On Thursday, 28 February 2013 at 18:23:04 UTC, H. S. Teoh wrote:
On Thu, Feb 28, 2013 at 04:31:59PM +0100, Andrea Fontana wrote:
I see cartesianProduct in std.algorithm. I read:
auto N = sequence!"n"(0); // the range of natural numbers
auto N2 = cartesianProduct(N, N); // the range of all pairs of
natural numbers
So it gives (0,0) (0,1) (1,0) ... and so on.
Is there a way to generate only tuple:
a[0] > a[1] (0,1) (0,2) ... (1,2) (1,3) .. (2,3) (2,4)
[...]
auto triangularProduct(R1,R2)(R1 r1, R2 r2)
if (isForwardRange!R1)
{
return map!(function(a) => zip(take(a[1].save, a[0]),
repeat(a[2])))(
zip(sequence!"n"(0), repeat(r1.save), r2)
)
.joiner();
}
Both ranges can be infinite, only the first one needs to be a
forward
range.
a[0] >= a[1] (0,0) (0,1) (0,2) ... (1,1) (1,2) (1,3) .. (2,2)
(2,3)
(2,4)
[...]
auto triangularProduct2(R1,R2)(R1 r1, R2 r2)
if (isForwardRange!R1)
{
return map!(function(a) => zip(take(a[1].save, a[0]),
repeat(a[2])))(
zip(sequence!"n"(1), repeat(r1.save), r2)
)
.joiner();
}
As above, both ranges can be infinite, only the first one needs
to be a
forward range.
Hope this helps. ;-)
T
triangulaProduct2 for [0,1,2,3] and [0,1,2,3] doesn't give (0,0)
(1,1) (2,2) (3,3)
:)
