Math Fun

Incorrect explanation to the chess board. The coloring has nothing to do with it.

If that’s all you have to say about it, just that it’s incorrect without any explanation of what’s incorrect about it, while everyone else gets it and thinks it’s a good explanation (and even mathematicians agree – this is a well-solved problem), then I’ll just assume you don’t get it.

If you’d like to expound, however, please do.

I’ve always found this puzzle quite interesting:

There are 5 pirates, of different ages. They are all, as luck would have it, perfect logicians. They also have just found some treasure: 100 coins. They are deciding how to split it. The rules for deciding the splitting are as follows:

The oldest proposes how to split the treasure.
Everyone except him votes on if it’s a good split.
Each pirate, having a strong sense of self-interest, bases their vote on the question, “Would I get less if this proposal failed?”
If the answer to that question is “yes,” they vote for it, else they vote against it.
If at least 50% of voting pirates vote for it, it passes, else it fails and the oldest is thrown over board, and the new oldest makes a new proposal.

So, again, they’re perfect logicians, and each ones has the explicit goals of
a) maximizing the money that gets split to them
b) not dying

So, what happens?

There’s an alternate version of the puzzle, in which all pirates vote, including the one who made the proposal, which has a different solution methinks.

[tab]So, when they get to the last two pirates, the youngest pirate will reject anything, because then she’ll end up with all the treasure.

When there are three, then, the second youngest will accept anything, possibly even zero, assuming being thrown overboard is less than just getting nothing*. The third youngest will then offer something like “I get everything, and neither of you get anything.” The vote will split, and the proposal will pass.

So the fourth youngest, knowing this, then knows that anything better than nothing will be accepted by the two youngest. He’ll propose something like “1 for the youngest, 1 for the second youngest, none for the third oldest, and the rest for me” The third will vote against, but the others will vote for, and it will pass.

The fifth, then, only has to give the youngest two 2 gold, the third youngest 1 gold, the fourth youngest 0 gold, and keep the rest to himself. The fourth will vote against, and the other three will vote for.

*If this assumption is bad, then the second youngest pirate gets 1 additional gold each round.[/tab]

here we go again #-o

My solution (I’ve tabbed each step because I’ve made it very detailed - you have been warned - answer is underlined in the final tab):

[tab]The only pirate who has no risk of dying (b) is the youngest pirate.
In the most extreme case, where he was the only one left, he would get the maximum of 100 coins all for himself, fulfilling (a) perfectly as well as (b).

In the next most extreme case, with 2 pirates left, the youngest would be the only voter, resulting in either 100% vote in favour of the proposal made by the oldest or 0%, which would get the oldest thrown overboard.
So to survive at all, the oldest has to offer an acceptable proposal for the youngest, who has nothing to lose. The only proposal that the youngest will accept is all 100 coins, or he will vote against it, be the only pirate left, and get all 100 coins anyway, due to (a).
So due to (b), the oldest of the two would offer 100 coins to the other pirate and have to put up with nothing for himself.[/tab]
[tab]With 3 pirates left, there are two voters. The oldest only has to satisfy 1 of them to get “at least 50%” of the vote, so he has to work out which one to satisfy and how much will satisfy them. The other doesn’t need to be offered anything, in order to maximise how much the other 2 get, as with (a) - it’s fine if just one of them votes against the proposal.
He can actually offer just 1 coin to the 2nd youngest, and nothing to the youngest, and take 99 for himself.

