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freedumb writes "From this article in Nature: "Two mathematicians have now taken the first step towards proving that pi contains not a single message but every conceivable message, meaningful or not."" Actually, it's a discussion concerning whether "that all strings of the same length appear in pi with the same frequency: 87,435 appears as often as 30,752, and 451 as often as 862, a property known as normality."
An amazing finding...
...next thing you know, they'll be finding secret messages hidden in the Bible
using some sort of letter skipping technique.
What is even more interesting is that NetBSD is a subset of this
"The interesting question is, must it also have all infinite strings embedded in it? I suspect that would lead to a contradiction, but this goes beyond my mathematical competence."
Sure would. Take pi, and add 1. Remove the decimal point so that you just have the infinite string 414159265358979... Suppose that's embedded in pi somewhere. The string 414159... must therefore also be embedded in itself (that is, aside from the trivial embedding, where you just look at the entire string). But wait! That means that the string 414159... must repeat, and thus, so must pi. Since pi is irrational, we have a contradiction.
(Sorry for the rather unclear explanation, it's 4 A.M. here and I can barely think straight.)
Soo.. if pi contained every possible message (ie, was truly random), couldn't you in theory find a specific position where pi prints out say, the Max Payne ISO, and distribute that position to friends?
Then, said friends, start calculating pi from that offset (wasn't there a story on slashdot about calculating any N digit of pi without having to calculate the first N-1 digits). Voila, kickass compression.
Of course, the small snags here are:
But once you get over these boring details, pi-based-compression can make for some very neat applications
It is more random, however it is not useful for most things as it is very predictable. Think of it as if you are using rand but everyone is using the same seed.
Erlang Developer and podcaster
> If we could just get enough of the message contained in Pi, maybe some order would magically appear.
I'm sure once it's calculated to an infinite number of digits, the true meaning will become clear.
You must have missed the story where they zipped DeCSS code and came out with a long prime number. The algorithm to get DeCSS code from this number is unzip (and it's available everywhere). So if the number itself is not illegal, then neither is the decimal displacement from pi.
___
___
If you think big enough, you'll never have to do it.
No it won't. First, the behaviour of rand() is implementation-dependent. Second, most implementations produce all their possible outputs (typically 2**16) in a fixed cyclic order.
No point other than karma whoring off of lame jokes about DeCSS (repeated at least 50 times per Pi story), no.
Oddly, my ICQ number is at position 19724810 counting from the first digit after the decimal point.
What're the odds of *that*?
;-)
Cheers,
Jim in Tokyo
Have no clue about firewalls?
-- My Weblog.
Is using the PI digit generators more random than using rand()?
> It will start a whole new branch of numerology dedicated to finding entire new holy books...
Ah, yes. Bible Codes for mathematicians.
I once seriously considered buying the Bible Codes program, just to see if it could find the message "bible codes, big lie". I wonder whether pi reveals its own naughty secrets.
At the very least we can expect helpful messages like "You can just round it off now, you moron."
--
Sheesh, evil *and* a jerk. -- Jade
> you should know one egregious example of funny strings in Pi at funny positions:
> 42424242 at position 242424.
Incredible! I have just discovered that it also lists out all the digits of pi, starting at offset zero!
Now instead of calculating the digits of pi, we can just look them up in the digits of pi!!!
On a serious note, observe that if pi does indeed have all possible strings embedded in it, then it must have all possible strings embedded in it twice. (And thrice, 4x, 5x, etc. The proof is left as an exercise to the reader.) Thus if it does embed all possible strings, it follows that the first n digits of pi must appear in it somewhere other than at offset zero, for any positive n.
The interesting question is, must it also have all infinite strings embedded in it? I suspect that would lead to a contradiction, but this goes beyond my mathematical competence.
--
Sheesh, evil *and* a jerk. -- Jade
Turing machines with attached "oracles" are used for proofs in theoretical computer science, but it's important to keep in mind that they are purely theoretical - no one will ever build an oracle. You don't think about how an "oracle" works. It's just a concept, like imagining that magic works. So why is this useful?
:-)
Well... For instance, you might be able to prove that such-and-such a problem with input size "n" could be solved in polynomial time (i.e. "fast") if you just had a magical oracle to supply you with only log(n)correct, one-bit answers.
The point of the proof would be that you don't need more than log(n) magic bits from the oracle. So what good is that? Well... If you can get the number of magic bits small enough, while still keeping the algorithm fast, it may provide a way to do a randomized algorithm where instead of using the magical oracle (which doesn't exist, remember?) you just use a random number generator, or maybe just try everything. Since a random number generator isn't as accurate as a magical oracle, you run the algorithm a lot of times with different random bits. Maybe you'll be lucky soon, and maybe that's good enough on average.
