WARNING: Do NOT calculate Pi in binary

[Gayle29uk]

It is conjectured that thisnumber is normal, meaning that it contains ALL finite bit strings.

If you compute it, you will be guilty of:

* Copyright infringement (of all books, all short stories, all newspapers, all magazines, all web sites, all music, all movies, and all software, including the complete Windows source code)
* Trademark infringement
* Possession of child pornography
* Espionage (unauthorized possession of top secret information)
* Possession of DVD-cracking software
* Possession of threats to the President
* Possession of everyone's SSN, everyone's credit card numbers, everyone's PIN numbers, everyone's unlisted phone numbers, and everyone's passwords
* Defaming Islam. Not technically illegal, but you'll have to go into hiding along with Salman Rushdie.
* Defaming Scientology. Which IS illegal -- just ask Keith Henson.

Also, your computer will contain all of the nastiest known computer viruses. In fact, all of the nastiest POSSIBLE computer viruses.

Some of the files on my PC are intensely personal, and I for one don't want you snooping through a copy of them.

You might get away with computing just a few digits, but why risk it? There's no telling how far into Pi you can go without finding the secret documents about the JFK assassination, a photograph of your neighbor's six year old daughter doing the nasty with the family dog, or a complete copy of
the not-yet-released Return of the King movie. So just don't do it.

The same warning applies to e, the square root of 2, Euler's constant, Phi, the cosine of any non-zero algebraic number, and the vast majority of all other real numbers.

There's a reason why these numbers are always computed and shown in decimal, after all.

[Embroglio]

Originally Posted by Gayle

of the not-yet-released Pearl Harbor movie.

You could have at least edited that part out.

ps. I don't do maths and numbers and shit, so don't 'get' this.

[Gayle29uk]

Originally Posted by Embroglio

You could have at least edited that part out.

Tum te tum te tum...edited

[JetLinus]

Is this some sort of variation of the old "Hey, l put some random bits into a file and hope it makes sense*" idea?



* worldovertaking virus; unreleased LOTR movie in DVD quality, preferrably XVid codec?


Also, why should PI be used in a credit card validation algorithm? Just because it's not rational? I think that kind of numbers is called "transcendent" or something...

[JetLinus]

There was once a nice news article about something similar.

Someone had "discovered" a crack for DVD protection, and put it into a zip file. The resulting hex-string of this, put into a decimal number, was (I think intentionally made to a) prime.
And people then claimed it would be "legal" and stuff coz noone could own the copyright on a prime.
Something like that.......

[queball]

First "Cantor rocks" before W.

Originally Posted by JetLinus

There was once a nice news article about something similar.

Someone had "discovered" a crack for DVD protection, and put it into a zip file. The resulting hex-string of this, put into a decimal number, was (I think intentionally made to a) prime.
And people then claimed it would be "legal" and stuff coz noone could own the copyright on a prime.
Something like that.......

Correct: http://www-2.cs.cmu.edu/~dst/DeCSS/G...ego/index.html
Though it's all rubbish, really. It's not the "number" but its context that determines its meaning as a picture or whatever. Shannon himself emphasised the distinction between "information" as used in the original post and information as in bitstrings.

[Nodrog]

intuitionism would disagree

[Radical Edward]

someone tell the daily mail. they will be baying for the banning of pi and other irrational numbers before you can see sense and buy the guardian.

[Radical Edward]

incidentally, wouldn't this make for excellent compression if it was true:

go to the 5 billionth digit of pi. take a string 2 million digits long. convert to binary, and read like so....

[queball]

Originally Posted by Radical Edward

incidentally, wouldn't this make for excellent compression if it was true:

go to the 5 billionth digit of pi. take a string 2 million digits long. convert to binary, and read like so....

Obviously it wouldn't be "compression", but http://pi.nersc.gov/

[Nodrog]

Originally Posted by Radical Edward

incidentally, wouldn't this make for excellent compression if it was true:

go to the 5 billionth digit of pi. take a string 2 million digits long. convert to binary, and read like so....

the number representing the start of the subsequence within pi would probably be longer than the number representing the data

[W]

May I be the first to say "old"?

