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Can someone help me with this problem? I think that it is something with the Dirichlet.

Suppose that a group of voters is to elect a mayor. There are $a$ voters that intend to vote for candidate A, $b$ voters for candidate B and $c$ voters for candidate C, where $0 <a <b <c$.

During the pre-election silence it is not allowed to meet with more than 2 voters to discuss candidates for mayor. In a meeting of 2 voters from different "camps" both will change their beliefs and vote for the only non-present camp. (So for example if voters for candidates $A$ and $B$ meet, they will leave the meeting wanting to vote for $C$)

If subsequently one of them would meet with another voter who voted for $A$ these two after the conversation would vote for candidate C.

The question is whether it can happen that all of $a + b + c$ voters will become convinced to vote for the same candidate?

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closed as not a real question by Michael Albanese, DonAntonio, Santosh Linkha, TMM, vonbrand Mar 31 '13 at 13:22

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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please complete the question –  Steven-Owen Mar 31 '13 at 12:12
    
I think read this as (and make some edits as the current description contradicts itself): is there any order of bilateral meetings such that after all such meetings all voters will vote for the same candidate? Is this always possible, or are there restrictions on the relative prior vote preference for $A, B, C$ in magnitudes $0 < a < b < c$? –  gnometorule Mar 31 '13 at 12:51
    
The two meeting rules are: (1) if voters currently voting for $A$ and $B$ meet, they will favor $C$ after (and symmetrical for the other cases), so $a= a -1, b = b - 1, c = c + 2$); and (2) if voters both currently for $A$ ($B, C$) voting meet, they intend to still vote for $A$ ($B, C$) after. This is a nice puzzle. –  gnometorule Mar 31 '13 at 12:51

1 Answer 1

This would be possible.

Easiest way to show is by an example: if you have $a=1$, $b=3$ and $c=4$.

First make $a$ and $c$ meet this yields $a=0$, $b=5$ and $c=3$.

Then make $b$ and $c$ meet to get $a=2$, $b=4$ and $c=2$.

Now make $a$ and $c$ meet twice and you have $b=8$ with $a=c=0$

In response to the comment (at the question) of @gnometorule

There are definitely restrictions on the starting values of $a$, $b$ and $c$. For one it is not possible with $a=1$, $b=2$ and $c=3$ because you will always be left with 1 rebellious voter.

[EDIT] Thinking of the problem a bit more. One condition for this to work is that $d+3=c$, where $d$ is $a$ or $b$ because in that case you can have $c$ and $e$ (where $e$ is $a$ or $b$ with $e\neq d$) meet such that $d=d+2$ and $c=c-1$ which makes sure that $c=d$ which can then continuously meet to have all votes at $e$. This is probably not the only condition, but it is at least a condition.

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Upvoted, but is there a general solution? –  gnometorule Mar 31 '13 at 12:55
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Pretty sure that there should be, but that is a bit too tough a nut to crack for me. –  Michiel Mar 31 '13 at 12:59
    
Agreed, and I checked these two cases when I first saw this. I just think it can be expressed as some relation between $a, b, c$ as a general case, and was wondering if someone might see which one. –  gnometorule Mar 31 '13 at 13:01

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