Prime

I do not see the answer to my dilemma. The premise for the correct system/logic is that if pirate does not follow it, he will end up worse off. In my previous post, I demonstrated clearly by example that the pirates who did not follow the purported best strategy, ended better off, and some of those who did follow ended worse. Whereas, each pirate thinks how to maximize his own situation -- not the captain’s.

Now your proof is complete. (I made the same omission, see my earlier posts.) I suggested P2 + 40 to Royal to shake the apparent notion that P2 + X is likely to be non-prime.

It's fascinating. I had no idea there was so much theory on the subject. But as far as this puzzle go, it meant to use a "handicap" language without all those possibilities. (Use of a loop was the whole trick.) Your first example uses pointers. And the #TEMPLATE in the second example, appears to be a built in language device, which just solves the problem directly. Another easy way would be just to read the file with the source code and write it out. But then it is not a puzzle.

Try P2 + 40. Where P is a prime and P > 5.

3 is a prime, yet it is a multiplier of 3. (3-1)*(3+1) is not a multiplier of 3. (3-1)*(3+1) + 27 is also not a multiplier of 3.

Oh, let's return from space onto our ship. We are pirates -- not astronauts. Tell me what's wrong with the following example: Given a situation where captain ( C ) and his first mate (P1) believe into the same logic as you as strongly as you. Whereas, P2 and P5 have equal faith in the alternative system. Then the alternative system wins. The ex-captain and P1 both lost (walked the board) due to their actions. P2 moved to captain's position and eventually became pirate king. P5 moved into P3 position, which became a potentially paying position under the new captain. (Under alternative system, captain is due to hand out 6 coins for the remainder of the game. Which makes it 1.5 coin payoff expectancy per pirate. Whereas, originally P5's payoff expectancy was 0.)

It does not have to be P2 who initiates the shift. It could be P4 or P6, or someone down the line if there are more pirates. My point is, the odd-man-out solution relies on faith. Within its framework, even-numbered pirates vote with the captain to avoid "dead end" odd positions. However, if pirates do not vote with the captain, the odd-numbered positions are no longer dead end. There exists another scheme of bribery, whereby captain pays more money and any of the remaining pirates have a chance to get some of it. Seeing that the odd-man-solution has not worked for the late captain, the new one may switch to that other scheme; and, given a sufficient number of pirates remaining, pirates get paid more. To make a specific example: In odd-man-out scheme, Captain pays total of 3 coins in the course of the entire game -- to P6, P4, and P2 at different stages. In the Captain-to-make-sure-to-stay-alive scheme, captains pays 12 coins for the duration of the game, which can go to anyone, except P1. Now consider P4 position in the very first round. If he votes Aye without pay, then he is voting for the odd-man-out scheme and stands to earn one gold coin. If he votes Nay, then he votes for the alternative scheme, where captain is still to pay 9 coins to be split in some random fashion between 5 pirates. This way P4's (who became P3) earning expectancy is 1.8 gold coins. Granted, the alternative scheme is also faith based. Still, when new captain sees that pirates do not wish to stick to the money saving odd-man-out scheme, he may be compelled to try and settle for something more expensive. This is pirates' way force the agenda, even though, captain moves first. Perhaps, there is a way to settle this issue definitively. We could try to apply the old mini-max algorithm here modified to work for multiple-player game. Also, we may need to set a limit for each individual bribe; otherwise, the problem may not have a definitive solution. Sorry about misspelling your name, Yoruichi-san. My keboad does not repond to key-press from time to tme. Perhaps, for that reason I don't attempt to write books, or researh papers.

I shall post my solution to my puzzle, before I forget what it was. Like in the original version of the weight problem dating back to 17th century, the powers of 3 proliferate. E.g., n reference weights seem to give (3n - 1)/2 combination reference weights. And the range which could possibly be solved with that 2*3n. That's as far as I went with research of this problem. I did not find any actual solutions for 18 lb range with two reference weights. With 3 weights, or more we must switch from pounds to grams, or milligrams. It is possible even for some modest weight ranges, we would have to transport tiny rocks in railroad cars and then use those cars and locomotive as reference weights.

But I am not convinced that solution is correct! So the greedy captain makes a bad offer putting his neck on the line, and half the pirates must go with it just because the captain is commited and cannot take his offer back? This whole system relies on the assumption that everyone believes such solution is the correct one. But I don't believe it. Now, let's say Youichi-san is the captain and Unreality is in the position of the first mate (P1). Let's further assume Unreality is of the same frame of mind with Y-s (captain) with respect to that problem. I am in the P2 position and there is a whole bunch of greedy disagreeable pirates behind me. Won't I prove the solution wrong by becoming a pirate king? Note that the pirates in the even positions will end up two steps closer to the captain in the even positions again.

I think benevolence/malevolence introduces an extra condition, which does not make problem more deterministic, but just adds another variable to tackle. Benevolence succumbs to greed. And it seems to go against financial interest here. There is much open to dispute without that extra factor. Let's save it for later. I'm guessing, what Y-S is going for is something as following: It seems like captain stands to save a whole lot of money. However, I am not convinced by such solution.

I think this wouldn't violate the conditions:

Clarification for my previous post. The assumption I'm making that the dime is standing vertically as it is rolled around the quarter. If OP meant that the dime is flat on the table, then the question is: revolutions from dime's perspective or from an outside observers'?

Great animation! But it looks like the dime is slipping to me... I think the trick here is...

