What a Math Party Game Tells Us About Graph Theory | Hacker News Reader

What a Math Party Game Tells Us About Graph Theory

quantamagazine.org

48 pointsindogooner19 comments

Seems like a introduction to the 'Handshaking lemma'?

https://en.wikipedia.org/wiki/Handshaking_lemma

> 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.

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

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.

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.

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.

> 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.

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

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.

The solution, when dealing with an odd number of folks, is to do a 'three-way' handshake. Some times you have to think outside the box.

That's 2 per person, or an impossible shake, since a hand is half a pair.

Nah, its half a handshake with each person, so only one total handshake per person :)

Isn't the simplest solution that everybody shakes hand with Byron?

Everybody but Byron has one handshake and Byron has n-1, with n the number of people

Depending on your definition of simpler, everyone could just pair off and shake exactly one hand.

It's the simplest option when n is even (which it is in this case), but fails when n is odd.

The article about how this problem is trivial for N even but impossible for N odd.

The article is talking about the biggest odd subgraph within the even subgraph solution.

I think the problem with the suggestion is that the solution to the even subgraph may not be under your control? Not sure.

While the title is entertaining this doesn’t seem to be a very good candidate for a game at a party.

Title is backwards. graph theory tells us about the party game.

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