Close Topic Options

Daishain's 5e D&D Rules, Q&A, And Support - Page 920

If you weight 4 against 4 and get balance, all 8 are good coins. You then weigh 2 known good against 2 unknowns. If you get a balance then the last 2 unknowns contain the bad coin, if not the two on the scale are one bad. So you weight one good against one of those and if it balances the one left out is bad.
If the 4 vs 4 is unbalanced, you know the last 4 are good. I’m trying to work out what two weighting’s will show the bad one.
Edit. Okay, if the two unknown 4s are 1234 and abcd, and you then weigh 12a against 3bc and they balance the bad coin is either 4 or d and you can weigh a good coin against either to tell which is bad.
If the 12a 3bc is light on the 12a side, then either a is a light bad coin, or b or c is a heavy bad coin. Vice versa if it’s light on the other side. Then you weight whatever pair is heavy (B against c in the heavy case) if it balances, the bad coin is the single one. If it doesn’t, then the heavy side is the bad one. Check my work please. I don’t want to miss it like the second one.

Edited: Gknightbc on 26th Sep, 2018 - 4:12am

Support QampA Rules Dragons and Dungeons e Daishains

Ok if we do not know if lighter or heavier you are weighing four at a time If the first weight is same all are good and the different coin is in the last four That makes it easy. If the first weight is different then it gets harder.

Daishain's 5e D&D Rules, Q&A, And Support Archive Pathfinder / D&D

Gknight, I think I have followed your reasoning, and I think it is sound. The worst that can happen is if we're wrong, we don't get as much intel from the sphinx. I say post it, as best you can. But maybe let one or two others chime in, first. I might be missing something as well.

Page 920 Support QampA Rules Dragons and Dungeons e Daishains

Well shoot…

Sorry, should have double checked it a bit more. Guess I was lax due to Knight having stated familiarity with the riddle's origin.

I'll let it stand.

Support QampA Rules Dragons and Dungeons e Daishains

I think you are on the right track, Gknight. Break them up into 3 sets of 4 coins. Then weigh the first 1-4 against 5-8. However, If they balance, you don't break up the last one into sets of two. Instead, you take 3 of the known good coins like 1-3 and compare them against 3 untouched coins, 9-11. If the new two stacks are still the same, then you know the weird coin is 12. If not, then you need to check and see which side was heavier. That's the key to this puzzle. If 9-11 is heavier than 1-3, then you know the odd coin is heavier than normal. But if it's lighter, then the 1-3 stack will be heavier. Now, you've identified that the strange coin is lighter or heavier. With that new piece of info, NOW you can test even numbers of coins. So test 9 versus 10 and put 11 to the side. If they match, then you know 11 is the odd coin. If 9 weighs more than 10, then refer back to whether or not you are looking for a lighter coin versus a heavier coin. If looking for a lighter coin, the answer in this example is 10. If you are looking for a heavier coin, then the answer is 9.

The solution above is pretty exact only if the first two stacks of 4 balance. If they don't, then the rest becomes a bit more complicated, but the key to the puzzle stays the same. At that point you will need to mix and match one coin from other sets into the sets of three I laid out above, and continue testing for lighter versus heavier.

EDIT Here are the other two scenarios broken down:

1. 1-4 is heavier than 5-8. This means either 1-4 contains a heavy coin OR 5-8 contains a light coin. The goal is to get back to testing groups of 3. And our secret weapon is knowing that the 9-12 stack are all normal coins (Because the unbalanced one is somewhere in 1-8). Now we break things up into groups of three and mix a few up. Let's do 2,6,8 and 3,7,9. We already know 9 is a normal coin and the only coins left are the potentially abnormal coins. This leaves 1,4,5 off to the side. If 2,6,8 and 3,7,9 balance, then you know the problem is somewhere in 1,4,5 and can do the 1 versus 4 test. Now, here's where it can get confusing and weight matters a lot. If 3,7,9 is heavier than 2,6,8, then either 3 is the heavy coin (Because we determined earlier that 1-4 is heavier than 5-8) OR 6 or 8 are light (Remember that only 1 coin can be the odd one). Now, compare 6 and 8. If they balance, then 3 is abnormal and heavy. If one is lighter, then that's the abnormal light coin.

I won't bother explaining the reverse because it's pretty similar. Now how the heck we get Lawrance to explain all this complicated mess to the sphinx in character is interesting. Puzzles in general are a bit meta for this reason. I could have Perry do it but it would be so out of character that I'd rather have Lawrance do it since he was on the right track.

Edited: seath on 26th Sep, 2018 - 1:34pm

Daishain's 5e D&D Rules, Q&A, And Support

Seath, you must have missed my edit. With regards to your version, you don't really need to know if the coin is lighter or heavier in the first scenario when the 4v4 is the same, just that one doesn't match. As far as your second part, that's the same as what I stated.
For laziness sake, Lawrance will be demonstrating. With illustrations. Does anyone have a solution to Riddle #2, because it has been bugging me. Familiarity with the riddle doesn't mean I can remember the solution from 30 years ago. .

Daishain's 5e D&D Rules Q&A Support - Page 920

The solution I am familiar with is to ask A the following. Though there are a couple other ways to do it.

"Is exactly one of the following two statements true?
-You are the knight
-B is the joker"

If the answer is yes, then B is either the knight or the knave, but cannot be the joker.
If the answer is no, then C is either the knight or the knave, but cannot be the joker.

Direct your following questions to the one you have eliminated as maybe being the joker. Now that you've eliminated the one who is inconsistent with the truth, the responses become more predictable.

Edited: daishain on 26th Sep, 2018 - 2:23pm

Daishain's 5e D&D Rules Q&A Support D&D / Pathfinder Archive - Page 920

Gknight, you do need to know lighter versus heavier in the first scenario because once you get down to the last three coins you still don't know if the coin you are looking for is lighter or heavier since the first stack balanced out. You only find that bit of info out in weigh #2 which let's you make the correct decision on the final weigh-in.

Your description of mine also WAS NOT the same in the second part. In my version of the latter two scenarios, I'm adding a coin from the third stack into the mix because it is 100% known as a good coin since one of the leftover two stacks of four definitely has a bad coin. You only take the ones from the potentially bad stacks and thus leave two coins (An even number, doesn't work) leftover.

Edited: seath on 26th Sep, 2018 - 2:38pm