decider for a question not clear This question was asked and answered but I cannot understand the solution.


Why is it sufficient to test all strings of |Q| + 1 length?
Why should special state q be found?


the original question:
Show that the set of all TMs that move only to the right and loop for some input is decidable

L2={ M | M is a TM and there exists an input w such that in the computation of M(w) the head only moves right and M never stops}
The pigeonhole principle. If you go through $|Q|+1$ states, then there must be a state you have been in twice already. This means that because the input is $\sqcup$ almost all of the time, then we are stuck in a loop and the machine wont halt