too much information!
John Baez
They say we live in the "information age", but how much information
is there, actually? Guess how much information there is in:
1) all the words ever spoken by human beings.
2) the capacity of all the hard disks produced in 2002.
3) the information in the genomes of all the people living in the world.
4) all the photographs taken in 2002.
5) the capacity of all the audiotapes produced in 2002.
6) the information required to precisely describe the position/momentum
of all the atoms in a red blood cell up to the limits imposed by
quantum mechanics.
Rank them! Which is the biggest?
(For serious scholars: we're not taking advantage of information
compression in any of these; that would make the figures harder to
compute. So, for example, we don't care about the fact that everyone
has DNA that's very similar to everyone else's.)
If you give up, try this:
Michael Jørgensen
"John Baez" <ba...@galaxy.ucr.edu> wrote in message
news:crt513$rjo$1...@glue.ucr.edu...
> http://math.ucr.edu/home/baez/information.html
The complete works of Shakespeare is only 5 megabytes? That corresponds to
five short novels. I'm surprised.
-Michael.
Androcles
> is there, actually? Guess how much information there is in:
>
> 1) all the words ever spoken by human beings.
> 2) the capacity of all the hard disks produced in 2002.
> 3) the information in the genomes of all the people living in the
> world.
> 4) all the photographs taken in 2002.
> 5) the capacity of all the audiotapes produced in 2002.
> 6) the information required to precisely describe the
> position/momentum
> of all the atoms in a red blood cell up to the limits imposed by
> quantum mechanics.
>
> Rank them! Which is the biggest?
7) Pixels on a computer screen. 16,000,000 colors for each pixel.
How many different pictures are possible?
Is it more or less than a google?
(only for serious scolars)
Androcles.
Uncle Al
>
> "John Baez" <ba...@galaxy.ucr.edu> wrote in message
> news:crt513$rjo$1...@glue.ucr.edu...
> >
> > They say we live in the "information age", but how much information
> > is there, actually? Guess how much information there is in:
> >
> > 1) all the words ever spoken by human beings.
> > 2) the capacity of all the hard disks produced in 2002.
> > 3) the information in the genomes of all the people living in the
> > world.
> > 4) all the photographs taken in 2002.
> > 5) the capacity of all the audiotapes produced in 2002.
> > 6) the information required to precisely describe the
> > position/momentum
> > of all the atoms in a red blood cell up to the limits imposed by
> > quantum mechanics.
> >
> > Rank them! Which is the biggest?
>
> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
> How many different pictures are possible?
> Is it more or less than a google?
>
> (only for serious scolars)
> Androcles.
Only for serious idiots.
(1.6x10^6)^(number of pixels).
Only for uber-idiots who would have to study to pass the qualifiers
for being certified as a card-carrying idiot by the International
Union of Pure and Applied Idiocy. The original headquarters were at
Santorini, Greece. IUPAI then sequentially moved to
Year Headqaurters
----------------------
50 Ambrym, Vanuatu
186 Taupo, New Zealand
260 Ilopango, El Salvador
1452 Kuwae, Vanuatu
1884 Laki, Iceland
1815 Tambora, Indonesia
1883 Krakatoa, Java
1902 Pelee, Carribean
1912 Katmai, Alaska
1989 Alaid, Northern Kurinel Islands
1991 Pinatubo, Philippines
UPAI is currently headquartered in Long Valley, California.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
ba...@galaxy.ucr.edu
Michael Jørgensen <in...@ukendt.dk> wrote:
>> http://math.ucr.edu/home/baez/information.html
I'm holding the complete works of Shakespeare in my hand;
not counting the index it's 1280 pages, which is indeed about
5 short novels. If I used the figure that a typewritten page
is 2 kilobytes, this would be just 2.6 megabytes. But in fact
the print on these pages is pretty small, so 5 megabytes seems
reasonable.
Hmm. These pages have 60 lines of text on them each, and
about 75 characters per line. That's 4500 characters, and
a character is about 5 bits, so we have about 22,550 bits
per page. Times 1280 pages, that's 28,800,000 bits, or
5,760,000 bytes. In other words, 5.49 megabytes.
So yeah, that figure was right, up to the one significant
figure given.
It goes to show that impact is not proportional to the number
of bits one spews.
John Baez
>In article <41e24162$0$256$edfa...@dread12.news.tele.dk>,
>Michael Jørgensen <in...@ukendt.dk> wrote:
>>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
>>news:crt513$rjo$1...@glue.ucr.edu...
>>> http://math.ucr.edu/home/baez/information.html
>>The complete works of Shakespeare is only 5 megabytes? That corresponds to
>>five short novels. I'm surprised.
>I'm holding the complete works of Shakespeare in my hand;
>not counting the index it's 1280 pages, which is indeed about
>5 short novels.
>Hmm. These pages have 60 lines of text on them each, and
>about 75 characters per line. That's 4500 characters, and
>a character is about 5 bits, so we have about 22,550 bits
>per page. Times 1280 pages, that's 28,800,000 bits, or
>5,760,000 bytes. In other words, 5.49 megabytes.
>It goes to show that impact is not proportional to the number
>of bits one spews.
It also goes to show that the chance of a monkey randomly
typing the complete works of Shakespeare is about
2^{-28,800,000}
or in other words, roughly
10^{-10,000,000}
Not surprisingly, writing this number out in decimal form
would require a book whose size is longer than the complete
works of Shakespeare: about log(30)/log(10) times longer.
Ken S. Tucker
This (my) post is a bit kooky...
Could it be that Baez is looking for intelligent
algorithms in random data, as said,
Shakespeare will appear.
Next, an algorithm to *recognize* intelligence
embedded in random data, ((A Beautiful Mind)).
Next, what's the algorithm?
Simple post from an interested lurker...
Ken S. Tucker
oÄŸin
>>>to
>>>five short novels. I'm surprised.
>
>>I'm holding the complete works of Shakespeare in my hand;
>>not counting the index it's 1280 pages, which is indeed about
>>5 short novels.
>
>>Hmm. These pages have 60 lines of text on them each, and
>>about 75 characters per line. That's 4500 characters, and
>>a character is about 5 bits, so we have about 22,550 bits
>>per page. Times 1280 pages, that's 28,800,000 bits, or
>>5,760,000 bytes. In other words, 5.49 megabytes.
>
>>It goes to show that impact is not proportional to the number
>>of bits one spews.
>
> It also goes to show that the chance of a monkey randomly
> typing the complete works of Shakespeare is about
>
> 2^{-28,800,000}
>
> or in other words, roughly
>
> 10^{-10,000,000}
>
> Not surprisingly, writing this number out in decimal form
> would require a book whose size is longer than the complete
> works of Shakespeare: about log(30)/log(10) times longer.
OK, lets say that you have 2^{-28,800,000} monkeys that were all patient
enough to type for long enough to type 5.49 megabytes of text (would a
monkey live long enough?). What is the probability of getting at least one
good copy. And lets say that the generated text was random (it would not be
but lets say it is). So what is the average mass and volume of a monkey and
a typewriter and paper, together with the required food and beverage for the
duration of the project? And what would be the minimum entropy generated by
this primate steno pool? Would it create a black hole? How many human
readers would be required to locate at least one good copy? How long would
that take. How did Shakespeare manage to do it all by himself, is the human
brain 2^{-28,800,000} more powerful than a monkeys brain? How many
Shakespeares would it take to write what just one of those 2^{-28,800,000}
monkeys wrote? What the Simian Society of America go ape-shit about it?
David Bernier
> In article <41e24162$0$256$edfa...@dread12.news.tele.dk>,
> Michael Jørgensen <in...@ukendt.dk> wrote:
>
>
>>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
>>news:crt513$rjo$1...@glue.ucr.edu...
>
>
>>>http://math.ucr.edu/home/baez/information.html
>
>
>>The complete works of Shakespeare is only 5 megabytes? That corresponds to
>>five short novels. I'm surprised.
>
>
> I'm holding the complete works of Shakespeare in my hand;
> not counting the index it's 1280 pages, which is indeed about
> 5 short novels. If I used the figure that a typewritten page
> is 2 kilobytes, this would be just 2.6 megabytes. But in fact
> the print on these pages is pretty small, so 5 megabytes seems
> reasonable.
In the 1950's, RAND produced a table of a million
random digits, and another one with
"100,000 Normal Deviates". ( link is below)
Having downloaded the 1 million digits (line numbers included),
I wrote a program to save them as a file with
12,500 lines at 80 digits per line.
This got compressed to a 484,760-byte *.zip file.
David Bernier
RAND link:
http://www.rand.org/publications/classics/randomdigits/randomdata.html
John Baez
>6) the information required to precisely describe the position/momentum
> of all the atoms in a red blood cell up to the limits imposed by
> quantum mechanics.
I didn't actually work this one out, but I worked out the answer for
a 1-milligram raindrop, and you can find it here:
http://math.ucr.edu/home/baez/information.html
Since the diameter of a red blood cell is about 10^{-2} centimeters,
I'd guesstimate that its entropy is about 10^{-6} times that of a
cubic centimeter of water, which is about 5 x 10^{23} bytes.
That's 5 x 10^{17} bytes, or 500 petabytes. By comparison, all the
photographs taken in 2002 held about 400 petabytes.
John Baez
oÄŸin <oÄŸi...@ragnarok.com> wrote:
>OK, lets say that you have 2^{-28,800,000} monkeys that were all patient
>enough to type for long enough to type 5.49 megabytes of text (would a
>monkey live long enough?). What is the probability of getting at least one
>good copy.
That's not much of a monkey there.
Androcles
"Uncle Al" <Uncl...@hate.spam.net> wrote in message
news:41E2AEDD...@hate.spam.net...
You are a member, huh? I wouldn't know.
Androcles.
Phil Carmody
> ba...@galaxy.ucr.edu wrote:
> > In article <41e24162$0$256$edfa...@dread12.news.tele.dk>,
> > Michael Jørgensen <in...@ukendt.dk> wrote:
> >
> >>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
> >>news:crt513$rjo$1...@glue.ucr.edu...
> >
> >>>http://math.ucr.edu/home/baez/information.html
> >
> >>The complete works of Shakespeare is only 5 megabytes? That corresponds to
> >>five short novels. I'm surprised.
> > I'm holding the complete works of Shakespeare in my hand; not
> > counting the index it's 1280 pages, which is indeed about
> > 5 short novels. If I used the figure that a typewritten page
> > is 2 kilobytes, this would be just 2.6 megabytes. But in fact the
> > print on these pages is pretty small, so 5 megabytes seems
> > reasonable.
> [...]
>
> In the 1950's, RAND produced a table of a million
> random digits, and another one with
> "100,000 Normal Deviates". ( link is below)
Be warned. RAND's data is complete crap. It's trivially
compressible by a non-negligible fraction of a percent due
to incompetant unbiassing. See a comp.compression archive
for estimates how non-random it is.
Phil
--
The gun is good. The penis is evil... Go forth and kill.
mensa...@aol.com
Androcles wrote:
> "John Baez" <ba...@galaxy.ucr.edu> wrote in message
> news:crt513$rjo$1...@glue.ucr.edu...
> >
> > They say we live in the "information age", but how much information
> > is there, actually? Guess how much information there is in:
> >
> > 1) all the words ever spoken by human beings.
> > 2) the capacity of all the hard disks produced in 2002.
> > 3) the information in the genomes of all the people living in the
> > world.
> > 4) all the photographs taken in 2002.
