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Topic: HARD: GREEDY PIRATES (Read 41146 times) |
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #100 on: Nov 30th, 2004, 9:20am » |
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To which problem do you claim this to be an optimised solution? If the original, then pirate 5 can get more than 332.
If the revised version, where proposals are only accepted provisionally until pirate 1 has had his say (with subsequent rounds of proposals if there's any gold left over), then I doubt your solution is correct because pirates 1 and 2 each stand to gain 500 gold if 5 dies, so 5's proposal must give pirate 3 a chance of ending up with some money (and pirate 4 a chance of survival)
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vicky
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Re: HARD: GREEDY PIRATES
« Reply #101 on: Dec 1st, 2004, 12:55am » |
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clarify some points:
1. Whats majority? 3 out of 5 or 2 out of reaming 4 (one excluded is the one who proposes) or 3 out of remaining four?
2. The above solution for original problem not the provisional one.
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #102 on: Dec 1st, 2004, 10:05am » |
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The "majority" most people seem to accept is:
1 pirate has to vote for his own proposal in order for it to pass.
With 2 pirates, both have to vote for the proposal in order for it to pass.
With 3 pirates, 2 have to vote for the proposal in order for it to pass.
With 4 pirates, 3 have to vote for the proposal in order for it to pass.
With 5 pirates, 3 have to vote for the proposal in order for it to pass.
In the original problem, if pirate 5 were to propose 334 each to pirates 3 and 4, and the remaining 332 for himself, pirates 1, 2 and 4 would all vote him down because pirate 4 could then propose to give 1 each to pirates 1 and 2, nothing to pirate 3, and take the remaining 998 himself, leaving all 3 of them better off than if they'd accepted pirate 5's proposal. Even if pirate 4 proposed to take all the gold himself (and got accepted), pirates 1 and 2 would still be better off than under 5's proposal because, while they'd still not have any gold, they would have got to watch pirate 5 drown.
To work out what the actual solution is, it may help to look at what happens with fewer pirates to start with - 2 pirates is easy, 3 may take a little thought, and if you can get 4, 5 should follow easily.
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vicki
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Re: HARD: GREEDY PIRATES
« Reply #103 on: Dec 2nd, 2004, 12:19am » |
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Ok ...
Lets move forward then ... what if it is a deadlock and each one gives a proposal that majority accepts??
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #104 on: Dec 2nd, 2004, 8:06am » |
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The conditions for the original version are:
1) Each pirate values survival over any amount of gold, gold over any number of corpses, and seeing other pirates die above any unlisted considerations.
2) There are 1000 standardised gold pieces to be divided among 5 pirates.
3) Pirate 5 makes the first proposal. If he is rejeted, pirate 4 has a turn, then pirate 3, pirate 2 and finally pirate 1 who only gets to make a proposal if everyone else is executed.
4) A proposal consists of an ordered list of non-negative integers, one for each surviving pirate, summing to 1000.
5) A proposal is accepted if a strict majority of the surviving pirates vote for it, and rejected otherwise.
6) If a proposal is rejected, the proposing pirate is executed.
7) If a proposal is accepted, it is enacted immediately, each pirate recieving the appropriate quantity of gold (this condition is changed for the harder version)
 All pirates know all these conditions, and are perfect reasoners, who do whatever gives them the highest expected value.
For the extension, 7) becomes:
7) If a proposal is accepted, it becomes the "current proposal" and the nesxt pirate takes their turn. Once all pirates have taken their turn, the current proposal is enacted. Any ownerless gold must then be shared out by this same process.
In neither situation does a deadlock arise (except in the impossible situation where the last surviving pirate votes against his own proposal, throws himself overboard and the 1000 gold is left ownerless)
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Icarus
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Re: HARD: GREEDY PIRATES
« Reply #105 on: Dec 2nd, 2004, 3:30pm » |
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on Dec 2nd, 2004, 8:06am, rmsgrey wrote:| If a proposal is accepted, it becomes the "current proposal" and the nesxt pirate takes their turn. Once all pirates have taken their turn, the current proposal is enacted. Any ownerless gold must then be shared out by this same process. |
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I didn't realize that it had been established that the 2nd round was only for leftover gold. There are other possibilities as well. There is also a question of what to do when two possible proposals are equally favorable under the conditions described. I believe this is being analyzed under the idea that the Pirates pick one at random.
I have decided (seeing as I came up with the variation in the first place) that I will offer other versions as separate puzzles in another thread (something I wish now I had done for the variation in first place), and leave this thread for the "leftover gold is distributed/random picks between equal positions" version.
