Fully animated view of a Turing machine on solving a particular input an check valid and invalid cases.

1. G r o u p - - - - - - - 4
S o u m i k M a i t y ( 1 0 )
Shyamal Krishna Jana(14)

2. M={Q, ∑, ┌, δ, q0, B,F} Q = The finite set of state.
∑ = The finite set of input symbol.
┌ = The complete set of tape symbol. ∑ is always a subset of ┌. δ = The transition function.
q0 = The start state. It is member of Q.
B = The blank symbol. The symbol is in ┌ but not in ∑. IT is not input symbol.
F = The final or accepting state, a subset of Q.
Q = {S, q1, q2, q3, q4, q5, q6, q7} ∑ = {a, b, c} ┌ = {a, b, c, X, Y, Z, $} δ = {(Q, ┌) → (Q, ┌, L/R)} q0 = {S}
F = {q7}
Design and Validate Turing Machine (Single Tape) in Three parts.
Three parts are (i) Algorithm (ii) Explanation of the machine through
transition diagram (iii) Verification of the design with non-trivial valid and
invalid input cases.
[ L = aibjck | i × j = k ; i, j, k ≥ 1 ]

3. M={Q, ∑, ┌, δ, q0, B,F} Q = The finite set of state.
∑ = The finite set of input symbol.
┌ = The complete set of tape symbol. ∑ is always a subset of ┌. δ = The transition function.
q0 = The start state. It is member of Q.
B = The blank symbol. The symbol is in ┌ but not in ∑. IT is not input symbol.
F = The final or accepting state, a subset of Q.
Q = {S, q1, q2, q3, q4, q5, q6, q7} ∑ = {a, b, c} ┌ = {a, b, c, X, Y, Z, $} δ = {(Q, ┌) → (Q, ┌, L/R)} q0 = {S}
F = {q7}
M = {Q, ∑, ┌, δ, q0, B,F}
Q = The finite set of state.
∑ = The finite set of input symbol.
┌ = The complete set of tape symbol. ∑ is always a subset of ┌.
δ = The transition function.
q0 = The start state. It is member of Q.
B = The blank symbol. The symbol is in ┌ but not in ∑. It is not input symbol. We
take $ as the blank symbol in the input tape.
F = The final or accepting state, a subset of Q.

4. M={Q, ∑, ┌, δ, q0, B,F} Q = The finite set of state.
∑ = The finite set of input symbol.
┌ = The complete set of tape symbol. ∑ is always a subset of ┌. δ = The transition function.
q0 = The start state. It is member of Q.
B = The blank symbol. The symbol is in ┌ but not in ∑. IT is not input symbol.
F = The final or accepting state, a subset of Q.
Q = {S, q1, q2, q3, q4, q5, q6, q7} ∑ = {a, b, c} ┌ = {a, b, c, X, Y, Z, $} δ = {(Q, ┌) → (Q, ┌, L/R)} q0 = {S}
F = {q7}
1. Scan the input from left to right to be sure that it is a member of a+b+c+ ;
reject if it is not.
2. Return the head at the left-hand end of the tape.
3. Cross off an ‘a’ and scan to the right until a ‘b’ occurs. Shuttle between the
b’s and c’s crossing of one of each until all b’s are gone. If all c’s have been
crossed of and some b’s remain reject.
4. Restore the crossed of b’s repeat stage 3 if there is another a to cross off. If all
a’s are crossed off, determine whether all c’s are crossed off. If yes accept,
otherwise reject.

5. b → Y, R
$ → $, R
a → a, R
q5
q7
q6
a → X, R
c → Z, L
Y → Y, R
Z → Z, L
Z → Z, R
b → b, L
X → X, R
Y → b, L
q4 q3
S q2
q1
b → b, R
Z → Z, L
a → a, L
Q = {S, q1, q2, q3, q4, q5, q6, q7}
∑ = {a, b, c}
┌ = {a, b, c, X, Y, Z, $}
δ = {(Q, ┌) → (Q, ┌, L/R)}
q0 = {S}
F = {q7}

6. M={Q, ∑, ┌, δ, q0, B,F} Q = The finite set of state.
∑ = The finite set of input symbol.
┌ = The complete set of tape symbol. ∑ is always a subset of ┌. δ = The transition function.
q0 = The start state. It is member of Q.
B = The blank symbol. The symbol is in ┌ but not in ∑. IT is not input symbol.
F = The final or accepting state, a subset of Q.
Q = {S, q1, q2, q3, q4, q5, q6, q7} ∑ = {a, b, c} ┌ = {a, b, c, X, Y, Z, $} δ = {(Q, ┌) → (Q, ┌, L/R)} q0 = {S}
F = {q7}

7. b → Y, R
$ → $, R
a → a, R
q5
q7
q6
a → X, R
c → Z, L
Y → Y, R
Z → Z, L
Z → Z, R
b → b, L
X → X, R
Y → b, L
q4 q3
S q2
q1
b → b, R
Z → Z, L
a → a, L
X a a Y b Z
Y Z
b
X Z Z
X Z Z $

8. b → Y, R
$ → $, R
a → a, R
q5
q7
q6
a → X, R
c → Z, L
Y → Y, R
Z → Z, L
Z → Z, R
b → b, L
X → X, R
Y → b, L
q4 q3
S q2
q1
b → b, R
Z → Z, L
a → a, L
X a a Y Z
b
X Z
X

9. THANK YOU