> > 5) the capacity of all the audiotapes produced in 2002.
> > 6) the information required to precisely describe the
> > position/momentum
> > of all the atoms in a red blood cell up to the limits imposed by
> > quantum mechanics.
> >
> > Rank them! Which is the biggest?
>
> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
> How many different pictures are possible?
> Is it more or less than a google?
For a 640x480 screen, it's 10^2219433, slightly larger than a googol.
John Baez
David Bernier <ez...@yahoo.com> wrote:
>In the 1950's, RAND produced a table of a million
>random digits, and another one with
>"100,000 Normal Deviates". ( link is below)
>
>Having downloaded the 1 million digits (line numbers included),
>I wrote a program to save them as a file with
>12,500 lines at 80 digits per line.
>
>This got compressed to a 484,760-byte *.zip file.
Someone else claimed that these "random" digits were
complete crap and could be significantly compressed; let's see
if your zip file reflects that.
1 million decimal digits should be about 3.3 million bits;
divide by 8 and get about 410,000 bytes.
Wait a minute - that's LESS than your supposedly compressed
file! What's going on? It can't be those line numbers.
Am I making some dumb mistake?
John Baez
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
>Be warned. RAND's data is complete crap. It's trivially
>compressible by a non-negligible fraction of a percent due
>to incompetant unbiassing. See a comp.compression archive
>for estimates how non-random it is.
Interesting. I haven't found those estimates yet, and would like
a more specific reference, but this reminds me of Stanislaw Lem's
SF comedy "His Master's Voice", where alien radio transmissions
go unnoticed at first and the punch cards holding the data get
thrown out... then used to compile a table of random numbers...
but then, by analyzing the data in the table, they realize it's
a fascinating communication! I don't want to give the rest of the
story away, but it's a wacky reflection on the concepts of randomness
and information.
...................................................................
"Generation of random numbers is too important to be left
to chance..."
Donald Knuth
How many exabytes of information are there in a raindrop?
See http://math.ucr.edu/home/baez/information.html
Ken S. Tucker
A bit OT, but would it be true that a *perfect* random
sequence has no compression algorithm?
or OTOH
Any random sequence is subject to compression?
Ken
David Bernier
[...]
> In the 1950's, RAND produced a table of a million
> random digits, and another one with
> "100,000 Normal Deviates". ( link is below)
>
> Having downloaded the 1 million digits (line numbers included),
> I wrote a program to save them as a file with
> 12,500 lines at 80 digits per line.
>
> This got compressed to a 484,760-byte *.zip file.
In reply to John Baez's question, I did another
experiment...
Well, I tried again with a (just digits) 1,000,000-byte file,
and compressed it with Winzip 9 using the
"Maximum (enhanced deflate)" method,
and the *.zip file size is 471,002 bytes,
compared to the 415,241.0 bytes (using info. theory).
David Bernier
Lewis Mammel
Yeah, you're ignoring the 3.3 bit -> 8 bit expansion of the
ascii encoding ... or you're PRETENDING to ignore it!
This shows the compression to be pretty good, actually.
I think you can foil LZ with,
0123456789
then
00010203040506070809101212141516171819 ...
000001002003 ...
because I think it remembers new tokens, then new sequences starting
with known tokens, then new extensions of those sequences, etc.
Of course, it doesn't globally analyse the sequences for regularity.
It's amazingly good. I messed around with it once and tried "seeding"
it and different things, but it never helped. Lempel and Ziv are my heros!
Lew Mammel, Jr.
David Bernier
[...]
> Be warned. RAND's data is complete crap. It's trivially
> compressible by a non-negligible fraction of a percent due
> to incompetant unbiassing. See a comp.compression archive
> for estimates how non-random it is.
Thanks for that.
In the RAND table, as some know, the 1 million
digits were printed at 50 digits per line.
In fact, 20,000 punch cards held 50 digits each...
As is explained on the RAND site, to
debias the 20,000 cards,
card n was added to card (n+1) (mod 10)
and this finally got printed.
In the article in comp.compression by
Matt Mahoney dated 2004-06-10 18:56:49 PST,
"Re: Compressing million random digits file",
he writes in part:
"It gets worse. Add up every 50'th digit starting at any position
from 0 to 49. The result is always even.
That's 50 more bits of entropy."
For 0<=j<50, serving as index for the 50 columns,
let S(j) be the sum of the 20,000 digits in
that column in the 400 pages of 50 rows
of the table. Then I get the following,
which agrees with what he said:
David Bernier
j S(j)
----------------
0 90164
1 90594
2 89668
3 89700
4 89346
5 89852
6 90068
7 89674
8 89968
9 90020
10 89330
11 89994
12 90140
13 90304
14 90198
15 90290
16 90172
17 89804
18 90212
19 89912
20 89182
21 89698
22 89654
23 89766
24 89160
25 89876
26 89444
27 90262
28 90104
29 89556
30 90266
31 89492
32 89804
33 89854
34 90032
35 89224
36 89668
37 90118
38 90402
39 89764
40 89832
41 89526
42 90678
43 89682
44 89742
45 90046
46 90378
47 90100
48 89968
49 89962
Phil Carmody
> In article <87u0po3...@nonospaz.fatphil.org>,
> Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
>
> >Be warned. RAND's data is complete crap. It's trivially
> >compressible by a non-negligible fraction of a percent due
> >to incompetant unbiassing. See a comp.compression archive
> >for estimates how non-random it is.
>
> Interesting. I haven't found those estimates yet, and would like
> a more specific reference,
Matt was the mastermind behind the analysis and I pulled out some
explicit entropy rate figures based on his 50-columns theory in
Message-ID: <87vfhyv...@nonospaz.fatphil.org>
It appears that there's maybe only ~3.3213 bits per digit rather
than 3.3219, perhaps less. That could correspond to about 180
characters of redundancy in the whole million digit file. I consider
that to be quite a lot.
The analysis was not taken particularly deep, the thread contains
enough clues how to progress (I was hoping Matt would continue the
investigation, as the idea was his baby).
Phil Carmody
> Someone else claimed that these "random" digits were
> complete crap and could be significantly compressed;
We need some context, for supposedly "random" data, even compressing
by a dozen bytes is statistically significant. It won't make a change
to your hard disk usage though, alas.
Gregory L. Hansen
>> In the 1950's, RAND produced a table of a million
>> random digits, and another one with
>> "100,000 Normal Deviates". ( link is below)
>
>Be warned. RAND's data is complete crap. It's trivially
>compressible by a non-negligible fraction of a percent due
>to incompetant unbiassing. See a comp.compression archive
>for estimates how non-random it is.
Hmm... I remember reading some experiments in parapsychology, and
correlations were found here and there. I'm a bit fuzzy on the details,
but I remember a common technique for finding random numbers was to take a
column or a row from a book of random number tables.
--
Irony: "Small businesses want relief from the flood of spam clogging their
in-boxes, but they fear a proposed national 'Do Not Spam' registry will
make it impossible to use e-mail as a marketing tool."
http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
jmfb...@aol.com
ba...@galaxy.ucr.edu (John Baez) wrote:
Yay!
>In article <crumni$sla$1...@glue.ucr.edu>, <ba...@galaxy.ucr.edu> wrote:
>
>>In article <41e24162$0$256$edfa...@dread12.news.tele.dk>,
>>Michael Jørgensen <in...@ukendt.dk> wrote:
>
>>>"John Baez" <ba...@galaxy.ucr.edu> wrote in message
>>>news:crt513$rjo$1...@glue.ucr.edu...
>
>>>> http://math.ucr.edu/home/baez/information.html
>
>>>The complete works of Shakespeare is only 5 megabytes? That corresponds
to
>>>five short novels. I'm surprised.
>
>>I'm holding the complete works of Shakespeare in my hand;
>>not counting the index it's 1280 pages, which is indeed about
>>5 short novels.
>
>>Hmm. These pages have 60 lines of text on them each, and
>>about 75 characters per line. That's 4500 characters, and
>>a character is about 5 bits,
Nitpick....round up to 10 which will take care of overhead
storage, white spacespace, hidden characters, shit like that.
> ..so we have about 22,550 bits
>>per page. Times 1280 pages, that's 28,800,000 bits, or
>>5,760,000 bytes. In other words, 5.49 megabytes.
>
>>It goes to show that impact is not proportional to the number
>>of bits one spews.
>
>It also goes to show that the chance of a monkey randomly
>typing the complete works of Shakespeare is about
>
>2^{-28,800,000}
>
>or in other words, roughly
>
>10^{-10,000,000}
>
>Not surprisingly, writing this number out in decimal form
>would require a book whose size is longer than the complete
>works of Shakespeare: about log(30)/log(10) times longer.
This used to be a very nice mental game to play. You are only
dealing with final output. If you begin to think of small
things that can generate that output (in sort of automatic
fashion), you'll begin to see why I was attracted to category
theory. Take two clumps of thingies; combine them in a prescribed
manner, and you can generate oodles of thingie-children.
1. I'd like to be able to have a control that ensures the
thingie-children are useful; a tweak control is ideal, where
tweak implies a little bit of my time without paperwork.
2. It would be nice to have the tweak be a semiautomatic
application of the last thingie-child. The IT biz is still
just shoveling virtual paperwork around.
But, I haven't done any thinking about this for years.
/BAH
Subtract a hundred and four for e-mail.
jmfb...@aol.com
Are you still assuming there's 5 bits/char?
Also the data was on cards. There's a different encoding with
cards that has to do with two fields and one hole punch in each.
Geography of the holes was how characters were defined.
Also 20,000 cards ain't very many cards. It's only 10 boxes.
jmfb...@aol.com
"mensa...@aol.compost" <mensa...@aol.com> wrote:
>
>Androcles wrote:
<snip>
>> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
>> How many different pictures are possible?
>> Is it more or less than a google?
>
>For a 640x480 screen, it's 10^2219433, slightly larger than a googol.
But none of it is information.
matma...@yahoo.com
I missed the beginning of this thread but I assume you mean these
random digits: http://www.rand.org/publications/classics/randomdigits/
The article gives some clues to how this data could be compressed. The
original data had some statistical biases that were removed by adding
pairs of adjacent cards modulo 10. I discovered last year that if you
add all the cards (i.e. every 50'th digit) then all the sums are even.
This means that each of the 20,000 original cards were used exactly
twice (or at least an even number of times).
I thought that if I could reproduce the original set of 20,000 biased
cards that were used to produce the table, that some signficant
compression would be possible, but I got stuck on how to do that. If
you could guess one of the original cards, then you could produce all
the others by alternately adding and subtracting cards. However it
seems that the final set of cards was reordered, perhaps shuffled. The
RAND paper does not mention this step, but I know this because if you
alternately add and subtract the final cards in their original order
they should sum to 0 modulo 10, but they do not.
It might be that the final cards were rearranged by some simple rule.
You could test a proposed rule by alternately adding and subtracting
cards, but if the cards were shuffled this would not work, even given
infinite computing resources. That's because there are 20,000!/5^50 >>
1 possible shuffles that would result in an alternating sum of 0 mod 10
in all 50 digits.
I don't believe that the cards were shuffled thororghly because there
is a slight bias in the distribution of differences of digits spaced 1
or 2 cards apart, but not further. So it is possible that a rule will
still be discovered.
David Bernier
> Phil Carmody wrote:
> [...]
>
>> Be warned. RAND's data is complete crap. It's trivially compressible
>> by a non-negligible fraction of a percent due
>> to incompetant unbiassing. See a comp.compression archive
>> for estimates how non-random it is.
