what is the maximum number of pawn moves?
Kilo, there is a maximum potential for 6 moves per pawn, and there are 16 pawns, so there is potential for 96 pawn moves. eight of these pawn moves must also be captures though, so there are 96+22=118 irreversable moves. To see a more detailed explanation of this and how to arrive at the correct maximum number of moves in a game(instant draw by 50 moves), check out pages 2 and 3 of this forum.
As for constructible angles:
Since 11 is not a power of two or a product of a power of two and distinct Fermat primes, 2 * Pi / 11 is not constructible. ( The first Fermat primes are 3, 5, 17. So a pizza of 17 equal slices would be constructible for example. )
I wonder though, can this be done with straight-edge and compass geometry?
Yes, Lewis Chen's construction that I just posted is straight-edge and compass. Just the slices don't have identical shape only identical area.
Yeah, I saw that a little after I posted. I originally ment by cutting it in the canonical way(as wedges whose points meet perfectly in the center. I thought there might be some problems with that idea(as you showed above) since regular 11-gons are not constructable.
How do you cut a round pizza into 11 equal slices?
I simply ask my pizzeria man to do it and if he can't who can do it?
How do you cut a round pizza into 11 equal slices?
I simply ask my pizzeria man to do it and if he can't who can do it?
number the pirates #1-5 with #1 being youngest and #5 being oldest.
solution: pirates #1 and #3 get 1 coin each, pirates #2 and #4 get no coins, and pirate #5 gets 98 coins.
To see this, lets work backwards. If there is only pirate #1 and pirate #2 left, pirate #2 gets 100 coins(because his vote is 50%, all that is needed), so pirate 1 does not want it to get down to 2 people, so when pirate #1, #2 and #3 are all around, pirate #3 can take 99 coins and leave 1 for pirate #1 to win with 2/3 vote., so pirate #2 does not want it to get down to 3 people. if #1-4 are still around, pirate #4 can get away with giving 1 coin to #2 and 99 for himself(#2 gets nothing if it goes down to 3 people, so must accept the 1 coin to give 50% vote), so pirate #5 realizing that #1 and #3 are screwed over with 4 people offers 1 coin to each of them, and they must accept, leaving 98 coins for himself with a 60% vote for it.
What if STRICTLY more than 50% of the vote is needed for the money to be divided up(so if there is a tie in votes, the pirates err to their bloodthirsty side and throw the oldest overboard), how does the answer change and why?
@262, similiar method as #261, but that has dual.
@262, similiar method as #261, but that has dual.
dual what? if you mean dual solutions, can you explain? because I see only one.
@264,
A,B,C,D,E youngest to oldest. yes voters are in bold
A -> A(100)
B -> doesn't matter
C -> C(100), B(0), A(0)
D -> D(98), C(0), B(1), A(1)
E (proposal 1) -> E(97), D(0), C(1), B(2), A(0)
E (proposal 2) -> E(97), D(0), C(1), B(0), A(2)
@264,
A,B,C,D,E youngest to oldest. yes voters are in bold
A -> A(100)
B -> doesn't matter
C -> C(100), B(0), A(0)
D -> D(98), C(0), B(1), A(1)
E (proposal 1) -> E(97), D(0), C(1), B(2), A(0)
E (proposal 2) -> E(97), D(0), C(1), B(0), A(2)
your line of reasoning for C doesn't make sense. B wouldn't vote yes for zero coins.(pirates are bloodthirsty) that changes everything afterwards.
>50% vote: Oldest to youngest, 97-0-1-0-2. Just a little extra bribe to the youngest.
@264,
A,B,C,D,E youngest to oldest. yes voters are in bold
A -> A(100)
B -> doesn't matter
C -> C(100), B(0), A(0)
D -> D(98), C(0), B(1), A(1)
E (proposal 1) -> E(97), D(0), C(1), B(2), A(0)
E (proposal 2) -> E(97), D(0), C(1), B(0), A(2)
your line of reasoning for C doesn't make sense. B wouldn't vote yes for zero coins.(pirates are bloodthirsty) that changes everything afterwards.
if B doesn't agree with C , then he(B) will die, becauce A will decline B's any proposal. so its better to live with 0 coin rather than die with zero coin. they are bloodthristy - but their own life has precedence over that.
>50% vote: Oldest to youngest, 97-0-1-0-2. Just a little extra bribe to the youngest.
97-0-1-0-2 is for 50+% condition (SeansEnglish's modification)
98-0-1-0-1 for 50% condition (original post)
Kilo, in your original stipulations "if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against" the pirate would get zero coins if he lives or dies, so he would vote against it regardless of his life. (better to die as a bloodthirsty pirate than live as a poor one I guess...)
Kilo, in your original stipulations "if a pirate would get the same number of coins if he voted for or against a proposal, he will vote against" the pirate would get zero coins if he lives or dies, so he would vote against it regardless of his life. (better to die as a bloodthirsty pirate than live as a poor one I guess...)
->
Lexicographic preferences or lexicographic orderings (lexicographical order based on the order of amount of each good) describe comparative preferences where an economic agent prefers an amount of one good (X) to any amount of another (Y). More generally, if offered several bundles of goods, the agent will choose the bundle that offers the most X, no matter how much Y there is. Only when there is a tie of Xs between bundles will the agent start comparing Ys. It represents the same generalization of utility theory as nonstandard infinitesimals extend the real numbers. With lexicographic preference, the utility of certain goods is infinitesimal in comparison to others.
The watch thing is well known but cheesy..
Cheesy, may be. But well known?
Judging from the responses here, I'm not sure...