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dragomanjk
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Re: HARD: GREEDY PIRATES
« Reply #106 on: Mar 17th, 2006, 2:41am » |
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The only way I can think of 5 getting 999 gold is for an infinitely smart pirate to not know everything that will happen.
P1 will kill P2 for lack of 'majority'. He then thinks he can say no to everyone and get all the money, greed clouds the mind.
P2 knows that if it gets down to it he will die so he will vote for 3 no matter what.
P3 gets P2's vote b/c he wants to live and so P3 gets all 1,000 gold.
P4 gives P2 one piece, but it doesn't make a difference, he can't get P1's or P3's vote b/c they believe they will get 1,000.
P5 gets P4's vote b/c P4 is going to die. P2 gets one piece of gold. P5 gets the 999.
Lets say P1 and P2 split if it gets down to it, 500 each.
P3 has to offer one of them more than 500 to get their vote. Giving only 500 will get him killed as the killing is seen as more money. P1=501/P3=499
P4 really has no option. he would have to offer each one more money and that's impossible. P4's only option is to try to convice P2 he's not getting anything if he passes this offer up and then give him a piece, and give P3 500 as well. P2=1/P3=500/P4=499
P5 must now convince P1 that he will be getting nothing if passes this offer and that P2 will only be getting 1. So he gives P1=1/P2=2/P5=997.
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #107 on: Mar 17th, 2006, 7:34am » |
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on Mar 17th, 2006, 2:41am, dragomanjk wrote:The only way I can think of 5 getting 999 gold is for an infinitely smart pirate to not know everything that will happen.
P1 will kill P2 for lack of 'majority'. He then thinks he can say no to everyone and get all the money, greed clouds the mind.
P2 knows that if it gets down to it he will die so he will vote for 3 no matter what.
P3 gets P2's vote b/c he wants to live and so P3 gets all 1,000 gold.
P4 gives P2 one piece, but it doesn't make a difference, he can't get P1's or P3's vote b/c they believe they will get 1,000.
P5 gets P4's vote b/c P4 is going to die. P2 gets one piece of gold. P5 gets the 999.
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I'm not sure where your analysis is coming from, but it doesn't seem to work under the standard assumptions.
If there are three pirates, P2 will vote for P3 through survival, so P3 gets everything. This means that P1 will get nothing if he lets it get down to 3 pirates, and should know that (if we can figure it out, so can he) so he'd vote for anything P4 offers that gives him any gold. So would P2 for the same reason, so P4 would offer 1,1,0,998... I'll let you figure out what P5 would offer for yourself.
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dragomanjk
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Re: HARD: GREEDY PIRATES
« Reply #108 on: Mar 17th, 2006, 10:35am » |
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on Mar 17th, 2006, 7:34am, rmsgrey wrote:
I'm not sure where your analysis is coming from, but it doesn't seem to work under the standard assumptions.
If there are three pirates, P2 will vote for P3 through survival, so P3 gets everything. This means that P1 will get nothing if he lets it get down to 3 pirates, and should know that (if we can figure it out, so can he) so he'd vote for anything P4 offers that gives him any gold. So would P2 for the same reason, so P4 would offer 1,1,0,998... I'll let you figure out what P5 would offer for yourself. |
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I realize that, but unless it happens that way, there is no other way to get 999 then like Icarus says should be possible and the riddle is solved at 997 for P5.
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #109 on: Mar 17th, 2006, 11:12am » |
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on Mar 17th, 2006, 10:35am, dragomanjk wrote:
I realize that, but unless it happens that way, there is no other way to get 999 then like Icarus says should be possible and the riddle is solved at 997 for P5. |
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There is no way to give P5 999 if proposals solely consist of division of money in 1 gold increments.
What Icarus had in mind was to allow proposals to include throwing people overboard as well - so, for instance, in the three pirate case, P3 suggests throwing P1 overboard and keeping all the gold for himself, and P2 will agree to avoid getting killed by P1.
This solution avoids the situation with very large numbers of pirates where some pirates just can't buy enough votes (though survivial votes will build up until you have enough at a higher number of pirates to make up the shortfall) but it does roughly halve the size of the crew each time (each proposal either buys a given pirate's vote, or throws that pirate overboard)
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Icarus
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Re: HARD: GREEDY PIRATES
« Reply #110 on: Mar 17th, 2006, 3:06pm » |
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Yes. If pirates are allowed to make any proposal which will be acted upon if agreed to, then pirates can buy votes with the life of the voter, not just with money.