It is in the interests of the 2nd youngest pirate to vote for the oldest pirate of 3 to be kept alive (accepting his proposal) as long as the proposal involves 1 coin or more for him, because even accepting 1 coin would be better than the 0 that he would get if the oldest pirate of 3 is thrown overboard (as in the scenario above with only the 2 youngest pirates). To ensure the life of the oldest of 3, the 2nd youngest needs to either be sure that the youngest will vote for him (in order for him to get at least 50% of the vote) whether he himself votes for the proposal or not, or he needs to vote for him himself (in order to give at least 50% of the vote) whether the youngest follows suit or not.
As pointed out, the youngest is ensured (b) and at least 0 coins, so only has to worry about getting as many more coins as he can more than 0, as with (a). He will have worked out the dilemma of the 2nd youngest, and that he’s in danger of getting 0 coins if he votes for any proposal. His concern is to get at least 1 coin, and the best scenario is that both vote against whatever the proposal is, so he can secure 100 coins - but that’s only a possibility if he can be sure that the 2nd youngest will vote against the proposal, then he can too, and he’s in the money.

At this point, he can’t actually be sure that the 2nd youngest will vote against the proposal if 0 coins are offered to him, because he’ll get 0 coins whether he votes for or against. Needing to be sure in order to secure his 100 coins by also voting against any proposal, this possibility goes out the window. So he will also accept anything more than 0 coins, even 1 coin, in order to escape the 0 coins he’s in danger of being offered.

Having worked out this, the oldest of 3 could actually offer either both of them 1, or either one of them 1 and the other 0. The former option only leaves him with 98 instead of 99 as in the latter 2 options. He can be more sure of his proposal being accepted with at least 1 vote if he offers the 1 to the 2nd youngest who is in the worse position - having nothing to lose.[/tab]
[tab]With 4 pirates left, there are 3 voters. The eldest needs to please 2 of these 3 to get at least 50% of their vote.
The 2nd youngest would do better to ensure the survival of the oldest of 4 pirates if he offers him 2 or more coins, else if he is thrown overboard, he’s in danger of getting 1 or less (as in the above scenario with only 3 pirates). Being offered 2 coins or more will get the oldest pirate 1 of the 2 votes he needs to stay alive.
Assuming the 2nd youngest is offered 2 coins, this would leave a maximum of 98 for the 3rd youngest, which is less than he would get if the oldest of 4 pirates was thrown overboard. It’s in his best interest to vote against the eldest. He can ensure this if the eldest offers less than 2 coins to the 2nd youngest, by being the 2nd vote against any proposal, or at least if he can be sure that the youngest will vote against the oldest pirate whether 2 coins are going to the 2nd youngest or not. The youngest is going to vote for whatever he is offered, as long as it is more than the 0 he’s going to get if the oldest of 4 pirates is thrown overboard.
So if the oldest of 4 proposes 2 to the 2nd youngest, 1 to the youngest, and nothing to the 3rd youngest, he can take 97 for himself and still have his proposal accepted.
However, since the 3rd youngest will have worked this out, he knows he is in danger of getting nothing, so will also accept as little as 1 coin (or more). The oldest of 4 will deduce this and question giving 2 to the 2nd youngest. If he offers him nothing, losing his vote, can he guarantee the 2 votes of the other two? No, because if the other two work out that the 2nd youngest won’t vote for the proposal, the 3rd youngest stands to get 98 coins if he also votes against the proposal and gets the oldest of 4 thrown overboard, which he will prefer.
So the oldest of 4 has to give 2 or more to the 2nd youngest, and 1 can go to the youngest and nothing to the 3rd youngest. This is because the youngest will fear the possibility of the oldest of 4 pirates being thrown overboard and getting 0 coins. He is loss averse and the 2nd youngest will be swayed by the gain of 1 coin. The 3rd youngest will vote against but this won’t matter with 2/3 of the vote passing the proposal. Without giving the 1 coin to the youngest, the 3rd youngest has a chance of his vote being combined with the vote of the youngest to move them over to the 3 pirate scenario where the 3rd youngest gets 98 coins. So the 1 coin to the youngest secures that this possibility doesn’t happen. The oldest of 4 gets his 97 coins.[/tab]
[tab]Now the Grande Finale:
With all 5 pirates
, there are 4 voters, and so only 2 of these 4 need to be satisfied.
According to 4 pirate scenario above, the pirates most easily pleased will be the 3rd youngest who would stand to get no coins, and the youngest who will get only 1 coin. Offer the youngest 2, the 3rd youngest 1, and they’ll have to accept. The other 2 are going to vote against getting nothing, but that doesn’t affect the proposal of 2 coins to the youngest, 1 coin to the 3rd youngest, nothing for the other two, and the remaining 97 coins for the oldest pirate. It will still pass with at least 50% of the vote in favour.