>
> [...]
>
> Thanks for that.
>
> In the RAND table, as some know, the 1 million
> digits were printed at 50 digits per line.
> In fact, 20,000 punch cards held 50 digits each...
>
> As is explained on the RAND site, to
> debias the 20,000 cards,
> card n was added to card (n+1) (mod 10)
> and this finally got printed.
>
> In the article in comp.compression by
> Matt Mahoney dated 2004-06-10 18:56:49 PST,
> "Re: Compressing million random digits file",
> he writes in part:
>
> "It gets worse. Add up every 50'th digit starting at
> any position from 0 to 49. The result is always even.
> That's 50 more bits of entropy."
If card 20,000 was added to card 1 (mod 10) to
get one row in the printed table (50 digits), the sum of
all the 20,000 digits in column j, 0<=j<50, should be a
multiple of 10...
The sums are all even, but they aren't all multiples
of 10...
David Bernier
See:
http://www.rand.org/publications/classics/randomdigits/
Section: "Production of the Random Digits"
Paragraph 4 beginning with:
"The random digits in this book were produced by rerandomization
of a basic table generated by an electronic roulette wheel."
David Bernier
Oops... Wrong on that!
(a1+a2) + (a2+a3) + ... (a_20000 + a1)
= 2*(a1+a2+ ... a_20000) == 0 (mod 2)
mensa...@aol.com
jmfb...@aol.com wrote:
> In article <1105420562.6...@c13g2000cwb.googlegroups.com>,
> "mensa...@aol.compost" <mensa...@aol.com> wrote:
> >
> >Androcles wrote:
> <snip>
>
> >> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
> >> How many different pictures are possible?
> >> Is it more or less than a google?
> >
> >For a 640x480 screen, it's 10^2219433, slightly larger than a
googol.
> But none of it is information.
I wouldn't say none. A lot of it would be useless. But just because
you can't tell the difference between a picture of a polar bear in
a blizzard and the inside of a ping pong ball doesn't mean the
picture of the inside of a ping pong ball isn't an accurate
representation.
If I had all 10^2219433 possible images, I would have every picture
taken by my digital camera, every picture that ever will be taken and
every possible picture my camera is capable of. This would include
closeups of every star in the universe. The thing is, a single image
could represent every star in the universe and be indistinguishable
from the inside of a ping pong ball.
So it's not a question of whether it's information but whether it's
usefull information.
John Baez
>>Androcles wrote:
>>> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
>>> How many different pictures are possible?
>>> Is it more or less than a google?
>>For a 640x480 screen, it's 10^2219433, slightly larger than a googol.
I'd say you're engaging in understatement, but that would be an
understatement. "Slightly" larger than 10^100?? This number makes
a googol look like one. That is, 1.
John Baez
>In article <crvr1o$r2n$1...@glue.ucr.edu>,
> ba...@galaxy.ucr.edu (John Baez) wrote:
>>In article <iUBEd.36478$SB6.1...@wagner.videotron.net>,
>>David Bernier <ez...@yahoo.com> wrote:
>>>In the 1950's, RAND produced a table of a million
>>>random digits, and another one with
>>>"100,000 Normal Deviates". ( link is below)
I keep on reading this as "100,000 normal deviants", which seems
like a description of usenet news. Apparently the librarians thought
similarly, since this book was first filed under psychology.
>>>Having downloaded the 1 million digits (line numbers included),
>>>I wrote a program to save them as a file with
>>>12,500 lines at 80 digits per line.
>>>
>>>This got compressed to a 484,760-byte *.zip file.
>>Someone else claimed that these "random" digits were
>>complete crap and could be significantly compressed; let's see
>>if your zip file reflects that.
>>
>>1 million decimal digits should be about 3.3 million bits;
>>divide by 8 and get about 410,000 bytes.
>>
>>Wait a minute - that's LESS than your supposedly compressed
>>file! What's going on? It can't be those line numbers.
>>Am I making some dumb mistake?
>Are you still assuming there's 5 bits/char?
No, since the data was decimal digits, I was assuming
ln(10)/ln(2) or about 3.3 bits per character.
But maybe you've solved the puzzle: maybe whatever data
compression Bernier used to generate his .zip file
was not smart enough to take full advantage of the
fact that the data was just numbers! Seems dumb, but...
At 5 bits/char we'd get 5 million bits or 625,000 bytes;
then some not-terribly-smart program might compress that
down to 484,760 bytes and feel proud of itself.
>Also the data was on cards. There's a different encoding with
>cards that has to do with two fields and one hole punch in each.
>Geography of the holes was how characters were defined.
Oh-oh - now this is getting too complicated for me to understand.
>Also 20,000 cards ain't very many cards. It's only 10 boxes.
That's irrelevant, your honor! What's at stake here is the
number of bits, and whether the defendent compressed the data
in an ill-advised manner.
John Baez
Gregory L. Hansen <glha...@steel.ucs.indiana.edu> wrote:
>In article <87u0po3...@nonospaz.fatphil.org>,
>Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
>>David Bernier <davi...@videotron.ca> writes:
>>> In the 1950's, RAND produced a table of a million
>>> random digits, and another one with
>>> "100,000 Normal Deviates". ( link is below)
>>Be warned. RAND's data is complete crap. It's trivially
>>compressible by a non-negligible fraction of a percent due
>>to incompetant unbiassing.
>Hmm... I remember reading some experiments in parapsychology, and
>correlations were found here and there.
Hi! You mean they used psychics to unearth correlations in the
RAND table of random digits? :-)
>Irony: "Small businesses want relief from the flood of spam clogging their
>in-boxes, but they fear a proposed national 'Do Not Spam' registry will
>make it impossible to use e-mail as a marketing tool."
>http://www.bizjournals.com/houston/stories/2003/11/10/newscolumn6.html
Grrr! One student here estimates he gets a gigabyte of spam per
year, which in information terms equals a pickup truck full of books.
Torbjorn Larsson
it much more.
But it get's worse. An efficiently compressed file is an example of
efficient communication, since the redundancy is removed. (Thus jpeg
and zip files.)
Redundancy is needed of you have a communication channel with loss.
(Shannon.) It's not great of you have an efficient channel.
A new efficient communication form is Ultra Wide Band radio. It's power
spectra looks like white noise on top of the background noise.
Vic Drastik
This is a quote from Coveyou, not Knuth. See the following or google
http://www.quotationspage.com/quote/461.html
What Knuth *did* say is
"Random numbers should not be generated with a method chosen at random."
優onald E. Knuth
Vic
Willem
)>>Someone else claimed that these "random" digits were
)>>complete crap and could be significantly compressed; let's see
)>>if your zip file reflects that.
)>>
)>>1 million decimal digits should be about 3.3 million bits;
)>>divide by 8 and get about 410,000 bytes.
415,241 bytes and change, if my calculator is accurate enough.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
Androcles
Not even close, there are 32 bits for EACH pixel.
32 * 32 for 2 pixels
32* 32 * 32 for 3, and 32 ^ (1024 * 768) for a screenful, and that's
just
a 17 inch monitor. I have used 1200 * 1600 pixels.
Total number of possible pictures = "Invalid input for function"
Never underestimate yourself, let others do it for you. - "Inspector
Morse", Colin Dexter.
Androcles.
mensa...@aol.com
John Baez wrote:
> In article <G5mdnSk4P5e...@rcn.net>, <jmfb...@aol.com>
wrote:
>
> >>Androcles wrote:
>
> >>> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
> >>> How many different pictures are possible?
> >>> Is it more or less than a google?
>
> >>For a 640x480 screen, it's 10^2219433, slightly larger than a
googol.
>
> I'd say you're engaging in understatement,
Deliberately.
> but that would be an
> understatement. "Slightly" larger than 10^100?? This number makes
> a googol look like one. That is, 1.
At least I spelled googol correctly.
gerard46
|> John Baez wrote:
|>> jmfbahciv wrote:
|>>>Androcles wrote:
|>>>> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
|>>>> How many different pictures are possible?
|>>>> Is it more or less than a google?
|>>>For a 640x480 screen, it's 10^2219433, slightly larger than a googol.
|> I'd say you're engaging in understatement, but that would be an
|> understatement. "Slightly" larger than 10^100?? This number makes
|> a googol look like one. That is, 1.
| Not even close, there are 32 bits for EACH pixel.
| 32 * 32 for 2 pixels
| 32* 32 * 32 for 3, and 32 ^ (1024 * 768) for a screenful, and that's
| just
| a 17 inch monitor. I have used 1200 * 1600 pixels.
| Total number of possible pictures = "Invalid input for function"
It doesn't matter what the size of the monitor is (how many inches).
What matters is the screen resolution.
My screen has 2048 x 1536 pixels. ___________________________Gerard S.
Timo Nieminen
Well, bits of storage, anyway.
2048 x 1536 x 32 is about 10^8 bits. Doesn't take much of a hard drive to
outdo that these days.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
mensa...@aol.com
Androcles wrote:
> "John Baez" <ba...@galaxy.ucr.edu> wrote in message
> news:cs1een$28j$1...@glue.ucr.edu...
> > In article <G5mdnSk4P5e...@rcn.net>, <jmfb...@aol.com>
> > wrote:
> >
> >>>Androcles wrote:
> >
> >>>> 7) Pixels on a computer screen. 16,000,000 colors for each
pixel.
> >>>> How many different pictures are possible?
> >>>> Is it more or less than a google?
> >
> >>>For a 640x480 screen, it's 10^2219433, slightly larger than a
googol.
> >
> > I'd say you're engaging in understatement, but that would be an
> > understatement. "Slightly" larger than 10^100?? This number makes
> > a googol look like one. That is, 1.
>
> Not even close, there are 32 bits for EACH pixel.
For 16 million colors it's 24 bits, not 32. Actually, it's
16777216 = 2^24
> 32 * 32 for 2 pixels
> 32* 32 * 32 for 3, and 32 ^ (1024 * 768) for a screenful, and that's
> just a 17 inch monitor.
You said _a_ computer screen. You didn't say _your_ computer screen.
And 640x480x24 is a valid screen setting. In fact, my old digital
camera
just happens to take 640x480 pictures.
> I have used 1200 * 1600 pixels.
So it's 10^18495282 instead. It's still bigger than a googol.
> Total number of possible pictures = "Invalid input for function"
What's the matter, don't know how to use logarithms?
>
>
> Never underestimate yourself, let others do it for you. - "Inspector
> Morse", Colin Dexter.
"Here's a nickel, kid. Get yourself a better computer."
- the unix guru from Dilbert
>
> Androcles.
mensa...@aol.com
Well, yeah, if you just want to store a single picture.
<understatement>
But if you want to store _all_ of them, you'll find today's hard drives
not quite up to the task.
</understatement>
Timo Nieminen
> Timo Nieminen wrote:
> >
> > 2048 x 1536 x 32 is about 10^8 bits. Doesn't take much of a hard
> drive to
> > outdo that these days.
>
> Well, yeah, if you just want to store a single picture.
>
> <understatement>
> But if you want to store _all_ of them, you'll find today's hard drives
> not quite up to the task.
> </understatement>
Of course, but the original question was which of the choices contained
the most information (ie stored at one time). The added choice 7) was a
red herring.
Today's hard rives won't be up to the task of storing _all_ of the
possible arrangements of bits that could possibly be stored on even a
mediocre hard drive.
David Bernier
[...]
>>Are you still assuming there's 5 bits/char?