[Nodrog]

Originally Posted by Gayle29uk

You might get away with computing just a few digits, but why risk it? There's no telling how far into Pi you can go without finding the secret documents about the JFK assassination, a photograph of your neighbor's six year old daughter doing the nasty with the family dog, or a complete copy of
the not-yet-released Return of the King movie. So just don't do it.

Originally Posted by Gayle29uk

a photograph of your neighbor's six year old daughter

Originally Posted by Gayle29uk

six year old daughter

Originally Posted by Gayle29uk

old

beaten

[W]

Originally Posted by Nodrog

beaten

I actually searched the thread for the word "old", only found that one reference, thought "who'd nitpick on that anyway?" and immediately thought of you.

[Radical Edward]

Originally Posted by queball

Obviously it wouldn't be "compression", but http://pi.nersc.gov/

1676941060
1676941060
1676941060
1676941060
1676941060

go go gadget censor evasion.

I see what you mean Nodrog.

[acropolis]

Originally Posted by Nodrog

the number representing the start of the subsequence within pi would probably be longer than the number representing the data

i'm suddenly curious about this.

offhand i'd guess they'd be about the same size, and i'm always right, so it seems likely.

but i'll check in class.

[W]

Originally Posted by acropolis

i'm suddenly curious about this.

offhand i'd guess they'd be about the same size, and i'm always right, so it seems likely.

but i'll check in class.

Given that any finite substring of pi is repeated infinite number of times, it's given that the average position of bitstrings within pi will be greated that the size of the string.

[acropolis]

Originally Posted by W

Given that any finite substring of pi is repeated infinite number of times, it's given that the average position of bitstrings within pi will be greated that the size of the string.

if it takes a million digits into pi (binary) to get a twenty bit sequence you want, it will take about 20 bits to state it's location.

[W]

Originally Posted by acropolis

if it takes a million digits into pi (binary) to get a twenty bit sequence you want, it will take about 20 bits to state it's location.

I think you missed my point.

If you have a bitstring containing all bitstring of 20 bits lenght, it's impossible that this bitstring be SHORTER than 2^20+20-1 bits long. So to give any subposition within this (optimal) bitstring, you'd need 20 bits. This is if none of the 20-bit strings within it is repeated. In pi, they are, pi has no such guarantee of optimality (as can be shown by noting that the first 10 decimals in decimal are not the 10 different digits)

Even if it was optimal, there'd be no average compression (as with ANY compression algorithm. That's right, no compression algorithm has an average gain on a random file)

[Scoot951]

Originally Posted by Gayle29uk

It is conjectured that thisnumber is normal, meaning that it contains ALL finite bit strings.

If you compute it, you will be guilty of:

* Copyright infringement (of all books, all short stories, all newspapers, all magazines, all web sites, all music, all movies, and all software, including the complete Windows source code)
* Trademark infringement
* Possession of child pornography
* Espionage (unauthorized possession of top secret information)
* Possession of DVD-cracking software
* Possession of threats to the President
* Possession of everyone's SSN, everyone's credit card numbers, everyone's PIN numbers, everyone's unlisted phone numbers, and everyone's passwords
* Defaming Islam. Not technically illegal, but you'll have to go into hiding along with Salman Rushdie.
* Defaming Scientology. Which IS illegal -- just ask Keith Henson.

Also, your computer will contain all of the nastiest known computer viruses. In fact, all of the nastiest POSSIBLE computer viruses.

Some of the files on my PC are intensely personal, and I for one don't want you snooping through a copy of them.

You might get away with computing just a few digits, but why risk it? There's no telling how far into Pi you can go without finding the secret documents about the JFK assassination, a photograph of your neighbor's six year old daughter doing the nasty with the family dog, or a complete copy of
the not-yet-released Return of the King movie. So just don't do it.

The same warning applies to e, the square root of 2, Euler's constant, Phi, the cosine of any non-zero algebraic number, and the vast majority of all other real numbers.

There's a reason why these numbers are always computed and shown in decimal, after all.

