In today’s class we worked through the support enumeration algorithm. This involved some discussions about what the algorithm is based on but also a bunch of tedious linear equations.

A recording of the class is available here.

We applied the algorithm described here to the zero sum game defined by:

\[A = \begin{pmatrix} 0 & -1 & 1\\ 1 & 0 & -1\\ -1 & 1 & 0 \end{pmatrix}\]The basic idea behind the algorithm is as follows:

- Assume what actions are played by both players (this is called the supports of the strategies
- Identify the strategies that ensure that indeed all actions of the chosen supports will be played: this only happens if all actions in the support itself have the same expected utility.
- Check that there is no better outside of the supports chosen.

**Note** We did not actually get to part 3 today.

We considered the supports of size 1: there are no single pairs of actions that are pairs of best responses to each other.

We considered 2 pairs of supports of size 2:

- \(I = \{R, P\}\) (the row player only using Rock and Paper) and \(J=\{R, P\}\) (the column player also only using Rock and Paper). This lead to a contradiction (at step 2 of the above general idea) where there would be no probabilities that work.
- \(I = \{R, P\}\) (the row player only using Rock and Paper) and \(J=\{P, S\}\) (the column player only using Paper and Scissors). We did not finish this in class but it leads to a contradiction (at step 3 of the above general idea) the row player would benefit from playing Scissors. I will finish this in class on Monday.

That is 2 of the 9 possible pairs of supports of size 2.

Source code: @drvinceknight Powered by: Jekyll Github pages Bootsrap css