Help me, I'm not Turing 🤣

[Tim O Rasso]

https://boardgamearena.com/3/turingmach ... =459488318

Could someone explain to me how it is possible give the right solution in this match using 2 verificators?
Thanks!

[poptasticboy]

Hi! So a few initial deductions:

E has to be "Sum is a multiple of 3" otherwise B would be redundant.
*EDIT: this is not actually true! In trying to recreate my reasoning quickly I actually forgot about the case where Sum=10!! For now, we'll *assume* the Sum is a multiple of 3, then I will add a section at the end dealing with why the Sum can't be a multiple of 5!*

F has to be saying that a certain colour is >1.
This is because, if it were saying that a colour is =1, then D would have to be referring to a different colour (otherwise it would be redundant). Then C must be referring to the final unused colour, (otherwise it would be redundant). For a multiple of 3, this only leaves 111 and the other Verifiers would all be redundant.

My two questions told me that:
A) Bl=Y
B) Sum is Odd.

For an Odd Sum multiple of 3 with Bl=Y, the only possible solutions are 111, 225, 441, 333, 555.

We still have C, D and F to do work for us.

F rules out 111. C rules out 555, but we'll keep them in mind for consideration of redundancy.

If C refers to P, we have 111, 441, 333 left.
D then must refer to Bl or Y to not be redundant, ruling out 441.
F rules out 111, leaving 333.

On the other hand, if C refers to Bl or Y, we have 111, 225, 333 left.
D must then refer to P to not be redundant, ruling out 225.
F rules out 111, leaving 333.

Either way, this leaves just 333 as the solution.

*EDIT: Now for the "Sum is a multiple of 5" case!*

So if the Sum is a multiple of 5, and from my two tests Bl=Y and Sum is Odd, then the solution can only be 113, 221 or 555, and again we have C, D and F to work with.

C just rules out 555, no matter which colour it refers to. D then chooses a unique solution between 113 and 221, leaving F redundant.

Therefore the Sum can not be a multiple of 5, and the previous solution 333 stands!

Sorry for the initial error!

[Tim O Rasso]

Thank you! I'll study your notes tonight. I had never considered the aspect of verifier redundancy. yet it is clearly written in the rules!

[Tim O Rasso]

Hi! So a few initial deductions:

E has to be "Sum is a multiple of 3" otherwise B would be redundant.
*EDIT: this is not actually true! In trying to recreate my reasoning quickly I actually forgot about the case where Sum=10!! For now, we'll *assume* the Sum is a multiple of 3, then I will add a section at the end dealing with why the Sum can't be a multiple of 5!*

F has to be saying that a certain colour is >1.
This is because, if it were saying that a colour is =1, then D would have to be referring to a different colour (otherwise it would be redundant). Then C must be referring to the final unused colour, (otherwise it would be redundant). For a multiple of 3, this only leaves 111 and the other Verifiers would all be redundant.

My two questions told me that:
A) Bl=Y
B) Sum is Odd.

For an Odd Sum multiple of 3 with Bl=Y, the only possible solutions are 111, 225, 441, 333, 555.

We still have C, D and F to do work for us.

F rules out 111. C rules out 555, but we'll keep them in mind for consideration of redundancy.

If C refers to P, we have 111, 441, 333 left.
D then must refer to Bl or Y to not be redundant, ruling out 441.
F rules out 111, leaving 333.

On the other hand, if C refers to Bl or Y, we have 111, 225, 333 left.
D must then refer to P to not be redundant, ruling out 225.
F rules out 111, leaving 333.

Either way, this leaves just 333 as the solution.

*EDIT: Now for the "Sum is a multiple of 5" case!*

So if the Sum is a multiple of 5, and from my two tests Bl=Y and Sum is Odd, then the solution can only be 113, 221 or 555, and again we have C, D and F to work with.

C just rules out 555, no matter which colour it refers to. D then chooses a unique solution between 113 and 221, leaving F redundant.

Therefore the Sum can not be a multiple of 5, and the previous solution 333 stands!

Sorry for the initial error!

First of all I apologize about my english. I used also google translate.
Thanks for explanation!
You are able to change my point of view.
You was able to find the code because your first attempt had blue=Yellow?
If you had inserted a code with blue not equal yellow, would you have had to ask other verificator in the next round?

Summing up, if I understood the method:
1) look verificators and see which combination are possible
2) insert the best code to prune the tree
3) analize the answer to reduce again the number of candidates

[poptasticboy]

Yes, this particular solution relied heavily on me asking "Blue=Yellow" and getting a .

I was writing this from the perspective of "given that I asked those questions, this is how I reached the solution". In reality of course, I remember spending a *very* long time deciding which questions to ask before settling on that, and at this stage I would need to go back and solve the puzzle from scratch to remind myself why I decided those would be good tests!