The stuff about knowing secrets wouldn't work, the truth is in there but also every possible lie. Otherwise pretty interesting tho...it is hard to imagine that if you just sprouted random ones and zeros long enough you'd get the Mona Lisa.

And the 'pi-compression' is also interesting if (a) we could figure out where exactly in pi the file we need is, and (b) there was some simple way of storing zillions of digits of pi. Perhaps if some government server stored it and everyone could access it as needed.

[acropolis]

Originally Posted by W

I think you missed my point.

If you have a bitstring containing all bitstring of 20 bits lenght, it's impossible that this bitstring be SHORTER than 2^20+20-1 bits long. So to give any subposition within this (optimal) bitstring, you'd need 20 bits. This is if none of the 20-bit strings within it is repeated. In pi, they are, pi has no such guarantee of optimality (as can be shown by noting that the first 10 decimals in decimal are not the 10 different digits)

Even if it was optimal, there'd be no average compression (as with ANY compression algorithm. That's right, no compression algorithm has an average gain on a random file)

000
001
010
011
100
101
110
111
1100010111
heh. that's neat.

[Intrepid00]

001101010100011010110000010101

[W]

Originally Posted by Scoot951

And the 'pi-compression' is also interesting if (a) we could figure out where exactly in pi the file we need is, and (b) there was some simple way of storing zillions of digits of pi. Perhaps if some government server stored it and everyone could access it as needed.

I think you missed the posts where we point out that this doesn't work?

[Phang]

Originally Posted by Radical Edward

1676941060
1676941060
1676941060
1676941060
1676941060

go go gadget censor evasion.

I see what you mean Nodrog.

banplz

[Scoot951]

Originally Posted by W

I think you missed the posts where we point out that this doesn't work?

Yea, well, that's what they said about flying cars!



...

[Wo2]

ummm...yea...altough if you fired a car out of a cannon you may get to fly for certain distance.

Any how back on topic.

True you would get all the above named thinks thanks to the random spoutings of binary string, you would first have to dig through the thousands and thousands, pieces of random text saying stuff like "Crlfroda" or "poot", the picts of random cloured blobs and programs that do nothing but flicker and go "Bing"

Yes I know my grammer is...lacking...but you still get what i'm saying.

[cnaw]

Originally Posted by Wo2

ummm...yea...altough if you fired a car out of a cannon you may get to fly for certain distance. I KNOW I KNOW. No relavence to the thread at all.

stick a Jato pack on a car and u'd get it to fly.

























(and before nodrog post the snopes link, i'll do it my self. Right here)

[acropolis]

anyway.

So, to check out compression of a random file, i took some 4-digit binary sequences,
like
0101,
1100,
0001,
0111,
etc.
Then i programmed my calculator to spit out 1s and 0s randomly, and counted how many digits were put out before getting to my 4-bit string of choice. Obviously it showed compression as impossible for a random file.

But fun question:
which 4-bit strings were likely to occur quickly (on average)?
which ones took longer?
why?

[AllFather|away]

you rock gayl

[W]

Originally Posted by acropolis

But fun question:
which 4-bit strings were likely to occur quickly (on average)?
which ones took longer?
why?

Random.
Random.
Pure coincidence.

[JetLinus]

The randomness 0wn°rz all of us.

Ok, here are some farb-rausch demos:

• 16 bytes
• 11 bytes
• 5 bytes

Now, even for the 5 byte one, you know how ridiculously small the probability of getting this by "random chance" is...

[JetLinus]

Another thought:
What about real bad luck?

Think about CRC error detection, parity checks and stuff. What, if for some sensitive data, it doesn't work? Nowadays we're all causing terrabytes of traffic, can't there be one tiny bit wrong?

(well, the answer is: it's quite unlikely ^^)


I just remember a true story, where someone working with quite some large dbs in his job (accounting, tourist information system, sales data, whatever) suddenly had a problem: Somewhere in a record, there was a negative number, that just COULDN'T be negative...