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LaCiTy
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Re: HARD: GREEDY PIRATES
« Reply #111 on: Sep 21st, 2006, 8:53am » |
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Solution
I didn't read all of your considerations, but it looks like the goal to reach is 999 for #5 (which is easily shown maximal as 1000 doesn't fit). Nervetheless, with rather sensible backward induction, one come only to 997. This is because one can not, (a priori) share a coin between 2 pirates or more. Or else, the solution will obviously be 1000 minus 3*epsilon, with epsilon fixed to be the minimal size for which a part of a coin is still valuable.
In fact, the correct answer lies between 1000 and 999, being very close to 1000.
As soon as i'll reveal my point, you'll immediately see a solution, so stop reading here if you want to figure out by yourself. Knowing the answer which is 999+ is actually quite good a hint. Or maybe you can keep on reading, i'll try to act like socrates and make the reader discover by himself.
There are coins right ? Many of them ! What do you do with coins ? You toss them ! That is you can introduce randomness. Still don't get it ? You can, as an illustration, suggest to give P1 1 coin with probability 1/2, which is tantamount to give him 0.5 coin ! (if he is risk neutral, or a little less if not, but still positive).
To finish the problem properly, on has to be sure that a lower bound exists for the minimum non zero value one can pledge. If it is arbirarily small, things get confused and one can not apply what has been used so far without randomness in a straightaway manner.
A pirate can not toss infinitely many coins which is clear given his life expectancy is finite and the speed of his moves is bounded by, say the speed of light  . Thus, the minimal probability he can obtain is 1/2^N, where N is the maximal number of coins he can toss. Let not assume all pirate are the same for the sake of useless complexity. We denote by epsilon1, epsilon2, epsilon3, epsilon4, epsilon5 the minimum probability they can draw.
Things work the same by backward induction, it's has been shown that in the situation there are 3 pirates left, 3 can grab all the money, yielding :
(0,0,1000)
Knowing that, 4 needing 2 favorable votes will give strictly more to 1 and 2 :
(epsilon4, epsilon4, 0, 1000 - 2*epsilon4)
[He does only 1 tossing for 1 and 2, so as to respect his drawing possibilities]
Now, #5 has to get 2 favorable votes. He chooses to give more (strictly) to those who would be dispriviledged by #4's proposal, that is 1 and 2.
One solution is :
(epsilon5, epsilon5, 0, 0, 1000-2*epsilon5) if epsilon5 > epsilon4
and
(n*epsilon5, 0, epsilon5, 1000-(n+1)*epsilon5) if
epsilon4 > epsilon5 and n := inf { t : t*epsilon5 > epsilon4 }
Note that if epsilon5 = epsilon4, a solution is provided by (0, 2*eps, eps, 0, 1000 - 3*eps).
In anycase, only epsilon4 and epsilon5 are relevant and well, if epsilon is very small, I virtually reach 1000 !
Can anyone be of better advice to #5 ?
Thank you for your reading.
P.S.: I am sorry if this solution was found before. But i feel quite proud of it and in need to communicate it !
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| « Last Edit: Sep 21st, 2006, 9:15am by LaCiTy » |
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Icarus
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Re: HARD: GREEDY PIRATES
« Reply #112 on: Sep 21st, 2006, 3:37pm » |
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Good idea! I believe you are the first to suggest probabilistic bribing.
But I suggest you skim through the thread anyway. There is an alternative interpretation of the problem which led to some interesting discussions.
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #113 on: Sep 22nd, 2006, 11:20am » |
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I agree that it's a new and worthwhile solution.
One minor quibble: I'm confident that, if the pirates are willing to trust each other to toss coins fairly, Pirate 4 would nominate the fastest surviving tosser (if the number of coin tosses is limited by life span then you can't toss for longer than the minumum of your and the potential recipient's life-spans - no self-respecting logical pirate is going to accept a proposal that will only pay him after he's dead! - therefore the difference in [epsilon]s had better arise from tossing speeds rather than life-spans) - so Pirate 5 will only have a smaller effective [epsilon] value if he's the fastest tosser - if not then he loses 3*(mini=1 to 4{[epsilon]i}; if he is, then he loses (t+1)*[epsilon]5 where t is the minimum integer for which t*[epsilon]5>mini=1 to 4{[epsilon]i}
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LaCiTy
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Re: HARD: GREEDY PIRATES
« Reply #114 on: Sep 23rd, 2006, 4:13pm » |
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Indeed.
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ThudnBlunder
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Re: HARD: GREEDY PIRATES
« Reply #115 on: Jan 8th, 2010, 10:19am » |
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What if there are a lot more pirates (say 2300) so that the bribe money runs out?
