G'day all.
Quoting Dan Weston :
Ouch. That's what happens when you let a machine do the translation.
How about:
"Once your good name is trashed, you can live unabashed."
"Until you've lost your reputation, you never realize what a burden it was."
-- Margaret Mitchell
Cheers,
Andrew Brom
Ouch. That's what happens when you let a machine do the translation. How
about:
"Once your good name is trashed, you can live unabashed."
David Virebayre wrote:
On Wed, Dec 9, 2009 at 11:47 AM, Henning Thielemann
wrote:
Ist der Ruf erst ruiniert, lebt es sich ganz ungeniert. 8-]
Is ther
On Wed, Dec 9, 2009 at 11:47 AM, Henning Thielemann
wrote:
> Ist der Ruf erst ruiniert, lebt es sich ganz ungeniert. 8-]
> Is there an English translation of it?
Google translate says : "If the reputation is ruined, one can live
quite openly."
David.
___
Am Mittwoch 09 Dezember 2009 11:47:49 schrieb Henning Thielemann:
> On Wed, 9 Dec 2009, Daniel Fischer wrote:
> > Am Mittwoch 09 Dezember 2009 00:02:30 schrieb Lennart Augustsson:
> >> And if you use quotRem it's faster (unless you're running on some
> >> exotic hardware like NS32K).
> >
> > Yes, b
On Wed, 9 Dec 2009, Daniel Fischer wrote:
Am Mittwoch 09 Dezember 2009 00:02:30 schrieb Lennart Augustsson:
And if you use quotRem it's faster (unless you're running on some
exotic hardware like NS32K).
Yes, but Henning Thielemann was busy in the exception vs. error thread, so I
didn't want
Am Mittwoch 09 Dezember 2009 00:02:30 schrieb Lennart Augustsson:
> And if you use quotRem it's faster (unless you're running on some
> exotic hardware like NS32K).
Yes, but Henning Thielemann was busy in the exception vs. error thread, so I
didn't want
to distract him by using quotRem :D
__
And if you use quotRem it's faster (unless you're running on some
exotic hardware like NS32K).
On Tue, Dec 8, 2009 at 10:19 PM, Richard O'Keefe wrote:
>
> On Dec 9, 2009, at 1:15 AM, Daniel Fischer wrote:
>
>> Am Dienstag 08 Dezember 2009 08:44:52 schrieb Ketil Malde:
>>>
>>> "Richard O'Keefe" w
On Dec 9, 2009, at 1:15 AM, Daniel Fischer wrote:
Am Dienstag 08 Dezember 2009 08:44:52 schrieb Ketil Malde:
"Richard O'Keefe" writes:
factors n = [m | m <- [1..n], mod n m == 0]
-- saves about 10% time, seems to give the same result:
factors n = [m | m <- [1..n `div` 2], mod n m == 0]++
On Dec 8, 2009, at 8:44 PM, Ketil Malde wrote:
"Richard O'Keefe" writes:
factors n = [m | m <- [1..n], mod n m == 0]
I should remark that I wasn't *trying* to write fast code.
I was trying to code as directly as I could, fully expecting
to have to rewrite later. I was pleasantly surprised
Am Dienstag 08 Dezember 2009 08:44:52 schrieb Ketil Malde:
> "Richard O'Keefe" writes:
> > factors n = [m | m <- [1..n], mod n m == 0]
>
> -- saves about 10% time, seems to give the same result:
> factors n = [m | m <- [1..n `div` 2], mod n m == 0]++[n]
Even faster (for large enough n):
fact
"Richard O'Keefe" writes:
> factors n = [m | m <- [1..n], mod n m == 0]
-- saves about 10% time, seems to give the same result:
factors n = [m | m <- [1..n `div` 2], mod n m == 0]++[n]
(But checking against primes is even faster, it seems)
-k
--
If I haven't seen further, it is by standin
Am Dienstag 08 Dezember 2009 01:54:12 schrieb a...@spamcop.net:
> G'day all.
>
> Quoting Richard O'Keefe :
> > These lines of Haskell code find the Zumkeller numbers up to 5000
> > in 5 seconds on a 2.2GHz intel Mac. The equivalent in SML took
> > 1.1 seconds. Note that this just finds whether a
G'day all.
Quoting Richard O'Keefe :
These lines of Haskell code find the Zumkeller numbers up to 5000
in 5 seconds on a 2.2GHz intel Mac. The equivalent in SML took
1.1 seconds. Note that this just finds whether a suitable
partition exists; it does not report the partition.
This takes 0.1
On Tue, 8 Dec 2009, Richard O'Keefe wrote:
is_Zumkeller :: Int -> Bool
is_Zumkeller n =
let facs = factors n
fsum = sum facs
in mod fsum 2 == 0 &&
I see this test is essential. I didn't do it and thus my program did not
find that 1800 is not a Zumkeller number within an hour. With
> From: Richard O'Keefe [mailto:o...@cs.otago.ac.nz]
>
> These lines of Haskell code find the Zumkeller numbers up to 5000
> in 5 seconds on a 2.2GHz intel Mac. The equivalent in SML took
> 1.1 seconds. Note that this just finds whether a suitable
> partition exists; it does not report the parti
> From: daniel.is.fisc...@web.de [mailto:daniel.is.fisc...@web.de]
> > Not related to Haskell, but do you think semi-Zumkeller numbers are
> > semi-perfect numbers?
>
> The site you linked to says so. I've not investigated.
Peter Luschny posted the link in a discussion in a German newsgroup:
h
On Dec 8, 2009, at 10:33 AM, Frank Buss wrote:
Anyone interested in writing some lines of Haskell code for
generating the Zumkeller numbers?
http://www.luschny.de/math/seq/ZumkellerNumbers.html
These lines of Haskell code find the Zumkeller numbers up to 5000
in 5 seconds on a 2.2GHz intel
Am Montag 07 Dezember 2009 22:33:32 schrieb Frank Buss:
> Anyone interested in writing some lines of Haskell code for generating the
> Zumkeller numbers?
>
> http://www.luschny.de/math/seq/ZumkellerNumbers.html
>
> My C/C# solutions looks clumsy (but is fast). I think this can be done much
> more e
On Mon, 7 Dec 2009, Frank Buss wrote:
Anyone interested in writing some lines of Haskell code for generating
the Zumkeller numbers?
http://www.luschny.de/math/seq/ZumkellerNumbers.html
My C/C# solutions looks clumsy (but is fast). I think this can be done
much more elegant in Haskell with l
Here's a completely naive implementation, it's slow as cold molasses
going uphill during a blizzard, but it doesn't seem to be wrong. I let
it run in the interpreter for the last 3 minutes or so and it's
reproduced the given list up to 126 (and hasn't crapped out yet).
I imagine there's pro
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