When two voters meet, they switch allegiance; might they all ally with the same candidate?

Question

Let's assume that there are three candidates running in an election. Right before the elections (when there is no more propaganda), it is forbidden to gather in groups of more than two people to discuss the candidates regarding the election.

There are basically three groups of people: group a which would vote for candidate A, group b which would vote for candidate B, and group c which would vote for candidate C, in such a way that the following is true: 0 < |a| < |b| < |c| (If we assume that the groups are sets). Each time two people from different groups meet, they change their opinions and vote for the third candidate. (So if a person which would vote for A meets a person which would vote for B, then after their conversation, they both would vote for C).

The question is: can all of the voters (the union of groups a, b, c) be persuaded to vote for the same candidate?

Answer 1

Let $s_0 = \newcommand{\tuple}[1]{\left\langle #1 \right\rangle}\tuple{a_0,b_0,c_0}$ be the initial state and $(s_k)_k$ be any sequence of following states. Set $$f(\tuple{a,b,c}) = \tuple{a-b, b-c, c-a} \bmod 3,$$ then $f(s_k) = f(s_0)$ for any $k$.

Proof.

Without loss of generality we can assume that

$$s_k = \tuple{a, b, c} \leadsto \tuple{a-1, b-1, c+2} = s_{k+1}.$$

Then:

\begin{align} (a-1)-(b-1) &= a-b &&= a-b \pmod 3, \\ (b-1)-(c+2) &= b-c-3 &&= b-c \pmod 3, \\ (c+2)-(a-1) &= c-a+3 &&= c-a \pmod 3, \end{align}

hence, $f(s_k) = f(s_{k+1})$.

Concluding, it is possible to get $a = b = 0$ only if $a_0-b_0 = 0 \pmod 3$. Moreover, there is a winning strategy if there are at least two different voters: WLOG we have $a \leq b$ (otherwise switch $a$ and $b$). First we obtain $c \geq b$ (trivial). Next we generate: $\tuple{0, b_0-a_0, c_0+2a_0}$, then $\tuple{2\frac{b_0-a_0}{3},2\frac{b_0-a_0}{3}, c_0+2a_0-1\frac{b_0-a_0}{3}}$, and finally $\tuple{0, 0, a_0+b_0+c_0}$.

I hope this helps ;-)

Answer 2

No. Let group a have 1 person, group b have 2, and group c have 3. It's not difficult to go through the different possible meetings of people from the groups to see that (1,2,3) (0,2,4) and (0,1,5) are the only possible distributions of votes (up to permutations of the groups.)

Answer 3

It can be done whenever b-a is a multiple of 3,when c-a is a multiple of 3 and a+b is greater than or equal to c, and when c-b is a multiple of 3 and a+b is greater than or equal to c. (3,6,7),(5,5,6),...(0,0,16). (5,6,9),(4,8,8),...(20,0,0) are examples.

Answer 4

If $2*$|a|+ |b| = |c| - |a|, then there is a solution by having everyone in a meet someone from c in an initial round and then pair up b and c in the second round so that then everyone is for a as there isn't anything mentioned for how many conversations could happen.

If one needs an example for the above, consider if $a=1, b=2, c=5$. Then after a and c meet, the values are $a=0,b=4,c=4$ and then each of b and c meet to produce the result of $a=8,b=0,c=0$.

The key is to get 2 of the 3 sets to have the same cardinality as then it is easy to move each of those people into the third set.

Answer 5

It may also worth noting that for there to be an initial configuration $(a,b,c)$ where no two of $a$, $b$, $c$ are congruent mod $3$ (and therefore, as others have shown, it is impossible to reach unanimity), it is necessary and sufficient that $a+b+c$ be divisible by $3$.