Perhaps interestingly, this may all change considering the possibility of finding coins in future. This might turn the pirates against each other, especially the youngest one against the others since he will never be thrown overboard if they keep the same rules, and especially since the 2nd oldest and the 2nd youngest are going to want more than nothing in future, and their primary target is going to be the oldest pirate who stands to rip them off in future in a similar way, and he’s the first in line to be thrown overboard.[/tab]

I think, between Carleas and Sil, my original answer was closest to Sil’s. But, Carleas said something interesting that I hadn’t thought of.
[tab]My first Solution Attempt (proposer can’t vote problem):

If there is only 1, he will take 100 coins.
If there are 2, the only way the oldest will survive is if he offers 1 100 coins. (remember this step, this is the step that I think is very likely wrong in my original reasoning, based on something Carleas said)
So, if there are 2, 1 gets 100 and 2 gets 0.
If there are 3, the oldest has to only earn 1 vote. He gives 1 coin to #2 and 0 to #1, and 99 for himself.
If there are 4, the oldest has to earn 2 votes.
He gives 1 to #1, 2 to #2, and 0 to #3, leaving 97 for himself.
If there are 5, the oldest has to earn 2 votes.
He gives 1 to #3, 2 to #1, and 0 to #2 and #4, leaving him with 97

see the second tab for my revised reasoning[/tab]
[tab]If there is only 1, he takes 100 coins. This is still true.
If there are 2, I thought before that 2 could offer 100 and survive. However…this isn’t the case, according to the way I’ve worded the problem.
“Each pirate, having a strong sense of self-interest, bases their vote on the question, “Would I get less if this proposal failed?”
If the answer to that question is “yes,” they vote for it, else they vote against it.”

If 2 offers 1 100 coins, the answer to the question is “no”, so thus he votes against it.
2 dies no matter what if there are only 2.

And Carleas was right, then:
If there are only 3, the oldest doesn’t have to offer 2 anything, 2 will still vote for the proposal in order to not die.
So if there are 3, 3 offers 2 0 and offers 1 0, gets 2’s vote and keeps 100 for himself.
If there are 4, then, the oldest needs 2 votes.
He can give 1 to #1 and 1 to #2, 0 to #3 and keep 98 for himself.
If there are 5, oldest needs 2 votes.
He gives 2 to (either of #1 or #2) and 1 to #3, and keeps 97 for himself.[/tab]

Aren’t our solutions the same?

I don’t think this needs to be tabbed: it’s a good observation, but are we to give it priority over “the explicit goal” of “b) not dying” when they conflict?

[tab]Perhaps it doesn’t matter hugely, since the solutions are identical if the revised version ended up with “#1” being offered the 2 coins instead of “#2” rather than the other way around.

However, my solution that has disregarded the stalemate procedure in favour of considering “b)” and certainty, would mean that “the other way around” would make all the difference. Perhaps I’ll re-evaluate my solution in light of this procedure, but give “not dying” precedence when they conflict - see if it makes a difference.[/tab]

" So, if there are 2, 1 gets 100 and 2 gets 0.
If there are 3, the oldest has to only earn 1 vote. He gives 1 coin to #2 and 0 to #1, and 99 for himself.
If there are 4, the oldest has to earn 2 votes.
He gives 1 to #1, 2 to #2, and 0 to #3, leaving 97 for himself. "

Why does he have to give #2 more than 1 coin?
I don’t see a problem with your solution of the chessboard, it seems indefeatable. Not so with the aryan guru and her so called logical necessity. Obviousness doesn’t figure into logic, it’s too far short of certainty. But the riddle is ‘ambiguous’ - it’s not meant to be taken literally.

