The logic of mathematics sometimes leads to the conclusion that it seems very weird. The general rule is that if the logic reasoning has no vulnerability, then the conclusion must stand the foot, even if it is contradictory with your intuition. In September 1998, Stephen M. Oomohundro of Palo Alto, California sent me a problem, it just belongs to this. This problem has been circulated for at least ten years, but omohundro has changed it, making its logic problems have been complex. Let's take a look at the original shape of this problem. 10 pirates robbed 100 gold, and intended to divide these trophy. This is some pirates that tell the democracy (of course, their own unique democracy), their habits are allocated in the following way: the most powerful pirate proposes allocation programs, then all pirates (including the proposal) The program is voted. If 50% or more pirates agree with this program, this program is passed and dispensed accordingly. Otherwise, the pirate of the proposal will be thrown into the sea, then the nominated pirate is repeated the above process. All pirates are willing to see that they are thrown into the sea, but if they choose, they are not to get a cash. They are of course not willing to be thrown into the sea. All pirates are rational and knowing other pirates are also rational. In addition, no two pirates are equally powerful - these pirates raise their seats in accordance with the top-down level, and everyone knows themselves and other owners. These gold blocks can not be divided, nor allowing a few pirates to have a gold block, because any pirates don't believe that his association will comply with the arrangement of shared gold blocks. This is a pirate that everyone is only for yourself. What kind of distribution plan should make him get the most gold for the most fierce pirate? For convenience, we give them a number in accordance with these pirates. The most embarrassing pirate is a pirate of No. 1, and the pirate of the second is 2 pirates, so so. This is the most powerful pirate should get the maximum number, and the proposal of the program will fall from top to bottom. Analysis of all the mystery of all such strategy games is that it should be reversed from the end. At the end of the game, you can easily know what decisions are beneficial to make it unfavorable. After you have determined this, you can use it on the second decision of the countdown, so so. If it is analyzed from the beginning of the game, it can't get far. The reason is that all strategic decisions are to be sure: "If I do this, what will the next person do?" So the decision made by the following pirates is important to you, but before you The decision made by the pirate is not important, because you have no power to these decisions anyway. Remember this, you can know that our starting point should be the game to leverage only two pirates - the first and No. 2 - time. At this time, the most powerful pirate is No. 2, and his best distribution plan is a look: 100 gold is all all of them, and the pirates of the 1st don't get it. Because he certainly specifically in favor of this program, it accounts for 50% of the total, so the program is passed. Now add 3 pirates. The 1st pirate knows that if the No. 3 program is veto, then there will be only 2 pirates left, and No. 1 will be affirmative. - In addition, the No. 3 is also understanding this situation. Therefore, as long as the distribution plan of No. 3 gives him a little sweet to return, then no matter what allocation scheme No. 3, No. 1 will vote. So No. 3 needs to separate as little gold as possible to bribe 1 pirates, this has the following distribution plan: No. 3 pirates to get 99 gold, No. 2 pirates do nothing, No. 1 sea stolen 1 gold. The strategy of Pirate 4 is also similar. He needs 50% of support tickets, so you need to find someone to do the same party as the same No. 3. He can give 1 gold for the minimum bribe of the party, and he can use this gold to collect 2 pirates.
Because if the No. 4 is veto, No. 3 is passed, the 2nd will be unameful. Therefore, the distribution plan of the 4th should be: 99 gold belongs to himself, and the No. 3 will not be found. No. 2 is 1 gold, and the No. 1 is also a piece. The strategy of the 5th pirate is slightly different. He needs to buy another two pirates, so at least 2 gold to bribe can be used to adopt its own scheme. His allocation plan should be: 98 gold belongs to yourself, 1 gold gives 3, 1 gold gives 1. This analysis process can continue to be done according to the above ideas. Each allocation scheme is unique, it can make the pirates of the program to achieve as much gold as possible, while ensuring that the program will definitely pass. According to this mode, the solution proposed by the 10th pirate will be that 96 gold belongs to his all, and the other numbers have a piece of pirate each to have a gold, and the numbered podium is not available. This solves the allocation challenge of 10 pirates. Omohundro's contribution is that he expanded this issue to a situation of 500 pirates, ie 500 pirate pumpys, 100 pieces of gold. Obviously, similar laws are still established - at least within a certain range. In fact, the rules described earlier until the pirates of No. 200 were established. The 2008 pirates will be: all odd pirates from 1 to 199 will be unpredictable, and all even pirates from 2 to 198 will have 1 gold, the remaining gold belongs to 200 Pirates all. At first glance, this demonstration method will no longer apply after 200, because 201 is can't get more gold to collect other pirates. But even if you don't have a golden, 201 will also hope that you will not be thrown into the sea, so he can assign: 1 to all odd numbers 1 to 199, 1 gold, you don't want it. The pirates of Queen 202 also don't have a choice. It can only be used in a gold - he must use all 100 gold to collect 100 pirates, and this 100 pirates must also be a person who will be unworthy in accordance with 201. Because such pirates have 101, the No. 202 program will no longer be the only - 101 bribery programs. The pirates of Question 203 must receive 102 votes, but he obviously not enough gold to pick up 101 accomplices. Therefore, he is destined to be thrown into the sea. However, although I have been destined to death in the 203th, it is not to say that he does not work in the game process. On the contrary, 204 now knows that in order to keep your life, No. 203 must avoid such a situation in which it will make a distribution plan. Therefore, no matter what kind of plan for piracy 204, 203 will be approved. Such No. 204, the pirate finally went to a life: he can get his own 1 vote, 1 ticket of 203, and the pirates of the other 100 pirates, just achieve 50% of the life. Getting the pirates of gold must belong to the No. 202 program will definitely will have the 101 pirates of the 101 pirates. What is the fate of the pirate 205? He can not be lucky. He can't expect $ 203 and 204 to support his program, because if they vote against the 205 program, they can see the 205th thrown into the sea to feed the fish, and their own life is still guaranteed. In this way, there is no doubt that there is no matter what the programs of the pirates 205 are. The 206 pirate is also the case - he must get support from Queen 205, but this is not enough to save him. Similarly, the pirates of 207 need 104 votes - except for the 100 Issue Tickets he bought and his own than 1 ticket, he still needs 3 votes to avoid death. He can get support from 205 and 206, but a ticket is still unable to get it, so the fate of the pirate 207 is also the sea feeding fish. 208 is still working again.
