Tips turingmachine

Unicity

All verificators are useful. You can gain a lot of information from it. Let's take an example, the verificators are:

A) blue is odd | blue is even

B) blue >1 | blue = 1

C) no 4 | one 4 | two 4 | three 4

D) blue <4 | blue >3

So firstly, if the answer is '444', that would mean that the verificators A, B, and D will be useless which is not possible, so it's not 444.

You can see that we have really few information about yellow and purple. If the answer is 123, then no verificator would tell the difference with the code 133. So that mean yellow can't be 2 or 3.

The only information we could have for yellow and purple is the number of 4. So we can deduce yellow and purple are equal to 4.

Furthermore, if blue was equal to 1, then B&C woud be enough to tell the combinaison is 144. So blue isn't equal to 1.

If blue was 4 or 5, then B will be useless compared to D. Say diferently, A&C&D are enough to tell if it's 444 or 544, and B would be useless, which isn't possible. So B isn't 4 or 5.

So, the two remaining combinaisons possible are 244 and 344.

But now, if you look at 244, and verificators ACD, we got two 4s, blue equal to 1, 2, or 3, and blue even... Which mean, it's already 244 without using B. So B is useless. So it's not that, and it's not 244.

And so it's exactly 344 with 0 check.

And we can verify it:

- with those answers (blue odd, blue>1, two 4s, blue<4), there is only one possibility: blue is forced to be at 3, and so y & p are forced to be at 4

- if you remove verificator A, there is 2+ solutions: 244 & 344

- if you remove verificator B, there is 2+ solutions: 144 & 344

- if you remove verificator C, there is 2+ solutions (25): per example 325 & 314

- if you remove verificator D, there is 2+ solutions: 344 & 544

So every verificator is useful, and there is only one possibility at the end. Without the reasonning above, that just mean it's a valid solution, not that it's the only one.