Theoretical computer science is fun. It's a little crazy and non-intuitive somtimes.
Torrey Hoffman (Azog)
Torrey Hoffman (Azog)
"HTML needs a rant tag" - Alan Cox
Wow, you just sparked a real idea, mathematitions say its impossible to encode a large truly random sequences of bits, into something that the outcome plus the decomressor is smaller than the first bytes. But given enough computing power you could find the large random bits in pit, somewhere and simple have the decompressor know the how to compute pi, and the start and stop points of the random number?
Why concentrate on just pi? If they show it's true for all trancendental numbers, they've got it for pi, e, etc.
;)
I'd be happy with just pi for starters...
Furthermore, it is not true for all trancendental numbers: for example, 1/n^(1!)+1/n^(2!)+1/n^(3!)+1/n^(4!)+1/n^(5!)+...
are trancendental, but with n=10 that number has only 1's and 0's, so it's not normal.
Can pi appear in pi anywhere? I guess not, since that would mean that pi repeats. Could e be in pi? I suppose if e was in pi, and pi was in e, then pi would be in pi, which I guessed earlier it couldn't be. But, maybe I'm wrong and there's a loophole since if pi contains itself, there's an infinite recursion going on.
Pi can't appear in pi, because that would make it repeat and make it rational, which it isn't (at least the way I understand "pi in pi"). Is e in pi? Not neccessarily. These guys are trying to prove that any finite number series can be found in pi, and e is infinitely long. If it's true, then you can choose any n and you can find n digits of e in pi, but not infinitely many.
Of course, e might be in pi (though I consider that unlikely - this would mean that pi=p/q+e*10^(-n) where p, q and n are integers, which seems quite weird). But what these guys are trying to prove doesn't show that.
I doubt, therefore I may be.
Why concentrate on just pi? If they show it's true for all trancendental numbers, they've got it for pi, e, etc.
Can pi appear in pi anywhere? I guess not, since that would mean that pi repeats. Could e be in pi? I suppose if e was in pi, and pi was in e, then pi would be in pi, which I guessed earlier it couldn't be. But, maybe I'm wrong and there's a loophole since if pi contains itself, there's an infinite recursion going on.
If there were an infinite number of monkeys typing on an infinite number of keyboards would they eventually produce all the works of Shakespeare? Not exactly - they would produce them immediately (as quickly as a monkey can type). That's infinity for you. The lazy 8. It goes on and on...
BigKahuna
Why ban it. Can't we just change it to someting simple like 3.000?
Comment removed based on user account deletion
So, does it contain the 7 line DeCSS implementation? how about just the ascii "Natalie Portman"? goatse.cx?
What am I trying to say? So what!
Well if it contains all conceivable messages, that must mean it contains all conceivable circumvention software! Down with Pi!
It's 10 PM. Do you know if you're un-American?
> and seven 7's is found at position 3346228, six 6's is found
> at position 252499, five 5's is found at 24466, and so on...
"and so on..." ?!?
That's not quite true. There exist numbers which have a PATTERN, but don't actually repeat.
.101001000100001000001...
For instance,
There's certainly a pattern (1 zero then 2 zeros then 3...) but the number never repeats.
Read: you're both pretty much right- it is a ratio of two numbers, and it is irrational (as odd as that sounds)
-bugg
this is the latest : microsoft sues pi for containing the complete source code to windoze.
btw, the code starts at position 4200394298 (in the binary expansion of pi), and continues for well, as long as anyone ccan read the stuff...
Don't Panic
All finite strings.
Say no to software patents.
Indeed. Better pick something like 42424242, which not only occurs way early (at position 242424), but for which not only the search string but also the position is an interesting pattern....
Probaility of it occurring so early should be less than 1% (we would expect it below 100000000, not below 1000000), and probability of the position being a permutation of the string is...well...amazingly small.
Small note for nitpickers: I counted the 3. as digits; the search engine does not. Hence the position shown is 242422 rather than 242424.
Say no to software patents.
Not necessarily so. If you define a (decimal) number as follows:
- it starts with 1.
after the dot, each digit at a prime number position is 1, and 0 otherwise
The number would be 1.01101010001010001...- The number would have no periodicity (because prime numbers become rarer and rarer), so it woule not be a rational number.
- It never ends.
- But still, by construction, it would not contain a single 8, much less a series of 5,646,498,765 eights.
