The original question and answer of pirate gold

Ian Stewart)

The logic of mathematics sometimes leads to seemingly weird conclusions. The general rule is that if the logic
There are no loopholes in reasoning, so the conclusion will surely survive, even if it is in conflict with your intuition. 1998
In September, Stephen M. omohundro from paloadto, Calif., sent me a question,
It Exactly belongs to this category. This problem has been around for at least ten years, but omohundro has made
The logic problem is made extra complicated.

Let's take a look at the original shape of this puzzle. Ten hackers snatched 100 gold from the cellar and planned
Share these benefits. This is a democracy-speaking pirate (of course, their own unique democracy ),
Their habit is to distribute in the following way: the most powerful pirate proposed the allocation scheme, however
All the following pirates (including those who propose the solution) will vote on the solution. If 50% or more
Pirates agree with this scheme, and the scheme will be approved and the trophy will be allocated accordingly. Otherwise, the Sea of proposed solutions
Hackers will be thrown into the sea, and the most nominated pirates will repeat the above process.

All the pirates are happy to see one of their associates thrown into the sea, but if they
If they choose, they would rather get a cash order. Of course they do not want to be thrown into the sea.
All pirates are rational, and others are rational. In addition, no
The two pirates are equally powerful-these pirates have arranged their seats in full order,
And everyone knows the levels of themselves and everyone else. These gold blocks cannot be further divided or used.
Several pirates have a total of gold coins, because no pirate believes that his associates will abide by
Schedule. This is a group of pirates who only plan for themselves.

What kind of allocation scheme should the most fierce pirate propose to obtain the most gold?

For convenience, we will number these pirates based on their severity. The most cowardly sea
The number of thieves is the number 1, and the number of cuts is the number 2, and so on. In this way, the most powerful pirate should
When the maximum number is obtained, the proposal of the solution will be carried out from top to bottom.

The secret to analyzing all such strategy games is that they should be pushed back from the end. Game conclusion
You can easily know which decision is favorable and which is unfavorable. After confirming this, you can
To use it for the last 2nd decisions, and so on. If the analysis starts from the beginning of the game,
That cannot go far. The reason is that all strategic decisions are to be determined: "If I do
What will the next person do ?" So what are the decisions made by the following pirates for you?
Important, but the decisions made by the pirates before you are not important, because you are also responsible for these decisions.
No.

With this in mind, we can know that our starting point should be that there are only two oceans in the game.
When stealing -- that is, number 1 and number 2. At this time, the most powerful pirate is the 2nd, and his best sub-Formula
The case is clear: 100 pieces of gold are owned by him, and no one gets anything from pirate 1. Because
He must have voted in favor of the solution, which accounts for 50% of the total, so the solution was approved.
Now add pirate 3. Pirate 1 knows that if the solution on the 3rd is rejected, there will be only 2
One pirate, and one will surely get nothing-In addition, the third understands the situation. Therefore,
As long as the allocation scheme on the 3rd is used to give the 1st a sweet spot so that he will not return empty-handed, no matter whether the proposal on the 3rd
For any allocation scheme, I will vote in favor. Therefore, the 3rd must be given as little gold as possible.
To bribe the first pirate, so that the following allocation scheme is available: the third pirate gets 99 gold, and the second sea
Nothing was stolen. Pirate 1 got 1 piece of gold. The same is true for pirate 4. He needs 50% yuan.
If you hold the ticket, you need to contact another person as the third party. The minimum bribe he can give to his peers is one gold.
Son, and he can use this gold to buy pirate 2. Because if No. 4 is rejected and No. 3 is passed,
Then the second part will be broke. Therefore, the allocation scheme of No. 4 should be: 99 yuan should belong to itself, and the allocation scheme of No. 3 should also be
No. I got one gold on the 2nd and one gold on the 1st. The strategy of pirate 5 is slightly different. He
Two other pirates need to be bought, so at least two pieces of gold must be used for bribery in order to get their solutions
Adopted. His allocation plan should be: 98 pieces of gold belong to himself, 1 piece of gold is given to 3, and 1 piece of gold is given to 1.

This analysis process can be continued based on the above ideas. Each allocation scheme is unique
It can make the pirate who proposed the scheme obtain as much gold as possible, and at the same time ensure that the scheme is willing
Yes. According to this pattern, the solution proposed by pirate 10 will be owned by 96 yuan,
The other even-numbered Pirates get one gold each, and the odd-numbered Pirates get nothing.
This solves the allocation problem of 10 pirates.

