A Turing Machine for the language \(\mathrm{L}=\left\{\mathrm{a}^{\mathrm{n}}…
2022
A Turing Machine for the language \(\mathrm{L}=\left\{\mathrm{a}^{\mathrm{n}} \mathrm{b}^{\mathrm{m}} \mathrm{c}^{\mathrm{n}} \mathrm{d}^{\mathrm{m}} \mid \mathrm{n} \geq 1, \mathrm{~m} \geq 1\right\}\) is designed. The resultant model is \(M =\)\(\left(\left\{\mathrm{q}_{0}, \mathrm{q}_{1}, \mathrm{q}_{2}, \mathrm{q}_{3}, \mathrm{q}_{4}, \mathrm{q}_{5}, \mathrm{q}_{6}, \mathrm{q}_{7}, \mathrm{q}_{\mathrm{f}}\right\},\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\},\left\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{X}_{1}, \mathrm{X}_{2}, \mathrm{Y}_{1}, \mathrm{Y}_{2}\right\}, \delta, \mathrm{q}_{0}, \mathrm{~B},\left\{\mathrm{q}_{\mathrm{f}}\right\}\right)\) and part of \(' 𝛿 '\) is given in the transition table. You need to write the following questions based on design of Turing Machine for the given language. Note that, while designing the Turing Machine \(X_1\) and \(X_2\) are used to work with \(′𝑎′𝑠\) and \(′𝑐′𝑠\) and \(Y_1\) and \(Y_2\) are used to handle \(′𝑏′𝑠\) and \(′𝑑′𝑠\) of the given string.
\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & \mathrm{a} & \mathrm{b} & \mathrm{c} & \mathrm{d} & \mathrm{X}_{1} & \mathrm{X}_{2} & \mathrm{Y}_{1} & \mathrm{Y}_{2} & \mathrm{~B} \\ \hline \mathrm{q}_{0} & \left(\mathrm{q}_{1}, \mathrm{X}_{1}, \mathrm{R}\right) & & & & \mathrm{M} 2 & & & & \\ \hline \mathrm{q}_{1} & \left(\mathrm{q}_{1}, \mathrm{a}, \mathrm{R}\right) & \left(\mathrm{q}_{1}, \mathrm{~b}, \mathrm{R}\right) & \mathrm{M} 1 & & & \left(\mathrm{q}_{1}, \mathrm{X}_{2}, \mathrm{R}\right) & & & \\ \hline \mathrm{q}_{2} & \left(\mathrm{q}_{2}, \mathrm{a}, \mathrm{L}\right) & \left(\mathrm{q}_{2}, \mathrm{~b}, \mathrm{~L}\right) & & & \left(\mathrm{q}_{2}, \mathrm{X}_{1}, \mathrm{R}\right) & \left(\mathrm{q}_{2}, \mathrm{X}_{2}, \mathrm{~L}\right) & & & \\ \hline \mathrm{q}_{3} & \mathrm{M} 3 & \left(\mathrm{q}_{4}, \mathrm{Y}_{1}, \mathrm{R}\right) & & & & \left(\mathrm{q}_{6} \mathrm{X}_{2}, \mathrm{R}\right) & & & \\ \hline \mathrm{q}_{4} & & \left(\mathrm{q}_{4}, \mathrm{~b}, \mathrm{R}\right) & & \left(\mathrm{q}_{5}, \mathrm{Y}_{2}, \mathrm{~L}\right) & & \mathrm{M} 5 & & \left(\mathrm{q}_{4}, \mathrm{Y}_{2}, \mathrm{R}\right) & \\ \hline \mathrm{q}_{5} & & \left(\mathrm{q}_{5}, \mathrm{~b}, \mathrm{~L}\right) & & & & \left(\mathrm{q}_{5}, \mathrm{X}_{2}, \mathrm{~L}\right) & \mathrm{M} 4 & \left(\mathrm{q}_{5}, \mathrm{Y}_{2}, \mathrm{~L}\right) & \\ \hline \mathrm{q}_{6} & & & & & & \left(\mathrm{q}_{6}, \mathrm{X}_{2}, \mathrm{R}\right) & & \left(\mathrm{q}_{7}, \mathrm{Y}_{2}, \mathrm{R}\right) & \\ \hline \mathrm{q}_{7} & & & & & & & & \left(\mathrm{q}_{7}, \mathrm{Y}_{2}, \mathrm{R}\right) & \left(\mathrm{q}_{6}, \mathrm{~B}, \mathrm{R}\right) \\ \hline \end{array}\)
What is the Move in the cell with number \(' M1 '\) of the resultant Table?
- A.
\(\left(\mathrm{q}_{2}, \mathrm{X}_{2}, \mathrm{R}\right)\) - B.
\(\left(\mathrm{q}_{2}, \mathrm{X}_{2}, \mathrm{~L}\right)\) - C.
\(\left(\mathrm{q}_{3}, \mathrm{X}_{2}, \mathrm{~L}\right)\) - D.
Error Entry
Attempted by 26 students.
Show answer & explanation
Correct answer: B
The machine must check the structure:
Mark one unmatched ‘a’ → X₁
Match it with one ‘c’ → X₂
And separately
Match one ‘b’ → Y₁
with one ‘d’ → Y₂
So the machine repeatedly performs two matching loops:
Loop 1: Match a ↔ c
Find next unmarked a, convert it to X₁, go right.
Find the matching c, convert it to X₂, go left to return.
Loop 2: Match b ↔ d
Find next unmarked b, convert it to Y₁, go right.
Find the matching d, convert it to Y₂, go left to return.
Where M1 Occurs
Look at the table row:
State = q₁
Symbol = c
Entry = M1
This means:
δ(q1,c)=M1\delta(q_1, c) = M1δ(q1,c)=M1
Now we must understand:
What should the TM do when it is in state q₁ and reading a ‘c’?
Understanding state q₁
State q₁ appears after:
Machine has just marked an ‘a’ as X₁
And is now moving right to find:
first b (for Y₁ marking), OR
first c (for X₂ marking)
In language anbmcndma^n b^m c^n d^manbmcndm, the order is:
anbmcndma^n \quad b^m \quad c^n \quad d^manbmcndm
So when the machine reaches the c block, it must:
✔️ Mark one c → X₂
✔️ Then return to the left to find the next a or b loop
This is exactly what M1 does.
Why M1 = (q₂, X₂, L)?
Let’s check step-by-step.
✦ The TM is in q₁
It has already matched an a
Now it is searching for a corresponding c
✦ It reads symbol c
This means it has reached the c-block.
✦ It should mark this c as processed:
c→X2c \rightarrow X_2c→X2
✦ After marking, the machine must return left
Why? Because it must go back to:
the b-marking phase, or
the next a-marking phase
That is why the move direction = L.
✦ It must now go to state q₂
State q₂ is designed for moving left until the machine reaches the boundary zone (the left side of the alphabet blocks).
So:
M1=(q2, X2, L)\boxed{M1 = (q_2,\; X_2,\; L)}M1=(q2,X2,L)
is exactly correct for both logic and direction control.
M1=(q2, X2, L)\boxed{M1 = (q_2,\; X_2,\; L)}M1=(q2,X2,L)
This marks a “c” as X₂, then moves left in state q₂ to continue matching.