Also, there’s been correspondence in this forum over the years since Sudokus emerged around ‘05. Many solutions were discussed. Roger Hui presented elegant compact solvers; John Scholes produced a nice one using “direct definition” in Dyalog APL, I forget where it’s published. My own is typically verbose, probably not publicised.
As for generation, I often wonder how the London Times manages to construct its puzzles; they’re always symmetric, sometimes mirror, sometimes 180 deg rotational. Cheers, Mike Sent from my iPad > On 14 Oct 2021, at 03:19, Devon McCormick <devon...@gmail.com> wrote: > > 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 >> > > > -- > > Devon McCormick, CFA > > Quantitative Consultant > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