>
>
> No, since the data was decimal digits, I was assuming
> ln(10)/ln(2) or about 3.3 bits per character.
>
> But maybe you've solved the puzzle: maybe whatever data
> compression Bernier used to generate his .zip file
> was not smart enough to take full advantage of the
> fact that the data was just numbers! Seems dumb, but...
>
> At 5 bits/char we'd get 5 million bits or 625,000 bytes;
> then some not-terribly-smart program might compress that
> down to 484,760 bytes and feel proud of itself.
[...]
Using WinRK 2.0, the 1,000,000-byte file
of 1,000,000 digits is compressed to a file
with an amazing 419,732 bytes!
David Bernier
P.S. It took about 15 seconds with CPU being Athlon 2200+
Androcles
"Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0501120928540.8948-100000@localhost...
ROFLMAO!
Androcles
Androcles
<mensa...@aol.com> wrote in message
news:1105487950.5...@f14g2000cwb.googlegroups.com...
Sure do, bubba.
What's the matter, don't know how to raise 32 to the power (1024 *
768)?
Androcles.
mensa...@aol.com
I'm not the one who got "Invalid input for function".
mensa...@aol.com
Furthermore, why would I want to raise 32 to the power of 1024*768?
That only gives you a number with 1,183,698 decimal digits, whereas
the _correct_ answer (2^(32*1024*768)) has 7,575,667 decimal digits.
Androcles
<mensa...@aol.com> wrote in message
news:1105494712....@c13g2000cwb.googlegroups.com...
You would if you used a calculator.
How many bits do you imagine they hold as an exponent anyway?
The Ghost In The Machine
<mensa...@aol.com>
wrote
on 11 Jan 2005 10:40:59 -0800
<1105468858....@z14g2000cwz.googlegroups.com>:
>
> jmfb...@aol.com wrote:
>> In article <1105420562.6...@c13g2000cwb.googlegroups.com>,
>> "mensa...@aol.compost" <mensa...@aol.com> wrote:
>> >
>> >Androcles wrote:
>> <snip>
>>
>> >> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
>> >> How many different pictures are possible?
>> >> Is it more or less than a google?
>> >
>> >For a 640x480 screen, it's 10^2219433, slightly larger than a
> googol.
>> But none of it is information.
>
> I wouldn't say none. A lot of it would be useless. But just because
> you can't tell the difference between a picture of a polar bear in
> a blizzard and the inside of a ping pong ball doesn't mean the
> picture of the inside of a ping pong ball isn't an accurate
> representation.
>
> If I had all 10^2219433 possible images, I would have every picture
> taken by my digital camera, every picture that ever will be taken and
> every possible picture my camera is capable of.
Worse than that. You'd have an awful lot of pictures of just about
everything, of varying quality; after all, for every "perfect"
picture, there are about 640x480x(2^24 - 1) = 5,153,960,448,000
which differ from that picture by just changing the color of 1 pixel.
That's an awful lot of copies of one's favorite Swimsuit model... :-)
> This would include
> closeups of every star in the universe. The thing is, a single image
> could represent every star in the universe and be indistinguishable
> from the inside of a ping pong ball.
>
> So it's not a question of whether it's information but whether it's
> usefull information.
>>
>> /BAH
>>
>> Subtract a hundred and four for e-mail.
>
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Androcles
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
message news:7dgeb2-...@sirius.athghost7038suus.net...
Wrong.
(2^24-1) colours for ONE pixel.
For two pixels, (2^24-1) * (2^24-1) .
For three pixels, (2^24-1) * (2^24-1) * (2^24-1)
For 640 pixels, (2^24-1) raised to the POWER of 640
Now calculate (2^24-1) ^ (640x480).
Discard all pictures of black cats in coal cellars at midnight.
Discard all pictures of polar bears in blizzards.
Discard all pictures of a cloudless daylight sky.
Discard all pictures of cherries on a bed of tulips.
Discard all pictures of blades of grass in closeup.
How many do you have left?
Androcles.
mensa...@aol.com
Maybe you would, but _I_ know how to use logs.
On Windows calculator:
[2][4]
[*]
[6][4][0]
[*]
[4][8][0]
[*]
[2]
[ln]
[/]
[1][0]
[ln]
[=]
2219433.9520314005560718541230247
> How many bits do you imagine they hold as an exponent anyway?
Doesn't matter. I only said there were 2219433 digits. I never
said what the digits were.
mensa...@aol.com
Timo Nieminen wrote:
> On Wed, 11 Jan 2005, mensa...@aol.compost wrote:
>
> > Timo Nieminen wrote:
> > >
> > > 2048 x 1536 x 32 is about 10^8 bits. Doesn't take much of a hard
> > drive to
> > > outdo that these days.
> >
> > Well, yeah, if you just want to store a single picture.
> >
> > <understatement>
> > But if you want to store _all_ of them, you'll find today's hard
drives
> > not quite up to the task.
> > </understatement>
>
> Of course, but the original question was which of the choices
contained
> the most information (ie stored at one time). The added choice 7) was
a
> red herring.
Oh, well I never did like tartar sauce. I was just trying to answer
the question of how the number of pictures displayable on your
computer compares to a googol (I did manage to avoid the red herring
of comparing it to a google).
tj Frazir
nice boat ,,
hongkong polar giant " snow dragon"
xue01.jpg
Address:http://www.gakei.com/xue/xue01.jpg Changed:8:34 AM on Sunday,
October 31, 2004
You can see sony in hong kong all the way from si dock ..look to the
left of the big ice breaker ,,yea thats a polar icebreaker in HK..so
what
fdb93fd9bcdfc63348916efc1e802570
Address:http://www.fotofinity.com/link/full/fdb93fd9bcdfc63348916efc1e802570
Lewis Mammel
John Baez wrote:
>
> In article <crt513$rjo$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu> wrote:
>
> >6) the information required to precisely describe the position/momentum
> > of all the atoms in a red blood cell up to the limits imposed by
> > quantum mechanics.
>
> I didn't actually work this one out, but I worked out the answer for
> a 1-milligram raindrop, and you can find it here:
>
> http://math.ucr.edu/home/baez/information.html
It seems a little strange to me that you remark, "Boltzmann, Shannon
and others figured out how entropy and information are related" but say
nothing whatsoever about the nature of this relationship. You make
remarks like, "To understand this ..." but you might better have
said, "You can't understand this, so I'm not going to explain"
I know you put a lot of effort into this stuff, and it's all
very well intentioned, but all that breezy stuff about the
fuzzy moles? I dunno. I was always ambivalent about Sagan's
COSMOS show, too, so maybe it's just me. I dunno.
Also, I'm peevish about the idea of specifying the position
and velocity of water molecules in a liquid state. In a gas,
we have a phase plane, px * x, divided into cells of size h squared,
so there's a nice segue from the classically specified state of
motion and the enumeration of quantum states. I don't think you
have that in liquid water.
Note that you end up with 4E24 bits to specify the momentum and
position of 6E23/18 molecules, so that's 120 bits per molecule.
Dividing six ways, you have 20 bits per phase space axis, which
brings you down to 1E-6 cm per displacement space axis. But this
is 200 times the Bohr radius. What kind of information is this?
"By my calculation, they're all in there somewhere." It's particularly
egregious to say that the water drop "holds" this information.
Well, it's all very thought provoking, anyway, so I'm not
really complaining, just ... commenting.
Lew Mammel, Jr.
Timothy Little
> Note that you end up with 4E24 bits to specify the momentum and
> position of 6E23/18 molecules,
Your calculation assumes that permutations of molecules are distinct.
Specifying a permutation of 3.3e22 molecules requires an extra 75 bits
per molecule, for a total of 195 bits per distinguished molecule.
- Tim
jmfb...@aol.com
"mensa...@aol.compost" <mensa...@aol.com> wrote:
>
>jmfb...@aol.com wrote:
>> In article <1105420562.6...@c13g2000cwb.googlegroups.com>,
>> "mensa...@aol.compost" <mensa...@aol.com> wrote:
>> >
>> >Androcles wrote:
>> <snip>
>>
>> >> How many different pictures are possible?
>> >> Is it more or less than a google?
>> >
>> >For a 640x480 screen, it's 10^2219433, slightly larger than a
>googol.
>
>I wouldn't say none. A lot of it would be useless. But just because
>you can't tell the difference between a picture of a polar bear in
>a blizzard and the inside of a ping pong ball doesn't mean the
>picture of the inside of a ping pong ball isn't an accurate
>representation.
It isn't an accurate representation. The picture is only in
your imagination. At no time would you be able to compare
the picture of the ping pong ball insides with a the polar
bear and be able to accurately specify which was which.
>
>If I had all 10^2219433 possible images, I would have every picture
>taken by my digital camera, every picture that ever will be taken and
>every possible picture my camera is capable of. This would include
>closeups of every star in the universe. The thing is, a single image
>could represent every star in the universe and be indistinguishable
>from the inside of a ping pong ball.
>
>So it's not a question of whether it's information but whether it's
>usefull information.
It is not information. Take an area of your disk; set all bits
to 1. You now have all possible programs. Now execute it.
Is the CPU generating all the work of all programs? NO.
You get a fault.
Androcles
<mensa...@aol.com> wrote in message
news:1105505378.9...@c13g2000cwb.googlegroups.com...
24*640*480*2 = 14745600
Why do you want the ln() of that?
When I use logarithms to calculate x^n, I take the log of x, multiply
that by the exponent n, and take the anti-log of the result.
You have attempted to calculate.... hmm...
ln( ln(4951200)/10).
How strange... what does it represent?
What's the matter, don't know how to use logarithms?
Androcles.
jmfb...@aol.com
ba...@galaxy.ucr.edu (John Baez) wrote:
>In article <G5mdnSk4P5e...@rcn.net>, <jmfb...@aol.com> wrote:
>
>>>Androcles wrote:
>
>>>> 7) Pixels on a computer screen. 16,000,000 colors for each pixel.
>>>> How many different pictures are possible?
>>>> Is it more or less than a google?
>
>>>For a 640x480 screen, it's 10^2219433, slightly larger than a googol.
>
>I'd say you're engaging in understatement, but that would be an
>understatement. "Slightly" larger than 10^100?? This number makes
>a googol look like one. That is, 1.
And that's (1) is all it is. Without a process, all of those
combinations are as useful as a single bit. And you can't
do much with only one solitary bit.
jmfb...@aol.com
ba...@galaxy.ucr.edu (John Baez) wrote:
>In article <G5mdnS44P5f...@rcn.net>, <jmfb...@aol.com> wrote:
>
>>In article <crvr1o$r2n$1...@glue.ucr.edu>,
>> ba...@galaxy.ucr.edu (John Baez) wrote:
>
>>>In article <iUBEd.36478$SB6.1...@wagner.videotron.net>,
>>>David Bernier <ez...@yahoo.com> wrote:
>
>>>>In the 1950's, RAND produced a table of a million
>>>>random digits, and another one with
>>>>"100,000 Normal Deviates". ( link is below)
>
>I keep on reading this as "100,000 normal deviants", which seems
>like a description of usenet news. Apparently the librarians thought
>similarly, since this book was first filed under psychology.
It's a good place for it. Whenever we talked about this stuff
we went nuts.
>
>>>>Having downloaded the 1 million digits (line numbers included),
>>>>I wrote a program to save them as a file with
>>>>12,500 lines at 80 digits per line.
>>>>
>>>>This got compressed to a 484,760-byte *.zip file.
>
>>>Someone else claimed that these "random" digits were
>>>complete crap and could be significantly compressed; let's see
>>>if your zip file reflects that.
