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 \('M3'\) of the resultant Table?
- A.
(q1,X1,L)
- B.
(q4,X1,R)
- C.
(q1,X1,R)
- D.
Error Entry
Attempted by 27 students.
Show answer & explanation
Correct answer: D
Note: All the provided options are incorrect. The right ans is D
Reason (Step-by-Step TM Logic):
q₀: Start, on first 'a' → mark X₁, R to q₁ (a/c phase begin).
q₁: Skip remaining a's/b's (R), reach first 'c' → to q₃? (Table: q₁ on c = M1? But partial shows q₃ on c undefined—M1 likely reject for early c).
Assume standard: After marking X₁'s, go right to c's, mark first c with X₂, L to q₃ (scan left for matching X₁).
In q₃:
On X₁: Match (erase or shift, R/L to continue).
On unmarked 'a': Extra 'a' (n > #c's) → no transition defined (error/reject). This is M3.
Table placeholders (M1/M2) are rejects; M3 follows pattern.
If match complete (no extra), shift to q₄ for b/d phase (Y₁ on b → (q₅, Y₁, R)).
Simulating "aabbccdd": Accepts (all match). "aaabbccdd" (extra a): Rejects at q₃ on extra 'a' → error.
Correct Option: D (Error Entry)
This is a "trap state" for decidability—TM halts and rejects on mismatch. If options were open-ended, it's "undefined" or "(reject, a, N)" but per table, D fits.