Think before you claim

[Jellby]

In the row that showed 114, there was a cell with an undefined reference. I "fixed" that to I10 and the 114 went away.

[Jellby]

Here's another challenge, with 10 verifiers (even if the game allows only up to 6):

A: 01: Blue = 1 | Blue > 1
B: 07: Purple even | Purple odd
C: 09: 0 threes | 1 three | 2 threes | 3 threes
D: 16: even > odd | even < odd
E: 21: no pairs | one pair
F: 22: ascending order | descending order | no order
G: 24: 3 numbers in ascending sequence | 2 numbers in ascending sequence | no ascending sequence
H: 27: Blue < 4 | Yellow < 4 | Purple < 4
I: 31: Blue > 1 | Yellow > 1 | Purple > 1
J: 39: Blue = 1 | Blue > 1 | Yellow = 1 | Yellow > 1 | Purple = 1 | Purple > 1

I believe there's no possible challenge with more than 10 verifiers (and this is the only one with 10).

[Murr_Rockstroh]

As I said above, I know what people are doing. Or at least I know how this could be done, answering with no questions asked.

Someone, without using a "solver," walk through the logic of this set of verifiers and tell me how you deduce the code without asking a single question.

A: Blue < 3 | Blue = 3 | Blue > 3
B: Blue < Yellow | Blue = Yellow | Blue > Yellow
C: Blue < Yellow & Purple | Yellow < Blue & Purple | Purple < Blue & Yellow
D: Even > Odd | Even < Odd

[Sunfall]

As I said above, I know what people are doing. Or at least I know how this could be done, answering with no questions asked.

Someone, without using a "solver," walk through the logic of this set of verifiers and tell me how you deduce the code without asking a single question.

A: Blue < 3 | Blue = 3 | Blue > 3
B: Blue < Yellow | Blue = Yellow | Blue > Yellow
C: Blue < Yellow & Purple | Yellow < Blue & Purple | Purple < Blue & Yellow
D: Even > Odd | Even < Odd

BIG EDIT: I'm pretty sure this is a 50/50, not an unambiguous solve. I still haven't run it through a solver to prove it, though.

I'll take a swing at this. Note that I'm doing it live, so apologies if it's a little rambly.

First, we can eliminate Blue = Yellow as a possibility. If it's a criteria, and Blue = 3, then verifier D is pointless. Both other choices for criteria A require criteria C to be Purple < BY (since neither Blue nor Yellow can be "the least"); if Blue < 3 then that forces 221 and verifier D is unnecessary, and if Blue > 3 then the solution is ambiguous, as 44x and 55x both force criteria D but provide no clue as to what value Purple has.

Now, consider Blue > 3. If Blue > Yellow, we again enter the realm of foggy ambiguity. Neither Purple < BY nor Yellow < BP provide nearly enough constraints on the values, as Blue is in {4, 5} and Yellow and Purple are swimming around in, at minimum, {1, 2, 3}, allowing for any mix of evens and odds verifier D could ask us for.

If Blue < Yellow, then we have 45x. If Blue < YP, then it's 455 and D is again unnecessary. So if it's anything it's Purple < BY. There are two odd numbers less than 4, but only one even number, so D would have to be Even > Odd and the answer is 452.

OK, that's great, so we need to prove now that it can't be Blue < 3. If it is, and Blue > Yellow, then it's 21x and we have to set C to Yellow < BP. But either choice for D is ambiguous, since purple can be any of {2, 3, 4, 5}, leaving it two choices whether there are more even or odd numbers. So nope. Our last option is Blue < Yellow. Let's go through the three options for C now:

• Blue < YP: Blue is {1, 2}, and Yellow and Purple are {2, 3, 4, 5} or {3, 4, 5} depending on Blue. Neither of the choices for D can possibly disambiguate this, as there are simply too many options.

• Yellow < BP: Literally impossible, since we already chose Blue < Yellow.

• Purple < BY: We have 2x1. Yellow can be {3, 4, 5}, so we have to pick Even > Odd as the choice, so that the result can be 241 rather than the ambiguous odd choices.

So it seems like the answer can be either 452 or 241, and it all hinges on whether Blue < 3 or Blue > 3 (since both of them end up with the same values for the other three criteria). That's a 50/50, though, so I could see someone taking a swing at it with zero verifiers. It's also possible they just didn't see the second alternative and thought there was only one option--as I did the first time I wrote this out--and got lucky.

P

[Romain672]

A: Blue < 3 | Blue = 3 | Blue > 3
B: Blue < Yellow | Blue = Yellow | Blue > Yellow
C: Blue < Yellow & Purple | Yellow < Blue & Purple | Purple < Blue & Yellow
D: Even > Odd | Even < Odd

So from B&C, it's a commun thing people not see. If b<yp, then B is useless. If y<bp then B is useless.
So C said that p<by.
Then A&D will most of the time fix the value of blue. So most of the time you will need to do some checks for blue.

Let's check b=y two seconds. So we still got p<(b=y), then if b=3 then 331 and 332 are potential solutions. But both have more odd than evens, so can't be dissociated.
Now if b<3, then it's 221 and D is useless: impossible.
If b>3, then it's 441/442/443/551/552/553/554. But you don't have information to separate them so it's impossible.
So b != y.