>>>
>>>1 million decimal digits should be about 3.3 million bits;
>>>divide by 8 and get about 410,000 bytes.
>>>
>>>Wait a minute - that's LESS than your supposedly compressed
>>>file! What's going on? It can't be those line numbers.
>>>Am I making some dumb mistake?
>
>>Are you still assuming there's 5 bits/char?
>
>No, since the data was decimal digits, I was assuming
>ln(10)/ln(2) or about 3.3 bits per character.
Which gives you four, not five, for max storage needed for _one_
decimal number.
>
>But maybe you've solved the puzzle: maybe whatever data
>compression Bernier used to generate his .zip file
>was not smart enough to take full advantage of the
>fact that the data was just numbers! Seems dumb, but...
Not really. It's a common error for those who don't
breathe binary or octal. If the guy only thought in hex,
he'ld have other problems. There are lots of "formulas"
for extracting characters out of packed bits. RADIX-50
was commonly used. This repeated reference smells of Bardot..
I just felt my brain spaz so this word is incorrect.
>
>At 5 bits/char we'd get 5 million bits or 625,000 bytes;
>then some not-terribly-smart program might compress that
>down to 484,760 bytes and feel proud of itself.
And, if I could write a 2-line FORTRAN program that generated
those 5 million bits, I'd have "compressed" the data down
to a maximum of 144 characters. I won't complicate your thinking
with mumblings about how many extra characters it takes to store
these 5 million numbers on media based on electricity.
>
>>Also the data was on cards. There's a different encoding with
>>cards that has to do with two fields and one hole punch in each.
>>Geography of the holes was how characters were defined.
>
>Oh-oh - now this is getting too complicated for me to understand.
Yep. That's how the conversation usually goes as soon as we
have to get practical and descent the ivory towers ;-).
>
>>Also 20,000 cards ain't very many cards. It's only 10 boxes.
>
>That's irrelevant, your honor! What's at stake here is the
>number of bits,
You are counting the bits which are place holders.
> ...and whether the defendent compressed the data
>in an ill-advised manner.
I would need to know the method of storage; the method of measuring
the size before _and_ after; and I usually want to know how the
original was manufactured. It has been my observation that each
pass at data modifies it; that's just how people work.
Lewis Mammel
Lewis Mammel wrote:
> we have a phase plane, px * x, divided into cells of size h squared,
Of course, px *x has units of h, not h squared.
Tim Arheit
Not surprising at all given that the million digits are all characters
0-9. (ie., lots of wasted bits to make it human readable).
In binary form the original million digits file is only 415,251
bytes. It can be downloaded from
http://www.datacompression.info/Miscellaneous/AMillionRandomDigits.bin
(at least for now).
-Tim
matma...@yahoo.com
That should be 415,241 bytes. Anyway there is a long standing
challenge to compress this file. The compressed size has to include
the size of the decompression program, but this may packed into an
archive along with the compressed file, along the lines of the rules of
the Calgary corpus challenge.
http://mailcom.com/challenge/ (but without the prize money).
So far we know that the sum of every 50th digit is even (50 bits of
redundancy) and there is a slight bias in the distribution of pairs of
digits spaced 50 or 100 places apart. This bias is very small, perhaps
in the tens of bits, not enough to write a decompressor. I am pretty
sure that there are larger biases, enough to meet the challenge,
waiting to be discovered.
Lewis Mammel
I was following the classical prescription, based on the suggestion
that the entropy prescribes the amount of information required to
do this. Of course, you make a good point, but the extra 75 bits
per molecule aren't there! ( That's N log N / N / log 2 . )
So what could you do with the bits you have?
If you want to glom indistinguishability onto a classical
enumeration, you still have N different positions, but the list
could be ordered, for example, so you might compress the information
by specifying successive differences, or something like that.
I think it's a moot point.
Lew Mammel, Jr.
mensa...@aol.com
I'm not, you parsed it wrong. The above keystrokes parse as
(24*640*480) * ((ln(2)/(ln(10))
>
> When I use logarithms to calculate x^n, I take the log of x, multiply
> that by the exponent n, and take the anti-log of the result.
> You have attempted to calculate.... hmm...
> ln( ln(4951200)/10).
> How strange... what does it represent?
Parsed incorrectly, it doesn't mean anything. Parsed correctly,
it represents:
(24*640*480) is the number of bits in the image (7372800).
If you consider that blob as a single base 2 number, it has
7372800 digits. To figure out how mnay base 10 digits that would
require, you multiply the number of base 2 digits by ln(2)/ln(10)
0.30102999566398119521373889472449 or simply 0.3 if you want a
rough estimate.
> What's the matter, don't know how to use logarithms?
So, using logarithms, it is possible to tell the magnitude of the
answer without actually having to calculate it.
I'm surprised that a refugee from sci.physics doesn't understand
orders of magnitude calculations.
>
>
>
> Androcles.
Androcles
<mensa...@aol.com> wrote in message
news:1105551443.8...@f14g2000cwb.googlegroups.com...
Yeah... Hopeless.
>
>>
>> When I use logarithms to calculate x^n, I take the log of x, multiply
>> that by the exponent n, and take the anti-log of the result.
>> You have attempted to calculate.... hmm...
>> ln( ln(4951200)/10).
>> How strange... what does it represent?
>
> Parsed incorrectly, it doesn't mean anything. Parsed correctly,
> it represents:
>
> (24*640*480) is the number of bits in the image (7372800).
Wrong.
24 bits per pixel, so 24^2 bits for two pixels.
24^3, so
24 ^ (640*480) ,
not
24 * (640*480)
So you know how to use logs, huh?
I wonder if you know how to raise to a power....
Androcles.
mensa...@aol.com
Why don't you ponder this a little further.
> 24 bits per pixel, so 24^2 bits for two pixels.
So what about the Macintosh I used to have that had 1 bit per pixel?
> 24^3, so
> 24 ^ (640*480) ,
So you're saying that my Macintosh could only display
1^(640*480)
images? Try punching that up in your calculator to see exactly
how wrong you are.
> not
> 24 * (640*480)
>
> So you know how to use logs, huh?
> I wonder if you know how to raise to a power....
All the know-how in the world does you no good if you can't
correctly apply it.
Timothy Little
> I was following the classical prescription, based on the suggestion
> that the entropy prescribes the amount of information required to do
> this.
Classical or not, swapping water molecules does not create a new
physical state. Let phase space be divided into M bins, with N
molecules occupying (some of) those bins. Two states are distinct if
there exists a bin with different number of molecules between the
states.
How many possible states are there? Your calculation assumes that the
answer is M^N.
- Tim
Lewis Mammel
Timothy Little wrote:
>
> Lewis Mammel wrote:
> > I was following the classical prescription, based on the suggestion
> > that the entropy prescribes the amount of information required to do
> > this.
>
> Classical or not, swapping water molecules does not create a new
> physical state. Let phase space be divided into M bins, with N
> molecules occupying (some of) those bins. Two states are distinct if
> there exists a bin with different number of molecules between the
> states.
Consulting Reif, I note he does say, "Note that in a STRICTLY
classical description it would be permissible to consider every
particle distinguishable." But this is just backfilling against
the already made decision to adopt quasi-indistinguishability.
> How many possible states are there? Your calculation assumes that the
> answer is M^N.
On John Baez's page the entropy of an object was stated as being,
"how much information it takes to completely specify all the positions
and velocities of all the particles composing it."
To me, this indicates the STRICTLY classical viewpoint, Just as
though one had said "all the rocks" or "all the marbles" instead
of "all the particles".
However, this runs up against the third law, that the entropy
be 0 at absolute 0. So from a classical point of view we would
have to suppose that the stated entropy value is the difference
of the entropy at the given temperature and the entropy at
absolute zero, where it would just be log( N! ) - we require
the uncertainty principle to make our positional box have a
single bin! Like Reif says, "The Gibbs paradox thus conceptually
foreshadowed already in the last century [ sic ] conceptual
difficulties that were resolved satisfactorily only by the
advent of quantum mechanics."
So the reasonable thing to do is to add in the 75 bits
and say that THIS is the information required for a complete
specification, which is what you suggested of course, and the
answer it gives is a lot more reasonable too. With an extra
12 bits per axis, we go from 200 a0 to 1/20 a0, which is still
a fair amount of uncertainty.
Actually, I'm sloughing off the issue of how to divide the
bits between momentum and position. I'm just assuming there's
a "momentum box" of side px and taking:
d_px / px * d_x / x = h/px/x
then taking d_px / px = d_x / x = sqrt( h/px/x )
but this is rather arbitrary and falls apart when px -> 0 .
Anyway, with this level of uncertainty, all this information
is completely lost in a picosecond or so, if we try to reason
semi-classically. How are we even supposed to imagine a classically
specified configuration of molecules in a liquid with this
uncertainty? To me it certainly contradicts the notion that
the water "holds" the information, which I see as a throwback
to determinism.
The philospher Hilary Putnam, for one, devotes a lot of time to
discussing brain states as though they can be treated deterministically,
and this was shocking to me when I saw it. There is a general view
that quantum uncertainty somehow only applies to the arcane,
and the classical world rolls on as far as we are concerned.
I most emphatically assert that this is not the case.
Lew Mammel, Jr.
John Baez
Lewis Mammel <l.ma...@worldnet.att.net> wrote:
>On John Baez's page
http://math.ucr.edu/home/baez/information.html
>the entropy of an object was stated as being,
>"how much information it takes to completely specify all the positions
>and velocities of all the particles composing it."
You may have read a preliminary version. Later I decided
this sort of statement would make people wonder if I'd
forgotten about the uncertainty principle. So, I decided
to be a wee bit clearer, and say:
A gram of water (at room temperature and pressure) holds
2.81 x 10^{24} nits
of information.
In other words, this is how much information it takes to completely
specify all the positions and velocities of all the particles composing
it, up to the limits imposed by the uncertainty principle! (If the
uncertainty principle didn't cut in, the information would be infinite -
that's one reason we need quantum mechanics.)
Even this is oversimplified, because I'm not trying to give a
course on entropy here. I'm just trying to give the reader a rough
sense of how entropy is related to information.
If I were trying to be more precise, I might have said that in
quantum theory the entropy is the amount of information it takes
to specify the microstate of a system, given the macrostate.
>To me, this indicates the STRICTLY classical viewpoint, Just as
>though one had said "all the rocks" or "all the marbles" instead
>of "all the particles".
No, we really want and *need* to take quantum theory into account
to properly treat issues like the entropy of water starting from
first principles.
But, I wasn't trying to explain this stuff in detail - just
compute the amount of information in a raindrop! So, I just
wanted to say the bare minimum about entropy, information and
quantum mechanics needed to give the reader a vague idea of
what the computation actually meant.
Lewis Mammel
John Baez wrote:
>
> In article <41E5F13D...@worldnet.att.net>,
> Lewis Mammel <l.ma...@worldnet.att.net> wrote:
>
> >On John Baez's page
>
> http://math.ucr.edu/home/baez/information.html
>
> >the entropy of an object was stated as being,
> >"how much information it takes to completely specify all the positions
> >and velocities of all the particles composing it."
>
> You may have read a preliminary version. Later I decided
> this sort of statement would make people wonder if I'd
> forgotten about the uncertainty principle. So, I decided
> to be a wee bit clearer, and say:
>
> A gram of water (at room temperature and pressure) holds
>
> 2.81 x 10^{24} nits
>
> of information.
