John Meacham wrote:
1. one really does logically derive from the other, Eq and Ord are like
this, the rules of Eq says it must be an equivalance relation and that
Ord defines a total order over that equivalance relation. this is a good
thing, as it lets you write code that depends on these
Hello, Bulat
I'm currently working on some kind of program for analysing FAT partitions.
Don't ask why did I chose to implement it in Haskell :) Just for fun.
My program needs to read scattered chunks of binary data from a huge file and to
do a good amount of deserialisation.
I implemented
Hello Daniel,
Friday, April 7, 2006, 2:05:03 AM, you wrote:
I've cleaned up my solver, removed a lot of redundant inference steps and made
the board a DiffArray (is that really faster?).
btw, DiffArray implementation can be made significantly faster by
using IORefs instead of MVars
--
Best
akamaus:
Hello, Bulat
I'm currently working on some kind of program for analysing FAT partitions.
Don't ask why did I chose to implement it in Haskell :) Just for fun.
My program needs to read scattered chunks of binary data from a huge file and
to
do a good amount of deserialisation.
Robert Dockins wrote:
The behaviour of NaN actually makes perfect sense when you realise that
it is Not a Number. Things that are not numbers are incomparable with
things that are.
Yes, NaN can be of type Float. But it's not a Float.
If you take that tack, then you have to concede that
Just out of curiosity, speed was not the objective when I wrote my solver, I
wanted to avoid guesswork (as much as possible), but in comparison with Cale
Gibbard's and Alson Kemp's solvers (which are much more beautifully coded),
it turned out that mine is blazingly fast, so are there faster
On Apr 7, 2006, at 9:43 AM, Jacques Carette wrote:
Robert Dockins wrote:
The behaviour of NaN actually makes perfect sense when you
realise that
it is Not a Number. Things that are not numbers are incomparable
with
things that are.
Yes, NaN can be of type Float. But it's not a Float.
given an Ord instance (for a type T) a corresponding Eq instance can be
given by:
instance Eq T where
a == b = compare a b == EQ
where did this second -^ == come from? (I guess if if Ordering
derives Eq :-) I think you meant
instance (Ord T) = Eq T where
a == b = case compare
Robert Dockins wrote:
On Apr 7, 2006, at 9:43 AM, Jacques Carette wrote:
[Lots of stuff about IEEE 754]
Is this an H' worthy item?
It is worth taking a serious look in conjunction with completely redoing
the Num class. Minor tweaking of the behaviour on NaNs (which requires
a large amount of
After reading Daniel Fischer's message about trying to use EnumSet in
his Sudoku.hs and finding that it was slower when processing a large
data set, I decided to do some profiling. I rewrote his solver to
use EnumSets and profiled it. The culprit turns out to be the
following function
Am Freitag, 7. April 2006 01:50 schrieben Sie:
On Apr 6, 2006, at 6:05 PM, Daniel Fischer wrote:
I've also written a version using David F. Place's EnumSet instead
of [Int],
that takes less MUT time, but more GC time, so is slower on the
36,628 test,
but faster for a single puzzle.
Robert Dockins writes:
On Thursday 06 April 2006 06:44 pm, John Meacham wrote:
On Thu, Apr 06, 2006 at 10:52:52PM +0100, Brian Hulley wrote:
[snip a question about Eq and Ord classes]
well, there are a few reasons you would want to use inheritance in
haskell, some good, some bad.
Am Freitag, 7. April 2006 17:33 schrieben Sie:
Just out of curiosity, speed was not the objective when I wrote my
solver, I wanted to avoid guesswork (as much as possible), but in
comparison with Cale Gibbard's and Alson Kemp's solvers (which are much
more beautifully coded), it turned out
Am Freitag, 7. April 2006 22:57 schrieb David F. Place:
After reading Daniel Fischer's message about trying to use EnumSet in
his Sudoku.hs and finding that it was slower when processing a large
data set, I decided to do some profiling. I rewrote his solver to
use EnumSets and profiled it.
On 4/7/06, Jared Updike [EMAIL PROTECTED] wrote:
given an Ord instance (for a type T) a corresponding Eq instance can be
given by:
instance Eq T where
a == b = compare a b == EQ
where did this second -^ == come from? (I guess if if Ordering
derives Eq :-) I think you meant
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