Projective Set - Revision history

Anton: Reverted edits by 83.26.212.107 (talk) to last revision by Anton

2020-02-07T03:53:10Z

Reverted edits by 83.26.212.107 (talk) to last revision by Anton

← Older revision		Revision as of 19:53, 6 February 2020
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	==Are there more games like this?==		==Are there more games like this?==
	[http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf This paper] of Davis and Maclagan claims on page 18 that Set and Projective Set are the only two "abstract SET games".		[http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf 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 [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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.		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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.

83.26.212.107: This is not true. you can find bigger non-sets in Z_3^4 is 20 not 16. Proof in https://en.wikipedia.org/wiki/Set_(card_game)

2019-10-16T12:03:34Z

This is not true. you can find bigger non-sets in Z_3^4 is 20 not 16. Proof in https://en.wikipedia.org/wiki/Set_(card_game)

← Older revision		Revision as of 04:03, 16 October 2019
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	==Are there more games like this?==		==Are there more games like this?==
	[http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf This paper] of Davis and Maclagan claims on page 18 that Set and Projective Set are the only two "abstract SET games".		[http://www.warwick.ac.uk/staff/D.Maclagan/papers/set.pdf 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 [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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.		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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.

Anton at 21:46, 16 July 2012

2012-07-16T21:46:12Z

← Older revision		Revision as of 13:46, 16 July 2012
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	[[Category:Blog]]		[[Category:Blog]]
			[[Category:Projective_Set]]

Anton: /* Are there always cards left over at the end? */

2011-11-30T16:48:30Z

Are there always cards left over at the end?

← Older revision		Revision as of 08:48, 30 November 2011
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	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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.		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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.

	First note that $\mathbb P^1_{\mathbb F_2}$ can be covered by disjoint lines since it is a single line. Now we induct on dimension. Suppose $\{(a_{i1},a_{i2},a_{i3})\}_i$ is a collection of triples of points in $\mathbb P^{2n-1}_{\mathbb F_2}$ so that each point appears in exactly one triple, and each triple forms a line. Then we have the following decomposition of $\mathbb P^{2n+1}_{\mathbb F_2}$ (in standard projective notation):		First note that $\mathbb P^1_{\mathbb F_2}$ can be covered by disjoint lines since it is a single line. Now we induct on dimension. Suppose $\{(a_{i,1},a_{i,2},a_{i,3})\}_i$ is a collection of triples of points in $\mathbb P^{2n-1}_{\mathbb F_2}$ so that each point appears in exactly one triple, and each triple forms a line. Then we have the following decomposition of $\mathbb P^{2n+1}_{\mathbb F_2}$ into lines (in standard projective notation):
	: $([0:\cdots:0:0:1], [0:\cdots:0:1:0], [0:\cdots:0:1:1])$		: $([0:\cdots:0:0:1], [0:\cdots:0:1:0], [0:\cdots:0:1:1])$
	: $([a_{i1}:0:0], [a_{i2}:0:0], [a_{i3}:0:0])$ for each $i$		: $([a_{i,1}:0:0], [a_{i,2}:0:0], [a_{i,3}:0:0])$ for each $i$
	: $([a_{i1}:0:1], [a_{i2}:1:0], [a_{i3}:1:1])$ for each $i$		: $([a_{i,1}:0:1], [a_{i,2}:1:0], [a_{i,3}:1:1])$ for each $i$

	=I don't like your cards=		=I don't like your cards=

Anton: /* Are there always cards left over at the end? */

2011-11-30T15:00:18Z

Are there always cards left over at the end?

← Older revision		Revision as of 07:00, 30 November 2011
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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 Thorngren came up with a nice proof.		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 Bejleri for a bit. Dori's friend Ryan Thorngren came up with a nice proof.

	First note that $\mathbb P^1_{\mathbb F_2}$ can be covered by disjoint lines since it is a single line. Now we induct on dimension. Suppose $\{(a_{i1},a_{i2},a_{i3})\}_i$ is a collection of triples of points in $\mathbb P^{2n-1}_{\mathbb F_2}$ so that each point appears in exactly one triple, and each triple forms a line. Then we have the following decomposition of $\mathbb P^{2n+1}_{\mathbb F_2}$ (in standard projective notation):		First note that $\mathbb P^1_{\mathbb F_2}$ can be covered by disjoint lines since it is a single line. Now we induct on dimension. Suppose $\{(a_{i1},a_{i2},a_{i3})\}_i$ is a collection of triples of points in $\mathbb P^{2n-1}_{\mathbb F_2}$ so that each point appears in exactly one triple, and each triple forms a line. Then we have the following decomposition of $\mathbb P^{2n+1}_{\mathbb F_2}$ (in standard projective notation):

Anton: Undo revision 760 by 213.220.224.254 (talk)

2011-11-22T18:57:08Z

Undo revision 760 by 213.220.224.254 (talk)

Show changes

Anton: Reverted edits by 219.106.157.196 (talk) to last revision by 213.220.224.254

2011-11-22T18:56:36Z

Reverted edits by 219.106.157.196 (talk) to last revision by 213.220.224.254

← Older revision		Revision as of 10:56, 22 November 2011
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	~~A piece of erudition unilke any other!~~		Ppl like you get all the brinas. I just get to say thanks for he answer.

219.106.157.196: THzoYzrpBPRggVS

2011-11-22T15:29:49Z

THzoYzrpBPRggVS

← Older revision		Revision as of 07:29, 22 November 2011
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	~~Ppl like you get all the brinas. I just get to say thanks for he answer.~~		A piece of erudition unilke any other!

213.220.224.254: kFXEyLZqJcPKfIN

2011-11-22T12:23:24Z

kFXEyLZqJcPKfIN

Show changes

Anton: /* Play */

2011-11-13T08:59:27Z

Play

← Older revision		Revision as of 00:59, 13 November 2011
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	=Play=		=Play=
	Play is essentially the same as for the game Set~~, but using 15 cards instead of 12~~.		Play is essentially the same as for the game Set.

	The dealer shuffles the cards and lays 15 cards (in a rectangle) 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.		The dealer shuffles the cards and lays 12 cards (in a rectangle) 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 18) are laid face up. These cards are not replaced when the next "set" is picked up, reducing the number to 15 again.		If all players agree that there is no "set" in the cards showing, three more cards (making a total of 15) are laid face up. These cards are not replaced when the next "set" is picked up, reducing the number to 12 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 play continues until the deck is depleted. At the end of the play there may be some cards which do not include a "set".