Consider a Turing machine M$1$ on alphabet $=\{0,1\}$ with three

states aside from qaccept and qreject. The start state is q$1.$ Here is the transition

function in table form:

q$1$ q$2$ q$3$

$$ accept, R accept, R $1,$ q$3,$ R

$00,$ q$2,$ R $1,$ q$3,$ R $1,$ q$2,$ R

$11,$ q$3,$ R $0,$ q$2,$ R $0,$ q$3,$ R

$($A$)$ Find a three letter word that this machine accepts.

$($B$)$ Find a three letter word that causes the machine to a infinite loop.

Note: There are only $8$ words of length $3!$ If you are unsure how to

proceed, just test a few.

$($C$)$ This version has no access to qreject. Modify this to make a new

machine M$2$ which accepts when M$1$ accepts and rejects when M$1$

would go into an infinite loop.