plasmid

Right on, PG, and glad you like it. This one I think I'll keep in a Sundae Best collection.

Now that you've solved the second part, what's the answer to the first part?

Well that gives new meaning to "pardon my French." And that takes care of the question. But wait, I made a mistake in the OP. You didn't throw out N+1 coins, you actually threw out N/2 coins. And if the cat were any lighter weight, denoted M, then you would have had to throw out more than M/2 coins. Now how heavy is the cat?

I got a different answer.

When I tried that, I got the following, which didn't seem to help as far as I could tell. Edit: fixed spoilers

I don't remember where the last Puzzle Land left off, so I'll pretend that you were stuck in jail with a warden who used to be a villain in a James Bond movie. He gives each prisoner a black or white hat and forbids anyone to communicate by any means other than his pet homing pigeon, which is trained to fly around the prison counterclockwise while counting each cell that it encounters and stopping whenever it reaches a prime number. After reading the remainder of the rules and leaving everyone in their cells to work out their hat color with the pigeon, you're the first to get the pigeon. Realizing that the warden never specifically forbade taking off your own hat and looking at it, you do so. You let the pigeon loose so that any other prisoners who have enough sense to look at their own hats will go free, and good riddance to those who would depend on some cockamamie bird-brained scheme to escape. After being released from prison, you then visit the castle of Puzzle Land and are called in service to the king. One of his knights takes you to a room with a pile of 27 coins, and explains that they are all gold and of equal weight except for one counterfeit that is slightly heavier (with the difference in weight between the counterfeit and a real coin being much smaller than the weight of a real coin). Your job, using only three weighings with the Royal Balance, is to identify the counterfeit and throw it away so the knight can bring the real coins to the king. Knowing full well how to handle this, you place nine coins on one pan and nine on the other. Seeing that one of the pans is heavier, you say "All right, looks like the counterfeit is among these nine." The knight, sounding as though something is amiss, says “Uh, you do know that's the Royal Balance, right?" You: Yes, what of it? Knight: Don't be too sure that the counterfeit is among those nine. You: What? Why not? They were in the heavier pan compared to another with an equal number of coins, so the counterfeit must be in there. Knight: Like I said, this is the Royal Balance. The one we use to trade with merchants. The pan that was lowest is the one that always weighs one coin heavier than the other. You: ... Knight: Well how do you think the king built such an empire? Magical royal elves? You: ... Knight: That counts as one of your three weighings, by the way. You: ... now you know there's no way for me to identify the counterfeit in two more weighings, right? Knight: Well then just find as many as you can that are genuine. As long as we don't give him a fake, maybe he won't bother to count them and figure out that you screwed up. You: … Knight: Look, I'll even make it easy for you by de-rigging the balance so it works like normal. As the knight finishes messing with the balance and you're about to get started, the royal kitteh comes along, hops up onto one pan of the balance, and (because it's a kitteh) promptly falls asleep. The knight warns you that aggravating the royal kitteh in an attempt to get it to move is punishable by instant slaying, so the cat will stay where it lays. But he tells you that the kitteh weighs exactly the same as N real gold coins. Using that fact, and the fact that the kitteh is not completely blocking you from putting some coins on the same pan with it, with two more weighings you get the worst possible luck as far as narrowing down where the counterfeit lies, but you toss out N+1 coins and give the rest to the king without being beheaded for giving him a counterfeit. As it turns out, if the kitteh were any heavier, you would have ended up tossing out N or fewer coins instead of N+1. To clarify that last bit, if the kitteh did not weigh N coins but instead weighed M = N+X coins for any positive integer X, then you would have tossed out M or fewer coins rather than M+1 coins in the worst case scenario. How much does the kitteh weigh?

Odd. If you type stuff, then highlight it, then bold or super/sub-script or change color or whatever for the highlighted area, it works fine. If you don't have anything highlighted and do any of the above in anticipation of typing something that should have that effect, then it adds characters.

More likely when dealing with Y-san, and follow up on Omega Scales:

Heaven sent with company Clad in cloaks of four or three Straight and narrow I'll not be Thus my fate is cast Sawn asunder as you're fain In coffin uninterred lie slain 'neath ice with blood-stained ball and chain Gouged until the last

Oh, if we're allowed to send two mice in a sort of staggered fashion like that - letting one mouse go in one direction and then letting the next mouse go in the same direction several seconds or a minute later to send a signal that's clearly distinct from, say, one mouse making two trips around the prison - then it can be done even faster.

lol @ everyone answering at once

I could mentor. Just include me when you PM the role. For the questions you mentioned, the lynch votes during the day phases generally just require a plurality: whoever has the most votes gets lynched -- killed with their role publicly revealed. While there are only 5 goodies and 2 non-goodies, some of those goodies have abilities that can give them information to identify other players as definite or probable baddies. And there's a sort of intangible aspect to the game of figuring out who seems to be acting weird and likely baddie based more on reading someone as if they were playing poker rather than pure logic. And of course, there's a reasonable chance that even if the goodies have no idea what they're doing they might lynch the baddie or indy with pure random luck.

That might be the most efficient way to do it; I can't think of a better one.

One thing I can say so far, which might or might not be pertinent, is that

Getting things started by solving for cases where {r1, r2, r3} are all rational numbers, but not solved for cases where they're irrational.

That should work. I had a different answer in mind.