 

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The contribution of Omohundro is that he has extended this issue to the situation where there are 500 pirates, that is, 500 seas.
100 pieces of gold are stolen. Obviously, similar rules are still valid, at least within a certain range.
As a matter of fact, the rules described above have been established until the 200th S. The solution for pirate 200 will be:
From 1 to 199 all the odd number of pirates will get nothing, and from 2 to 198 all the even number
Each pirate gets one gold, and the remaining one gold is owned by the pirate 200.

At first glance, this demonstration method will no longer apply after the 1949th, because the 1949th cannot get more
More gold to buy other pirates. But even if you cannot find the gold, at least you still want
Thrown into the sea, so he can allocate 1 gold to each of the odd-numbered pirates from 1 to 199.
Son, you don't need one. Pirate 202 has no choice but to drop a single piece of gold --
He must use all the 100 pieces of gold to buy 100 pirates, and the 100 must be
Some people who will find nothing according to the 201 plan.

Since there are 101 such pirates, the 202 plan will no longer be the only one-the bribery plan
There are 101 types. Pirate 203 must receive 102 votes, but apparently he does not have enough gold to buy
There were 101 associates. Therefore, no matter what allocation scheme is proposed, he is destined to be thrown into the sea to feed
Fish. However, although 203 is destined to be a dead end, it does not mean that he is not able to afford the game.
What is the function. On the contrary, 204 now knows that in order to save his life, 203 must avoid
We have already proposed a distribution scheme, so no matter what kind of scheme the 204 pirate proposed
Will definitely vote in favor. In this way, the man of the 9th pirate finally got a life: he can get his own
One vote, one vote on the 1949th, and one vote for the other 203 purchased pirates, just to save their lives.
50%. A pirate who obtains gold will belong to the 202 who will surely find nothing under solution 101.
List of pirates.

What is the fate of the 205 pirate? He is not so lucky. He can't count on
204 supports his scheme, because if they vote against the 205 scheme, they will be able to gloat
Seeing that 205 was thrown into the sea to feed the fish, their own lives can still be preserved. In this way, none
No doubt that no solution can be proposed by pirate 205.

The same is true for pirate 206-he can certainly get support from pirate 205, but this is not enough to save him.
Life. Similarly, pirate 207 requires 104 votes in favor-except for the 100 votes he had bought
In addition to one of his own votes, he still needs three votes to survive the death. He can get 205
No. And no. 206 support, but there is still a difference in a ticket, but it cannot be obtained in any way, so no. 207 pirate
The fate is to feed fish in the sea.

208 is running again. He needs 104 votes, and 205, 206, and 207 will support
He, with his own pass and his 100 pass, was able to survive. Whoever gets his bribe will belong
Those who will certainly find nothing under the 204 Plan (candidates include all the even numbers in the 2 to 200)
And 201, 203, and 204 ).

Now we can see a new rule that will always be effective later: those who can pass through the solution
(All of their allocation plans are to use gold to buy 100 associates, but none of them can .)
The distance between them is getting farther and farther, and no matter what solution they propose, the pirates will be thrown into the sea.
-So to save their lives, they will vote for any sharding formula proposed by a pirate who is better than them.
Case. Piracy that can avoid the loss of fish, including 201, 202, 204, 208, 216, 232, 264,

328 and 456, that is, a pirate whose number is equal to the power of one of 200 and 2.

Now let's take a look at which pirates are lucky enough to get bribes. The method for allocating bribes is not unique.
One of the ways is to let the 201 pirates distribute bribes to all the odd-numbered pirates from 1 to 199,
Let the number 202 be assigned to all the even-numbered pirates from 2 to 200, and then bribe the number of The number 204
A pirate, who Brid even numbers on the 1949th, and so on, that is, take turns to bribe odd numbers and even numbers.
Number of pirates.

The conclusion is: when 500 pirates use the best policy to divide gold, the first 44 are no doubt dead,
The 456 issue is solved by dividing each of the odd-numbered pirates from 1 to 199 into one gold.
Yes. Because of the democratic system that these pirates have implemented, their affairs have become the most amazing.
Most of the pirates feed the fish off the sea, but sometimes they may feel lucky-although they cannot
The stolen gold can always be avoided. Only the shortest 200 pirates may have a dirty copy.
And only half of them can really get a piece of gold. It is indeed the cowardly who inherit wealth.