Mike Orrison, of Harvey Mudd, visited the REU on Wednesday and talked about the generalized Condorcet criterion and the underlying linear algebra used to prove a surprising result.

The Condorcet criterion is said to hold for a voting system if it guarantees that, if a candidate would win each head-to-head election, then that candidate would win the election.

In our usual voting system, we only ask voters to vote for their first choices. But you could imagine having voters list their preferences. For instance, if I had to rank my cats, I’d rank them M > T > S (Moo’s my favorite, then Tipper, and lastly, there’s Sophie).

Now, I could ask my friends to rank my cats, and maybe I’d come up with:

- M > T > S – ten ballots
- T > M > S – four ballots
- T > M > S – five ballots
- S > T > M – two ballots

We can take away a few things from these rankings:

- No one likes Sophie (poor Sophie!).
- Moo wins the election!
- I have a lot of friends (okay, okay: it’s a fictional example).
- Tipper beats Moo (11 to 10); Tipper beats Sophie (19 to 2); Moo also beats Sophie (19 to 2).

Item 4 means that Tipper is the Condorcet winner. However, if we use our usual “most first place votes” election, we’d see that Moo is the winner.*

If you have more than three candidates, the results get weirder. You can have someone who wins every head-to-head competition (call the 2-winner Bob), but someone else could win in every three-way battle (call the 3-winner Alice).

The surprising and cool result is that once Bob’s 2-winningness (my term) is overshadowed by Alice’s 3-winningness, Bob can’t go on to be a 4-winner. He can’t be any n-winner for n > 3. And the proof, my friends, rests in linear algebra. You can read all about the work Mike did with his REU students in their paper: Generalized Condorcet winners.

*In reality, Moo would win by any metric because she’s fantastic.