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SMQ
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Re: HARD: GREEDY PIRATES
« Reply #116 on: Jan 8th, 2010, 1:05pm » |
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on Jan 8th, 2010, 10:19am, ThudanBlunder wrote:| What if there are a lot more pirates (say 2300) so that the bribe money runs out? |
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Using the following extension of rmsgrey's formalism to N pirates:
1) Each pirate values survival over any amount of gold, gold over any number of corpses, and seeing other pirates die above any unlisted considerations.
2) There are 1000 standardised gold pieces to be divided among N pirates.
3) Pirate N makes the first proposal. If he is rejeted, pirate N-1 has a turn, then pirate N-2, ..., and finally pirate 1 who only gets to make a proposal if everyone else is executed.
4) A proposal consists of an ordered list of non-negative integers, one for each surviving pirate, summing to 1000.
5) A proposal is accepted if a strict majority of the surviving pirates vote for it, and rejected otherwise.
6) If a proposal is rejected, the proposing pirate is executed.
7) If a proposal is accepted, it is enacted immediately, each pirate recieving the appropriate quantity of gold
8) All pirates know all these conditions, are perfect reasoners, and are risk-neutral, who do whatever gives them the highest expected value.
If I've managed the induction properly,
- for 5 <= N <= 1000, pirate N can make a proposal in which he receives 999 - 2 floor((N-2)/2) coins by offering 2 coins to any floor((N-2)/2) of the first N-3 pirates, 1 coin to pirate N-2, and keeping the rest. This given an expected value to each of the first N-3 pirates of (2 floor((N-2)/2)/(N-3) coins which is >= 1.
- for N = 1001, pirate 1001 can offer 2 coins to any 499 of (the first 998 pirates OR pirate 1000 (who only receives 1 coin by his best proposal)), 1 coin to pirate 999, and keep one coin himself. This gives an expected value to pirates 1-998,1000 of 998/999 coins which is < 1.
- for N = 1002, pirate 1002 can offer 1 coin to any 501 of pirates 1-998,1000 and keeping the other 499 coins for himself.
- for 1003 <= N <= 2000, pirate N can make a proposal in which he receives 1000 - floor(N/2) coins by offering 1 coin to any floor(N/2) of the first N-2 pirates and keeping the rest.
- for N = 2001, pirate 2001, can offer 1 coin to any 1000 of the first 2000 pirates, keeping none for himself.
- for N = 2002, there is no proposal pirate 2002 can make which will be accepted by 1001 pirates other than himself.
- for N = 2003, pirate 2003 can offer 1 coin to any 1000 of the first 2001 pirates, leaving nothing for pirate 2002 or himself but still receiving the needed 1002 votes.
In general:
- for N > 2000, n != 1999 + 2k, there is no proposal pirate N can make which will be accepted.
- for N = 1999 + 2k, k > 0, pirate N can offer 1 coin to any 1000 of the first 1999 + 2k-1 pirates and receive the needed 1000 + 2k-1 votes.
So for 2300 pirates, pirates 2300 down through 2256 are executed. Pirate 2255 offers one coin to any 1000 of the first 2127 pirates and nothing to pirates 2128 through 2255 and his proposal is accepted.
--SMQ
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ThudnBlunder
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Re: HARD: GREEDY PIRATES
« Reply #117 on: Jan 9th, 2010, 10:00am » |
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SMQ, why can't 2002 rely on 2001's vote?
2001 knows he will get nothing or be executed if 2002 is executed.
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SMQ
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Re: HARD: GREEDY PIRATES
« Reply #118 on: Jan 9th, 2010, 10:54am » |
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If 2002 makes a proposal in which 2001 receives no gold, 2001 will prefer his own proposal (in which he still receives no gold, but gets to see 2002 executed), and so will vote against.
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Bumble_Bee_Tuna
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Re: HARD: GREEDY PIRATES
« Reply #119 on: Aug 9th, 2010, 7:21am » |
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I believe that I have a better solution for the problem as originally stated. I agree with the solutions already posted if proposals are restricted only to how to distribute the treasure. However, if they're allowed to make ANY proposal, even Icarus's solution of 999 gold isn't good enough! Pirate 5 can in fact get away with taking ALL the gold, and killing ALL the other pirates. (or 3 out of 4, if ties don't pass) EDIT: I goofed, I guess #5 can't kill them all in the ties pass scenario 
I prefer using the assumption that ties pass, as it results in the most fun solution. I'll list what each pirate gets like this:
aye [257, 2]
where aye is what they vote, 257 is how much gold they get from the proposal (or "dead" if they're killed) and 2 is just a listing of how many pirates besides themselves they get the satisfaction of knowing die.
