Majority and Minority games
Let G be a graph with all degrees odd. Each vertex is initially
randomly assigned a colour (black or white), and at each time
step every vertex updates to the majority colour among its
neighbours.
This always reaches a stable state or a 2-cycle (Goles and Olivos;
Poljak and Surâ) and the number of steps required to do so is
less than the number of edges (Poljak and Turzík).
Suppose each vertex is aiming to be the same colour as most of
its neighbours. Say a vertex is happy at time t if this is the case.
The process must reach a fixed point or 2-cycle, so for t
sufficiently large each vertex is always happy or always unhappy.
What proportion of vertices are ultimately happy on average?
Call this h(G).
If G is bipartite, h(G)=½. This is the worst case; in fact for any G
every vertex has probability at least ½ of being happy.
We have necessary and sufficient conditions for equality.
If G is 3-regular and contains a 3- or 5-cycle then h(G)>½.
This is not true for longer odd cycles...
... or for higher regularity.
What if some vertices use the opposite (minority) rule?
If all vertices do this then the process will evolve in exactly the
same way, except the colours are swapped when t is odd.
In particular, again we always reach a fixed point or 2-cycle.
If G is bipartite with one part playing the majority rule and the
other part playing the minority rule, we always reach a 4-cycle.
Suppose that there is some set of special vertices X with no two
vertices at distance less than 4, such that no triangle of G meets
X. If every vertex not in X plays the majority rule and the vertices
in X do anything that depends only on the colours of their
neighbours, then we must eventually reach a 1-, 2- or 4-cycle.
The same result applies if G is bipartite, and every vertex not in X
plays the majority rule if it is in the first part and the minority
rule if it is in the second. (We have the same conditions on X, but
now the triangle condition is automatically satisfied.)
In particular, if all of the vertices play the majority rule apart
from one vertex, v, which plays the minority rule, and v is not in
a triangle, then we always reach a 1-, 2- or 4-cycle.
The requirement that no triangle contains v is necessary.
What if vertices start off playing the majority rule but swap if
that isn’t working well?
One possible scheme is for every vertex which is unhappy at
time t to swap rule. If the swap happens before the rule is
applied to get the colour at t+1 we run through the same states
but more slowly. So we should swap afterwards.
This gives us a mapping from the colours and rules at time t to
colours and rules at time t+1. We can show this is a bijection, so
we will eventually return to the starting state. This makes it easy
to compute the long-run average happiness for a given graph.
The time taken can be surprisingly long (up to 2916 for one 10vertex cubic graph).
This seems to generally do worse than the simple majority rule,
but better than 50% for non-bipartite graphs. It does sometimes
do better than simple majority. If G is bipartite we will still get
exactly 50%; does it always do better if not? Can we prove any
good upper bound?
An alternative is to do the same thing with the restriction that no
vertex swaps rule twice in a row. This seems to do much better,
even for bipartite graphs – we can show that for Kr,r there will be
3r/2 – O(√r) happy vertices on average. Again, can we prove that
it always does better?
We also consider what happens with random errors. There are
two obvious settings: vertices may make observation errors or
play errors independently with some probability p<½.
These both fit into a general scheme where a vertex v with k
black neighbours plays black with probability π(v,k), and this is
increasing in k for each v. White h(t) for the expected proportion
of happy vertices at time t. We can show that in such a scheme
h(t)≥½ for every t.
These probabilities give a Markov chain on the possible states of
the graph, and h(t) tends to a limit given by the expected
happiness of the limit distribution, h∞(p).
Results for the Wagner graph generated using PRISM.
Perhaps surprisingly, h(t) does not have to be increasing.
For play errors it is not true that h∞(p)→1 as p→0. The graph
shown is a counterexample.
The majority game on this graph has six essentially different
configurations which arise as fixed points or 2-cycles of the
majority game.
2p2
2p
2p
2p
6p2
7p2
3p2
8p2
6p2
6p
4p
For the same graph with observation errors, h∞(p)→1 as p→0. Is
this always the case? Is it always the case for cubic graphs?
Balister, Bollobás, Johnson and Walters studied “Random
Majority Percolation”, play errors on the (4-regular) discrete
torus. Using the current colour of a vertex to break ties gives
strong links to Bootstrap Percolation.
Bootstrap Percolation on a lattice: each vertex is on or off.
Vertices switch on if enough of their neighbours are already on,
never switch off.
They showed that for very small p the torus spends almost all
the time in a state with all vertices the same colour.
What if we limit the number of observation errors a vertex
makes at any single time step?
If a vertex of degree 2r+1 may make r errors then there is a
positive probability of making either play unless its neighbours
are unanimous.
If G is not bipartite, that condition guarantees that it will
eventually reach a state with all colours the same, then stay
there.
For cubic G there is just one parameter: the probability of
playing incorrectly if the neighbours are split 2:1, p.
This can take a long time to reach a monochromatic state if p is
close to 0 – it is of order p–k for some k, but k is unbounded.
If vertices of degree 2r+1 may only make r–1 errors then we may
get stuck in a 2-cycle where some vertices are unhappy.
If we only make errors when the neighbours are split 3:2, we
seem to be better off with more errors.