Author |
Topic: ELEGANTLY GREEDY PIRATES (Read 2326 times) |
|
casual_kumar
Newbie
Gender: 
Posts: 21
|
 |
ELEGANTLY GREEDY PIRATES
« on: Aug 8th, 2006, 4:23am » |
 Quote  Modify
|
Lets start backwards...
On the day when only 2 are left, the treasure will get split for sure as 50% vote of the 2nd pirate will be in favour. So, on the day when 3 are left, 1 and 2 will try killing the 3rd, knowing this, 3rd would avoid the situation by voting in favour of a split on the day when 4 are alive. Similarily, 5th 6th 7th and 8th pirates would go for a split on the day when 8 are alive. Otherwise, 1st, 2nd, 3rd and 4th would vote for a kill till only they are left.
Calculating similarily, on the day when 512 pirates are left alive they will go for a split. Infact, a split is only possible on days when alive pirates are exactly 2^n (i.e. 1, 2, 4, 8, 16, 32, 64, 128, 256, 512).
So heres the strategy. 512 pirates from the top will vote for killing till they have only 512 left. On this day, pirates 1 to 256 will vote for killing and 257 to 512 will go for a split. Having 50% votes for a split, the treasure would get split among 512 pirates. None, of the pirates from 257 to 512 will dare risking voting against the split, because if today (when 512 are left) they kill the 512th pirate then the pirates 1 to 256 will always vote for killing and none of the later half will survive.
|
| « Last Edit: Aug 8th, 2006, 4:25am by casual_kumar » |
 IP Logged |
|
|
|
c
Newbie
Posts: 6
|
|
Re: ELEGANTLY GREEDY PIRATES
« Reply #1 on: Mar 25th, 2012, 11:48am » |
Quote Modify
|
For the supermajority cases:
* Pirate 2 will vote with P3, or else he gets nothing;
* Pirate 4 will vote for a split in a two-thirds majority with pirates 5-9 (but not sooner than day 9);
* Pirate 10 will vote with 11-27, etc;
* Eventually, the split will be successfully voted for by pirates 244-729, against P1-P243.
The third case will be the most bloody, with the last vote required being that of P65-P256 against P1-P64.
In a three-fifths majority, the "split" days would be 3, 8, 20, 50, 125, 313, and 783. Thus this would be the least bloody scenario so far. 
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print