I see... You mean 5th round when there are captain and three crewmen left (C, P1, P2, P3). Indeed, it appears, in that case P2 has a tangible reason to vote with the captain to kick off P3 even without any pay in order to avoid P1 position in the next round. If you project same reasoning to the rounds before, it would seem always be the case for P2. That leads to a bit of a paradox though, as I mentioned before. For if P2 sticks to that logic, it encourages the captain not to pay for his position. So in a way his position becomes like P1 for earlier rounds. That in turn leads to further regression to P3, P4, etc.. On the other hand, if P2 breaks this strategy, the line moves along, and in 2 turns P2 can end up in the position of the captain. Also, note that there is hardly a reason for P2 to reject payoff at any time. And that would be the way for captain to ensure own position. So the real question here is can Captain risk his position and not pay P2? This problem has the element of famous "Unexpected Hanging" paradox. Which has been posted on the forum under "Drop Quiz" title. I'm still leaning towards the idea that Captain cannot save any money by counting on P2's vote in any round, even the 5th.

Square root has two solutions -- positive and negative. x2 = y2 does not mean x = y. Case in point, your equation before taking square root: ((a-b)/2)2 = ((b-a)/2)2 perfectly okay at that point. Whereas, (a-b)/2 = (b-a)/2 is no longer true, unless a was equal to b to start with.

You assume I understand your intent. But I honestly don't. And my question was does the man who's designated by the captain to go off the ship vote? And I still don't know what is the "original pirate problem" you're refering to. At any rate, that does not change the problem significantly. My answer is still the same. Please, don't take it personally, if this solution does not match your intended one. For what's it worth, I think it's a good problem and that's the reason I posted my solutions here -- not to pick on your problem statement. I consider myself voted off this ship.

Now you change the rules yet again. So the man who is being voted off the ship does participate in voting? P4 becomes P3 one way or another in your scenario. He should not refuse any pay off offer, even the minimum one.

Aha, a bribery attempt! I love bribes! However, my solution to the problem as stated, is simply a correct one. Not being insightful, I cannot help. So let's restate the conditions, as I understand them now: Each round, captain announces a proposal of whom to vote off the ship, which comes together with money offering to some crew members. Captain cannot back off his proposal or change money offering.

Small addition: In the third step when 3, 4, 5, and 6 are left -- pirate 3 has a choice of bribing either 4 or 5. Since if it goes into the next round neither of them stands to get the coin. And money takes precedence. So in the beginning, the captain also has a choice between bribing 6 who can't get any money otherwise, and 4 or 5 who only have 50% chance of getting coin if hey voted against. And the whole coin beats 50% chance plus whatever fun they get watching 2 men walking the plank.

The way I uderstood bribing: Every round, after announcing who is the candidate to be off the ship, the captain can walk up to any pirate and confidentially offer a bribe -- one gold coin. The pirate may accept or decline. Then captain walks to another pirate if he still needs to influence more votes, and so on. When captain feels secure he has enough men on his side, the voting commences. If the captain approached every pirate in turn and still didn't secure enough votes, the voting shall commence, anyway. Another assumption is that the pirates can calculate all possible outcomes and follow the strategy which gives highest return. I don't see how announcing bribery attempt to a group of pirates in the open changes things. Make a link to the "original" pirate problem, as there are number of duplications of pirate problems in this forum. You should also be able to show where you think my analysis is wrong. That is a different strategy, where captain paid less money, and didn't walk the plank, or where a pirate followed a different pattern and ended up richer or king in all possible outcomes. We, pirates are mercyless.

New puzzles do not necessarilly come out right the first time around. Like computer programs they need testing and debugging. I am not sure what you meant the solution should be, but it's apparent at this point the statement of the problem needs work. There seems to be a contradiction with the notion that pirates possess perfect reasoning ability. There is a test you can apply to your problem/solution: Try all variations and see if any of the pirates made a wrong decision when acting according to your scheme. If that's the case, then pirate made a mistake, which contradicts the notion of perfect logic ability in pirates, which we know, all of them possess. We pretty much zeroed in on the discrepancy here. Test the case where captain attempts to bribe his next successor, before moving on to bribe next pirate. Make best effort on the part of captain. See if the first mate ended up better refusing the bribe vs. taking it. When testing cases where you chose first mate to be voted off the ship, start by offering bribe to the next man. When enough men accepted bribes to secure the vote, captain need not offer more bribes. Check for all cases when any pirate refused bribe, see if that pirate ended up better off.

Or are we refering to different puzzles. There are so many copies of those. I am talking about 5 pirates dividing 100 gold coins. Where first pirate offers how to split the loot and votes with the others on whether to accept his proposal.

You refered me to the original pirates' puzzle, and I read and solved it before returning here.

Then the first mate is dumb. He is not versed in game theory. His decision will only lose him money, but not gain any position. Note, unlike the original pirate puzzle, where captain divides money and puts it up for voting, without any chance of changing anything if it does not go his way, here captain can make rounds offerening bribes to anyone. If you don't believe me, conduct a trial game. Be a captain, and as far as pirates are concerned stick to this strategy: vot Aye if you got a bribe, and Nay if you didn't. Try choosing different men to vote off the ship. See how much pirates earn collectively if they refuse reasonable bribe offer, versus when they don't. Remember as a captain you can offer bribe to anyone. Try offering it to the first man in succession first before going to the next pirate. See if the first man ends up better off when refusing the bribe.