Thus proving that never ending irrational number does not necessarily contain all strings. Btw, irrational numbers are always never ending, or else they would be a fraction whose denominator would be a large power of ten. Think about it.Say no to software patents.
If you did any math in your youth, you'd know that this is a perfectly valid way to do a proof. It's called "coming up with a counterexample". If somebody claimed that all prime numbers were odd, it would be perfectly valid to point out that 2 is both prime and even. Discarding the proof because "you purposefully picked 2 to show me wrong" is invalid, as this is the whole point of the proof.
Likewise, in this case DNS-and-BIND claimed that all infinitely long irrational numbers have necessary a long sequence of 8's in them. I refuted his claim by showing him a number which had no 8 at all inside. Now, what exactly is your problem with my refutation of that claim?
> But that doesn't tell us anything about wether or not there is a string of 5,646,498,765 8s in pi or any other irrational number in decimal.
You're right on that, but nobody claimed the contrary. Saying "not all irrational numbers have a long string of 8's in them" is not the same thing as "no irrational numbers have a long string of 8's in them".
It's like in real life: "not all pies are cream pies" (or expressed differently "it is a pie, so it has to be a cream pie"): indeed, there are also apple pies...
But that doesn't mean that "no pies are cream pies" (i.e. cream pies don't exist): indeed Bill Gates was hit by one in the face...
Say no to software patents.
42424242 at position 242424.
Oddly enough, according to the pi search page, the same string can be found again at position 1404114, which is also below 100000000. On a normal pi, you'd expect a single occurrance of 42424242 below 100000000, and at a completely random position...
Say no to software patents.
Cannot be, or else it would repeat itself an infinite number of times, cyclically, which would make it a rational number.
Say no to software patents.
There's a signifigant difference between random and irrational. Random means you'll get a totall random value every time you look at it. Irrational means there's no pattern to it, while the number is most definately set and unchanging.
The second application should be the start digit of a Metallica MP3.
Anyone want to start up a distributed network to look for these?
I'm trying to teach myself to set people on fire with my mind... Is it hot in here?
I'm sure there's a lot of cool stuff in the first 2^128 digits. If 128 bits is a long on an itanium system, I'm sure we could have a lot of fun searching the first 2^128 digits of pi for stuff without even breaking out of the long address space. If 2^128 seems small, how about 2^256? 2^1024? 2^65536? That's not a lot of bytes, but it's a hell of a lot of space to search. Probably more than modern computers will be able to handle for years (Even if we do start up a distributed net type search engine to look for things.) Who knows. My next computer might have to include a pi coprocessor...
I'm trying to teach myself to set people on fire with my mind... Is it hot in here?
So, yes, trivially, all strings appear an infinite number of times. Or are we talking about another measure of frequency (number of appearances in a substring of pi's digits of a given length?)
There's 10 types of people in this world, those who understand binary and those who don't.
It will start a whole new branch of numerology dedicated to finding entire new holy books... the Book of the Damned, I Microsoft, II Microsoft, the letter of BOFH to the Great Unwashed, and, of course, the source code to Office (which will take up the space between 2^8 and 2^40906) ...
Zaphod B
Zaphod B
When duplication is outlawed, only outlaws will have
So if anything, they are proving normality to base 2^n, NOT base 10^n. And it may actually be that their proof is general enough to show normality in all bases - the article is not clear on that point.
Ok, if an average MP3 is 5MB (5,242,880 bytes) then the odds of finding a specific MP3 using a sequence of random numbers is 1/256**5,242,880 (since 8-bit bytes have 256 possibilities). This is about the same as 1/10**12,626,113 (since our base 10 numbering system gives each digit only 10 possibilities)
P(MP3)=1/10**12,626,113 (really close to 0)
Thus the odds of not finding a specific 5Mb MP3 using a sequence of random digits is 1-P(MP3)
~P(MP3)=1-1/10**12,626,113 (much closer to 1)
Since the longest known expansion of Pi is about 500 Billion digits (500,000,000,000) there are 499,987,373,888 consecutive strings of 12,626,113 digits contained in the known expansion. So the question is, what is the probabilty that at least one of these is the specific 5MB MP3 we are looking for.
An easier question is to ask what is the probabilty that we won't find the specific 5MB MP3 in a 500 Billion digit expansion of Pi.
The probability that any specific 12,626,113 digit substring of the 500 Billion digit expansion of Pi is not the 5MB MP3 we are looking for is ~P(MP3). So the probability that every one of the 499,987,373,888 possible 12,626,113 digit expansions is not the 5MB MP3 we are looking for is ~P(MP3)**499,987,373,888.