>
> In other words, this is how much information it takes to completely
> specify all the positions and velocities of all the particles composing
> it, up to the limits imposed by the uncertainty principle! (If the
> uncertainty principle didn't cut in, the information would be infinite -
> that's one reason we need quantum mechanics.)
I saw this, but it leaves the classical view intact, merely placing
a limit on it, which is necessary, as you say, if one is to pursue
the quantification of the information involved.
> Even this is oversimplified, because I'm not trying to give a
> course on entropy here. I'm just trying to give the reader a rough
> sense of how entropy is related to information.
You never did this. You merely asserted a proportional relation
with no justification except that it had been "figured out".
Since this assertion is a paraphrase of the immortal S = k log W,
I guess that's something, but this is the point that requires
explanation. It seemed to me you glided over it and chose to
dwell unduly on superficial arithmetic.
> If I were trying to be more precise, I might have said that in
> quantum theory the entropy is the amount of information it takes
> to specify the microstate of a system, given the macrostate.
There you go. Quantum theory is necessary to have any grasp
of this. Classical physics is utterly incompetent to reveal
the secrets of entropy.
> >To me, this indicates the STRICTLY classical viewpoint, Just as
> >though one had said "all the rocks" or "all the marbles" instead
> >of "all the particles".
>
> No, we really want and *need* to take quantum theory into account
> to properly treat issues like the entropy of water starting from
> first principles.
Yes, we need to chuck classical physics overboard when speaking
of the microstate of water, so why did you adhere to the idea
that its state can be specified as a classical microcanonical
ensemble? You're pandering to classical revanchism!
I'm looking at Schroedinger's booklet, Statistical Thermodynamics,
from 1944 - "In principle we have always in mind a quantum-mechanical
system" - Go for it!
> But, I wasn't trying to explain this stuff in detail - just
> compute the amount of information in a raindrop! So, I just
> wanted to say the bare minimum about entropy, information and
> quantum mechanics needed to give the reader a vague idea of
> what the computation actually meant.
But you didn't compute it, in any meaningful sense. You looked
up the entropy in a table and asserted that it measured the
information required to specify etc. etc. ... which really, it
doesn't.
Well, I'm being way to critical, as I keep saying, and I can
reduce all my objections to quibbles if I work at it. But these
were my thoughts.
Lew Mammel, Jr.
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:odeFd.112084$48.3...@fe1.news.blueyonder.co.uk...
[snip]
> > Parsed incorrectly, it doesn't mean anything. Parsed correctly,
> > it represents:
> >
> > (24*640*480) is the number of bits in the image (7372800).
>
> Wrong.
> 24 bits per pixel, so 24^2 bits for two pixels.
> 24^3, so
> 24 ^ (640*480) ,
> not
> 24 * (640*480)
>
> So you know how to use logs, huh?
> I wonder if you know how to raise to a power....
>
>
> Androcles.
Nice one!
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LogsHuh.html
"So you know how to use logs, huh?"
Androcles and vectors:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorLength.html
Androcles and limits:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegativeInfinity.html
Androcles and equations:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras2.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AndroDistri.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ToothlessBite.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Competent.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/UseTrans.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Sheesh.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/DivZero.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Think.html
Androcles and square roots:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/STILL.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CanSpecify.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GrowUp.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Material.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html
Androcles and exclusive ors:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html
Androcles and partial differential equations:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff2.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff3.html
Dirk Vdm
ste...@nomail.com
: <mensa...@aol.com> wrote in message
: Wrong.
: 24 bits per pixel, so 24^2 bits for two pixels.
: 24^3, so
: 24 ^ (640*480) ,
: not
: 24 * (640*480)
: So you know how to use logs, huh?
: I wonder if you know how to raise to a power....
: Androcles.
Lets try that with 2 bits per pixel, and three pixels.
How many images are there? Here's all of them.
00 00 00 00 01 00 00 10 00 00 11 00
00 00 01 00 01 01 00 10 01 00 11 01
00 00 10 00 01 10 00 10 10 00 11 10
00 00 11 00 01 11 00 10 11 00 11 11
01 00 00 01 01 00 01 10 00 01 11 00
01 00 01 01 01 01 01 10 01 01 11 01
01 00 10 01 01 10 01 10 10 01 11 10
01 00 11 01 01 11 01 10 11 01 11 11
10 00 00 10 01 00 10 10 00 10 11 00
10 00 01 10 01 01 10 10 01 10 11 01
10 00 10 10 01 10 10 10 10 10 11 10
10 00 11 10 01 11 10 10 11 10 11 11
11 00 00 11 01 00 11 10 00 11 11 00
11 00 01 11 01 01 11 10 01 11 11 01
11 00 10 11 01 10 11 10 10 11 11 10
11 00 11 11 01 11 11 10 11 11 11 11
Should be pretty clear that there are 2^6 = 64 images.
According to Androcles logic there should be be
2^(2^3) = 256 pictures. Where are the other 192 pictures?
Stephen
Randy Poe
> > Maybe you would, but _I_ know how to use logs.
> >
> > On Windows calculator:
> >
> > [2][4]
> > [*]
> > [6][4][0]
> > [*]
> > [4][8][0]
> > [*]
> > [2]
> > [ln]
> > [/]
> > [1][0]
> > [ln]
> > [=]
> > 2219433.9520314005560718541230247
>
> 24*640*480*2 = 14745600
> Why do you want the ln() of that?
log10(2^(24*640*480)) = ln(2^(24*640*480))/ln(10)
= ln(2)*24*640*480/ln(10)
That's what the above sequence of keys is
calculating.
The first [ln] gives you ln(2) on screen.
Hitting the / key does the multiplication of
all the terms to date (24*640*480*ln(2)) and gets
ready to divide by what comes next.
[1][0][ln] gives you ln(10).
The last [=] gives you 24*640*480*ln(2)/ln(10).
The point of calculating log10 of a large number is
to tell you how many digits it has, in base 10.
- Randy
Dirk Van de moortel
<ste...@nomail.com> wrote in message news:cs6b75$2aum$1...@msunews.cl.msu.edu...
Do not try to teach an Ape.
It does *not* work.
Dirk Vdm
Brad Guth
How about unauthorised copies of information, or even authorised copies?
As I tend to believe the perpetrated disinformation and vast
reproductions of such crapolla far outweighs the truth and nothing but
the truth, perhaps by at least a ten fold factor, and then some.
Most high resolution video is entirely artificial to start with, or
summarily modified to suit whatever the mainstream status quo desires,
as for that we're talking google tera bytes worth, and that's just the
smut portion. The digital music archives and vast copies distributed is
another appalling fiasco having no end in sight.
Therefore, how much honest information do we actually need?
How much global energy is going into the creation, conversions,
archiving, publishing, distributing and educational aspects of said
information?
The global internet and of it's online and off-line users are consuming
how many gigawatts/hr?
Regards, Brad Guth / GASA-IEIS
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
mensa...@aol.com
That's not quite right. Androcles logic actually implies
pictures = depth^area
so for your example, his logic would be
pictures = 2^(1*3) = 8
which is still wrong. He underestimates the count.
> Where are the other 192 pictures?
Since the actual way to count pictures is
pictures = 2^volume
the underestimate between
32^(1024*768) and 2^(32*1024*768) is a whopping
6000000+ orders of magnitude.
Apparently, he thinks that just because he got a
million digit number, it must be correct.
>
>
> Stephen
ste...@nomail.com
: ste...@nomail.com wrote:
:>
:>
:> Should be pretty clear that there are 2^6 = 64 images.
:>
:> According to Androcles logic there should be be
:> 2^(2^3) = 256 pictures.
: That's not quite right. Androcles logic actually implies
: pictures = depth^area
I actually was not sure exactly what his logic was, and
realized that was a possible interpretation. To me
it seemed like he might have been arguing about what the exponent
should, not what the total number of images was. Either way
he is wrong.
Stephen
Androcles
<ste...@nomail.com> wrote in message
news:cs77ta$53k$1...@msunews.cl.msu.edu...
Good grief....
Suppose each pixel is either black or white. How many monochrome
pictures can you display on a 640x800 screen?
Start with a 2x1 screen.
bb
bw
wb
ww
= 2^2 = 4
now a 3x1 screen
bbb
bbw
bwb
bww
wbb
wbw
wwb
www
= 2 ^ 3 = 8
That same 8 above with a
2x2 screen will have
b(xxx) and w(xxx) = 2^4 = 16 possible images.
so, its the number of colours raised to the power of the number
of pixels.
with 24 bits to each pixel, we have
2^24 possible colours per pixel. That's
(2^24) = (16,777,216) colours
raised to the power of 512000 = (a number too large for a calculator).
Androcles.
mensa...@aol.com
And surely you now realize by the law of exponents
(x^m)^n = x^(m*n)
that (2^24)^(640*800) = 2^(24*640*800) which is what I said.
Mike
John Baez wrote:
[snip]
>
> But, I wasn't trying to explain this stuff in detail - just
> compute the amount of information in a raindrop!
[snip]
Now that you "know" the amount of information, can you use it in any
way to replicate the rain drop?
If not, this means this is the apparent information, or information at
the phenomenal level. The information needed at a substance level to
replicate the exact same rain drop could be significantly higher or
even asymptotic to infinity.
I argue that even if you know the amount of information you claim that
is not nearly enough to replicate the rain drop.
Mike
Androcles
<mensa...@aol.com> wrote in message
news:1105683474....@c13g2000cwb.googlegroups.com...
If I recall correctly, you claimed to be able to evaluate the
number using Window's calculator and logarithms, jeering at me.
Crow is a favorite on the menu for you, is it?
Androcles.
jmfb...@aol.com
"Mike" <ele...@yahoo.gr> wrote:
>
>John Baez wrote:
>
>[snip]
>>
>> But, I wasn't trying to explain this stuff in detail - just
>> compute the amount of information in a raindrop!
>
>[snip]
>
>Now that you "know" the amount of information, can you use it in any
>way to replicate the rain drop?
That depends. However, now he knows how to recognize a raindrop.
Recognition is not a trivial problem. Another interesting poser
is how much of the base information can be changed and still
have the item recognized as a raindrop? This is what compression
is all about.
>
>If not, this means this is the apparent information, or information at
>the phenomenal level. The information needed at a substance level to
>replicate the exact same rain drop could be significantly higher or
>even asymptotic to infinity.
>
>I argue that even if you know the amount of information you claim that
>is not nearly enough to replicate the rain drop.
First things first. Before you can think about how to replicate it,
you have to learn how to recognize it.
Mike
essentially argue that the information needed for recognition of a
small subset of the information needed for replication.
Mike
Willem
) John Baez wrote:
)> But, I wasn't trying to explain this stuff in detail - just
)> compute the amount of information in a raindrop!
)
) [snip]
)
) Now that you "know" the amount of information, can you use it in any
) way to replicate the rain drop?
Obviously not. If I know the amount of information on a floppy disk, can I
recreate that floppy disk ?
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
ste...@nomail.com
: <ste...@nomail.com> wrote in message
: news:cs77ta$53k$1...@msunews.cl.msu.edu...
:> In sci.math mensa...@aol.compost <mensa...@aol.com> wrote:
:>
:> : ste...@nomail.com wrote:
:> :>
:> :>
:> :> Should be pretty clear that there are 2^6 = 64 images.
:> :>
:> :> According to Androcles logic there should be be
:> :> 2^(2^3) = 256 pictures.
