On Sun, May 30, 2010 at 19:01, Simen kjaeraas <[email protected]
<mailto:[email protected]>> wrote:
Andrei Alexandrescu <[email protected]
<mailto:[email protected]>> wrote:
This reminded me that Phobos lacks a combinatorial range,
taking two or
more (forward) ranges and giving all combinations thereof:
combine([0,1],[2,3])
=>
(0,2), (0,3), (1,2), (1,3)
Work has commenced on implementing just that.
Yah, that would be useful. If Philippe agrees to adapt his work,
maybe that would be the fastest route. And don't forget - the
gates of Phobos are open.
Too late for that, as I've already written this. :p
Hey, cool, lots of static foreach. Man, it's like I'm reading my own
code. I'm happy to see convergent evolution: this must be the D way to
do this.
Current problems: back and popBack not implemented. I'm not sure they
even should be. Doing so would be a tremendous pain the region below the
spine.
Ow.
I *guess* it's doable, but I for sure don't want to do it.
May very well be there are other problems, I don't know. If anyone finds
any, please let me know.
Well, I like the way you dealt with popFront. You length is more costly
than mine, but I cheated: I count the numbers of popFronts and substract
them from the original length.
Your empty use .length in the foreach loop. You should use .empty
instead, I think. And with an infinite range in the mix, the resulting
product is infinte also, so you can define your combine to be infinite.
Then I can give you something to munch over :-)
When one range is infinite, the resulting combination is infinite.
What's sad is that you'll never 'traverse' the infinite range:
auto c = combine([0,1,2], cycle([2,3,4]));
->
(0,2),(0,3),(0,4),(0,2),(0,3), ...
You never reach the (1,x) (2,x) parts, as it should be seeing how we
both defined combinations: iterating on the ranges as if they were
digits in a number.
But there is another way to iterate: diagonally, by cutting successive
diagonal slices:
c is:
(0,2),(0,3),(0,4),(0,2),...
(1,2),(1,3),(1,4),(1,2),...
(2,2),(2,3),(2,4),(2,2),...
->
(0,2),(0,3)(1,2),(0,4),(1,3),(2,2) ...
It's even better if all ranges are infinite.
I never coded this, but it seems doable for two ranges. Lot thougher for
any number of ranges... Pretty obscure, maybe?
btw, I think we should divert this discussion in another thread if you
wish to continue: this is bearophile's Huffman thread, after all.
