Set operation like cartesian product

[Andrea Fontana]

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)

:)