Projective Set: Difference between revisions

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{{todo|update page to reflect 63 card deck}}
[http://en.wikipedia.org/wiki/Set_(game) Set] (see the [http://www.setgame.com/set/rules_set.htm rules]) is the game of finding lines in $\mathbb F_3^4$. [[Projective Set]] is the game of finding 2-dimensional subspaces in $\mathbb F_2^5$, or if you prefer, the game of finding lines in $\mathbb P^4_{\mathbb F_2}$.
[http://en.wikipedia.org/wiki/Set_(game) Set] (see the [http://www.setgame.com/set/rules_set.htm rules]) is the game of finding lines in $\mathbb F_3^4$. [[Projective Set]] is the game of finding 2-dimensional subspaces in $\mathbb F_2^5$, or if you prefer, the game of finding lines in $\mathbb P^4_{\mathbb F_2}$.


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Also answered on page 18 of the [http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf Davis and Maclagan paper]. The maximum number of projective set cards without a set is $2^d$, where you're using a deck of $2^{d+1}-1$. (You can easily realize this by picking a property (e.g. squiggle) and taking all the cards which have that property.)
Also answered on page 18 of the [http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf Davis and Maclagan paper]. The maximum number of projective set cards without a set is $2^d$, where you're using a deck of $2^{d+1}-1$. (You can easily realize this by picking a property (e.g. squiggle) and taking all the cards which have that property.)
==Are there always cards left over at the end?==
==Are there always cards left over at the end?==
No. If you play with a deck of size $2^{d+1}-1$ where $d$ is odd, it is possible to break the deck up into disjoint sets. I thought about this problem with Dori for a bit. Dori's friend Ryan came up with a nice proof. [Anton:insert proof here]
No. If you play with a deck of size $2^{d+1}-1$ where $d$ is odd, it is possible to break the deck up into disjoint sets. I thought about this problem with Dori for a bit. Dori's friend Ryan came up with a nice proof. {{todo|insert proof here}}


=I don't like your cards=
=I don't like your cards=
I generated them with LaTeX as follows. Tweak to taste.
I generated them with LaTeX as follows. Tweak to taste.
<pre>
[Anton: insert code here]
</pre>


{{todo|insert code here}}


[[Category:Blog]]
[[Category:Blog]]

Revision as of 07:46, 4 November 2011

[Todo: update page to reflect 63 card deck]

Set (see the rules) is the game of finding lines in $\mathbb F_3^4$. Projective Set is the game of finding 2-dimensional subspaces in $\mathbb F_2^5$, or if you prefer, the game of finding lines in $\mathbb P^4_{\mathbb F_2}$.

Set is fun because any line in $\mathbb F_3^4$ contains three points, so any two cards (i.e. points in $\mathbb F_3^4$) uniquely determine a third. I realized that since the same is true in $\mathbb P^4_{\mathbb F_2}$, projective set would also be a fun game.

The Cards

There are five possible features a card can have: a squiggle, circle, vertical, horizontal, and smile:











Any given card must have at least one of these (i.e. must correspond to a nonzero element of $\mathbb F_2^5$), but can otherwise have any combination of them. There are therefore 31 cards in a complete deck of projective set.

To get your own deck of projective set, print out this pdf on some cardstock and cut out the cards. Or use the bigger deck of 63 cards. I recommend cutting off the corners as shown to make it easy to orient all the cards properly.

The Rules

Three cards form a set if two of them XOR to the third. Another way to put it is that three cards form a set if for any given property, either none of the cards have it, or exactly two of them have it. Here are some examples of sets. Note that if three cards form a set, any two of them XOR to the other. Note also that given any two cards, there is a unique third card that will form a set with those two.












Here are some examples of non-sets.






Play

Play is essentially the same as for the game Set, but using 9 cards instead of 12. Also, since the number of cards is not divisible by 3, the final time cards are added, only one card is added instead of 3. (Maybe this is a good argument that projective set should be played with six properties instead of five, so there would be 63 cards. For that, you can use this pdf.)

The dealer shuffles the cards and lays 9 cards (in a square) face up on the table so that they can be seen by all players. The players remove a "set" of three cards as they are seen. Each "set" is checked by the other players. If correct, the "set" is kept by the player and the dealer replaces the three cards with cards from the deck. Players do not take turns but pick up "sets" as soon as they see them. A player must call "set" before picking up the cards. After a player has called "set", no other player can pick up cards until the first player is finished. If a player calls "set" and does not have one, the player loses one point. The three cards are returned to the table.

If all players agree that there is no "set" in the cards showing, three more cards (making a total of 12) are laid face up. These cards are not replaced when the next "set" is picked up, reducing the number to 9 again.

The play continues until the deck is depleted. At the end of the play there may be some cards which do not include a "set".

The number of "sets" held by each player are then counted, one point is given for each and added to their score. The deal then passes to the person on the dealer's left and the play resumes with the deck being reshuffled.

When all players have dealt, the game ends; the highest score wins.

Questions

Are there more games like this?

This paper of Davis and Maclagan claims on page 18 that Set and Projective Set are the only two "abstract SET games".

How many cards can there be without a set?

Also answered on page 18 of the Davis and Maclagan paper. The maximum number of projective set cards without a set is $2^d$, where you're using a deck of $2^{d+1}-1$. (You can easily realize this by picking a property (e.g. squiggle) and taking all the cards which have that property.)

Are there always cards left over at the end?

No. If you play with a deck of size $2^{d+1}-1$ where $d$ is odd, it is possible to break the deck up into disjoint sets. I thought about this problem with Dori for a bit. Dori's friend Ryan came up with a nice proof. [Todo: insert proof here]

I don't like your cards

I generated them with LaTeX as follows. Tweak to taste.

[Todo: insert code here]