How my opponent could know the answer ?

[biwebof]

"4 solutions: 125-135-521-531"

I don't feel like going through this logic again because I remember it being more difficult than the 212 one.
But these 4 solutions feel familiar, if I remember right, I was in this situation multiple times yesterday.
So after the first time, I use my solver AFTER the game, because I'm curious what the answers are.
(I think I should be allowed to do that, I can get the same result by listing all the answers on paper + I made the solver myself, so you cannot say I used someone else's work.)

And then, if I see the cards again, I remember the solve.

I think this is unfortunate situation, I see the same puzzles sometimes. (I think I saw the 542/142/524/124 one like six times already.) Sometimes I joke in chat "I have psychic powers, I can get this one in 2 questions."
Now of course me having seen the question already gives me perhaps unfair advantage of someone seeing it for the first time.
But first of all, it's like innovation, if you know the cards and combos, you will have advantage over those who don't.
And second of all, any unfair advantage is game's fault, not mine. They should make more cards so that seeing the same puzzle happens less often.

[mdpeterson42]

I am really confused by a couple of points of this discussion. First, it seems that you are saying there could be multiple solutions, but that is not what the rules say. Can you clarify?

Second, I disagree that having seen a puzzle before gives you any less advantage than knowing the cards in Innovation. This is supposed to be a deduction game, where each puzzle has one answer. If you've seen the puzzle, you know the answer. It's not a game at that point. And for that reason, I don't see any reason to play this competitively on BGA.

Third, with all the discussion of one question (or even zero question) solves, it seems like this isn't even a game in the best of circumstances - it's just a math question. Whoever has "solved" the game will get it right on the first turn - maybe both will. In the immortal words of Jim's dad from American Pie - "It's not a game. It can be fun, but it's not a game." Am I missing something?

[Jellby]

I am really confused by a couple of points of this discussion. First, it seems that you are saying there could be multiple solutions, but that is not what the rules say. Can you clarify?

My guess is that there are several answers compatible with the criteria cards selected. The actual answer depends on the specific verifier selected for each criterion. Some of them may be shown to be incompatible (according to the rules of the game), but at some point you can't do any more deduction and have to actually ask some question to proceed.

If you've seen the puzzle, you know the answer. It's not a game at that point. And for that reason, I don't see any reason to play this competitively on BGA.

You don't know if you have seen the same puzzle. A given selection of criteria cards could be compatible with different choices of verifiers, so a puzzle that looks the same on the surface could have a different answer at the end, because the verifiers are different. Of course, all the deductions that let you rule out possibilities before asking any question will be the same, and you won't need to redo them (if you remember the conclusion).

Third, with all the discussion of one question (or even zero question) solves, it seems like this isn't even a game in the best of circumstances - it's just a math question.

You can call that a game, but it's a mostly solved game. It's the same with some abstract games. Is tic-tac-toe a game? Are pure-luck games real games? You could say they're just a coin flip.

[The Jester]

Same topic, different game.
I am curious how we get to a 0 check solve on this one:
https://boardgamearena.com/table?table=459148726

I could only eliminate the third option of verifier C (straightforward), the second option for verifier E (it makes verifier F useless). After that I am blocked. I got quite lucky with my first checks and set of numbers but not as fast as the winner just from the problem statement.

I guess we don't have the luxury to look at the logic of every single problem but I am curious to know how to improve my game so feel free to have a go at it and educate me and others.

Thanks in advance

[Romain672]

https://boardgamearena.com/table?table=459148726

E&F are fun.
If there is a pair, then F is useless, so there is no pair.
If there is an order, than E is useless, so there is no order.
You obviously can't have three 4s.
And after, you are stuck. Let's see what we could do.

Purple is hard to have a single solution.
If p=2, then in most cases p can be equal to 4. So if that's the case, either: b=4 or y=4 or (y=3 & b=1).
If p=4, that's the opposite of above.
If p is odd, then you want to differenciate p=1, p=3, and p=5: If y=3, then purple will always have two solutions. If y=1 or y=5, only solutions are 153 315 351 513 by using only BEF. That makes C useless. So p isn't odd.

So we got 4x2 x42 132 / 2x4 x24 534 left.
By doing a second turn with E&F we got 452 142 342 132 214 254 324 524 534 left.
That leave 9 possibilities by only using B, E, and F. I will not continue since the rest look boring, but yeah that look pretty hard to me.

[poptasticboy]

Same topic, different game.
I am curious how we get to a 0 check solve on this one:
https://boardgamearena.com/table?table=459148726

I could only eliminate the third option of verifier C (straightforward), the second option for verifier E (it makes verifier F useless). After that I am blocked. I got quite lucky with my first checks and set of numbers but not as fast as the winner just from the problem statement.

I guess we don't have the luxury to look at the logic of every single problem but I am curious to know how to improve my game so feel free to have a go at it and educate me and others.

Thanks in advance

Hey! I've since discussed this game privately with The Jester, but I've noticed it show up here, and I'm never one to miss an opportunity to explain my reasoning!

E and F are the usual, they force each other to be "No pairs" and "No order".

C can not be two 4s or three 4s, because that would make F redundant

Bl+Y>6 would make A redundant.

If Bl+Y=6 and there are no 4s, F would be redundant.
If Bl+Y=6, there is a 4 and P is Odd, A would be redundant.
If B+Y=6, there is a 4 and P is Even, then No Pairs forces Bl & Y to be Odd, making F redundant.

Therefore we know Bl+Y<6. That's D, E and F solved.

If Bl>1, there is no 4 and P Odd, no unique solution (231/213, for example).
If Bl>1, there is no 4 and P Even, no unique solution (e.g. 312/222).
If Bl>1, one 4 and P Odd then F would be redundant.
If Bl>1, one 4 and P Even, no unique solution (e.g. 412/214).

Therefore Bl=1 and A is solved.

If there is No 4 and P Even, then F is redundant.
If there is No 4 and P Odd, only solution would be 111. Looking at it now, I'm not sure how I ruled 111 out. This might have been a mistake.

If there is a 4 and P Odd, Solution is 143 but D would be redundant.
If there is a 4 and P Even, solution is 142.

So looking at it now, unless I am missing something, the solution has to be 142 or 111, and I should have needed 1 question to C to solve it. I must have missed 111 as an option when I did it before and was lucky that was not correct. It's too easy to forget triples with card 21!