Well, debugging, everything didn't help, the error was not reproducable, but the minus was there! I imagine some confused and puzzled people standing around a screen. Well... End of story: In the evening, the db server or the job-processing machine, whatever, died down, memory fault. Apparently there had been ONE broken bit (or more?) before, and the parity check just didn't detect it.
In debug-mode, the program was on a different position in memory, so the error didn't occur twice. But well, parity check isn't as powerful as CRC for example, or ECC...



Where do we get if we can't trust our own RAM?

[MrL_JaKiri]

We can't trust our own ram, we can't trust our own hard drives. They're not totally accurate.

[acropolis]

Originally Posted by W

Random.
Random.
Pure coincidence.

well, of course it is random.

but just thinking about what you would logically expect to see.
it should take, on average, about as long to get to your first 111 series as your first 010, right?

here's the point:

comparing 1110 to 0101,

every 111 is guaranteed to terminate in a 1110 string. so whenever you see '111', 1110 isn't far behind.

every 010 only has a 50% chance of terminating in an 0101 string, and if it doesn't you start over from scratch.

so one could expect 1110 strings to appear, on average, sooner than 0101.

but there are obvious points of weakness in this argument (notably, how do you know that 111 shows up as quickly as 010?), so actually testing it out is always better form.

i get 1110 showing up 12-13 in,

and 0101 showing up 18-20 in.

i would love for someone to prove me wrong.

[W]

Originally Posted by acropolis

well, of course it is random.

but just thinking about what you would logically expect to see.
it should take, on average, about as long to get to your first 111 series as your first 010, right?

here's the point:

comparing 1110 to 0101,

every 111 is guaranteed to terminate in a 1110 string. so whenever you see '111', 1110 isn't far behind.

every 010 only has a 50% chance of terminating in an 0101 string, and if it doesn't you start over from scratch.

so one could expect 1110 strings to appear, on average, sooner than 0101.

but there are obvious points of weakness in this argument (notably, how do you know that 111 shows up as quickly as 010?), so actually testing it out is always better form.

i get 1110 showing up 12-13 in,

and 0101 showing up 18-20 in.

i would love for someone to prove me wrong.

You're dead wrong, and could prove it to yourself by simply doing it over with a different random seed.

Yes, the different bitstrings are dependant on eachother, so if you have 1001, you know you need atleast 4 bits more to find 0001, and if it's there you've already found 0010 0100 and 1000 as well. But for each of the different bitstrings, the probable position is entirely random and with equal distribution.

[Belgarath The Sorcerer]

This thread reminds me too much of one of my modules at Uni.

[Intrepid00]

Originally Posted by idimmu

this thread doesnt deliver

100110100001001101000000001000001010000100100

[acropolis]

Originally Posted by W

You're dead wrong, and could prove it to yourself by simply doing it over with a different random seed.

Yes, the different bitstrings are dependant on eachother, so if you have 1001, you know you need atleast 4 bits more to find 0001, and if it's there you've already found 0010 0100 and 1000 as well. But for each of the different bitstrings, the probable position is entirely random and with equal distribution.

i have a number ways of 'proving' to myself that they are equal, on average. my problem is i have what appear to be better proofs that they aren't, and i see no way of refuting them.

i've got one better now:
1111
vs.
1110

in any random string, we reach the first time we have three ones in a row (the prerequisite for a 1111 or a 1110).

00010111
k?
from here, how far is it to the last digit of 1110?
we can calculate.
1*1/2 + 2*1/4 + 3*1/8 + 4*1/16..... = ~1.64
a 0 is coming along pretty damn soon.

from the exact same point, how far on average to the last digit of 1111?
1*1/2 + (average distance from scratch to the last digit of 1111+1)*1/2
because there's 1 chance of the next digit being a 1 and 1 chance of a 0, and having to start over.

even if god intervened here, and everytime after that frustrating 0 you immediately got 1111,
that's 1*1/2 + (4+1)*1/2 = 3. which is still greater than 1.64.

in reality (average distance from scratch to last digit of 1111) is closer to 18 than 4, so 1111 takes ~ (9-1.64) ~ 7 longer than 1110.

counterintuitive =/= wrong

[MrL_JaKiri]

What the hell are you talking about?

