How my opponent could know the answer ?

[kostinen]

Hello everyone

I have to ask a question about this specific game (well, which I lost, obviously. If not I probably would not be here )

The game I talk about : https://boardgamearena.com/gamereview?table=453737742

My question is simple, how to find the blue number without checking the C verifier ?
Understand this could help me to be better at this game.

Thanks !

[kostinen]

Ok I got it. Given that violet is odd (from B verifier) and yellow is 3 (from A verifier), any other result than "no 4" would lead to a E verifier useless.

[Romain672]

That post is wrong. I still submit it to show my reasonning:

So from your game:
- if you take A & B & D, it's pretty interesting:
*) if y=3 then B&D will fix the value of purple which work (p=1245)
*) if y=12, then D must give purple higher, and you can't differenciate p3 from p5. So purple is even and p=24. But note that all possibilities look like to work
*) if y=45, then D must give purple lower, purple must be even (else you can't separate p1 from p3), so p=24

So potentials answers:
- x31 x32 x34 x35
- x12 x14 x24
- x42 x52 x54

But then C & E if they only give informations about b would tell blue even and would make b=2 or b=4.
It look really hard to advance from there.

Spoil ahead.
And my solver give only two answers: 211 and 231.
And yeah I forgot the case where y=p, I though it was impossible, but with more thinking that work.

[plotinus__]

Ok I got it. Given that violet is odd (from B verifier) and yellow is 3 (from A verifier), any other result than "no 4" would lead to a E verifier useless.

I'm new to this game so I'm sure I've misssed something, but how does the opponent know that violet is odd? From the replay it seems that the opponent did not use B verifier.

[Romain672]

I'm new to this game so I'm sure I've misssed something, but how does the opponent know that violet is odd? From the replay it seems that the opponent did not use B verifier.

I gave the only two possibilities in my answer above.
All others possibilities either:
- make one verificator useless (ie, if you take all the answers from the others verificator and remove that verificator you still got 1 solution left)
- have at least another solution (ie, two or more solutions give the same answers from all verificators)

[Falveryn]

Blue has to be 2.
Yellow can't be 1, 2, 4 or 5.
Y can't be the same as P.

So you only need to ask D if Y>P or P<Y, so it is doable with only 1 query.

[Romain672]

Blue has to be 2.
Yellow can't be 1, 2, 4 or 5.
Y can't be the same as P.

So you only need to ask D if Y>P or P<Y, so it is doable with only 1 query.

I redo my solver to be sure. But no, the end of my post is right, yours is wrong.
235 makes verificator B useless.


If no one change it:
https://docs.google.com/spreadsheets/d/ ... edit#gid=0

[GMT1]

I wonder the same with this game : https://boardgamearena.com/gamereview?table=457065005
And some other games I played...

If players don't solve the game by themselves, I don't see the point of playing.

[Romain672]

I wonder the same with this game : https://boardgamearena.com/gamereview?table=457065005
And some other games I played...

If players don't solve the game by themselves, I don't see the point of playing.

So instantly from B&C, you know that p<yb else B would be useless.
Then you don't have much information about purple, so that mean there is a blue or a yellow which equal 2 or 3 or 4 and purple below it (and maybe the parity of D) will fix the value of purple. But that doesn't look to help much.
So if b=y, then D tell the parity of both of them, and A tell the value which isn't 3. So 221 look like a valid solution, and the only one if b=y.
If b>y, then we got b>y>p. If b=3, then D is useless. If b=4, then it's either 432, 431, or 421, D would make 432 and 421 not be differenciated, so 431 is left.
If b<y, then we got y>b>p. If b=4, then y=5, and we got 452 as only valid solution (451 & 453 can't be differenciated). If b=3 then only solution is 342. If b=2 then only solution is 241.

So we got 221/431/452/342/241.
Let me look at my solver. Apparently 241 & 452 are only solutions. At least I got them.
221 makes verificator D useless.
431 has 5 others solutions like 521
and 342 makes verificator B useless.

But that's not too hard.
Note that if you play in 'easy 4' with very high elo players, you can expect them to be way better than me who have around 275 elo, and I reduced the possibilities to five pretty easily.


Note that it's doable in 1 question, and they used 2 for safety because you used 3 questions instantly.

[kostinen]

That post is wrong. I still submit it to show my reasonning:

So from your game:
- if you take A & B & D, it's pretty interesting:
*) if y=3 then B&D will fix the value of purple which work (p=1245)
*) if y=12, then D must give purple higher, and you can't differenciate p3 from p5. So purple is even and p=24. But note that all possibilities look like to work
*) if y=45, then D must give purple lower, purple must be even (else you can't separate p1 from p3), so p=24

So potentials answers:
- x31 x32 x34 x35
- x12 x14 x24
- x42 x52 x54

But then C & E if they only give informations about b would tell blue even and would make b=2 or b=4.
It look really hard to advance from there.

Spoil ahead.
And my solver give only two answers: 211 and 231.
And yeah I forgot the case where y=p, I though it was impossible, but with more thinking that work.

Thanks for the tips !