Several Turing Machines, building up to a TM that stops once it has found the nth prime
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README.md

I asked my programming teacher how to create a Turing Machine that reaches the nth prime.

He thought I was joking. He was WRONG.

I never make jokes :)

Anyways, to grok how Turing machines, as described in Automata and Computability, by Dexter C. Kozen, work, here are:

  • A Turing Machine that accepts if a number n doesn't divide another number m and rejects otherwise.
  • A Turing Machine that accepts if n doesn't divide m, or if n=m, and rejects otherwise.
  • A Turing Machine that accepts if n is prime, and rejects otherwise.
  • A Turing Machine that accepts once it has found the nth prime.

Early versions, deprecated, start with 0.

  • A Turing Machine that accepts if a number n doesn't divide another number m and rejects otherwise.
  • A Turing Machine that detects whether a number >=2 is prime.

Here is a walkthrough of how the TMs work. The states are gradually defined as the Turing machine moves, and then modified when it makes sense to do so. I think that this will be easier to understand than a table.

(divisor 1.0) Accepts if n doesn't divide m, rejects otherwise.

Gets as an input 0111...11822..229␣␣␣␣␣...., where:

  • 0 is the left endmarker
  • There are (n-1) '1's, and 1 '8'. The 8 signals the end of the '1's.
  • Similarly, there are (m-2) '2's and 1 '9'. The 9 signals the end of the '2's.
  • The ␣ are empty characters, and the TM has an infinite number of them to the left.

state 0:

  • The TM starts on this state.
  • Inmediatly, it moves to the right, to state 1.

state 1:

  • Looks for a 1 to replace by a three
  • if symbol = 1, write 3, move right, change to state 2.
  • otherwise: move right, keep state.

state 2:

  • it looks for a 2 to replace by a 4.
  • if symbol = 2, write 4, change to state 3.
  • else: move right, keep state

state 3:

  • if it doesn't find a 0, go to the left, keep state
  • if it finds a 0, write 0, move right, go to state 1.

// As I write this, I realize that if I replace state 0 by state 3, nothing happens.

// Excursus: After competing this project, I searched for similar ones, and found one by a William Bernoudy. My finding the nth prime TM had 14 states and 12 symbols, while his had 14 states and only 10 symbols. But if I replace state 0 by state 3, I have one state less! Anyways, from now on no state is state 3. We modify state 2, which refers to it.

state 0:

  • If symbol = 0, move right, change to state 1.
  • else: move left, keep state.

state 2:

  • it looks for a 2 to replace by a 4.
  • if symbol = 2, write 4, change to state 0.
  • else: move right, keep state

Now, once all the 1s are turned into 3s, state 1 would go on searching, so we want to modify it to notice that it has run out of threes.

state 1:

  • if symbol = 1, write 3, move right, change to state 2.
  • if symbol = 8, write 8, move right, change to state 4.
  • otherwise: move right.

If there were no 2s left to turn to 4s, then n|m (n divides m). But if there are, n can still divide m, so we keep on going.

state 4:

  • if symbol = 2, write 4, move to the right, change to state 5.
  • if symbol = 9, REJECT, n | m

We also notice that if state 2 finds no 2s to turn into 4s, then ¬ (n|m), so we add that option.

state 2:

  • if symbol = 2, write 4, change to state 0.
  • if symbol = 9, ACCEPT.
  • else: move to the right.

If state 4 does find a 2, then the situation looks, for example, like this; take n=3, m=6, then the tape cpuld have:

0338444229

3 '2's were turned into '4's. and we are in state 5, which turns the '3's into '1's and keeps going

state 5:

  • if symbol = 3, write 1, move to the left, keep state.
  • if symbol = 0, write 0, move to the right, change to state 1.
  • else: move to the left, keep state.

End.

(divisor 1.1) Accepts if n|m or if n=m.

Now the input is: 05111...11822..229 The 5 will be cahnged to a 6 once we change the 2 that corresponds to the 8. If there is no such 2, n = m. So we modify state 4 and 5 and create a state 6.

state 5:

  • if it reads a 3, write 1, move left, keep state.
  • if it reads a 5, write 6, keep state. // It has found such a 2.
  • if it reads a 0, write 0, change to state 1.

state 4:

  • if it reads a 2, write 4, move to the left, change to state 5
  • if it reads a 9, write 9, move to the left, change to state 6.
  • else: move to the right, keep state.

state 6:

  • if it reads a 5, accept.
  • if it reads a 6, reject.
  • else: move to the left.

End.

(is prime 1.1) Accepts if n is prime.

We start with: 0518777...77722..229

This reads: The initial n=2, after which there are enough 7s to increase n to find divisors of m.

Q: But, how many 7s?

A: Well, a priori at least sqrt(n), but up to n, if you one.

Q: But Nuño, if you put ceil(sqrt(n)) '7's, aren't you offloading some of the calculations to yourself instead of making the machine calculate it?

A: Yes, I am.

Anyways, right now the only state which can accept is state 2, which does so if ¬(n|m) for a given n. Instead of accepting, we want to increase n by 1. We modify state 2, and create 2 new states: state 7 and state 8.

state 2:

  • if it reads a 2: write 4, move left, change to state 0.
  • if it reads a 9: write 9, move left, change to state 7.
  • else: move to the right

state 7:

  • if it reads a 6: write 5, move left, keep state.
  • if it reads a 4: write 2, move left, keep state.
  • if it reads a 3: write 1, move left, keep state.
  • if it reads a 0: write 0, move right, change to state 8.

state 8: -if it reads an 8: write 1, move right, keep state. -if it reads a 7: write 8, move right, change to state 3.

  • if it reads a 2: ACCEPT. There is no space left, at least one of each pair of divisors has been tried

End.

(Find the nth prime 1.1)

Initial input: 0AA...AA51829␣␣␣...␣␣

It will replace an A by a B each time it finds a prime, so if n-1 is the number of 'A's, it will find the nth prime.

State 9 will change an A to a B. States 10, 11 and 12 initialize n to 2. 12, 13 and 14 move m one step to the right and increase it to m+1. By moving it one step to the right, n is bounded only by m+1.

The states that can accept are state 6 and state 8, and only state 6 can reject.

state 6

  • if it reads a 5, write 5, change to state 9.
  • if it reads a 6: write 6, change to state 10.
  • else: move to the left.

state 8: -if it reads an 8: write 1, move right, keep state. -if it reads a 7: write 8, move right, change to state 3. -if it reads a 2: write 2, move left, change to state 9.

state 9:

  • if it reads an A: write B, move right, change to state 10.
  • if it reads a 0: ACCEPT. There are no more As to change.
  • else: move left.

state 10:

  • if is reads a 1: write 1, move right, change to state 11.
  • if is reads a 3: write 1, move right, change to state 11.
  • else: move right.

state 11:

  • if it reads a 1: write 8, move right, change to state 12
  • if it reads a 3: write 8, move right, change to state 12
  • if it reads a X: write 8, move right, change to state 12
  • else: REJECT // Shouldn't be seeing anything else.

state 12:

  • if it reads a 1: write 7, move right, keep state
  • if it reads a 3: write 7, move right, keep state
  • if it reads a 8: write 7, move right, keep state
  • if it reads a 2: write 7, move right, change to state 13.
  • if it reads a 4: write 7, move right, change to state 13.
  • else: move right.

state 13:

  • if it reads a 9: write 2, move right, keep state
  • if it reads a 4: write 2, move right, keep state
  • if it reads a ␣: write 2, move right, change to state 14
  • else: move right.

state 14:

  • if it reads a ␣: write W, move left, change to state 0.