0-question solution to Hard/6 puzzle #C63 U9M 3
Posted: 25 January 2024, 21:20
I had another request from an opponent to go through my deductions after a game, so I wrote another BGG post with my solution. Hope it's useful!
https://boardgamegeek.com/thread/323411 ... -c63-u9m-3
Edit: Here's a link to this game on here, so you can see the criteria cards in the replay:
https://boardgamearena.com/gamereview?table=466339736
The puzzle is #C63 U9M 3
A(2): Blue compared to 3
B(14): Which colour is strictly smallest
C(24): Is there a sequence of (consecutive) ascending numbers
D(26): A specific colour is <3
E(39): A specific colour compared to 1
F(47): The number of 1s OR the number of 4s
E) Whichever colour B says is smallest, E must be verifying the same colour. This is because, if it were verifying a different colour then it would automatically be >1 (as it's not smallest).
Not verifying Bl=1, as A would be redundant.
If Y=1 then Y is smallest and D must say that P<3. This would make Y=1, P=2 and C would be redundant.
If P=1 then P is smallest and D must say that Y<3. This would make Y=2, P=1, and no matter what A says, C would be redundant.
Therefore this must be verifying that the smallest colour is >1.
D) This must also refer to the same colour as B & E, because if the smallest colour is at least 2, no other colour can be <3. Furthermore, the three verifiers (B, D and E) can not refer to Bl, as D would make A redundant. So we know that either Y=2 or P=2, and whichever it is is the smallest colour.
F) This can not be verifying the number of 1s, as we already know there are none. This can not be verifing Two 4s, as we would then already know from B, D & E that the solution is 424 or 442, leaving A and C redundant. So there must be No 4s or One 4.
A) Can not be Bl<3 as we already know Bl>2. Can not be Bl=3, because if Bl=3 and another colour is smallest, and we know that D has to refer to that same colour, then we would already know that it is <3 so D would be redundant. Therefore Bl>3.
C) There can not be an ascending sequence of 3 numbers, because either Y or P is smallest.
If there's an ascending sequence of 2 numbers, Bl>3 and Y<3 then Y would be forced to be smallest, making B redundant.
If there's an ascending sequence of 2 numbers, Bl>3 and P smallest then Bl=4, Y=5 and F would be redundant.
Therefore there is no ascending sequence.
B, D & E) If P is smallest and there is No 4, then Bl=5, so there could not be an ascending sequence and C would be redundant.
If P is smallest, there's One 4 and no ascending sequence, there's no unique solution between 432 and 542.
Therefore B, D & E refer to Y and we know Y=2.
F) If there were One 4, there'd be no unique solution between 425 and 524.
Therefore there are No 4s and the solution is 525 in 0 questions!
https://boardgamegeek.com/thread/323411 ... -c63-u9m-3
Edit: Here's a link to this game on here, so you can see the criteria cards in the replay:
https://boardgamearena.com/gamereview?table=466339736
The puzzle is #C63 U9M 3
A(2): Blue compared to 3
B(14): Which colour is strictly smallest
C(24): Is there a sequence of (consecutive) ascending numbers
D(26): A specific colour is <3
E(39): A specific colour compared to 1
F(47): The number of 1s OR the number of 4s
Solution:Before we begin, it's important to understand the two basic tools:
1) No verifier can be redundant.
2) There must be a unique solution.
Rule 1 means that every verifier must be necessary to find the solution. If any response or set of responses from verifiers would mean that another verifier no longer gives you any additional information, then that set of responses is not a valid combination.
Furthermore, if you think you have a possible solution, but removing any one of the verifiers would still lead you to the same unique solution, then that solution can not be the correct one.
Rule 2 means that if a particular combination of responses leads either to zero codes or to more than one code, then that combination can not be correct; only combinations of responses that lead to exactly one solution are valid.
It is also important to understand how the verifiers work, in particular in relation to the non-exclusive criteria cards (26–48). Each verifier knows that one of the criteria listed on its criteria card is true. However, it knows nothing about the other listed criteria; they may independently be true or false, but they will have no bearing on the responses that this verifier gives. The verifier will simply respondif your code obeys the one fact that it knows and
if your code does not obey it; the other listed criteria are irrelevant. Your job is to work out which one fact it knows.
Throughout this post I'll use Bl=Blue (Triangle), Y=Yellow (Square) and P=Purple (Circle).
E) Whichever colour B says is smallest, E must be verifying the same colour. This is because, if it were verifying a different colour then it would automatically be >1 (as it's not smallest).
Not verifying Bl=1, as A would be redundant.
If Y=1 then Y is smallest and D must say that P<3. This would make Y=1, P=2 and C would be redundant.
If P=1 then P is smallest and D must say that Y<3. This would make Y=2, P=1, and no matter what A says, C would be redundant.
Therefore this must be verifying that the smallest colour is >1.
D) This must also refer to the same colour as B & E, because if the smallest colour is at least 2, no other colour can be <3. Furthermore, the three verifiers (B, D and E) can not refer to Bl, as D would make A redundant. So we know that either Y=2 or P=2, and whichever it is is the smallest colour.
F) This can not be verifying the number of 1s, as we already know there are none. This can not be verifing Two 4s, as we would then already know from B, D & E that the solution is 424 or 442, leaving A and C redundant. So there must be No 4s or One 4.
A) Can not be Bl<3 as we already know Bl>2. Can not be Bl=3, because if Bl=3 and another colour is smallest, and we know that D has to refer to that same colour, then we would already know that it is <3 so D would be redundant. Therefore Bl>3.
C) There can not be an ascending sequence of 3 numbers, because either Y or P is smallest.
If there's an ascending sequence of 2 numbers, Bl>3 and Y<3 then Y would be forced to be smallest, making B redundant.
If there's an ascending sequence of 2 numbers, Bl>3 and P smallest then Bl=4, Y=5 and F would be redundant.
Therefore there is no ascending sequence.
B, D & E) If P is smallest and there is No 4, then Bl=5, so there could not be an ascending sequence and C would be redundant.
If P is smallest, there's One 4 and no ascending sequence, there's no unique solution between 432 and 542.
Therefore B, D & E refer to Y and we know Y=2.
F) If there were One 4, there'd be no unique solution between 425 and 524.
Therefore there are No 4s and the solution is 525 in 0 questions!