Um, if there’s 3, the oldest has to earn 2 votes to get at least 50%.

The eldest of 4 has to give the youngest 2 coins due to the clause: “Each pirate, having a strong sense of self-interest, bases their vote on the question, “Would I get less if this proposal failed?”
If the answer to that question is “yes,” they vote for it, else they vote against it.”
The youngest would ask himself that question, and answer “no”, he wouldn’t get less, he would get the same. So he would vote against it. The eldest needs his vote to both survive and get the most coins out of the proposal, so he has to change the proposal so that the youngest answers “yes” to that question, and thus votes FOR his proposal.

As for that old guru one, u clearly didn’t keep up - we all proved the need for her several times over in several ways. That one’s well and truly solved, explored and resolved.

So you figure that if you painted the board all the same color leaving only the grid, you would be able to solve it???
The solution involves odd and even numbers and symmetry.
Anything that you can’t do to one side of the board, you can’t do to the other.

On this one, I think James has a point. While the square coloring demonstrates the fact that it can’t be done, the explanation for why can’t rely on the color of the board. The coloring is a stand-in for odds/evens.

The first time I actually heard the explanation that I offered, I heard it in the context of a colorless board. The problem was just a grid of squares, all white, with 2 squares removed (the grid was not 8x8, it was a bit smaller I think, maybe 6x4 or something).

The solver first showed a couple diagrams of attempts to cover all the squares, all of them failing.

And then he said, “Instead of trying every possible arrangement of dominos, there’s this faster way of solving it: imagine the squares as colored black and white, like a chessboard.” And then he offered the same solution.

So, the solution, though it refers to the colors of the chessboard, doesn’t explicitly necessarily depend on them being those colors. Even if they weren’t those colors, even if they were all white, we could still, in our solution, artificially imagine them as colored and the solution would still hold. That our solution involves a checkered grid doesn’t mean that the board has to be a checkered grid. We merely have to imagine the board as checkered to see that it still holds, even on a fully white board.

“The solution involves odd and even numbers and symmetry.”

This could mean any number of things.
The board in question, with opposite corners removed, is symmetrical. It’s symmetrical across both diagonals.
There are symmetrical boards that have ways to tile the dominos to cover all squares, and asymmetrical boards that have ways to do the same.
There are symmetrical boards that are impossible to domino, and assymetrical boards that are also impossible to tile.

When you said “odd and even” – again, I’m not sure what you mean by that, but it’s potentially very closely related to the chessboard solution.
If you were to number the tiles in a spiraling fashion, 1-64, you would find that the black tiles are even and the white tiles are odd (or vice versa, depending on where you start, etc.). You can’t tile the board if you remove 2 even squares only, or 2 odd squares only, but this becomes synonymous then with 2 black squares only, or 2 white squares only. “Odd/even” is interchangable with “black/white” if you number them in a spiral order.

Carleas, do you not consider it an explanation for why? If I’ve used the coloring to prove to you that no possible arrangement could cover all the squares…isn’t that showing why? It’s a fully general solution, and if someone takes such a board up to me and asks “Why can’t I cover them all with dominos?” I don’t feel anything is wrong with the answer, “Well, notice that when you cover 2 squares with a domino, it must cover 1 black and 1 white square. So no matter how many dominos you put on, there will always be an equal number of black and white squares covered. The board you’ve brought to me has less whites than blacks (or vice versa), and so that’s why it can’t be done.” I don’t see how it fails at ‘why’.

The one problem, classically, with this solution that most people have is not that it doesn’t work as a solution, not that it doesn’t explain why, but that it’s creative. In the world of mathematics, logic and proofs, this proof was too…informal. It worked for them, it was a great explanation, made it clear why it’s impossible, but they wanted a less creative solution. They raced to come up with the most non-creative solution.

I wouldn’t think you guys would have a problem with creative solutions though…

James and Carleas - obviously the colors are only illustrative of the odd even thing. But a very solid illustration.

