On Fri, 2004-07-16 at 12:10 -0400, Jonadab the Unsightly One wrote:
Austin Hastings [EMAIL PROTECTED] writes:
Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
collapses to the range [Inf, Inf).
It's not that simple. By that reasoning, 10% of all numbers in
[0,Inf)
On Mon, 19 Jul 2004, Jonadab the Unsightly One wrote:
If I extend the natural numbers N with Inf to a new set NI (N with
Inf)
The problem is, NI is not a group with respect to addition for any
definition of addition of which I am aware. Translated from mathese
In other words, or more
Ph. Marek [EMAIL PROTECTED] writes:
This is obviously some new definition of Inf of which I was not
previously aware.
Well, after reading my sentence one more, I see what may have caused
some troubles. Inf is not in N; but *in my understanding* it fits
naturally as an extension to N, that
On Mon, Jul 19, 2004 at 09:55:14PM -0400, Jonadab the Unsightly One wrote:
: It is possible to construct a group that includes infinities, but NI
: isn't it, and for Perl purposes it doesn't seem necessary.
Though if someone wants to hack surreals into 6.1, that'd be cool. :-)
Larry
On Friday 16 July 2004 18:23, Jonadab the Unsightly One wrote:
Please take my words as my understanding, ie. with no connection to
mathmatics or number theory or whatever. I'll just say what I
believe is practical.
[...]
I'd believe that infinity can be integer, ie. has no numbers after
Austin Hastings [EMAIL PROTECTED] writes:
Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
collapses to the range [Inf, Inf).
It's not that simple. By that reasoning, 10% of all numbers in
[0,Inf) would be in [Inf/10,Inf), also reducing to the range
[Inf,Inf). For that
David Storrs [EMAIL PROTECTED] writes:
Does it even make sense to take the Infiniteth element of an array?
No. At best, it would be undefined, so we could define it to return
undef.
I think I would prefer if using Inf as an array index resulted in a
trappable error.
Or that, yeah.
--
Ph. Marek [EMAIL PROTECTED] writes:
Please take my words as my understanding, ie. with no connection to
mathmatics or number theory or whatever. I'll just say what I
believe is practical.
[...]
I'd believe that infinity can be integer, ie. has no numbers after
the comma; and infinity is in
[EMAIL PROTECTED] (David Storrs) writes:
Does it even make sense to take the Infiniteth element of an array?
You should have used a hash in the first place.
--
BASH is great, it dumps core and has clear documentation. -Ari Suntioinen
On Wed, 14 Jul 2004, Ph. Marek wrote:
Please take my words as my understanding, ie. with no connection to
mathmatics or number theory or whatever. I'll just say what I believe is
practical.
OT
As a side note, being what one would probably call a mathematically
oriented person, it is very
On Wednesday 14 July 2004 12:58 pm, Brent 'Dax' Royal-Gordon wrote:
Andrew Rodland wrote:
So if we have @x = [1, 3, 5, 6 .. 9, 10 .. Inf, 42];
...
42 is just one number, so questions of indexing
it are moot, but its distance from the left is Inf. So, there's no way
to access the 42 by
On Wed, Jul 14, 2004 at 07:40:33AM +0200, Ph. Marek wrote:
To repeat Dave and myself - if
@x = 1 .. Inf;
then
rand(@x)
should be Inf, and so
print $x[rand(@x)];
should give Inf, as the infinite element of @x is Inf.
Does it even make sense to take the Infiniteth element
On Wednesday 14 July 2004 08:39, David Storrs wrote:
To repeat Dave and myself - if
@x = 1 .. Inf;
then
rand(@x)
should be Inf, and so
print $x[rand(@x)];
should give Inf, as the infinite element of @x is Inf.
Please take my words as my understanding, ie. with no
On Wednesday 14 July 2004 04:55 am, Ph. Marek wrote:
On Wednesday 14 July 2004 08:39, David Storrs wrote:
To repeat Dave and myself - if
@x = 1 .. Inf;
then
rand(@x)
should be Inf, and so
print $x[rand(@x)];
should give Inf, as the infinite element of @x is Inf.
On 7/12/04, Austin Hastings wrote:
--- Larry Wall [EMAIL PROTECTED] wrote:
The hard part being to pick a random number in [0,Inf) uniformly. :-)
Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
collapses to the range [Inf, Inf). Returning Inf seems to satisfy the
uniform
David Green writes:
On 7/12/04, Austin Hastings wrote:
--- Larry Wall [EMAIL PROTECTED] wrote:
The hard part being to pick a random number in [0,Inf) uniformly. :-)
Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
collapses to the range [Inf, Inf). Returning Inf seems
--- Larry Wall [EMAIL PROTECTED] wrote:
The hard part being to pick a random number in [0,Inf) uniformly. :-)
Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
collapses to the range [Inf, Inf). Returning Inf seems to satisfy the
uniform distribution requirement: if you
On Thursday 08 July 2004 05:25, Larry Wall wrote:
: say @x[rand]; # how about now?
