CaptainEd

Just so that I'm sure of what you mean. When you say that any two of the sides can be used to form those sums, are you referring to this die as having 6 sides--4 of them have two faces, and 2 of them have just one face?

For that matter, does the carnivore get to take part in the deliberations--the decision on the proper strategy?

So, my solution, spread out over 38, 39 and 41 solves this, right?

Good point, Bushindo! Prisoner 2 is the first one who knows both colors, so he adopts a position that makes Prisoner 1's location correct as well. And, following TSLF's suggestion of NEWS-coding, Prisoner 1 tells all arrivals their own color. Very pretty!

Channeling other denizens, as we all thought this would constitute "communication", hence illegal.

OK, I think we are agreeing that we have to relax our interpretation of Wolfgang's proscription against communication. The first two prisoners can establish the locations of all 4 colored rows, before the third one enters the room. Then, the prisoners can inspect each arriving prisoner, rotate their positions to indicate the newcomer's color, and the newcomer can take his place in the appropriate row.

In addition to being apparently impossible, this puzzle has a number of details we haven't used yet: * there is an "unknown" number of prisoners. * the unknown number is greater than 14 * the prisoner can't see anything until he (alone) arrives in the room, at which point he is "unfolded", presumably now able to see the earlier arrivals. (From the standpoint of this puzzle, why bother? he can see the other rows, or at least where they are supposed to be, but he doesn't know what color he is. Why bother to let him see where the rows are?) * the prisoners can arrange a plan beforehand. So, I'm wondering, in the spirit of the slight cheating that various solutions have proposed, what other kind of cheating might be acceptable to Wolfgang and the Warden. Do we imagine that the prisoners themselves get to arrange the order in which they enter the room? (How would that help? I dunno...) Notice that having the "like colored prisoners fidget" has the flaw of annoying the warden, but it also has the flaw that it can't be sure of working for the first few prisoners, as not all the rows will be populated at first. We can see how the first two prisoners can arrange themselves so that all four rows' locations are unambiguously visible. But, short of opening and closing their eyes to encode the next arrival's color, I don't see how to cheat surreptitiously enough to get it past Warden Wolfgang.

Thanks, k-man. That makes perfect sense, and I don't see why I didn't get that from Y-san's exposition--I think I need more biscuits this morning...

I'm missing something even before I try to answer: I see actions of * cutting into rounds * rolling out * rolling into rectangles * requiring 1/4 inch thick rollout I think I hear that you cut them into circles, then roll them into 1/4 thickness, and then roll them into rectangular shapes. This will necessarily reduce the thickness. But maybe the 1/4 thickness is required AFTER rolling the rounds into rectangular shapes? Please, what is the sequence of events and constraints thereupon? Alternatively, what was the stuff about rectangular shapes about?

Handedness: 1) Mr. Green, when asked the time, looked at his watch, on his right wrist--generally the location for left-handers. He ALSO has a tan line on his left wrist! Does he usually wear two watches? 2) The suit of armor has stabbed Wellesley--with which hand? I'd like to ask Mr. Green (while lie detector is operational) * whether he wears two watches at once * what he wears on his left wrist I'd like to watch Mr. Green: * to see if he seems left-handed I'd like to remember the crime scene: * to see which hand the suit of armor used to hold the sword that allegedly killed Wellesley * to see which side of the body the suit of armor was kneeling at * to see whether Wellesley was stabbed through the front or through the back (as he was found face-down, I'm assuming through the back, but I'd like a more official statement)

Prime, your "disproof" notwithstanding, we should congratulate you on your insight that has identified the limit towards which any experiment to 137 should tend (IMH and Naive Opinion). I'm interested to understand about Bushindo's recursive code. My contribution was only to respond to Bushindo's numeric result, as it matches the cyclic stream of digits I learned in childhood.

Thanks K-man! I get it now! (blushing in shame at my obtusity).

Bonanova, as you say you believe it has no solution, I infer that you believe you know what the problem is! All I can tell is that there's a 16x16 grid (Thanks to your reformatting!). Perhaps it could be viewed as a 4x4 grid of 4x4 cells (without lines). I can't tell how it relates to Sudoku, which I think of as a 9x9 grid. What is your interpretation of the problem? At least I can learn from that :-) Of course, we could see each hex digit as a 2x2 bit matrix, but I don't see that that helps, either...

Nice puzzle! We are pretty good at weighing coins here, but when you have to pay per card and pay per round, it's a different story. I started with one more than Bonanova found, but now I've caught up with him.