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45 minutes ago, Jez said:

You can run the 'take my pile if you think it's bigger' method for powers of 2, but not 5.

OK, so the first step is to get the 1st pirate out of the way. 

1. Rough out 1/5 of the loot, and ask if anyone wants it, if no-one wants it then keep adding to the pile until someone takes it, or if everyone wants it, keep removing pieces until there is only one pirate asking for it, then they get that. A pirate considering whether to take it has to decide whether they are more likely to get more at this stage than if they go to the next, so there is a motivation to 'lock in the deal' early.

Then you can do this with any power of 2;

2. Roughly halve the remaining pile and the two pirates standing next to a given pile will spilt it like for two people. If there are 3 or more pirates standing next to a pile, move the treasure piece by piece to the other pile until each pile has two pirates who want to share that particular pile.

You're right. +1

Spoiler

They could just repeat the step 1 for 4 pirates then for 3 pirates, making 1/4 then 1/3 of the remaining loot, until only 2 pirates left. Then use the split-and-choose procedure for the last 2. This method is of course directly generalizable for any number of pirates.

I hope you keep going and solving other puzzles as well. Please, hide your answers under "Spoiler".

 

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1 hour ago, Genady said:

You're right. +1

  Reveal hidden contents

They could just repeat the step 1 for 4 pirates then for 3 pirates, making 1/4 then 1/3 of the remaining loot, until only 2 pirates left. Then use the split-and-choose procedure for the last 2. This method is of course directly generalizable for any number of pirates.

I hope you keep going and solving other puzzles as well. Please, hide your answers under "Spoiler".

 

Ah, sorry, I will hide my answer (if you want to) .. oh, OK, no editing later on .. meh.

Edited by Jez
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On 6/4/2023 at 4:26 AM, Jez said:

1. Rough out 1/5 of the loot, and ask if anyone wants it, if no-one wants it then keep adding to the pile until someone takes it, or if everyone wants it, keep removing pieces until there is only one pirate asking for it, then they get that.

What do you do if there are 2 pirates who want it, and then neither wants it after the adjustment, or vice versa?

On 6/4/2023 at 4:26 AM, Jez said:

2. Roughly halve the remaining pile and the two pirates standing next to a given pile will spilt it like for two people. If there are 3 or more pirates standing next to a pile, move the treasure piece by piece to the other pile until each pile has two pirates who want to share that particular pile.

What if there's 3 pirates on one side and 1 on the other, and you move some treasure and two (or three) pirates move to the other side? How many adjustments will it take to guarantee there are 2 on each side?

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4 minutes ago, md65536 said:

What do you do if there are 2 pirates who want it, and then neither wants it after the adjustment, or vice versa?

What if there's 3 pirates on one side and 1 on the other, and you move some treasure and two (or three) pirates move to the other side? How many adjustments will it take to guarantee there are 2 on each side?

Funnily enough I have been thinking that through, because the same sort of thoughts occurred to me.

In particular, if that was me and my loot, I don't like any step where I might have to judge 1/5th versus 4/5ths of the pile, I mean, we're geared to weigh up two equal sides, right?

I was thinking that the pile first needs to be roughed out into 1/5th and 4/5ths and the larger then split into 4, and one stands out. Once in 4 equal piles, according to the 4 pirates engaged in that split, items in the 'excess' pile can be used or added to so as to 'tune' the right number of pirates during the 4 way split.

Now we have 5 piles, 4 of which are equal according to 4 of them, and .... still trying to work out the next bit. ;)

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On 5/27/2023 at 2:11 PM, Genady said:

Five aging pirates decided to split their treasure and to retire. Each one will be satisfied if in his estimate his share is not smaller than one fifth of the total. Their estimates vary. How can they do it peacefully?

image.png.259834fa82ff475e551b51b63bab9c16.png

Spoiler

My apologies if this solution has already been offered:

If we have fifty thousand coins in the chest (the puzzle demands a treasure that can be equally divided into fifths, so I've chosen this treasure to make the math easier, lol).

The first pirate gets 10.000 coins. 

The second pirate: 8,000

The third: 6,400

The fourth: 5,120

The last: 4,096

 

 

Edited by S.T. Ranger
Seeing if there is a prompt to hide the answer.
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1 minute ago, S.T. Ranger said:

My apologies if this solution has already been offered:

If we have fifty thousand coins in the chest (the puzzle demands a treasure that can be equally divided into fifths, so I've chosen this treasure to make the math easier, lol).

The first pirate gets 10.000 coins. 