:>
:> : That's not quite right. Androcles logic actually implies
:>
:> : pictures = depth^area
:>
:> I actually was not sure exactly what his logic was, and
:> realized that was a possible interpretation. To me
:> it seemed like he might have been arguing about what the exponent
:> should, not what the total number of images was. Either way
:> he is wrong.
:>
:> Stephen
: Good grief....
Yes, good grief. To quote you
Wrong.
24 bits per pixel, so 24^2 bits for two pixels.
24^3, so
So according to you two 24 bit pixels contains 576 bits.
Here are two 24 bit pixels
000010101101000100100000
100010101101010100100011
There are 48 bits total, not 576 as you claimed.
For three 24 bit pixels there are 72 bits total,
not 24^3=13824 as you claimed.
Stephen
Mike
1. Get an enpty floppy disk
2. Transfer the information to the empty disk
3. Now, if you've done that carefully and these is no label on the
disks, you will not be able to distinguish between the two.
4. Do not repeat the process with a third disk because you will be
accused of pirating information:)
Mike
Willem
) Absolutely, follow the steps below:
)
) 1. Get an enpty floppy disk
) 2. Transfer the information to the empty disk
) 3. Now, if you've done that carefully and these is no label on the
) disks, you will not be able to distinguish between the two.
) 4. Do not repeat the process with a third disk because you will be
) accused of pirating information:)
Hmm, maybe I misunderstood what you meant by recreating a raindrop.
If I read this correctly, you're asking to recreate a raindrop, given
that you have access to the original ?
Androcles
<ste...@nomail.com> wrote in message
news:cs92rm$1oh1$1...@msunews.cl.msu.edu...
Here are 10 digits, 0,1,2,3,4,5,6,7,8,9
999 uses only 3 digits total, not a thousand minus one as you claim.
Here is a 2x2 screen, and all the monochrome pictures that you can
display.
__
|00| 01 10 11 00 01 10 11
|00| 00 00 00 01 01 01 01
__
00 01 10 11 00 01 10 11
10 10 10 10 11 11 11 11 total 16.
================================
Here is a 2x2 screen with some Red Green Blue either on or off
00 0R 0G 0B R0 G0 B0
00 00 00 00 00 00 00
How many possible pictures are there?
Androcles.
Mike
Willem wrote:
> Mike wrote:
> ) Absolutely, follow the steps below:
> )
> ) 1. Get an enpty floppy disk
> ) 2. Transfer the information to the empty disk
> ) 3. Now, if you've done that carefully and these is no label on the
> ) disks, you will not be able to distinguish between the two.
> ) 4. Do not repeat the process with a third disk because you will be
> ) accused of pirating information:)
>
> Hmm, maybe I misunderstood what you meant by recreating a raindrop.
>
> If I read this correctly, you're asking to recreate a raindrop, given
> that you have access to the original ?
>
Almost. Your example was a good one though. It seems that all the
information you can have is not enough if you do not have the proper
media to store it.
It seems that the type of information obtained by physics, such as
states of motion, is good only for a "mental" or virtual reproduction
of physical reality. hands on rerpoduction required much more
information than that, which physics does not know what it is.
Mike
Randy Poe
answer directly. Nevertheless I can't leave this
illogic uncommented.
Androcles wrote:
> <ste...@nomail.com> wrote in message
> news:cs92rm$1oh1$1...@msunews.cl.msu.edu...
> > Yes, good grief. To quote you
> > Wrong.
> > 24 bits per pixel, so 24^2 bits for two pixels.
> > 24^3, so
> >
> > So according to you two 24 bit pixels contains 576 bits.
> > Here are two 24 bit pixels
> > 000010101101000100100000
> >
> > 100010101101010100100011
> >
> > There are 48 bits total, not 576 as you claimed.
> > For three 24 bit pixels there are 72 bits total,
> > not 24^3=13824 as you claimed.
> >
> > Stephen
>
> Here are 10 digits, 0,1,2,3,4,5,6,7,8,9
>
> 999 uses only 3 digits total, not a thousand minus one as you claim.
There are a thousand possible 3-digit numbers
(or a thousand minus one, if 000 doesn't count).
> Here is a 2x2 screen, and all the monochrome pictures that you can
> display.
> __
> |00| 01 10 11 00 01 10 11
> |00| 00 00 00 01 01 01 01
> __
>
>
> 00 01 10 11 00 01 10 11
> 10 10 10 10 11 11 11 11 total 16.
>
Or 2^4 (or 2^4-1, not counting the all-black screen).
> ================================
>
> Here is a 2x2 screen with some Red Green Blue either on or off
>
> 00 0R 0G 0B R0 G0 B0
> 00 00 00 00 00 00 00
>
> How many possible pictures are there?
Are you allowing each pixel to only have 4 possible
values: 0, R, G, B? If so, then there are 4^4 = 256
possible pictures.
On the other hand if each pixel is represented with
3 bits (R on/off, G on/off, B on/off), then each
pixel can have 2^3 = 8 different values:
0, R, G, B, RG, RB, GB, RGB
and there are (2^3)^4 = 2^(3*4) = 4096 different
pictures.
- Randy
mensa...@aol.com
Not.
> you claimed to be able to evaluate the number
I never said I could evaluate the number, only evaluate
whether it was bigger than a googol.
> using Window's calculator and logarithms,
Which I did.
> jeering at me.
There was no jeering until you falsely claimed that I was
not even close and then justified it with false mathematics.
> Crow is a favorite on the menu for you, is it?
It's not on the menu. If you want to eat Crow you have to
ask for it. And you've been asking for it.
> Androcles.
Dirk Van de moortel
<mensa...@aol.com> wrote in message news:1105733672.1...@f14g2000cwb.googlegroups.com...
>
> Androcles wrote:
[snip]
> > If I recall correctly,
>
> Not.
Androcles has a very peculiar sense of recalling.
>
> > you claimed to be able to evaluate the number
>
> I never said I could evaluate the number, only evaluate
> whether it was bigger than a googol.
>
> > using Window's calculator and logarithms,
>
> Which I did.
>
> > jeering at me.
>
> There was no jeering until you falsely claimed that I was
> not even close and then justified it with false mathematics.
Androcles thrives on being jeered at.
>
> > Crow is a favorite on the menu for you, is it?
>
> It's not on the menu. If you want to eat Crow you have to
> ask for it. And you've been asking for it.
Androcles has turned asking to get Crow for dinner
into a fine art:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Androrgasm.html
Succulent, right?
Dirk Vdm
ste...@nomail.com
: I'm on Androcles pretend-to-ignore list, so he won't
: answer directly. Nevertheless I can't leave this
: illogic uncommented.
It is amazingly illogical.
: Androcles wrote:
:> <ste...@nomail.com> wrote in message
:> news:cs92rm$1oh1$1...@msunews.cl.msu.edu...
:> > Yes, good grief. To quote you
:> > Wrong.
:> > 24 bits per pixel, so 24^2 bits for two pixels.
:> > 24^3, so
:> >
:> > So according to you two 24 bit pixels contains 576 bits.
:> > Here are two 24 bit pixels
:> > 000010101101000100100000
:> >
:> > 100010101101010100100011
:> >
:> > There are 48 bits total, not 576 as you claimed.
:> > For three 24 bit pixels there are 72 bits total,
:> > not 24^3=13824 as you claimed.
:> >
:> > Stephen
:>
:> Here are 10 digits, 0,1,2,3,4,5,6,7,8,9
:>
:> 999 uses only 3 digits total, not a thousand minus one as you claim.
: There are a thousand possible 3-digit numbers
: (or a thousand minus one, if 000 doesn't count).
What possible point was Androcles trying to make here?
:> Here is a 2x2 screen, and all the monochrome pictures that you can
:> display.
:> __
:> |00| 01 10 11 00 01 10 11
:> |00| 00 00 00 01 01 01 01
:> __
:>
:>
:> 00 01 10 11 00 01 10 11
:> 10 10 10 10 11 11 11 11 total 16.
:>
: Or 2^4 (or 2^4-1, not counting the all-black screen).
Which is 2^(2*2) which is what he denied.
:> ================================
:>
:> Here is a 2x2 screen with some Red Green Blue either on or off
:>
:> 00 0R 0G 0B R0 G0 B0
:> 00 00 00 00 00 00 00
:>
:> How many possible pictures are there?
: Are you allowing each pixel to only have 4 possible
: values: 0, R, G, B? If so, then there are 4^4 = 256
: possible pictures.
Apparently he has no idea what RGB values are and is
just making something up. I wonder how much deeper he will
continue to dig.
Stephen
Androcles
<mensa...@aol.com> wrote in message
news:1105733672.1...@f14g2000cwb.googlegroups.com...
And got the wrong answer, 10^2219433, which is not correct,
being 1 followed by 2219433 zeros, an exact multiple of 10 and
not a power of 2.
>
>> jeering at me.
>
> There was no jeering until you falsely claimed that I was
> not even close and then justified it with false mathematics.
I refer to the trite comment
"What's the matter, don't know how to use logarithms?"
which was clearly intended to be insulting.
Eat crow.
Androcles.
Randy Poe
> > Which I did.
>
> And got the wrong answer, 10^2219433, which is not correct,
> being 1 followed by 2219433 zeros, an exact multiple of 10 and
> not a power of 2.
Why do you open yourself up to ridicule this way?
Is it fun?
I see in this thread that when the Windows calculator
method is explained in detail, the answer given is
10^2219433.9520314005560718541230247, which is
still not an exact power of 2, but a more precise
representation of the actual power of 2 (which we
already knew). The point is to find out
how many digits it has, compared to a google.
Have you NEVER heard of the concept "order of
magnitude" or "number of digits"?
You know that when computer literature talks about
"16 million colors", that 16 million isn't a power
of 2 either, right?
- Randy
Brad Guth
operational computers, and perhaps as many as 10% of those individual
PCs are on-line at any one time.
6.5e7 potentially active computers at 100 watts each = 6.5e9 watts, 6.5
gigawatts
10% of those PCs being on-line at 10 w/circuit requires another 65e6
watts, 65 megawatts
1:1000 internet/intranet servers and mainframes somewhat insures there's
another 65 megawatts
Avg. lighting, HVAC and transport related considerations should be worth
another 6.5 gigawatts
Thus far we've arrived at 13.130 gigawatts without involving printers,
scanners or copies of documents being reprinted and/or burned onto CDs
and DVDs. I believe global hardcopy publishing and distribution will
remain as yet another 10 fold over whatever digital aspects, thus we've
arrived at 133 gigawatts and climbing. That's 133/6.5 = 20.5 watts per
every soul upon Earth.
Of all that's being stored, transferred, shared, added to, converted
from hardcopy to digital format, of how much of all that is actually
necessary isn't 0.0001%. At least of what's on all my computers amounts
to perhaps 10 gigabytes, whereas I'd have to honestly say 100 megabytes
represents what truly matters out of all of what I've got to work with.
Unfortunately, I know a good number of folks that can't even manage
their PCs on less than 100 GB hard-drives, whereas the 200GB along with
another stash of 100+ DVDs and CDs to boot is becoming the current
minimum standard.
Going by the internet overdose of graphics, spam, smut, audio and video
sharing that's almost entirely bogus and/or unrequested, at least I've
never requested nor intentionally downloaded music or movies, and my
idea of whatever's smut would be certified by a Nun. I'd have to suggest
that perhaps at most 0.1% of the average internet traffic load is usable
and/or generally necessary for humanity to advance upon whatever's being
shared. The remainder is intellectual crapolla, disinformation or simply
bogus individuals and their governments taking every possible advantage
of snookering humanity by way of selectively sharing whatever needs to
be perpetrated, of whatever is not even theirs to share, much less
entitled.
