After reading what this is about,
I guess a1, a2 and a3 aren’t atomic
so maybe you want
> is&.>/ a1;a2;a3
instead
Am 14.10.21 um 01:00 schrieb Hauke Rehr:
Regarding your second question:
I think it should be
is/ a1, a2, a3
Only literal numbers may be joined into an array by juxtaposition.
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?
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