Fergus Henderson wrote:
| Yes, the correct Haskell syntax for this is somewhat different.
| You can use
|
| [(a, b-a) | b - [1..], a - [1..b-1]]
Or:
[ (a,b) | (a,b) - [1..] // [1..] ]
For a suitable definition of (//), for example:
(//) :: [a] - [b] - [(a,b)]
xs // ys =
Koen Claessen [EMAIL PROTECTED]
proposes the following diagonalisation function:
[ (a,b) | (a,b) - [1..] // [1..] ]
For a suitable definition of (//), for example:
(//) :: [a] - [b] - [(a,b)]
xs // ys = diagonalize 1 [[(x,y) | x - xs] | y - ys]
where
Marcin 'Qrczak' Kowalczyk wrote:
1. Is there any difference between
\a b - (a, b)
and
let f a b = (a, b) in f
?
No
2. Is there any difference between
case x of (a, b) - (b, a)
and
let (a, b) = x in (b, a)
?
Yes, if x is bottom.
-- Lennart
| To me, it seems unsatisfactory to have a solution to this pure
| list problem with auxiliary functions relying on integers.
| It turns out to be a nice exercise to implement
|
| diagonalise :: [[a]] - [a]
|
| without any reference to numbers.
Here's my definition of an integer free
1. Is there any difference between
\a b - (a, b)
and
let f a b = (a, b) in f
?
2. Is there any difference between
case x of (a, b) - (b, a)
and
let (a, b) = x in (b, a)
?
--
__("Marcin Kowalczyk * [EMAIL PROTECTED] http://kki.net.pl/qrczak/
\__/ GCS/M d- s+:--
