jack: (Default)
[personal profile] jack
I can't remember the name (edit: thank you Woodpijn) but at alextfish and woodpijn's we played a card game where the point was to select from a deck of cards with pictures on, the ones someone else would choose, corresponding to a certain word, and you score according to how identical the two selections are.

Specifically, there's a deck of cards with words (although you could choose random words in another way) and one is turned up. Each player has an identical deck and chooses 5 cards from it (in fact, 5 varies in a proprietary way as your avatars move along a scoring track).

It's quite fun guessing surprising interpretations or amusing choices of picture -- it's surprising how often people _do_ agree on them -- or any special knowledge of the person matching against.

The rules as given describe either being in a team and trying to match each other more than the opposing team members match each other, or a free-style each taking it in turns to be a designated "judge" each other player must match.

The rules give (iirc) a traditional mastermind-style matching score. Give five (or however many) cards in order and score 3pts for same-card-same-place as the matchee, and 1 pt for same-card-different-place. That works reasonably well. But the question is, what ought the rules to be?

* At first, I thought the idea of a rotating judge was odd, since it seemed just an arbitrary contrivance of the rule system. It still offends my aesthetics that you can't just match against all the other players. But OTOH, if you were omniscient and knew what everyone were going to play, you ought to get a perfect score, and you couldn't if two people were going to play different things and you couldn't match both of them. Is there are compromise?

* The three/one point match is acceptable, but I don't think it quite describes matching. Suppose the word were "black" and I played pictures crow, sloe, fishing boat, night, black dog, and someone else played another card in first place, followed by my four first place cards -- that feels like they matched me except for one datum, but the old scoring system would only give four points. Is there a better algorithm, maybe something that (in a simple way) includes a information theory metric of distance?

Date: 2008-10-20 04:22 pm (UTC)
simont: A picture of me in 2016 (Default)
From: [personal profile] simont
Oh, rats. I've just realised this system gives the same score to ABCDE/ACBDE as it does to ABCDE/EABCD. There's always something. Back to the drawing board, I fear; I have a nasty feeling that deals a death blow to the entire LCS strategy.

Perhaps I'll stop replying to myself now, and wait to see if anyone else is interested in this subthread at all before I try again :-)

Date: 2008-10-20 04:36 pm (UTC)
From: [identity profile] cartesiandaemon.livejournal.com
I tried to start by drawing which inequalities I thought there should be, but even then wasn't sure: ABCDE>ABCED>ABCDX?ABCXD... That sort of depends whether you think right place has an importance less than order, or none. I think it must be none, if you accept the XABCD example to show that decisions can be binary "is this of $foo importance, or not?"

My best guess was some metric based on "fewest moves of some sorts", including some sort of simple moves and replacements, but I don't know what would be appropriate if anything.

Date: 2008-10-20 04:47 pm (UTC)
simont: A picture of me in 2016 (Default)
From: [personal profile] simont
Yes. Hmm. Well, you could certainly do a simple enough one like that based on my original idea: if you score x for a deletion or insertion, and y for the interchange of adjacent elements, then it seems clear that the minimum edit distance between two lists is equal to 2x*(number of completely different elements) + y*(number of misordered pairs within the common subset of elements). (Based on the strategy "remove all elements that need to be absent, reorder the common subset, insert all new elements". Any other approach risks using additional interchanges to move things past elements that could instead have been removed beforehand or added afterwards.)

I'd be inclined to arrange that y is proportional to x/(n-1) as n varies, so that no matter what the length of the list you'd score the same fraction of full marks for getting exactly the right list in exactly the wrong order. Then it's just a matter of adjusting the remaining constant factor between x and y, and scaling so that full marks is some sort of nice round number (either a constant or n, as you please).

eta: oh no, I'm wrong again! In fact, my formula above is not quite right, because of the possibility that a single massively misordered element might be more cheaply dealt with as a deletion and insertion than as lots of interchanges to move it to the right place. This only comes into play if the constant factor is such that 2x can ever be less than (n-1)y, but if that's the case then working out the minimum edit distance might get fairly hairy. Perhaps it would be better to treat this as a sanity constraint on the constant factor :-)

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