Reputation: 1997
Is there an algorithm to lead all possible combinations of given amount of three-valued logic values?
For example, F(2)
should return this list:
t t
t u
t f
u t
u u
u f
f t
f u
f f
The function would look like this (in Haskell):
data Tril = FALSE | NULL | TRUE
all :: Int -> [[Tril]]
all amount = ???
all1 :: [Tril]
all1 = join (all 1)
all2 :: [(Tril, Tril)]
all2 = map (\[f, s] -> (f, s)) (all 2)
all3 :: [(Tril, Tril, Tril)]
all3 = map (\[f, s, t] -> (f, s, t)) (all 3)
Upvotes: 2
Views: 445
Reputation: 62818
You can do this very simply as a list comprehension:
all2 = [ (v1, v2) | v1 <- [FALSE, TRUE, NULL], v2 <- [FALSE, TRUE, NULL] ]
You can write it equivalently as a monadic do-block:
all2 = do
v1 <- [FALSE, TRUE, NULL]
v2 <- [FALSE, TRUE, NULL]
return (v1, v2)
And that gives us an idea for how we can write the variable-size one:
all 0 = [[]] -- Note: Empty list with one empty item.
all n = do
v <- [FALSE, TRUE, NULL]
vs <- all (n-1)
return (v:vs)
As it turns out — and this is slightly mind-bending — this is the net effect of the replicateM
function. It takes a monadic action, does it N times, and gathers the results together.
all n = replicateM n [FALSE, TRUE, NULL]
Upvotes: 7
Reputation: 116139
replicateM
does exactly that:
> import Control.Monad
> replicateM 2 [1,2,3]
[[1,1],[1,2],[1,3],[2,1],[2,2],[2,3],[3,1],[3,2],[3,3]]
Hence,
all :: Int -> [[Tril]]
all amount = replicateM amount [FALSE,NULL,TRUE]
I'd suggest to pick anouther name, since all
is already taken by Prelude.all
.
Upvotes: 4