If I roll 200 6's in a row, it doesn't affect the probability of me rolling a 6 next time (assuming it's a fair die)

[acropolis]

Originally Posted by MrL_JaKiri

What the hell are you talking about?

If I roll 200 6's in a row, it doesn't affect the probability of me rolling a 6 next time (assuming it's a fair die)

okay you've rolled 200 6s in a row.

which string is going to be completed sooner, on average:
200 6s and then a five,
or 201 6s?

it's 200 6s then a five.

if your next roll is a 1, 2, 3, or 4, it's going to take 3 gabillion rolls (on average) to get either one.

if your next roll is a five, that's one for 200 sixes then a 5 and 3 gabillion (on average) to get 201 sixes.

if you roll a six, it's one for 201 sixes and only about ~2.3 gabillion for 200 sixes then a five.

the very last part is the trick. because if you roll another six, you've still got another shot at rolling a five (if you are going for 200 sixes then a five), whereas if you roll a five (and you are going for 201 sixes), you have to start over.

[MrL_JaKiri]

You DON'T UNDERSTAND PROBABILITY.

ALL THAT HAS OCCURED BEFORE IN A RANDOM SEQUENCE DOESN'T AFFECT FUTURE HAPPENINGS.

THERE IS NO LAW OF AVERAGES.

[acropolis]

Originally Posted by MrL_JaKiri

You DON'T UNDERSTAND PROBABILITY.

ALL THAT HAS OCCURED BEFORE IN A RANDOM SEQUENCE DOESN'T AFFECT FUTURE HAPPENINGS.

THERE IS NO LAW OF AVERAGES.

heh.

if my question is "will this next roll be the last digit of 1111"

obviously things that have occurred before have an effect. if my last three rolls weren't all ones, i have 0 chance, if they were all ones, i have a 50-50 chance.

i'm not going to contend that 'i understand probability'

i'm just going to give everyone ample oppurtunity to find a flaw in my arguments, and watch that not happen.

[Gayle29uk]

Originally Posted by acropolis

understand probability'

i'm just going to give everyone ample oppurtunity to find a flaw in my arguments

You have no argument, you don't understand probability.

[queball]

If you get 011111110 then that's three 1111's.

[queball]

To elucidate slightly: say you get 111xxx at the end where xxx are still to be chosen.
xxx 1110's 1111's
000 1 0
001 1 0
010 1 0
011 1 0
100 1 1
101 1 1
110 1 2
111 0 3
total: 7 7

[Tactitus]

Originally Posted by Gayle29uk

You have no argument, you don't understand probability.

I think what he's trying to say is that if you already have a partially completed sequence then your probability of completing it is higher than if you don't have a partially completed sequence (e.g., the probability of getting one six after already getting N-1 sixes is higher than the probability of getting N sixes).

[W]

MrL, you're missunderstanding the question here.

from a throw of 200 dice (which accidentally all turned up 6's), the probability for the following results from the next 202 dice throws are:

of there being 0 or more 6's then a 5: >1/6
of there being any other dice, then 200 6's and one 5: <1/6^201


but this does not affect the average number of throws you need to throw to get 200 6's in a row and then a 5. (which is the same for the number of throws as for 201 6's)

[MrL_JaKiri]

I was going to state that as well, but I had to have a shower.

[W]

Originally Posted by queball

To elucidate slightly: say you get 111xxx at the end where xxx are still to be chosen.
xxx 1110's 1111's
000 1 0
001 1 0
010 1 0
011 1 0
100 1 1
101 1 1
110 1 2
111 0 3
total: 7 7

And for average distance:
xxx | 1110 distance | 1111 distance
000 | 1 | -
001 | 1 | -
010 | 1 | -
011 | 1 | -
100 | 2 | 1
101 | 2 | 1
110 | 3 | 1
111 | - | 1

Given that for a - in the 1111 column, the average distance to a 1111 is atleast 15, but for a - in 1110 column, the average distance is 1.6, should easilly show that GIVEN a 111 already, the average distance to a complete 1110 is shorter than 1111

[W]

My guess is acropolis is totally correct. Since 1111's tend to bunch up together, the average distance between the bunches has to be longer.