Why is the following language  decidable? $L_{one\ right\ and\ never\ stops}$ I can't understand how the following language can ever be decidable:
$L= \{ \langle M \rangle |  M \ is \ a \ TM \ and\ there\ exists\ an\ input\ that\ in\ the\ computation\ $ $of\ M(w)\ the\ head\ only\ moves\ right\ and\ M\ never\ stops \}$, 
but apparently it is. 
We need to approve that there's at least one input as requested, so even if run on all inputs in parallel, we can't say something as "if it never stops, accept", it's paradox, and even if we could do that it was showing that the language is in $RE$- recursive enumerable.
so how come it is recursive?
Thanks a lot

I think that Karolis's answer is almost correct.  I believe in order to deal with the possible non-determinism instead of constructing a graph whose vertices are the states of the TM $M$ in question, we should instead let the vertices be all nonempty sets of non-halting states.
Given any tape-symbol $\mathtt{a}$ we will draw a (directed) edge from vertex $A$ to vertex $B$ (labelled by $\mathtt{a}$) iff the following conditions are satisfied:


*

*for each state $q$ in $A$ the only instructions of $M$ with hypothesis $q,\mathtt{a}$ are of the form $q,\mathtt{a} \rightarrow q',\mathtt{b},R$ where $q'$ is a state in $B$ (in particular, no left moves are allowed, and no halting states can be entered);

*every state in $A$ has such an instruction (so every computation will continue at least one more step); and

*every state in $B$ is reachable from a state in $A$ by such an instruction.


We now analyse this directed, labelled graph.


*

*For each cycle in this graph consisting only of distinct blank-labelled edges we see if (some vertex in) this cycle is reachable from the vertex $\{ q_\text{start} \}$ using a path without cycles;

*If so, we accept $M$;

*If we run out of such blank-labelled cycles, we reject $M$.

