bill lam wrote:
> btw anyone knows how to do it in C or have a pointer? May be this serves as
> an
> example of "C for J Programmers" :-)
Here's a rough approximation, in J, of how it could be done in C:
zigzag=:3 :0
lim=.*/'X Y'=.y
'J K'=.y-1
r=.,y$n=.j=.k=.0
d=.1
while.n<lim do.
r=.n (k+Y*j)} r
n=.n+1
if.d>0 do.
if.k<K do.
k=.k+d
else.
d=._1
j=.j+1
continue.
end.
if.j do.
j=.j-d
else.
d=._1
end.
else.
if.j<J do.
j=.j-d
else.
d=.1
k=.k+1
continue.
end.
if.k do.
k=.k+d
else.
d=.1
end.
end.
end.
y$r
)
An advantage of this approach is that it's O(#r)
However, for example, if I had not included the variables J and K
and had used X and Y in their place, the code would be incorrect
and the reasons for that might have been difficult to determine.
FYI,
--
Raul
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