2 pirates:
any proposal will pass.
1: nay [dead, 3]
2: aye [1000, 4]
3 pirates:
pirate #1 will accept anything where he lives. Propose to kill #2 and take it all.
1: aye [0, 3]
2: nay [dead, 2]
3: aye [1000, 3]
4 pirates:
pirate #2 will accept anything where he lives, but he'll also accept death if 3 other pirates die with him instead of the two he'd see in the 3 pirate scenario. Propose to kill #1, #2, and #3 and take it all.
1: nay [dead, 3]
2: aye [dead, 3]
3: nay [dead, 3]
4: aye [1000, 4]
5 pirates:
Propose to kill #4 and 1 other pirate (doesn't matter which one). Assuming 1 and 3 are chosen to be spared:
1: aye [0, 3]
2: nay [dead, 3]
3: aye [0, 3]
4: nay [dead, 3]
5: aye [1000, 4]
It's almost the same if a true majority is needed to pass instead of a tie:
2 pirates:
Anything proposed fails.
1: nay [1000, 4]
2: aye [dead, 3]
3 pirates:
#3 needs to win #2's vote by letting him live. Kill #1, take 1000.
1: nay [dead, 2]
2: aye [0, 3]
3: aye [1000, 3]
4 pirates:
#4 needs to win both #1 and #2 over. He can let #1 live, but he'll have to give #2 1 gold. 3 gets killed.
1: aye [0, 2]
2: aye [1, 2]
3: nay [dead, 1]
4: aye [1000, 2]
5 pirates:
Killing #4 and #2 is enough to satisfy #3, who would otherwise only die alongside 1 pirate instead of 2. And killing 2, 3, and 4 is enough to satisfy #1, who would otherwise only get to see 2 pirates die instead of 3. Propose to kill 2, 3, and 4 and take all 1000.
1: aye [0, 3]
2: nay [dead, 2]
3: aye [dead, 2]
4: nay [dead, 2]
5: aye [1000, 3]
Of course, other assumptions could be thrown into the mix (if probabilistic stuff is allowed, now just a miniscule chance of survival is enough to convince a fellow voter instead of a 100% chance of survival. Or what if you could propose "I'll kill pirate #2...in 10 minutes". Would pirate #2 value that over being killed immediately? Then you have another silly avenue for making proposals. But those kinds of assumptions sort of kill the riddle.
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| « Last Edit: Aug 9th, 2010, 1:40pm by Bumble_Bee_Tuna » |
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rmsgrey
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Re: HARD: GREEDY PIRATES
« Reply #120 on: Aug 9th, 2010, 8:38am » |
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In your first scenario, I think you got #4's proposal wrong:
4 pirates:
pirate #2 will accept anything where he lives, or where there's only one survivor. Propose to kill #1, #2 and #3 and take it all.
1: nay [dead, 3]
2: aye [dead, 3]
3: nay [dead, 3]
4: aye [1000, 4]
making #5's proposal need the support of two other survivors. #6 then also needs to let someone live since only #4 is guaranteed death if #5 gets to bid.
Your analysis of the second variant seems to hold.
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Bumble_Bee_Tuna
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Re: HARD: GREEDY PIRATES
« Reply #121 on: Aug 9th, 2010, 1:42pm » |
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on Aug 9th, 2010, 8:38am, rmsgrey wrote:In your first scenario, I think you got #4's proposal wrong:
4 pirates:
pirate #2 will accept anything where he lives, or where there's only one survivor. Propose to kill #1, #2 and #3 and take it all.
1: nay [dead, 3]
2: aye [dead, 3]
3: nay [dead, 3]
4: aye [1000, 4]
making #5's proposal need the support of two other survivors. #6 then also needs to let someone live since only #4 is guaranteed death if #5 gets to bid.
Your analysis of the second variant seems to hold. |
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Noted, I have edited to make your correction. That's a shame, I really liked having the answer involve killing everyone I think if I ever tell this riddle it will be 4 pirates instead of 5.
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Hippo
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Re: HARD: GREEDY PIRATES
« Reply #122 on: Aug 9th, 2010, 4:13pm » |
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Just one thing ... I would rather use notation where last coordinate is minus the number of alive (other) pirates.
Advantage of this notation is that it is independent on the original number of pirates.
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| « Last Edit: Aug 9th, 2010, 4:16pm by Hippo » |
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