Question

Construct a Turing machine with tape symbols \(0,1\), and \(B\) that, when given a bit string as input, replaces the first \(0\) with a \(1\) and does not change any of the other symbols on the tape.

Short answer

The constructed Turing machine with tape symbols\(0,1\), and\(B\)that, when given a bit string as input, replaces the first\(0\)with a\(1\)and does not change any of the other symbols on the tape is

\(\begin{array}{c}{\bf{T = }}\left( {{\bf{S,I,f,}}{{\bf{s}}_{\bf{0}}}} \right)\\{\bf{S = }}\left\{ {{{\bf{s}}_{\bf{0}}}{\bf{,}}{{\bf{s}}_{\bf{1}}}} \right\}\\{\bf{I = \{ 0,1,B\} }}\\\left( {{{\bf{s}}_{\bf{0}}}{\bf{,0,}}{{\bf{s}}_{\bf{1}}}{\bf{,1,R}}} \right)\\\left( {{{\bf{s}}_{\bf{0}}}{\bf{,1,}}{{\bf{s}}_{\bf{0}}}{\bf{,1,R}}} \right)\\\left( {{{\bf{s}}_{\bf{0}}}{\bf{,B,}}{{\bf{s}}_{\bf{1}}}{\bf{,B,R}}} \right)\end{array}\)

Step-by-step solution

Definition

A Turing machine \({\bf{T = }}\left( {{\bf{S,I,f,}}{{\bf{s}}_{\bf{0}}}} \right)\) contains a finite set \({\bf{S}}\) of states, an alphabet \({\bf{I}}\) containing the blank symbol \({\bf{B}}\), a starting state \({{\bf{s}}_{\bf{0}}}\) and a partial function from \(S{\bf{ \times I}}\) to \({\bf{S \times I \times }}\{ R,L\} \). Note: The partial function \({\bf{f}}\) is often represented as \(5{\bf{ - }}\)tuples.

It will require \(2\) states \({{\bf{s}}_{\bf{1}}}\) and \({{\bf{s}}_{\bf{0}}}\) (\({{\bf{s}}_{\bf{0}}}\)is the start state, \({{\bf{s}}_{\bf{1}}}\) will be the final state), which will be contained in the set \({\bf{S}}\).

\({\bf{S = }}\left\{ {{{\bf{s}}_{\bf{0}}}{\bf{,}}{{\bf{s}}_{\bf{1}}}} \right\}\)

The alphabet \({\bf{I}}\) needs to contain the tape symbols. The tape symbols are given as \(0,1\)and\({\bf{B}}\).

\({\bf{I = \{ 0,1,B\} }}\)

Finding the partial function as five tuples

Next, you will define the partial function as five tuples.

As long as the input is a \(1\), it will keep the input unchanged and move one position to the right (as it only want to change the first \(0\)).

\(\left( {{{\bf{s}}_{\bf{0}}}{\bf{,0,}}{{\bf{s}}_{\bf{1}}}{\bf{,1,R}}} \right)\)

The first time the input is a \(0\), it moves to the state \({{\bf{s}}_{\bf{1}}}\) (as the machine can halt once the first \(0\) has been changed) and change the \(0\) to a \(1\). It then moves one position to the right.

\(\left( {{{\bf{s}}_{\bf{0}}}{\bf{,1,}}{{\bf{s}}_{\bf{0}}}{\bf{,1,R}}} \right)\)

Once it arrives at the first blank symbol (while not having read any \(0\)'s), it will move to the final state \({{\bf{s}}_{\bf{1}}}\) as it reads the entire string.

\(\left( {{{\bf{s}}_{\bf{0}}}{\bf{,B,}}{{\bf{s}}_{\bf{1}}}{\bf{,B,R}}} \right)\)

Finding the partial function as five tuples

It doesn't add any five-tuples that start from \({{\bf{s}}_{\bf{1}}}\) such that the algorithm ends once it reaches state \({{\bf{s}}_{\bf{1}}}\) (which is considered the final state).

The constructed Turing machine with tape symbols\(0,1\), and\(B\)that, when given a bit string as input, replaces the first\(0\)with a\(1\)and does not change any of the other symbols on the tape is

\(\begin{array}{c}\left( {{{\bf{s}}_{\bf{0}}}{\bf{,0,}}{{\bf{s}}_{\bf{1}}}{\bf{,1,R}}} \right)\\\left( {{{\bf{s}}_{\bf{0}}}{\bf{,1,}}{{\bf{s}}_{\bf{0}}}{\bf{,1,R}}} \right)\\\left( {{{\bf{s}}_{\bf{0}}}{\bf{,B,}}{{\bf{s}}_{\bf{1}}}{\bf{,B,R}}} \right)\end{array}\)