Of photographs and ongoing document page scannings capable of 50
megabytes per frame is another obscene waste of time as well as digital
technology that's having to be stored and subsequently shared over an
already overloaded internet/intranet, causing mainframes and secondary
servers to having those arrays of essentially mega-gigabyte drives that
are already operating in compression mode to being replaced by those
1e100 (google) hard-drives (what's next? a 100,000 rpm googleplex
hard-drive that operates in H2), so that the information stalkers such
as mostly dishonest corporations, GOOGLE and of course of almost
anything NSA/DoD/CIA/FBI, British MI6 and so forth that's related to
their global domination agenda can continue keeping tabs upon every
tidbit of whatever's out there.
Thus far the information being compiled and of all the energy
consumption and resulting pollution has been escalating by a factor of
roughly 10 fold per year, of which at most 0.0001% is actually
necessary.
BTW; It takes absolute loads of energy in order to obtain, store,
transfer, share and to continually upgrade the standards by which newer
and newer computers and of everything associated (such a massive digital
displays, theater class screens and eventually millions more of various
HDTVs) must follow suit, or else you're out of the technological loop.
Thus soon the global energy demand as for creating such and sustaining
this primarily luxury growth (most of which the lower 99.9% of humanity
doesn't require) will become the primary drain upon our limited global
energy resources. These days, no one fixes anything, they trash whatever
they have and upgrade.
I believe there's a breaking point of roughly 2 kw/soul. Surpassing 2
kw/soul is insuring the demise of the existing energy resources
sustaining life as we know it, and even that's based upon no further
aircraft smashing into tall buildings, and of our insane warlords going
after more of those stealth WMD. Perhaps they should have been going
after lunar He3 instead of WMD.
Digital files need to be drastically reduced, duplications need to
become minimized and/or at least centralized via some global intranet
data bank that's accessed via a Skyship optical network of extremely
energy efficient and thereby least polluting alternative to any surface
and/or satellite alternatives. The average computer and accessory loads
(including whatever personal environment aspects) will eventually need
to become limited to 100 watts/unit (with damn few exceptions),
especially if eventually (within a decade from now) 10% of humanity
becomes PC interfaced with 10% of those being on-line somewhere
throughout the global net. Thereby less graphical and more raw text
based data, whereas if the end-user desires to take each and every 100k
file of perfectly usable information and dress each of those up to
100mb, then so be it, but just don't share that sort of meaningless
garnishment over your intranet, much less over the internet.
Sorry this contribution became such a wall-of-words. Perhaps this
argument explains as to why our cold-war has refocused upon global
energy domination.
Uncle Al
>
> For this argument I've estimated there's at most 1% of humanity having
> operational computers,
6.5 billion -> 65 million. I peg it closer to 15% in terms of access
rather than individual ownership. Businesses, governments at all
levels, and schools/universities have an awful lot of hardware.
Upscale cell phones are bottom-feeding computers. Every car/bus/truck
is a network.
> and perhaps as many as 10% of those individual
> PCs are on-line at any one time.
6 million on-line. OK. Probably more given education, government,
and commerce.
> 6.5e7 potentially active computers at 100 watts each = 6.5e9 watts, 6.5
> gigawatts
At least double with CRTs, maybe 50% more with LCDs. A Pentium cooks
70 W scratching its butt.
> 10% of those PCs being on-line at 10 w/circuit requires another 65e6
> watts, 65 megawatts
>
> 1:1000 internet/intranet servers and mainframes somewhat insures there's
> another 65 megawatts
>
> Avg. lighting, HVAC and transport related considerations should be worth
> another 6.5 gigawatts
After all the pushing and shoving, not unreasonable as conservative
Fermi estimates.
> Thus far we've arrived at 13.130 gigawatts without involving printers,
> scanners or copies of documents being reprinted and/or burned onto CDs
> and DVDs. I believe global hardcopy publishing and distribution will
> remain as yet another 10 fold over whatever digital aspects, thus we've
> arrived at 133 gigawatts and climbing. That's 133/6.5 = 20.5 watts per
> every soul upon Earth.
1) We need more civilian power reactors if we wish to think and
play. A rational civilian fuel recycling plan, too.
2) Heat emitted from computers and support is approaching heat
emitted from brains and support. We're outsourcing the scut work of
intellect.
> Of all that's being stored, transferred, shared, added to, converted
> from hardcopy to digital format, of how much of all that is actually
> necessary isn't 0.0001%.
Sure - consider e-mail and spam alone. I would say 30% of everything
transmitted - not counting spam - is garbage beyond argument. Another
30% is marginally garbage. If you count music and video, 99% is
garbage without argument.
> At least of what's on all my computers amounts
> to perhaps 10 gigabytes, whereas I'd have to honestly say 100 megabytes
> represents what truly matters out of all of what I've got to work with.
> Unfortunately, I know a good number of folks that can't even manage
> their PCs on less than 100 GB hard-drives, whereas the 200GB along with
> another stash of 100+ DVDs and CDs to boot is becoming the current
> minimum standard.
A world bursting with crap. I agree. Vital programming and data
maybe 100 MB on my 20 GB drive. I just ordered a tricked out AMD
Athlon 55-FX 2.6 GHz with 120 GB HDD and 2 GB Corsair extreme DDR
RAM. Windows is a pig. Less the OS, maybe 4 GB of stuff total right
now. I'm assuming the future will be improved means to deteriorated
ends.
You can have my WordStar when you pry it from my cold dead hands.
Give it good hardware and it handles 90,000-line datafiles no
problem. Don't try that with anything from Microcrap unless you have
a copy of "War and Peace" to pass the time.
> Going by the internet overdose of graphics, spam, smut, audio and video
> sharing that's almost entirely bogus and/or unrequested, at least I've
> never requested nor intentionally downloaded music or movies, and my
> idea of whatever's smut would be certified by a Nun. I'd have to suggest
> that perhaps at most 0.1% of the average internet traffic load is usable
> and/or generally necessary for humanity to advance upon whatever's being
> shared. The remainder is intellectual crapolla, disinformation or simply
> bogus individuals and their governments taking every possible advantage
> of snookering humanity by way of selectively sharing whatever needs to
> be perpetrated, of whatever is not even theirs to share, much less
> entitled.
Makes one fondly pine for Sturgeon's Law. Sturgeon was an optimist.
Smut is good. It diverts social breakdown. Sex, drugs, hunt, or
homicide.
[snip]
> BTW; It takes absolute loads of energy in order to obtain, store,
> transfer, share and to continually upgrade the standards by which newer
> and newer computers and of everything associated (such a massive digital
> displays, theater class screens and eventually millions more of various
> HDTVs) must follow suit, or else you're out of the technological loop.
> Thus soon the global energy demand as for creating such and sustaining
> this primarily luxury growth (most of which the lower 99.9% of humanity
> doesn't require) will become the primary drain upon our limited global
> energy resources. These days, no one fixes anything, they trash whatever
> they have and upgrade.
Quantum limits are being approached. Multiple core CPU's and 64-bit
OS next. To do what, surf 7-bit Usenet? To type a multimedia memo?
To help Uncle Al calculate parity divergence of crystal lattices!
Screw NASA and its 10,000+ Itanium "Columbia" (being upgraded to
Itanium-2 - the original spec was crap). Should have used Opteron
800-series. Uncle Al has been afforded access to a very small piece
of CERN for number crunching. Nice folks - and it wasn't doing
anything anyway.
The working part of society amazes me. I submitted 60 lines of
executable processes, 5-8 hrs/line estimated. As the e-mail sends
another one comes in. Benchmarks from an overachiever who diddled the
executable dropped the time estimates by 8%. Output remained exact to
18 decimal places. If we cleaned up the world there would be mobs
demanding their right to squalor.
> I believe there's a breaking point of roughly 2 kw/soul. Surpassing 2
> kw/soul is insuring the demise of the existing energy resources
> sustaining life as we know it, and even that's based upon no further
> aircraft smashing into tall buildings, and of our insane warlords going
> after more of those stealth WMD. Perhaps they should have been going
> after lunar He3 instead of WMD.
When China comes fully on-line the world implodes. We cannot sustain
the US gorging on 40% of the planet's resources and China gorging on
another 100%. I expect the US will lose and the Third World will
end. Call it a push. Europe and russia will be interesting.
> Digital files need to be drastically reduced, duplications need to
> become minimized and/or at least centralized via some global intranet
> data bank that's accessed via a Skyship optical network of extremely
> energy efficient and thereby least polluting alternative to any surface
> and/or satellite alternatives.
Who watches the watchers? Those with wealth and power will not
sacrifice for the unwashed mob, nor should they. Death to the unable
by their own hands.
> The average computer and accessory loads
> (including whatever personal environment aspects) will eventually need
> to become limited to 100 watts/unit (with damn few exceptions),
No way. That won't even run the motherboard. Will you create a
computational underclass of users? Those who would benefit most would
be the first to be denied access. Management does not seek genius, it
hunts it to destruction in the name of Korporate Kulture
(self-survival).
> especially if eventually (within a decade from now) 10% of humanity
> becomes PC interfaced with 10% of those being on-line somewhere
> throughout the global net. Thereby less graphical and more raw text
> based data, whereas if the end-user desires to take each and every 100k
> file of perfectly usable information and dress each of those up to
> 100mb, then so be it, but just don't share that sort of meaningless
> garnishment over your intranet, much less over the internet.
>
> Sorry this contribution became such a wall-of-words. Perhaps this
> argument explains as to why our cold-war has refocused upon global
> energy domination.
We are here to conquer not to survive. War is economics spurred on by
birthrate. Absent a new open frontier it will make little difference
what we do. By 2050 First World civilization will be abolished,
smothered within its socialist Ponzi schemes and overrun by
reproductive warriors who value life beneath all else.
Uncle Al says, "Ownership in the 21st century is the ability to
destroy."
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
John Baez
>John Baez wrote:
>> But, I wasn't trying to explain this stuff in detail - just
>> compute the amount of information in a raindrop!
>Now that you "know" the amount of information, can you use it in any
>way to replicate the rain drop?
No - not me, anyway.
>If not, this means this is the apparent information, or information at
>the phenomenal level. The information needed at a substance level to
>replicate the exact same rain drop could be significantly higher or
>even asymptotic to infinity.
I'm using "information" in the usual sense of information theory -
see for example Shannon's book "A Mathematical Theory of Communication".
In this sense, information is what it takes to answer questions about
something. It's not all it takes to build something: for that you
need some raw materials as well.
For example, you can't build a brick house even if you have the
blueprints, if you don't have any bricks. Or, for that matter,
if you don't have the proper tools and knowhow!
The last point is also relevant: even if we had the complete information
describing a raindrop and we had enough water to make a raindrop, we
couldn't build an exactly identical raindrop, because we don't know
how to manipulate atoms that well.
>I argue that even if you know the amount of information you claim that
>is not nearly enough to replicate the rain drop.
You need something more - but what you need isn't more information
about the raindrop. The wonderful thing about statistical mechanics
is that it allows us to measure the total amount of information in
something like a raindrop, by measuring its entropy. Boltzmann, Gibbs
and Shannon - they were awfully clever.
John Baez
Brad Guth <brad...@yahoo.com> wrote:
>Terrific topic. However, does disinformation count?
Yes, but a negative amount. :-)