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Topic: Political slugfest (Read 5162 times) 

ecoist
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Political slugfest
« on: Nov 1^{st}, 2008, 9:22pm » 
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4 libertarians, 13 republicans, and 17 democrats gather to argue their political philosophy. They wander about and debate each other in pairs. When two of them of different political persuasions debate each other, they become so disallusioned that they both change to the third political persuasion. Show that it cannot happen that, after awhile, all of them acquire the same political philosophy.

« Last Edit: Nov 1^{st}, 2008, 9:39pm by ecoist » 
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towr
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Re: Political slugfest
« Reply #1 on: Nov 2^{nd}, 2008, 8:29am » 
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Reminds me of a type of lizard.. Let's see if I can find a link Ah, here we go, and more to the point, here.


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Hippo
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Re: Political slugfest
« Reply #2 on: Nov 2^{nd}, 2008, 2:09pm » 
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let = e^{2i/3} be primitive 3rd root of unity. Look at (l,r,d)=(l+r^{2}+d) Operations are additions of (1,1,2),(1,2,1) or (2,1,1) to (l,r,d). As addition of (1,1,1) does not change , additional (0,0,3),(0,3,0) or (3,0,0) does not change (mod 3). Starting position (4,13,17) is mod 3 equal (1,1,1) having = 1 (mod 3). (34,0,0) = . (0,34,0) = ^{2} (0,0,34) = 1 I am not able to finish the proof. So try to go to 34d. 1) (4,13,17)+=(6,3,3)=(10,10,14) 2) (10,10,14)+=(10,10,20)=(0,0,34). At least I have prooved noone else will be able . (12,13,14)=+ 1 = i3 (mod 3) and (39,0,0)=(0,39,0)=(0,0,39)=0 (mod 3). (13,15,17)=1=i3(mod 3) and (45,0,0)=(0,45,0)=(0,0,45)=0 (mod 3). If two of l,r,d are equal (mod 3), we can equalize them as in 1). Than convert to the remaining kind as in 2). If l,r,d are distinct mod 3 we get (l,r,d) = i3(mod 3) and (l+r+d,0,0)=(0,l+r+d,0)=(0,0,l+r+d)=0 (mod 3).

« Last Edit: Nov 2^{nd}, 2008, 3:02pm by Hippo » 
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towr
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Re: Political slugfest
« Reply #3 on: Nov 2^{nd}, 2008, 3:04pm » 
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on Nov 2^{nd}, 2008, 2:09pm, Hippo wrote:So try to go to 34d. 1) (4,13,17)+=(6,3,3)=(10,10,14) 2) (10,10,14)+=(10,10,20)=(0,0,34). 
 That bodes well for Obama

« Last Edit: Nov 2^{nd}, 2008, 3:05pm by towr » 
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Eigenray
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Re: Political slugfest
« Reply #4 on: Nov 2^{nd}, 2008, 3:53pm » 
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For the general case, the analysis is simpler if they're allowed to borrow people. For example, 3 libertarians left to themselves will stay libertarian. But after a couple rounds with a republican in the room (who then leaves, still a republican), they might just find themselves all democrats. Find a simple criterion to determine whether one state can turn into another without borrowing. Hippo: I also thought roots of unity were the right way to go. But they don't really work for a composite number of parties. That is, with n parties, and a given number of people, there are n^{n1} distinct states (allowing negative people), by looking at all the differences mod n. But []/(n) has size n^{(n)}, where = e^{2pi i/n} is an nth root of unity.

« Last Edit: Nov 2^{nd}, 2008, 4:08pm by Eigenray » 
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ecoist
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Re: Political slugfest
« Reply #5 on: Nov 2^{nd}, 2008, 5:27pm » 
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I screwed up, guys! There is an elementary solution if the total number of people involved is a multiple of 3. I should have said there are only 3 libertarians. Thinking about what Eigenray wrote, I came up with the following variation. Let there be 15 parties with the ith party having i members, for i=1,...,15. Whenever 14 of these guys meet, no two belonging to the same party, they all switch to the party none belong to. Show that it cannot happen that after awhile everyone belongs to the same party.


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Hippo
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Re: Political slugfest
« Reply #6 on: Nov 3^{rd}, 2008, 12:41am » 
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on Nov 2^{nd}, 2008, 3:53pm, Eigenray wrote:For the general case, the analysis is simpler if they're allowed to borrow people. For example, 3 libertarians left to themselves will stay libertarian. But after a couple rounds with a republican in the room (who then leaves, still a republican), they might just find themselves all democrats. Find a simple criterion to determine whether one state can turn into another without borrowing. Hippo: I also thought roots of unity were the right way to go. But they don't really work for a composite number of parties. That is, with n parties, and a given number of people, there are n^{n1} distinct states (allowing negative people), by looking at all the differences mod n. But []/(n) has size n^{(n)}, where = e^{2pi i/n} is an nth root of unity. 
 Actually I didn't thing about general case and the roots of unity was used in confusing way ... I actualy thought on triangular grid mod 3. I have used 1,, and ^{2} as a coordinates. You can as well use x,y,z space coordinates and project it to the plane perpendicular to main diagonal (t(1,1,1) line). ... There is 9 positions in the projection mod 3, 7 of them are on projections of axis. Mod 3 just projections of (1,1,0) and (1,1,0) are not. I don't think it can be easily generalised to higher number of parties. At least not from my point of view . ... Oh yes, it looks well for prime p number of parties, but I have had problems to imagine the mod p operation . BTW: The greasmonkey eats spaces after escape sequences as . How should it be avoided in posts?

« Last Edit: Nov 3^{rd}, 2008, 1:18am by Hippo » 
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ecoist
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Re: Political slugfest
« Reply #7 on: Nov 11^{th}, 2008, 4:27pm » 
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Ok, how about this less ambitious generalization of the chameleon problem? Suppose that there are n>2 political parties whose members satisfy two conditions. a) For any two parties, the numbers of members of each party are incongruent modulo n. b) If any n1 people meet, no two of them belonging to the same party, then all n1 switch to the party none belong to. Show that it can never happen that everyone belongs to the same party.


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ecoist
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Re: Political slugfest
« Reply #8 on: Nov 16^{th}, 2008, 11:44am » 
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I'm a complete idiot (thx for letting me find out for myself)! This problem, and its generalization, belongs in easy. Wowbagger suggested the easy solution, and Hippo noted what was needed to be assumed to make his solution work. The numbers of members of each party forms a multiset of residues modulo n. That multiset never changes when party switching occurs (see Hippo's example with (4,13,17). There are always exactly two parties with the same number of members modulo 3). The reason for this is that, when switching occurs, it amounts to subtracting 1 from each membership modulo n (adding n1 members to one party's membership is the same as subtracting 1 modulo n). Hence, since initially all memberships are incongruent modulo n, they must remain incongruent modulo n; and so the guys can never all belong to the same party. The only way such a thing could happen is if, initially, n1 memberships are congruent modulo n.


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