The length command is too greedy. It wants to check the entire length of
both lists, and only then compare. If you did this by hand, you would
know to quit at the first nonequal element. To do this in Haskell, you
should generally avoid "length". Lazier functions such as "and" stop as
soon as they can:
and (map isEqual xs)
doesn't really map to every xs. The "and" keeps asking map to check the
next element for equality, but stops asking when it gets a negative
answer. length doesn't know to stop asking.
Dan
Alexteslin wrote:
What wrong with my original solution?
allEqual2 :: [Int] -> Bool
allEqual2 xs = length xs == length (filter isEqual xs)
where
isEqual n = (head xs) == n
It looks simpler to me
Dan Weston wrote:
The real lesson here is that
1) no problem is "too easy" to cheat good software engineering practices.
2) no solution not accompanied by at least a QuickCheck result should be
viewed with suspicion
I obviously have to go back and relearn these lessons.
Dan
Jason Dagit wrote:
On 7/19/07, Jason Dagit <[EMAIL PROTECTED]> wrote:
On 7/19/07, Dan Weston <[EMAIL PROTECTED]> wrote:
Oops, you got me. I didn't even look at the third line, I just took it
from the previous post. My first instinct actually was to write:
allEqual x@(h:t) = and (zipWith (==) x t)
I prefer,
allEqual [] = True
allEqual xs = foldl1 (==) xs
But, unfortunately, it's not a one liner like yours (unless you allow
allEqual [] = undefined).
Oh and silly me, that only works for [Bool].
I'll hushup :)
Jason
but I don't think that zipWith is allowed in the question.
Dan
Antoine Latter wrote:
On 7/19/07, Dan Weston <[EMAIL PROTECTED]> wrote:
I would define:
allEqual [] = True
allEqual [_] = True
allEqual (x1:x2:xs) = (x1 == x2) && allEqual xs
What does this function do for "allEqual [1, 1, 2]" ?
Antoine
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