Oh, I see what you're thinking. In the solution you're proposing, one prisoner can send the mice to any other specific prisoner he wants, with the mice arriving from either the left or the right. In my mind, the food slots face out through the main door of each cell and into a common area like in the picture. I was thinking that if, for example, prisoner 4 were to send one mouse right and one mouse left, then the mouse he sends right would visit prisoners 3, 2, 1, 7, 6 etc in that order, and the mouse he sends left would visit prisoners 5, 6, 7, 1, etc in that order. The mouse wouldn't have any indication that it's intended to visit one particular prisoner along that route and would visit everyone along the way, and might be picked up by any prisoner along the way and either have its direction switched or be held onto while waiting for the other mouse to arrive. TSLF might be able to clarify what's intended in this problem.

If you were to attempt to use that strategy in a solution

Roster: 1) Phil1882 2) Framm18 3) Flamebirde 4) bonanova (that's with a lower case b) 5) 6) 7) Nana77 (or can back-up or mentor as needed) Back-ups: A) plasmid B) C) I'll switch from backup to player in a couple of days if signups haven't filled by then, but will leave a couple spots open for people new to mafia. Does an indy win end the game and make the goodies lose? For Knuckle's APR ability, does it start counting APR on the night that it's used -- meaning that the target will probably go into the next day phase with 2 APR (one due to the night phase passing and one due to their night action that night) and then die at the end of the upcoming day phase? Does Shoot's trap act like a block, or does it act more like what I typically think of as a trap by also saving and preventing the trapped player from voting and being voted for the next day? If the latter, is the indy prohibited from continually self-trapping to make themselves invincible and guarantee that they'll outlive their wincon targets?

If your game is ready to run and we still don't hear from Kikacat, how about go ahead and post the OP and we can have critique and signups at the same time? If it's a straightforward design then it probably won't need much if any tweaking. If you're not ready, I could go ahead and post one.

Looks like you've mostly solved your own puzzle And no one tried to argue that it's impossible because there are only 6*6*4*4 = 576 ways to orient the dice, while there are 6! = 720 ways to arrange the prisoners, so the dice can't encode all possible arrangements of prisoners. [spoiler=Well, here's my implementation]First, agree on a cycle to pass the dice. For example, if the guard will ask you to name a prisoner to give the dice, then agree on some order like prisoner A -> prisoner B -> prisoner C -> prisoner D -> prisoner E -> prisoner F -> prisoner A. Represent the orientation of the dice with four numbers as follows. The first number (from one to six) is the number facing the lion on the die that's on the side with the “D”, the second number (also from one to six) is the number facing the lion on the die that's on the side with the “E”. The third number (from one to four) is derived by considering the die on the side with the “D”, looking at the number facing the “DNE” and seeing where that is among the four numbers that do not face toward or away from the lion. For example, if face 2 were toward the lion and face 5 were away from the lion, then if face 4 is pointing toward the DME then 4 is the third number among {1, 3, 4, 6} that don't face toward or away from the lion, and represents a 3 as the third digit in the code. The fourth digit is obtained similarly by considering the die on the side with the “E”. The first prisoner to get the dice will consider his cell number, call it A, and pass the dice to the next prisoner arranged with the code [A A 1 1]. The fact that the first two numbers are the same and the second two numbers are both 1 will let the prisoner who receives the dice know that he's the second prisoner to get them, and will tell him that the prisoner before him in the cycle is in cell A. The second prisoner will consider his cell number to be B, and will pass the dice in the orientation [A A 1+int(B/3) 1+(Bmod3)]. The third prisoner to receive the dice will know that he's the third because the first two numbers are the same and the last two numbers are not both 1, and he will know the cell numbers of the two prisoners before him in the cycle. The third prisoner will consider his cell number to be C. He will calculate X as the number that C would be if A and B were excluded – for example, if C=5 and A=2 and B=6, then C=5 would be the fourth number out of the numbers left if A and B are removed {1, 3, 4, 5}, so X would be 4. The third prisoner will pass the dice in the orientation [A B X X], which will let the fourth prisoner know that he's the fourth to get the dice because the first two numbers are different and the last two numbers are the same. The fourth prisoner will similarly calculate Y based on his cell number using the same method that the third prisoner used to calculate X, and will pass the dice in the orientation [A B X Y]. The fact that the first and second numbers are different and the third and fourth numbers are different will let the next prisoner know that at least four other prisoners have seen the dice, and will tell him which cells each of the four prisoners preceding him are in, so knowing that in combination with knowing his own cell number will mean that the fifth prisoner will know everyone's cell number. Since the fifth prisoner knows everyone's cell number, he can pass the dice on to the sixth player with the same [A B X Y] orientation that he would have used if he were the fourth prisoner. It doesn't really matter if the sixth prisoner knows that he's sixth and not fifth, as long as he gets enough information to know that he's somewhere after the fourth and can figure out everyone else's cell number. The sixth prisoner then passes it back to the first prisoner, again using the orientation [A B X Y] that he would have used if he were the fourth prisoner. Now the fifth, sixth, and first prisoners to get the dice know everyone's cell number, and they can each free themselves and the prisoner that is three steps ahead of them in the cycle that was agreed upon at the beginning.

Ready to run Asylum, or still ironing things out? If not, maybe we could run a more-or-less standard mafia in the meantime to keep things going.

Ok, now we have enough information from what's been posted so far. I wonder if anyone will simultaneously give an algorithm that saves everyone, and prove that it's impossible to save everyone.