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Is setup #B52F2TL incorrect?
Posted: 22 August 2026, 10:46
by muharred
I'm confused by today's daily: #B52F2TL. It seems to me that one of the verifiers is always redundant.
The verifiers are:
A: Purple is even or odd
B: Number of 1s in the code 0/1/2/3
C: Sum of all numbers even or odd
D: Sum of yellow and blue compared to 6: <6/=6/>6
E: No pairs in the code / a pair in the code
Spoilers follow:
It seems to me that A or C is always redundant - you only need one of them.
My logic is as follows: the sum of any pair is always even. So, if there is a pair, then if it is yellow=blue, then the parity of purple (verifier A) will solely determine the parity of the sum of all three numbers including but not limited to the case where purple=yellow=blue making verifier C redundant (or vice versa). Therefore, all three numbers can't be equal to one another and if there is a pair present, it must be purple with another color.
But by looking at the verifiers we can notice that there is nothing distinguishing blue from yellow - any solution where yellow=x and blue=y will also work with yellow=y and blue=x. Therefore yellow must be equal to blue.
That's a contradiction.
The proposed correct answer was yellow=2, blue=2 and purple=4. Which is understandable, if you assume yellow=blue. By querying D you can tell if they are 2/3/(4 or 5). And by querying E you could tell if purple is the same with them. If it is, its parity could distinguish between 4 or 5 case. If it's not, then combining other verifiers we can find the value of purple just as well.
Am I wrong in my understanding of the rule that no verifier can be redundant? Sometimes I'm kinda struggling with it, but my definition is that any verifier has to be either queried or has to play a part in your deductions eliminating some of the answers just by the fact that the verifier is present. In this case verifier A or C breaks that rule.
Re: Is setup #B52F2TL incorrect?
Posted: 22 August 2026, 13:59
by Romain672
I did it on my solver:
https://docs.google.com/spreadsheets/d/ ... id=0#gid=0
Three possible answers: 222, 224, and 555.
muharred wrote: ↑22 August 2026, 10:46Am I wrong in my understanding of the rule that no verifier can be redundant? Sometimes I'm kinda struggling with it, but my definition is that any verifier has to be either queried or has to play a part in your deductions eliminating some of the answers just by the fact that the verifier is present. In this case verifier A or C breaks that rule.
Don't hesitate to pm me if you want that I check one of your games with my solver to help you understand, I can tell you all potential solutions with no question asked.
If you remove A:
- 222 isn't a solution anymore because 235 and 325 would answer the same.
- 224 isn't a solution anymore because 233 and 323 would answer the same.
- 555 isn't a solution anymore because 254 and five others trio would answer the same.
If you remove C:
- 222 isn't a solution anymore because 234 and 324 would answer the same.
- 224 isn't a solution anymore because 232 and 322 would answer the same.
- 555 isn't a solution anymore because 253 and five others trio would answer the same.
That's pretty weird and interresting how far from the real solutions those others solutions are, and that's hard to understand.
Re: Is setup #B52F2TL incorrect?
Posted: 22 August 2026, 14:29
by Jellby
https://turingmachinesolver.tiiny.site/
Enter the criteria cards (7 8 18 19 21) and you'll get a list of all 96 combinations. Click on "Test" for any of them and it will tell you if that's a possible solution or not. For example:
Code: Select all
34 ● even #1s = 2 ▲+■+● odd ▲+■ = 6 double Incompatible
(if there are two 1s and the sum of all is odd, no number can be even)
Code: Select all
53 ● odd #1s = 0 ▲+■+● even ▲+■ > 6 no double Not unique
(it doesn't result in a unique solution)
Code: Select all
91 ● odd #1s = 3 ▲+■+● odd ▲+■ < 6 no double A is redundant
(actually, if #1s = 3 everything else is redundant, and such verification card doesn't even exist)
And if you click on "Solve" you get a list of the 3 possible solutions.
I think your mistake may be that a code with 3 identical numbers (blue=yellow=purple) would give "no double" in card 21
Re: Is setup #B52F2TL incorrect?
Posted: 23 August 2026, 12:37
by muharred
Romain672 wrote: ↑22 August 2026, 13:59
If you remove A:
- 222 isn't a solution anymore because 235 and 325 would answer the same.
- 224 isn't a solution anymore because 233 and 323 would answer the same.
- 555 isn't a solution anymore because 254 and five others trio would answer the same.
If you remove C:
- 222 isn't a solution anymore because 234 and 324 would answer the same.
- 224 isn't a solution anymore because 232 and 322 would answer the same.
- 555 isn't a solution anymore because 253 and five others trio would answer the same.
That's pretty weird and interresting how far from the real solutions those others solutions are, and that's hard to understand.
That's the issue here - uncommon reasoning allows to eliminate all the solutions where blue is not equal to yellow based on the reasoning that verifiers that limit them only limit both at once. This means that if there is a solution where blue=2 and yellow=3, then blue=3 and yellow=2 will also be a solution.
But, writing this, I realised my mistake. The reasoning I mentioned means that somehow the answers given by verifiers must invalidate all combinations where blue≠yellow. Making these verifiers useful. Meanwhile I took condition that blue=yellow as if it was an axiom.
The condition is indeed always true with these verifiers, but A and C aren't redundant, because they are the verifiers that ensure that this condition is true.
Re: Is setup #B52F2TL incorrect?
Posted: 23 August 2026, 12:46
by muharred
Jellby wrote: ↑22 August 2026, 14:29
I think your mistake may be that a code with 3 identical numbers (blue=yellow=purple) would give "no double" in card 21
This I understood. This was the key to differentiate between 222 and 224 as a solution for the daily.
I realised my mistake writing a reply to another comment. Since I correctly assumed that blue must be equal to yellow in any solution, I didn't think that a combination of verifiers must ensure that every other combination is invalid. And that's why both A and C are present. But since I took blue=yellow as axiom, it made presence of both A and C redundant.
Re: Is setup #B52F2TL incorrect?
Posted: 23 August 2026, 12:58
by Jellby
It is an axiom because A and C are present, right?
Re: Is setup #B52F2TL incorrect?
Posted: 23 August 2026, 13:35
by muharred
Jellby wrote: ↑23 August 2026, 12:58
It is an axiom
because A and C are present, right?
Kinda. They play their part in this. With this combination of verifiers A and C will always give the same answer (both odd or both even) this way eliminating combinations where the parity of blue triangle is different from the parity of yellow square. With the specific challenge it eliminated codes 23? and 32?, I think. And it had to do so, or both codes would match every other verifier making the challenge invalid. But other verifier answers will have to eliminate codes like 245 (and 425) or 134 (and 314) for the puzzle to have a single solution.
The fact that blue must be equal to yellow follows from the fact that there is no verifier that gives a clue about just one of them. And all that give a clue about both of them only deal with their sum. And since a+b equals b+a in every respect, somehow all the verifiers working together must eliminate cases where blue≠yellow or we will end up with a situation where two codes match all the answers.