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As far as I know for Non-Deterministic Turing Machine (NTM) there are 4 kind of branches:

  1. An input is accepted if there is at least one node in the tree that is an accept.
  2. An input is rejected if all branches are rejected, which means that it is a finite tree.
  3. An input is not-accepted if there are loop nodes with reject nodes.
  4. An input is not-accepted if there are only loop nodes.

We know that for RE-R language, our NTM must not accept some inputs.

My question is on not-accepted branches.
For a specific NTM that accepts RE-R language, is option 4 (for all nodes looping) necessary to exist?
Means that only option 3 exists without option 4.

Thanks.

Yuval
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