The second pirate: 8,000

The third: 6,400

The fourth: 5,120

The last: 4,096

 

 

The chest has mix of many different items, which cannot be compared objectively. I.e., each pirate has only a subjective evaluation of what is worth what.

In your proposed solution, I don't understand in what sense each pirate gets at least one fifth of the total. And, how it adds up to the total.

Please use the "Spoiler" feature to hide your answers. It is an "eye" icon on the top of your reply window.

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22 minutes ago, Genady said:

The chest has mix of many different items, which cannot be compared objectively. I.e., each pirate has only a subjective evaluation of what is worth what.

In your proposed solution, I don't understand in what sense each pirate gets at least one fifth of the total. And, how it adds up to the total.

Please use the "Spoiler" feature to hide your answers. It is an "eye" icon on the top of your reply window.

Spoiler

Spoiler in place, lol. 

It is okay to have varying items as the treasure, we would simply replace the coins used in the example with those items. The puzzle in the OP asks how this treasure could be divided peacefully with each pirate receiving 1/5 share. I didn't see a requirement that nothing remained of the treasure (I'll go back and take a look).

This solution a peaceful division into the required 1/5 division for each pirate. This leaves a little left over to be reburied, lol. 

And, sorry about my confusion concerning the spoilers. Let me know if you are now talking about the second post or if I am still not using the right prompt (or if I am doing that incorrectly). I highlighted the text and hit the eye icon. Is that how it is done?

I have reached the limit on five posts for a new member's first day, so will have to make this apology in the edit.

 

Edited by S.T. Ranger
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2 minutes ago, S.T. Ranger said:

Spoiler in place, lol. 

It is okay to have varying items as the treasure, we would simply replace the coins used in the example with those items. The puzzle in the OP asks how this treasure could be divided peacefully with each pirate receiving 1/5 share. I didn't see a requirement that nothing remained of the treasure (I'll go back and take a look).

I don't see anything hidden. It should've looked like this:

Spoiler

This is how the "spoiler" hides the text.

I can clarify the puzzle as needed.

We cannot simply replace the coins with those items because, e.g., there are 50 thousand different items.

The pirates all know what is in the chest, and each one will be satisfied if he gets what he thinks is one fifth of it.

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Spoiler

Seems to me they could start a pile and keep adding to it until one of them will accept it as their 1/5 share and repeat the process until the last two where; "one cuts, the other chooses". (if 2 pirates want the same pile and there isn't equivalent items to make an identical pile, repeat process using different combinations of items until only 1 accepts)

 

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18 minutes ago, npts2020 said:

 

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Seems to me they could start a pile and keep adding to it until one of them will accept it as their 1/5 share and repeat the process until the last two where; "one cuts, the other chooses". (if 2 pirates want the same pile and there isn't equivalent items to make an identical pile, repeat process using different combinations of items until only 1 accepts)

 

Your solution is a bit different from the one mentioned before, but it seems to be the same in principle and will work as well. +1

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On 6/11/2023 at 1:52 PM, npts2020 said:

 

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Seems to me they could start a pile and keep adding to it until one of them will accept it as their 1/5 share and repeat the process until the last two where; "one cuts, the other chooses". (if 2 pirates want the same pile and there isn't equivalent items to make an identical pile, repeat process using different combinations of items until only 1 accepts)

 

Spoiler

This probably typically works but I think you have to make use of "at least one pirate must be willing both to take the pile, or to let someone else take it," unless you say that the pirates judgment are all strictly different by some known amount. Otherwise you could have multiple pirates who have the same opinion no matter how finely you adjust a pile.

You can't "make an identical pile" if you're relying on the pirates' judgment to determine their equality, as they won't necessarily agree on that. It would be the same reason that making 5 identical piles at the start, wouldn't work.

 

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On 6/12/2023 at 7:05 PM, md65536 said:
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This probably typically works but I think you have to make use of "at least one pirate must be willing both to take the pile, or to let someone else take it," unless you say that the pirates judgment are all strictly different by some known amount. Otherwise you could have multiple pirates who have the same opinion no matter how finely you adjust a pile.

You can't "make an identical pile" if you're relying on the pirates' judgment to determine their equality, as they won't necessarily agree on that. It would be the same reason that making 5 identical piles at the start, wouldn't work.

 

Spoiler

Well, it kind of assumes they all want to make an agreement. There is the possibility in virtually any group that there could be one or more members who will be unhappy no matter the outcome, so will always call at the same time as someone else. One solution would be to allow the first (then second, then third etc) to withdraw if they feel they can get a similar share later or force a recount. Eventually, the loot will get divided with no valid reason for any of them to complain.

 

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