Unless you're suggesting this: https://code.jsoftware.com/wiki/NYCJUG/2012-11-13#Simple_.E2.80.9CSudoku_N.E2.80.9D_Generation ?
On Thu, Oct 14, 2021 at 10:21 AM Devon McCormick <devon...@gmail.com> wrote: > Easier said than done. > > On Thu, Oct 14, 2021 at 6:56 AM Hauke Rehr <hauke.r...@uni-jena.de> wrote: > >> No, generation is easier. >> Take a random permutation along both axes of >> |."0 1~ >:i.9 >> and then erase numbers as long as they’re derivable >> by what remains. >> >> Am 14.10.21 um 04:19 schrieb Devon McCormick: >> > I just put up this page from a past NYCJUG meeting where we talked about >> > sudoku - >> https://code.jsoftware.com/wiki/NYCJUG/2012-11-13#Show-and-Tell - >> > and it only took me ten years! >> > Perhaps a harder puzzle is to generate sudoku. >> > >> > On Wed, Oct 13, 2021 at 7:59 PM Hauke Rehr <hauke.r...@uni-jena.de> >> wrote: >> > >> >> but since N2, N1 and N0 are boxed, >> >> you need (N2;N1;<N0) rather than (N2;N1;N0) >> >> or, even simpler since they’re already atomic >> >> boxes (only one item, empty shape): >> >> >> >> > is&.>/ (N2,N1,N0) (<@mn@:{)"0 _ g >> >> >> >> you may even get rid of the "0 _ >> >> (but read the vocabulary page for { carfully) >> >> >> >> >> >> Am 14.10.21 um 01:37 schrieb Hauke Rehr: >> >>> if I now understand your first question correctly, >> >>> you want to learn how to shorten a phrase with >> >>> much repetitive structure >> >>> >> >>> first of all, you can define 'is' this way >> >>> is =: e. # [ >> >>> >> >>> >> >>> your expression looks like >> >>> (v N2{g) is (v N1{g) is (v N0{g) >> >>> with v equal to mn etc. >> >>> >> >>> if you want to factor out g, you can say >> >>> ((N2 v@:{ ]) is (N1 v@:{ ]) is (N0 v@:{ ])) g >> >>> >> >>> … and if you define >> >>> MN =: mn@:{ >> >>> you get >> >>> ((N2 MN ]) is (N1 MN ]) is (N0 MN ])) g >> >>> >> >>> if what I wrote for a1, a2 and a3 is correct, >> >>> you may then do >> >>> >> >>> > is&.>/ ((N2 MN ]);(N1 MN ]);(N0 MN ])) g >> >>> >> >>> which may be further simplified to >> >>> >> >>> > is&.>/ (N2;N1;N0) (MN&.>)"0 _ g >> >>> >> >>> and we may substitute back >> >>> >> >>> > is&.>/ (N2;N1;N0) (mn@:{&.>)"0 _ g >> >>> >> >>> I hope this works; and maybe someone wants to >> >>> comment on different ways to reduce repetitive >> >>> expressions. >> >>> >> >>> >> >>> But if your main concern was solving the task >> >>> of writing a sudoku solver, take a look at the >> >>> wiki page Ric pointed to. >> >>> >> >>> >> >>> Am 14.10.21 um 01:04 schrieb Hauke Rehr: >> >>>> Concerning the first question: >> >>>> >> >>>> try adding ] where g used to be, >> >>>> or inserting an & before the { >> >>>> (but I didn’t study your code >> >>>> enough to /know/ this will work) >> >>>> >> >>>> Am 14.10.21 um 00:52 schrieb 'Viktor Grigorov' via Programming: >> >>>>> Hello, >> >>>>> >> >>>>> Recently I saw an article in lobste.rs >> >>>>> (https://www.hillelwayne.com/post/sudoku/) about sudoku solving, >> and >> >>>>> though t, "it'd be nice to try it J". (Didn't even bother reading >> it, >> >>>>> but later glancing at it found the author had used J in the end. :D) >> >>>>> >> >>>>> I reshape a list of integers of length 81 with blanks being 0s, >> >>>>> row-wise, from top, going left-to-right; into a 4-cube of length 3; >> >>>>> and define 3 auxiliary verbs. >> >>>>> >> >>>>> g=:3 3 3 3 $ ... >> >>>>> q=:i.3NB. has missing >> >>>>> hm=:0&=@(<./)@, >> >>>>> NB. intersection >> >>>>> is=:([e.])#[ >> >>>>> NB. missing numbers >> >>>>> mn=:13 :'((-.((>:@i.9)e.,y))#(>:@i.9))' >> >>>>> >> >>>>> Although one usually solves sudoku (I think) by thinking of >> >>>>> exclusions, as the article's first paragraph or so suggested. I >> >>>>> wanted to find symmetries or something similar, thinking of it as a >> >>>>> higher dimmensional thing, hence the 4-cube. The constraints, or >> >>>>> symmtetries, or whatever they may be are the 9 rows, columns, and >> >>>>> faces each containing 1--9 once. My idea is to try each of the 81 >> >>>>> cells until once with only one overlap is found, break, then repeat >> >>>>> until no change. >> >>>>> >> >>>>> The sudoku given as an example in the english wikipedia article for >> >>>>> sudoku has an 'easy' example, wherein the center of the center >> >>>>> resolves to 5 using the intersection of the missing numbers of the >> 5. >> >>>>> row, 5. column, and 5. face; or within the tesseract: >> >>>>> >> >>>>> (mn(<1;q;1;q){g)is(mn(<1;1;q;q){g)is(mn(<q;q;1;1){g) >> >>>>> ((mn(<1;q;1;q)&{)is(...)is(...))g NB. nope >> >>>>> >> >>>>> Taking out g and binding the from doesn't work, giving me just 1--9. >> >>>>> Trying to take out the coordinates doesn't fair well for me either. >> >>>>> >> >>>>> How can one shorten the former expression using bindings? >> >>>>> >> >>>>> My second question regards is/: if I define a1, a2, a3 as missing >> >>>>> numbers in the row, column, face of some cell, why does is/ a1 a2 >> a3 >> >>>>> give a syntax error, when a1 is a2 is a3 doesn't? >> >>>>> >> >>>>> >> >>>>> >> ---------------------------------------------------------------------- >> >>>>> For information about J forums see >> http://www.jsoftware.com/forums.htm >> >>>>> >> >>>> >> >>> >> >> >> >> -- >> >> ---------------------- >> >> mail written using NEO >> >> neo-layout.org >> >> >> >> ---------------------------------------------------------------------- >> >> For information about J forums see http://www.jsoftware.com/forums.htm >> >> >> > >> > >> >> -- >> ---------------------- >> mail written using NEO >> neo-layout.org >> >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm >> > > > -- > > Devon McCormick, CFA > > Quantitative Consultant > > -- Devon McCormick, CFA Quantitative Consultant ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
