I agree. Computation on the type level does not imply computation on the
value level.
On 8/18/07, Tim Chevalier [EMAIL PROTECTED] wrote:
On 8/17/07, Kim-Ee Yeoh [EMAIL PROTECTED] wrote:
Incidentally, GHC's type checker is Turing complete. You
already have as much static evaluation as is
On Thu, 2007-08-16 at 12:50 -0700, Kim-Ee Yeoh wrote:
Aaron Denney wrote:
On 2007-08-15, Pekka Karjalainen [EMAIL PROTECTED] wrote:
A little style issue here on the side, if I may. You don't need to use
(++) to join multiline string literals.
text = If you want to have multiline
Lennart Augustsson wrote:
On 8/16/07, Kim-Ee Yeoh [EMAIL PROTECTED] wrote:
'Course not. The (++) function like all Haskell functions is only a
/promise/ to do its job. What does assembling at compile time
mean here:
s = I will not write infinite loops ++ s
But if the strings are all
It's very hard to tell if it's worth it or not. Concatenating constant
strings will turn the string into WHNF, which might enable some other
transformation. By having lots of little transformations that on their own
look worthless you can make big improvements.
I'd like to see as much static
. (map
p011_toGroups) $ p011_input
--
L.Guo
2007-08-17
-
From: Ronald Guida
At: 2007-07-20 11:39:50
Subject: [Haskell-cafe] Hints for Euler Problem 11
To handle the diagonals, my plan is to try
On 8/17/07, Kim-Ee Yeoh [EMAIL PROTECTED] wrote:
Lennart Augustsson wrote:
And as a previous poster showed, ghc does concatenate strings.
And Haskell (as in the current language definition) does not.
I was talking about Haskell.
Haskell says nothing about compile time or run time in
On 8/17/07, Kim-Ee Yeoh [EMAIL PROTECTED] wrote:
Incidentally, GHC's type checker is Turing complete. You
already have as much static evaluation as is practically possible.
You already knew that.
I don't see how the first statement implies the second.
Cheers,
Tim
--
Tim Chevalier *
Aaron Denney wrote:
On 2007-08-15, Pekka Karjalainen [EMAIL PROTECTED] wrote:
A little style issue here on the side, if I may. You don't need to use
(++) to join multiline string literals.
text = If you want to have multiline string literals \
\in your source code, you can break
But if the strings are all constant it's perfectly feasible to concatenate
them at compile time.
On 8/16/07, Kim-Ee Yeoh [EMAIL PROTECTED] wrote:
Aaron Denney wrote:
On 2007-08-15, Pekka Karjalainen [EMAIL PROTECTED] wrote:
A little style issue here on the side, if I may. You don't need
I came up with this function to try to extract the main diagonal.
getDiag :: [[a]] - [a]
getDiag = map (head . head) . iterate (tail . map tail)
The problem is, this function doesn't work unless I have an infinite
grid.
Could anyone provide me with some hints to lead me in the right
On Thu, Jul 19, 2007 at 11:39:26PM -0400, Ronald Guida wrote:
In Problem 11, a 20x20 grid of numbers is given, and the problem is to
find the largest product of four numbers along a straight line in the
grid. The line can be horizontal, vertical, or diagonal.
I found it easier to work with
Ronald Guida wrote:
Hi, again.
I started looking at the Euler problems [1]. I had no trouble with
problems 1 through 10, but I'm stuck on problem 11. I am aware that
the solutions are available ([2]), but I would rather not look just
yet.
[...]
FWIW I used a 2D array and a
Jim Burton wrote:
Ronald Guida wrote:
Hi, again.
I started looking at the Euler problems [1]. I had no trouble with
problems 1 through 10, but I'm stuck on problem 11. I am aware that
the solutions are available ([2]), but I would rather not look just
yet.
[...]
FWIW I
Hi, again.
I started looking at the Euler problems [1]. I had no trouble with
problems 1 through 10, but I'm stuck on problem 11. I am aware that
the solutions are available ([2]), but I would rather not look just
yet.
In Problem 11, a 20x20 grid of numbers is given, and the problem is to
Here's my hint, FWIW.
Pick a data structure that makes your life easier, i.e. where horz,
vert, and diag lines are handled the same way. Instead of a 2D
structure, use a 1D structure.
Then,
data Dir = Horz | Vert | LL | LR
stride Horz = 1
stride Vert = rowLength
stride LL = rowLength - 1
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