[jstanley]

> At the start, everyone has shaken zero hands, so you all start at an even number. Let’s say Anna and Byron take the initiative and shake each other’s hands. Now they are at one handshake each.

> Caitlin needs to join in, so she shakes Byron’s hand, giving her one handshake, an odd number. But now Byron has two handshakes, and so he’s back to an even number.

The first blunder of the game. Byron was already in a winning position and was under no obligation to shake a hand, but he has allowed Caitlin to capitalise on his stupidity.

[sva_]

It isn't meant to be a competitive game, but a cooperative one

[tarentel]

We interpreted this much differently. I think everyone has to shake hands with an odd number of people not just any single person. It's everyone wins or no one does.

[tromp]

How is playing a losing move in a lost position a blunder? All moves are losing after all...

[bena]

The goal of the game is to end with everyone having an odd number of handshakes.

I'll also add that his comment about winning/losing is a bit off as well.

This isn't a competitive game, it's a cooperative one. Everyone needs to shake an odd number of hands. The game is won, when everyone in the room has shaken an odd number of hands. It's more a puzzle than a game.

There may be additional rules on if disconnected sub graphs are allowed or not, so with an even number of people, the puzzle is trivial to solve.

[tromp]

Cooperative games still allow a distinction between winning and losing moves. A winning move is one where the resulting subgame still has a solution. A losing move is one where the resulting subgame has no solution.

In this game, looking only at the parities, all moves are reversible, so every move preserves the property of having a solution. For an even number of nodes, all moves in all positions are winning. And for an even number of nodes, all moves in all positions are losing.

[bena]

Yes, but before that person had an even number of handshakes, so shaking their hand gets them to odd. But at the expense of putting him at even. At worst, it's a neutral move.

And like I said, it really depends on the other rules which are never explicitly stated.

But I have a sneaking suspicion that discussing the game is arguing the metaphor.

The game is just there to help us get to the realization that graphs with an even number of vertices can have an odd number of edges, but a graph with an odd number of vertices can't have an odd number of edges.

[tromp]

> a graph with an odd number of vertices can't have an odd number of edges.

A 3 node graph can have 1 edge. The article is about a different notion of oddness, namely that all nodes have an odd degree.