PDA
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