Silhouette - ah yes, thanks.
No, the riddle has not been solved at all in the way you claim. No one showed me how my demonstration of 4 blues and 4 browns fails. I understood that Sauwelios thinks my and James’ position were interchangeable, and that he therefore did not even try to understand my explanations. Maybe you’re doing the same. To be honest I did not read most of James’s stuff there because he was dealing with it in a very different manner and making much more complicated claims. I could not keep up with that at the same time as verifying/falsifying my own position.

Well but, they will not ever do that if they don’t understand how* they could start from the same number. And apparently, that isn’t obvious. Interestingly, even in the canonical solution it is required that all start from the same number. Maybe the whole problem is that none of them understands the logical requirements of the canonical solution.** This has crossed my mind before.

  • I am familiar with this phenomenon - I used to suffer from it when I was a child. Often I would not be able to work with a tool or toy if I did not understand how it worked.

** They would in the canonical case simply not know that they do not know. Which speaks for it being called canonical.

Yes and no. I think it’s a flawless proof that it is impossible. But my understanding is that for any chessboard from which 1 white and 1 black tile are removed, there is a solution. Because this seems like a generalization of the problem, I have the intuition that a truly general solution would solve both, i.e. a general solution should work for any chessboard from which coordinates (x1,y1) and (x2,y2) are removed. Color gives a good approximation for solutions to the narrow subset where (x1,y1) and (x2,y2) are both the same color, but tells us next to nothing about the situation where they are different colors.

So, maybe “why” is the wrong term, but I do think there is a more general solution, and that the hand-waivey odds/evens and symmetry explanation, while incomplete, hints at that more general solution

I think that’s also the case, and I also wouldn’t know how to prove that.
My explanation, you’re right, merely rules out the solution to any chessboard in which the removed pieces are the same color, it doesn’t prove that if they’re of different colour that there will necessarily be a solution.

I consider that a different problem from the problem being asked, though, as the problem that’s being asked is indeed only about a chessboard in which the pieces removed are of the same color, and why there’s no possible solution in that case.

Your “understanding” may be a misunderstanding. Where did you make a demonstration? Note that general claims about hypotheticals are not a demonstration. Post or link to your step-by-step solution, and I will explain to you which steps, if any, are false—just as I’ve done with James’.

Yes, I’m really saying “this solution isn’t the best solution to this problem, because there’s this other problem it doesn’t solve.” Which, I admit, is a bullshit criticism about a solution to this problem. But I still think a solution that addresses the whole class of problems would be the better solution.

I personally doubt that there’s any single proof which covers both. Even the mathematicians and logicians working on the problem are creating proofs for the independent problem of “Why is there no solution if the squares removed are of the same color?”, while not proving that there are always solutions for squares removed of different color.

You might find a proof for one, and in the same paper a proof for the other…but just putting it on the same paper doesn’t make it the same proof.

[tab]Due to the symmetry, it doesn’t matter if you begin with rows or columns, so let’s choose rows.
In the top row, there are an odd number of spaces to fill thus
There is no alternative but to share a domino between the first row and the second.
The domino could be placed on either the right or left side of the board but in either case
There is no alternative but to cause an odd number of spaces on the same side of our domino as the missing space was on
We now have the same situation of even on one side an odd on the other but on the second row thus
We have no alternative but to share a domino with the third row and restricted to the odd side which
Leads to the same situation with the fourth row with the odd number shifted opposite as the last
As we go from row to row, we cause the next row to be odd on opposite side of our domino depending on how we started
That pattern has no alternative but to continue down the rows. With each row, the odd number of spaces shifts from right to left.

In order to solve the problem, because the last row has a missing space opposite from the first row,
We must have the odd number of spaces shifted to the opposite side from where we began with the second row
But there are an odd number of rows from the second to the last yielding no alternative but to have
the odd side be the same on the last row as on the second row.

Thus there is no solution.[/tab]