From jrr@atml.co.uk Fri May 30 00:08:31 CDT 1997 Article: 123443 of sci.math Path: gannett.math.niu.edu!corn.cso.niu.edu!vixen.cso.uiuc.edu!howland.erols.net!news.maxwell.syr.edu!eerie.fr!cnusc.fr!ciril.fr!univ-angers.fr!jussieu.fr!oleane!weld.news.pipex.net!pipex!warm.news.pipex.net!pipex!tank.news.pipex.net!pipex!news-relay.atml.co.uk!not-for-mail From: jrr@atml.co.uk (John Rickard) Newsgroups: sci.math Subject: Re: Set (game, not collection) Date: 29 May 1997 19:18:58 +0100 (BST) Organization: ATM Ltd Lines: 42 Message-ID: References: <338D3559.7983@mindless.com> NNTP-Posting-Host: puffball.atml.co.uk X-Server-Date: 29 May 1997 18:19:00 GMT X-Newsreader: TIN [version 1.2 PL2] Xref: gannett.math.niu.edu sci.math:123443 S (ika@mindless.com) wrote: : So, the game Set has become very popular around my campus. If you don't : know how to play, it is very simple. [81 different cards, each with four attributes each of which has three values: Number: 1, 2, 3 Shape: Diamond, Oval, Squiggle Colour: Red, Green, Purple Filling: Filled, Shaded, Empty A set consists of three cards such that for each of the four attributes either all three cards have the same value for that attribute or they have three different values for that attribute (so not two the same and one different).] One can have 20 cards with no sets: 1 2 3 DOS DOS DOS F *.* ... .*. R S ... .*. *.* E *.* ... .*. F ... ... ... G S .*. ... .*. E ... ... ... F .*. ... *.* P S *.* .*. ... E .*. ... *.* I have a proof that every collection of 21 cards contains a set, but it's a bit messy; I'll post it later if nobody comes up with a pointer to a proof that's already been published. It starts with the observation that if one considers the set of 81 cards as a 4-dimensional affine space over the 3-element field then a set is simply a line. -- John Rickard From zare@cco.caltech.edu Mon Jun 2 11:29:28 CDT 1997 Article: 123792 of sci.math Path: gannett.math.niu.edu!corn.cso.niu.edu!vixen.cso.uiuc.edu!howland.erols.net!newsfeed.internetmci.com!news.ridgecrest.ca.us!nntp-server.caltech.edu!zare From: zare@cco.caltech.edu (Douglas J. Zare) Newsgroups: sci.math Subject: Re: Set (game, not collection) Date: 1 Jun 1997 02:33:34 GMT Organization: California Institute of Technology, Pasadena Lines: 51 Message-ID: <5mqn1u$cjp@gap.cco.caltech.edu> References: <338D3559.7983@mindless.com> NNTP-Posting-Host: morocco.cco.caltech.edu Xref: gannett.math.niu.edu sci.math:123792 In article , John Rickard wrote: >[...] >One can have 20 cards with no sets: > > 1 2 3 > DOS DOS DOS > F *.* ... .*. > R S ... .*. *.* > E *.* ... .*. > > F ... ... ... > G S .*. ... .*. > E ... ... ... > > F .*. ... *.* > P S *.* .*. ... > E .*. ... *.* > >I have a proof that every collection of 21 cards contains a set, but >it's a bit messy; I'll post it later if nobody comes up with a pointer >to a proof that's already been published. It starts with the >observation that if one considers the set of 81 cards as a >4-dimensional affine space over the 3-element field then a set is >simply a line. The instructions suggest that children under 6 might want to recognize lines in affine 3-space instead. I do not know of a written, published proof that 21 points must contain a line, though perhaps one was mentioned in the discussion of Set on sci.math.research . I believe Ken Smith "published" a proof in a talk in 1994 after we worked one out, but we found earlier claims. I gave this proof to Tom Magliery ( http://sdg.ncsa.uiuc.edu/~mag/Set/ ), but it doesn't seem to be there. I forget whether we proved that the above is unique up to isomorphism; does that follow from your argument? If you have only a few cases, please add it too the rec.puzzles archives, since the current entry is only empirical. The 3-dimensional