Say for example I want to construct a triplet, taking in every combination of numbers from 1..100 in a triplet; i.e:
[(0,0,0),(0,0,1),(0,1,1),(1,1,1),(0,0,2),(0,1,2),(0,2,2),(1,2,2)]
..etc etc, up until a given bound (i.e: 100: giving us a final triplet of (100,100,100)); is there any reasonable way of doing this within haskell, or would I be best off writing a method that in short held a boundary pointer, and recursively increased each number until it was equal to the number to its right?
I think your description best fits a list comprehension to express what you want to do:
[(a, b, c) | c <- [0..100],
b <- [0..c],
a <- [0..b] ]
You say you want the numbers 1..100, but mention 0 in the examples. Further, you say "every combination" but then do not mention (1,0,0).
[(a,b,c) | m<-[0..100], a<-[0..m], b<-[0..m], c<-[0..m] ]
to avoid duplicates, enumerate them in the order of their total sum:
let l = 100; in [(a,b,c) | m<-[0..3*l],
a<-[0..l], b<-[0..l], c<-[0..l],
a+b+c == m ]
The calculation can be accelerated (by keeping the same result), by narrowing down the possible ranges:
let l = 100; in [(a,b,m-a-b) | m<-[0..3*l],
a<-[(max 0 (m-2*l))..(min l m)],
b<-[(max 0 (m-a-l))..(min l (m-a))] ]
