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Problem: Given a start state q0, and a completely blank tape except for one square with a # symbol, find the # and halt on it.

Non-Deterministically:

This machine chooses to either search left or right of the start state, and keeps going in that direction until the next symbol is the # symbol, where it stays.

Deterministically: ?

How do I replicate this machine in a deterministic form? I've done some research and it seems this problem can be solved by addressing both possibilities/branches of the "tree", but I can't seem to connect the dots here...

Austin Amon
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1 Answers1

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You do not just run over the non-# cells, but you mark them as visited. Further you do the two branches simultaniously by alternating between them.

  1. Mark the current cell with "+" (unless it is the # already)
  2. Run to the right while you see +. When you see the first blank, mark it +.
  3. Run to the left while you see +. When you see the first blank, mark it +. Go to 2.

In this way you can process any fixed number of non-deterministic branches in a deterministic way. Of course, the running around takes much more time than the actual simulation.

Peter Leupold
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  • Welp, this is exactly how I ended up doing it, albeit I figured it out eventually. My hangup was that I was thinking you couldn't have two transitions out of a state in a deterministic machine... but alas you just can't have two transitions on the same input. That opened it up for me. Thanks for the reply. – Austin Amon Nov 29 '17 at 02:54