Well, that's always going to ask for @x[0], which isn't a problem.
However, if you say rand(@x), it has to calculate the number of
elements in @x, which could take a little while...
I'd expect to be rand(@x) =
Ph. Marek wrote:
On Thursday 08 July 2004 05:25, Larry Wall wrote:
: say @x[rand]; # how about now?
Well, that's always going to ask for @x[0], which isn't a problem.
However, if you say rand(@x), it has to calculate the number of
elements in @x, which could take a little while...
I'd expect to
Ph. Marek [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Thursday 08 July 2004 05:25, Larry Wall wrote:
: say @x[rand]; # how about now?
Well, that's always going to ask for @x[0], which isn't a problem.
However, if you say rand(@x), it has to calculate the number of
--- Dave Whipp [EMAIL PROTECTED] wrote:
rand(@x) == @x.rand == @x[ rand int @x ] == @x[ rand(1) * @x ]
guaranteeing a uniform distribution unless adverbial modifiers are
used.
Meaning I can do:
$avg_joe = rand @students :bell_curve;
?
=Austin
On Mon, Jul 12, 2004 at 10:12:03AM -0700, Austin Hastings wrote:
: --- Dave Whipp [EMAIL PROTECTED] wrote:
:
:rand(@x) == @x.rand == @x[ rand int @x ] == @x[ rand(1) * @x ]
:
: guaranteeing a uniform distribution unless adverbial modifiers are
: used.
The hard part being to pick a random
--- Larry Wall [EMAIL PROTECTED] wrote:
On Mon, Jul 12, 2004 at 10:12:03AM -0700, Austin Hastings wrote:
: --- Dave Whipp [EMAIL PROTECTED] wrote:
:
:rand(@x) == @x.rand == @x[ rand int @x ] == @x[ rand(1) * @x ]
:
: guaranteeing a uniform distribution unless adverbial modifiers
are
On Mon, Jul 12, 2004 at 11:11:58AM -0700, Austin Hastings wrote:
: --- Larry Wall [EMAIL PROTECTED] wrote:
: The hard part being to pick a random number in [0,Inf) uniformly. :-)
:
: Half of all numbers in [0, Inf) are in the range [Inf/2, Inf). Which
: collapses to the range [Inf, Inf).
On Wed, Jul 07, 2004 at 11:50:16PM -0400, JOSEPH RYAN wrote:
To answer the latter first, rand (with no arguments) returns a number
greater than or equal to 0 and less than 1 which when used as an index
into an array gets turned into a 0.
As to why the second pop would take forever, I'd
On Fri, Jul 02, 2004 at 09:32:07PM -0500, Dan Hursh wrote:
: how 'bout
:
: @x = gather{
: loop{
: take time
: }
: } # can this be @x = gather { take time loop }
: push @x, later;
: say pop @x;# later
Can probably be made to work right.
: say pop @x;# heat death?
Yes.
- Original Message -
From: Larry Wall [EMAIL PROTECTED]
Date: Wednesday, July 7, 2004 11:25 pm
Subject: Re: push with lazy lists
On Fri, Jul 02, 2004 at 09:32:07PM -0500, Dan
Hursh wrote:
: how 'bout
:
: @x = gather{
: loop{
: take time
: }
: } # can
On Wed, Jul 07, 2004 at 11:50:16PM -0400, JOSEPH RYAN wrote:
- Original Message -
From: Larry Wall [EMAIL PROTECTED]
Date: Wednesday, July 7, 2004 11:25 pm
Subject: Re: push with lazy lists
On Fri, Jul 02, 2004 at 09:32:07PM -0500, Dan
Hursh wrote:
: how 'bout
:
: @x
Joseph Ryan wrote:
The way I understand the magicness of lazy lists, I'd expect:
@x = 3..Inf;
say pop @x; # prints Inf
@x = 3..Inf;
push @x, 6; # an array with the first part being
# lazy, and then the element 6
say pop @x; # prints 6
say pop @x; # prints Inf
say pop @x; # prints Inf
- Original Message -
From: Dan Hursh [EMAIL PROTECTED]
Date: Friday, July 2, 2004 10:32 pm
Subject: Re: push with lazy lists
Joseph Ryan wrote:
I guess that's true with X..Y lazy lists. I
thought there were
other
ways to make lazy lists, like giving it a closure
that gets called
Hi,
If I can assume:
@x = 3..5;
say pop @x;# prints 5
@x = 3..5;
push @x, 6;
say pop @x;# prints 6
say pop @x;# prints 5
What should I expect for the following?
@x = 3..Inf;
say pop @x;# heat death?
- Original Message -
From: Dan Hursh [EMAIL PROTECTED]
Date: Friday, July 2, 2004 2:23 pm
Subject: push with lazy lists
Hi,
If I can assume:
@x = 3..5;
say pop @x;# prints 5
@x = 3..5;
push @x, 6;
say pop @x;# prints 6
say
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