bonanova

Prime, that analysis is convincing. Consider the following as well. I wonder, in light of your betting on "1" appearing in just four die rolls means we should be looking at median outcomes rather than average.

Someone claimed to get all sixteen. With only one odd one. I will look for the list.

Uhhh ... There! I just cut a king-size hole in the box so we can all leave.. Since we're looking for a proof and a symbol, I considered saying what the symbol is. Suffice it that the symbol would probably bring the search to an end, and specifying the proof most certainly would. So I'll just let that information itself be the clue. Relax, and let the solution come to you.

Yes. Simple concept that cuts thru all the computational complexity.

No. Sorry that's not it, either. Keep trying.

OK, I'll ask: why?

Ah, but there is one. No correct answers yet.

Yes. It does. Great solve. To clarify, the OP might better have read: Each letter has a value selected from just eight digits. The actual wording, and the title, suggests (wrongly) that eight digits are used.

Maybe in linguistics some notions stand in the same place that axioms do in geometry and math: they're not derivable from the rest of the stuff. Define and definition might be one such notion. It does have synonyms, like describe, expose, clarify, that have more satisfactory dictionary meanings. Define is what we say we're doing when we give the meaning of a word or idea. So the meaning of define has to come from some already agreed upon idea, or else, as you point out, it becomes a bit circular.

There are two constraints from a normal reading of the OP that, when combined, render the puzzle impossible: No communication. Enter and take position singly. We've now shown that if either is relaxed the puzzle has a solution.

Edit. Removed misleading information from the original post. I ran across this puzzle - it's part of a larger puzzle - a week ago. There are twelve letters, each of which corresponds to one of eight digits. The letters are juxtaposed into two-digit numbers, and they occur that way in six equations. Here are the equations - six, in twelve unknowns. Their integer properties provide additional constraints. VG = CV - QP JY = GP - ZY QG = VJ - MV XC = BC - XP XK = KX - QB ZP = CY - GM Can you find consistent values of the 12 letters B C G J K M P Q V X Y Z using just eight digits?

If only a binary signaling scheme is allowed, CaptainEd has shown that only prisoner 1 needs to change his mind. Even in the stricter NS view, there are some latitude to define a quadnary signaling scheme due to an implied necessary condition You're right.NS signaling is enough.Four easily distinguishable rotations: None, 180ocw, 180occw, and 360occw.And entering prisoners must be able to see the rotation. I think we spent most of the time on this puzzle with a poor understanding of the conditions.But it was fun to finally solve something possible.It's just that the problem became teasing out the constraints that admitted a solution!Bertrand was helpful in this case. Thanks, Wolfgang.I want to thank you all....I was thinking like this:After the first three prisoners took their right places,they all should be facing the door, the possible combinations would be:Yellow,Green,Red,........Yellow,Green,........,BlueYellow,.........,Red, Blue.........,Green, Red, Blueso when the 4th one enters, Let him to have X color..the X one standing in the raw will turn his face to the wall,and the new comer will stand in that raw facing the door....and when nobody turns to the wall so he should stand at the empty place facing the door.when the four raws are made, each new comer will know where he belongs when the man with the same color turns his face toward the wall. The first three may not all be different. What if all the prisoners are say red, except for the last three to make all the colors used? Edit: ok, that's not a problem I guess, it just delays the row formation.

This is great - beautiful, even.Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)I made a graph of the divergence point sometime back, I will look for it.When the records are written, however, this def gets Honorable Mention. Great proof as well. I have no beef with the power tower, but isn't the argument similar to this one Nicely debunked Bushindo. However, as I stand now the the company of two great logicians/mathematicians, I tread lightly regarding weak points, knowing the desired solution might likewise be so described. Its redeeming quality being only that it fits the given conditions so nicely.

Ah, the old dazzle 'em with double angles and parentheses ploy! Remember it well from my college days. Did you think I would actually evaluate all those functions? Unfortunately that is unnecessary. It came out wrong by a factor of 4. While the problem asked for a "proof" that 1=2; you ended up with 2=1; I.e., that 4=2. In a dyslexic world, where a skeptic is unsure there is such a thing as a dog, that might suffice. But I'm afraid the quest must go on. Efforts should be rewarded, nonetheless. Thus, a clue. If you get far enough out of the box, it might be the start of a relaxing hour after work.

This is great - beautiful, even. Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.) I made a graph of the divergence point sometime back, I will look for it. When the records are written, however, this def gets Honorable Mention.

Hi markdane, and welcome to the Den. That's the answer. Thanks for posting.

You throw m fair six-sided dice and bet that all numbers from 1 to 6 will show at least once. What is the smallest value of m that gives you a winning bet on average?

You are given n > 0 of each of the standard denomination US coins: 1¢, 5¢, 10¢, 25¢, 50¢, $1. Your task is then to select from them a set of n coins whose total value is exactly $1. Clearly, if n=1, you can do this by selecting the $1 coin. If n=2 you select the two half-dollar coins. But if n > 100, then every set of n coins (e.g. all the pennies) will have a total that is too large. What is the smallest n such that it is impossible to select n coins that make exactly a dollar?

Good observation Phaze. I hadn't thought of counting the word itself. RG caught the idea.

That's the answer I had in mind. The complete set. Did you notice the font change I used to emphasize it?

Actually not bad. But not the solution I had in mind.

Good thinking so far, but it's not as complicated as it looks. You'll just have to go a bit more outside the box.

If only a binary signaling scheme is allowed, CaptainEd has shown that only prisoner 1 needs to change his mind. Even in the stricter NS view, there are some latitude to define a quadnary signaling scheme due to an implied necessary condition You're right. NS signaling is enough. Four easily distinguishable rotations: None, 180ocw, 180occw, and 360occw. And entering prisoners must be able to see the rotation. I think we spent most of the time on this puzzle with a poor understanding of the conditions. But it was fun to finally solve something possible. It's just that the problem became teasing out the constraints that admitted a solution! Bertrand was helpful in this case. Thanks, Wolfgang.

OK so I feel obligated to make a puzzle that is not a duplicate of a previous one. We can "straighten" the situation out by requiring that the words made from a 3x3 array of letters not have bends or kinks. This reduces the maximum number to sixteen words. Thus, W D O A I D S Q N permits AID WIN WAS and SAW, but not DID DIN SIN WAD or ODD. Sixteen is perfect, and minimum for an entry is ten. You make the grid. Duplicate letters are permitted. Enjoy.

I found this puzzle in a book which attributed it to the British Monthly Games & Puzzles. I have adapted it somewhat. How many limericks [allowing for variations in form] are listed here? There was a young girl in Japan Whose limericks never would scan. ___ When someone asked why, ___ She said, with a sigh, "It's because I always attempt to get as many words into the last line as I possibly can." Another young poet in China Had a feeling for rhythm much fina. ___ His limericks tend ___ To come to an end Suddenly. There was a young lady of Crewe Whose Limericks stopped at line two. There was a young man of Verdun.