Games on Cellular Spaces: An Evolutionary Approach

P. R. Andrade, A. M. V. Monteiro, G. Camara

During the first four billion years or life on Earth, the primary means of information transfer was genetic. (M. Nowak)

S. E. Riechert

Games spiders play: behavioural variability in territorial disputes. Behavioural Ecology and Sociobiology 3:135-162. 1979.
Games spiders play II: Resource assessment strategies. Behavioural Ecology and Sociobiology 6:121-128. 1979.
The consequences of being territorial: spiders, a case of study. American Naturalist 117:871-892. 1981.
Spider interaction and strategies: communication vs. coercion. 281-315. In Spider communication, Mechanisms and ecological significance. Princeton University Press. 1982.

Adaptive dynamics in space

Lion, Sebastien

Over the last twenty years, the role of spatial self-structuring as a template for evolution has drawn much attention. Spatial structure can be an important component of the eco-evolutionary feedback loop: the evolution of a trait (e.g. migration) can shape the local structure of the population, which in turn creates new selective pressures on the evolving trait. As a consequence, the evolution of spatially structured populations often displays very different features from the evolution of well-mixed populations.

ESS for Chicken and Nash Equilibrium

I is stable iff:

E(I,I) > E(J,I)

or

E(I,I) = E(J,I) and E(I,J) > E(J,J)

	S0	S1
S0	0	-1
S1	1	-10

S0 is not ESS against S1:

E(I,I) > E(J,I)
E(S0,S0) > E(S1,S0)
0 > 1 (FALSE)

E(I,I) = E(J,I) and E(I,J) > E(J,J)
E(S0,S0) = E(S1,S0) and E(S0,S1) > E(S1,S1)
0 = 1 (FALSE)

S1 is not ESS against S0:

E(I,I) > E(J,I)
E(S1,S1) > E(S0,S1)
-10 > -1 (FALSE)

E(I,I) = E(J,I) and E(I,J) > E(J,J)
E(S1,S1) = E(S0,S1) and E(S1,S0) > E(S0,S0)
-10 = -1 (FALSE)

Adding Neq (S0.1) to the game, we have:

	S0	S0.1	S1
S0	0	-0.1	-1
S0.1	0.1	-0.1	-1.9
S1	1	-0.1	-10

S0.1 is ESS against S0:

E(I,I) > E(J,I)
E(S0.1,S0.1) > E(S0,S0.1)
-0.1 > -0.1 (FALSE)

E(I,I) = E(J,I) and E(I,J) > E(J,J)
E(S0.1,S0.1) = E(S0,S0.1) and E(S0.1,S0) > E(S0,S0)
-0.1 = -0.1 and +0.1 > 0 (TRUE)

S0.1 is ESS against S1:

E(I,I) > E(J,I)
E(S0.1,S0.1) > E(S1,S0.1)
-0.1 > -0.1 (FALSE)

E(I,I) = E(J,I) and E(I,J) > E(J,J)
E(S0.1,S0.1) = E(S1,S0.1) and E(S0.1,S1) > E(S1,S1)
-0.1 = -0.1 and -1.9 > -10 (TRUE)