case is particularly interesting. The configuration of 9 points containing no line is essentially unique, and can be specified by the solutions of a quadratic. The group of symmetries is 2-transitive on the set. There is a unique point in the complement of the 4-dimensional example which is the intersection of 10 lines containing two points each, and projectivising with respect to this point gives a quadratic in projective 3-space. An equivalent problem is as follows: How many points must be chosen from a d-dimensional lattice so that 3 of them must have a lattice point as their center of mass? An easy variation of this appeared on the Putnam. Douglas Zare From stadler@math.ohio-state.edu Fri May 30 00:08:20 CDT 1997 Article: 123430 of sci.math Path: gannett.math.niu.edu!corn.cso.niu.edu!vixen.cso.uiuc.edu!feeder.chicago.cic.net!chi-news.cic.net!math.ohio-state.edu!not-for-mail From: stadler@math.ohio-state.edu (Jonathan Stadler) Newsgroups: sci.math Subject: Re: Set (game, not collection) Date: 29 May 1997 17:18:37 GMT Organization: Department of Mathematics, The Ohio State University Lines: 59 Message-ID: <5mkdpd$fj0$1@mathserv.mps.ohio-state.edu> References: <338D3559.7983@mindless.com> NNTP-Posting-Host: math.mps.ohio-state.edu X-Newsreader: trn 4.0-test58 (13 May 97) Xref: gannett.math.niu.edu sci.math:123430 Status: R In article <338D3559.7983@mindless.com>, S wrote: >So, the game Set has become very popular around my campus. If you don't >know how to play, it is very simple. > >There are 81 cards. Each card is different. The cards have pictures of >1, 2, or 3 shapes on them. There are three shapes, diamon, oval, and >squigle. Also, they are three different colors, red green or purple and >can be filled, shaded or empty. A set consists of three cards for which >each attribute is either the same for all cards or different. For >example 1 red filled diamond, 2 filled green diamonds, and 3 filled >purple diamonds is a set. so is 1 empty grenn squigle, 2 shaded red >ovals, and 3 filled purple diamonds. The basic rule is, if two cards >match, but the third does not, for any given characteristic, the set is >not valid. > >It is normal to lay out 12 cards at a time from which to draw sets. It >is possible that there can be no sets in 12 cards. Then, you lay down 3 >more, so you have 15 out. Very rarely, this also fails to yeild a set. >Then you lay down 3 more (so there are 18 out) I have never seen this >fail. Could it? My intuition says that the minimum number of cards to >guarantee a set is somewhere between 15 and 20. Can anyone solve this >problem? No big rush or anything, I'm just curious. > >sara l mastros Sara, I, too, have been thinking about this recently. So, I tried to deal with the problem with one fewer dimension (by using only purple cards) and can give a collection of 9 (purple) cards with no set. This means that there is a collection of 18 with all 4 dimensions yielding no set. The way to get 18 is to take the green cards identical to the purple. Here is a collection of 18 cards with no set: 1 P D E 1 G D E 1 P S E 1 G S E 1 P O F 1 G O F 2 P O E 2 G O E 2 P D E 2 G D E 2 P S F 2 G S F 3 P S F 3 G S F 3 P O F 3 G O F 3 P D S 3 P D S Check out the web site "www.setgame.com". Follow the link to "Other pages devoted to SET" and on the next page, you should be able to find a link to the rec.puzzles.games FAQ, which gives a listing of 20 cards with no set. And if that's not enough for you.... On the SET web site, they give alternate rules. In the description of Set-Up, they indicate that it is impossible to have more than 21 cards without a set. I have e-mailed them to find out how they proved this, but have had no response. Good luck with your problem. Jon Stadler stadler@math.ohio-state.edu