Question: 4 . [ 4 pts ] Draw a Turing Machine that accepts the set of all bitstrings that have 0 in all the

4.[4 pts] Draw a Turing Machine that accepts the set of all bitstrings that have "0" in all the even-numbered positions starting with position 0 at the left. So the \(\mathbf{T M}\) should accept and reject strings such as the following (I show you all strings of length \(0-5\) bits). The table contains hints to help you (note bold underlining). You can do this in 4 states (maybe fewerl). But more states would be OK, too. Think of the diagrams above, one node is the starting state, one node is the Accept state, and the rest are working nodes.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline String & Result & String & Result & String & Result & String & Result \\
\hline Empty & reject & 0000 & Accept & 00000 & Accept & 10000 & reject \\
\hline 0 & Accept & 0001 & Accept & 00001 & reject & 10001 & reject \\
\hline 1 & reject & 0010 & reject & 00010 & Accept & 10010 & reject \\
\hline 00 & Accept & 0011 & reject & 00011 & reject & 10011 & reject \\
\hline 01 & Accept & 0100 & Accept & 00100 & reject & 10100 & reject \\
\hline 10 & reject & 0101 & Accept & 00101 & reject & 10101 & reject \\
\hline 11 & reject & 0110 & reject & 00110 & reject & 10110 & reject \\
\hline 000 & Accept & 0111 & reject & 00111 & reject & 10111 & reject \\
\hline 001 & reject & 1000 & reject & 01000 & Accept & 11000 & reject \\
\hline 010 & Accept & 1001 & reject & 01001 & reject & 11001 & reject \\
\hline 011 & reject & 1010 & reject & 01010 & Accept & 11010 & reject \\
\hline 100 & reject & 1011 & reject & 01011 & reject & 11011 & reject \\
\hline 101 & reject & 1100 & reject & 01100 & reject & 11100 & reject \\
\hline 110 & reject & 1101 & reject & 01101 & reject & 11101 & reject \\
\hline 111 & reject & 1110 & reject & 01110 & reject & 11110 & reject \\
\hline & & 1111 & reject & 01111 & reject & 11111 & reject \\
\hline
\end{tabular}
*No copy/paste please*
4 . [ 4 pts ] Draw a Turing Machine that accepts

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