I've just played with 8 Queens puzzle, and found out that it seems there is no _dvl operator (from k(v2)) in k(v4). Also I checked other k versions from ngn k impls and found ^ operator in k(v6), for example at JohnEarnest's impl:
l^a or l^l is except. Remove all instances of each of y from x.
k) 1 3 2 5 1 2 3^1 3 5 2 2
I really love SQL-style and would like to apply it in q. But is the following way idiomatic in q/k(v4) and is it a good solution? Or maybe there are shorter ways to do such list comparison/exclusion exists:
q)show s:til 8
0 1 2 3 4 5 6 7
q)s where not s in 2 4 6 /bother about this line, can it be shorter?
0 1 3 5 7
My version of q8 code is a bit longer then in nsl k2, without recursion and without conditions:
f:{raze {(x,) each (til 8) where not (til 8) in {x,(x-f),x+f:reverse 1+til count x} x} each x}
\ts:10 7 f/til 8 /248 100128
count 7 f/til 8 /92
first 7 f/til 8 /0 4 7 5 2 6 1 3
Upd: command I was looking for is except:
q)f:{raze {(x,) each (til 8) except {x,(x-f),x+f:reverse 1+til count x} x} each x}
Upd2: generalized 8 Queens solution in k(v4):
k){(x-1){,/{(x,)'(!y)@&~(!y)in{x,(x-f),x+f:|1+!#x}x}[;y]'x}[;x]/!x}8
Upd3: add 8 queens puzzle to blog
It is just keyword except.
How to find it: we know idiomatic k construction @&, so just search through .q namespace for it:
q)qfind:{([] q:k;k:.q k:key[.q] where (string value .q) like "*",x,"*")}
q)qfind "@&"
q k
-----------------------------------
inter k){x@&x in y}
except k){x@&~x in y}
xcols k){(x,f@&~(f:cols y)in x)#y}
q) (til 8) except 2 4 6
0 1 3 5 7