And if b>y. So we got an order p<y<b, so it's 321/421/521/431/531/541/432/532/542/543. We can separate odds and evens: 321/521/431/531/541/532/543 & 421/432/542. And finally for each group we need to separate the 3 cases of blue: 321 for odds and b=3, too many for odds and b>3, 421/432/542 for evens and b>3.
None of those work: 321 is solo b=3 (so D is useless), and 421/432/542 can't be separated.

Now if b<y, we got an order: p<b<y, so it's 231/241/251/341/351/451/342/352/452/453. So we separate odds and evens: 241/342/452 & 231/251/341/351/451/352/453. Then for 241/342/452 they got differents answers for blue so that seem to work. For the odds, we got one solution for b<3/b=3/b>3, and for odds, too many for each of those.
So we got 241 & 342 & 452 as solutions.

But now we can check if another verificator was useless for those.
I though there was none, but my solver tell that B is useless for 342. Which would mean we have b=3, p<by, and more evens. And it's true, since B is odd, that force the value of two others numbers as evens.

And my solver confirm the two others solutions.

But I have 200 to 250 elo. I still made lots of mistakes and is not as fast as others: often I think there is only one solution left while there is 2 or 3 remaining.
I suppose all of those is not possible for me in 4min, but I think in 6min (so after validating the code once) I can maybe have it but it's close. And there would be my mistake still.


edit: and you can have the feeling it's forced to be two evens with those verificators. If you commit on that and with p<by, you are already at 10 possibilities remaining (and two verificators not used): 421 241 221 441 / 342 442 542 432 452 / 443.

[Sunfall]

I just... totally forgot to test the B = 3 case (I eliminated Blue = Yellow, but not that). Oops! Thanks for catching that.

Working it through my way:

Given B = 3, if Blue > Yellow, Yellow is in {1, 2}. If Purple < BY, then it's 321 but we didn't need verifier D, so that's not valid. And if Yellow < BP, we didn't need the Blue > Yellow constraint, so that's invalid.

If Blue < Yellow, if Blue < YP, then we didn't need the first constraint at all, so that's an invalid choice. The only remaining option is that Purple < BY, which as always has too many options with the odd D constraint, but the even one forces 342. And, hey, that's the third one that Romain672 found.

Like Romain said, it's easy to think you've "got it" only to realize you don't, as I just ably demonstrated on my own behalf.

P

[Murr_Rockstroh]

As I said above, I know what people are doing. Or at least I know how this could be done, answering with no questions asked.

Someone, without using a "solver," walk through the logic of this set of verifiers and tell me how you deduce the code without asking a single question.

A: Blue < 3 | Blue = 3 | Blue > 3
B: Blue < Yellow | Blue = Yellow | Blue > Yellow
C: Blue < Yellow & Purple | Yellow < Blue & Purple | Purple < Blue & Yellow
D: Even > Odd | Even < Odd

I'll take a swing at this. Note that I'm doing it live, so apologies if it's a little rambly.

(snip, see solution walk through above)

So it seems like the answer can be either 452 or 241, and it all hinges on whether Blue < 3 or Blue > 3 (since both of them end up with the same values for the other three criteria). That's a 50/50, though, so I could see someone taking a swing at it with zero verifiers. It's also possible they just didn't see the second alternative and thought there was only one option--as I did the first time I wrote this out--and got lucky.

P

It is one of those two answers. Thank you for walking through that for me. And I appreciate that you did it "live" that is kind of what I wanted.

And I see what Romain above is doing, which is more a little deduction, then some brute force to check the deductions made.

None of this would be done at the table in live game and I suppose I was hoping that's how people might play here competitively, and most of the games I've joined, that's exactly what seems to be going on. But when I got into two different games and the winner solved it with zero questions asked, I just felt like that really wasn't in the spirit of the game, again since I was approaching it more like I was actually at the table playing it with other people, not taking a long time to make deductions and checks to see if I could do it in zero questions.

I'll just mark those that do that as "I'd like to avoid this person" or play it solo.

[Romain672]

None of this would be done at the table in live game and I suppose I was hoping that's how people might play here competitively, and most of the games I've joined, that's exactly what seems to be going on. But when I got into two different games and the winner solved it with zero questions asked, I just felt like that really wasn't in the spirit of the game, again since I was approaching it more like I was actually at the table playing it with other people, not taking a long time to make deductions and checks to see if I could do it in zero questions.

Yeah I think I would enjoy a specific timer for this game, a really short timer.
For easy, something around 1:00 for all 3 questions (+0:40 per extra turn). And no extra time between the turn or when you guess (or like +0:10).
For extreme/nightmare, maybe the double.
And a way to punish people out of time (maybe either instant loss if out of time, or -5% elo per extra second, capped as the normal loss).
All those things aren't possible on bga atm.

That change the game into forcing you to deduce things fast (kinda like the game Reflection), or to make you have a good feeling about the right questions to ask.

I suppose I can already do that by increasing the difficulty, but I like the simplicity of 'normal easy 4' or similar.