Chaddaï Fouché-2 wrote:
2008/3/30, Bulat Ziganshin [EMAIL PROTECTED]:
although the last alternative,
(Branch l r) = (Branch l' r') = l == l' r = r' || l = l'
seems suspicious to me. isn't it the same as
(Branch l r) = (Branch l' r') = l = l'
Yes, it should be :
(Branch l
2008/3/31, Simeon Mattes [EMAIL PROTECTED]:
why I should take as right
(a,b) = (a',b') iff (a a' or (a == a' and b = b'))
and not
(a,b) = (a',b') iff (a = a' or (a == a' and b = b'))
The latter seems more logical, doesn't it?
No, it doesn't, since in the latter (1,2) = (1,1) because 1
Hello everyone,
I would like to ask something that I found in the ebook a Gentle
Introduction to Haskell.
http://haskell.org/tutorial/stdclasses.html#sect8.4
data Tree a = Leaf a | Branch (Tree a) (Tree a)
instance (Ord a) = Ord (Tree a) where
(Leaf _) = (Branch _) = True
Although I have tried to make sense what lexicographic order means I haven't
figured out. Maybe an example with a simple application of this would be
helpful. To be honest I can't understand what the symbol = really means.
= means less than or equal to.
Normally, lexicograpic order is the
Hello Simeon,
Monday, March 31, 2008, 12:45:54 AM, you wrote:
The latter specifies a lexicographic order: Constructors are ordered by the
order of their appearance the data declaration, and the arguments of a
constructor are compared from left to right.
Although I have tried to make sense
Hello Bulat,
Monday, March 31, 2008, 1:16:35 AM, you wrote:
if you can compare chars and 'a' 'b', then *lists* of chars compared
in lexicographic order will be
aaa aab
aab aba
baa abb
as it was mentioned by Niklas Broberg, the last sentence should be reversed:
abb baa
sorry for +1
2008/3/30, Bulat Ziganshin [EMAIL PROTECTED]:
although the last alternative,
(Branch l r) = (Branch l' r') = l == l' r = r' || l = l'
seems suspicious to me. isn't it the same as
(Branch l r) = (Branch l' r') = l = l'
Yes, it should be :
(Branch l r) = (Branch l' r') = l l' ||
