Board Game Arena

Are there any setups by the Turing Machine manufacturer that we know of that use the fact that a verifier is required to complete the code BUT the verifier also does not ever produce a true for the given code? Example below:

For b = blue, y = yellow and p = purple you could have:

1. b = 2
2. y = 3
3. p = 3


You could also have the verifiers:

1. v1 = one colour is less than the other two
2. v2 = yellow is =/>/< 3
3. v3 = there are x even numbers
4. v4 = one colour is greater than the other two

Using that you can get the following results from the verifiers (eventually):

1. v1: blue is < yellow or purple
2. v2: yellow = 3
3. v3: there is 1 even number
4. v4: all results are negative

In this case BECAUSE all results on v4 are negative, the code is solvable and is b2 y3 p3

IF v4 had a positive for p > b/y then the result would also be meaningful: b2 y3 p5

Therefore in this case v4 is both: required for the solution AND provides only negative results for this code

I'm interested in whether the game setup provided by the manufacturer uses this feature to allow for such games. It seems 'legal' within the game rules, but it's not clear whether these cases are actually provided and I can imagine that for 'ease of understanding/play' they might have shied away from those setups.

It's relevant because knowing whether they are possible setups or not affects your number of guesses. If it's not possible, then you can assume the above case HAS TO BE the p5 solution and do (potentially) at least one less guess of the verifiers

No. The conditions are that there is 1 and only 1 code that yields positive for all verifiers, and that removing any single verifier would result in multiple codes yielding positive for all remaining verifiers.

So the mere presence of "one colour is greater than the other two" eliminates 2 3 3 or similar codes.

Unless this has been changed recently, no that not possible. v4 tell you that one of the three digits is greater than the others, so there can't be two 5s.
With help of my solver that give you 4 solutions: 124/245/421/542

235 isn't possible because the same answers gives 134 as another solution.


Interestingly, if you was allowing that, that would have given 16 solutions:
111/124/133/222/233/245/331/332/334/335/421/433/444/533/542/555

Thanks everyone, the clarification that the rules state that your code HAS TO pass all the verifiers answers the question - given that's the goal I understand why it would be an illegal setup. I disagree that in my example multiple answers were possible or that the all negative verifier wasn't pointing to a unique solution, but it's moot given that the code must specifically be able to tick every verifier. Thanks!

Tackster wrote: ↑12 July 2025, 21:26I disagree that in my example multiple answers were possible or that the all negative verifier wasn't pointing to a unique solution

b<yp, y=3, one even, p>by.
I didn't even verify since I'm confident on my solver, but 134 and 235 gives those 4 answers.
So both of them can't be the real solution.

And I guess you didn't understood what i mean by my solutions: it's the potential solutions with no questions asked. If we got the four answers you listed on that post, then there is no valid solution anymore.