P(~MP3)=~P(MP3)**499,987,373,888
So now that we know the probability that a specific 5MB MP3 file is not contained in the 500 Billion digit known expansion of Pi, we can calculate the probability that we can find at least one instance of a specific 5MB MP3 as ~P(~MP3)=1-P(~MP3).
~P(~MP3)=1-P(~MP3) =1-~P(MP3)**499,987,373,888 =1-(1-P(MP3))**499,987,373,888 =1-(1-1/10**12,626,113)**499,987,373,888
Hmmm... I think I'll go by a lottery ticket.
Other possible things that we could find -
PI IS THREE!!
froin laven, i didn't think i'd have to use that.
sig?
Then there are the 'dirty' irrational numbers like pi and e that seem to have random digits. The research mentioned has moved a big step closer to proving that the digits of pi don't just seem random, they truly are random (at least in the sense that all possible combinations occur).
The part that'll really blow your mind is that somebody found an equation that tells you any binary digit of pi you want, without having to calculate any of the other binary digits. (See here.) That is why people are excited by the conjectured normality of pi: if normal, it produces all possible strings of bits from a trivial deterministic equation. This mixture of randomness with order is at the heart of many interesting questions in chaos theory, computational theory, and cryptography.
-- ;-)
Kuro5hin.org: where the good times never end.
And that is a pain in the neck for everyone in comp.compression.
There is a frequent fallacy among those who almost understand how compression works, that works like this:
The assumption, of course, is that the number of digits and the offset can be encoded in a form that will be smaller than the original sequence. There is nothing to warrant that assumption. The fact is that the number of possible inputs that a lossless compression method can handle places lower bounds on the average length of its outputs. This means that no lossless compression method can achieve a lower average length for its outputs than would be achieved by simply numbering them all with the non-negative integers.
In fact, 'compressing' a sequence of digits into a (length, offset) pair will do substantially worse, since there are multiple (length, offset) pairs that will correspond to a given digit sequence; for instance, "1" could be encoded as (1,0) or (1,2). This duplication means that (1,2) is essentially wasted, since it could be representing a sequence that currently has a longer representation.
Lossless compression methods need to be used in conjunction with models: some criteria that separates the data we will want to compress from the vast majority of files, about which we do not care. The accuracy of this model affects how many of our inputs we can actually compress, and its precision affects the average compression ratio.
If people are to respect the law, perhaps the law should begin by respecting the people.
What if PI is a rational number, with infinitely large numerator and denominator? :)
The REAL sam_at_caveman_dot_org is user ID 13833.
Pi is not that unusual. Here is a simple number called (I believe) the Champernowe Constant
0.123456789101112131415161718192021222324...
After the decimal point we are, in effect, counting. Clearly, any string "35002134" will appear in the Champernowe constant, and infinitely often. (Anybody know where I got the sample string?)
Pi is an amazing number, clearly, but sometimes it is erroneously represented as the only number with the above property.
DJS
http://www.dougshaw.com
God is real unless declared integer
The middle mind speaks!
Do you have a constructive proof of this oracles existence? Constructive is the key word here. The problem is similar to finding the (Kolmogorov) complexity of a natural number. You can prove that every number does have a complexity (measured as the minimum size representation of that number), but you can't compute the complexities. The function exists, but you can't compute it.
Turing's theorem is much more interesting that you seem to give it credit for. Let B be a set of natural numbers. Define the "B-Halting Problem" as Given an arbitrary Turing machine, determine whether that Turing machine, when equipped with a B-oracle, halts after being started with empty input. Then Turing's proof shows that no Turing machine equipped with only a B-oracle can solve the B-Halting Problem. The usual, oracle-less, version is a special case of this: just take B = empty set. You're absolutely right. The Turing Theorem is interesting (as long as you don't consider the original paper. . .snore!). It's unfortunate that everyone just chooses to ignore it, instead of really investigating what the implications of the theorem are (along side with Godels Incompleteness Theorem). They're mostly held up as clever examples, then discretely swept under the table.
Randomness is not an essential part of the study of Turing machines-with-oracle, and the Halting Problem does not involve randomness at all.
Not at all? The fields share many similarities, and it's hard to find a modern discussion of one without the other.
Cheers!
The middle mind speaks!
A teacher, a physicist and a mathemetician are having drinks together in a Scottish pub when the teacher looks out the window and sees a white sheep. The teacher says, "There are white sheep in Scotland". The physicist looks out the window and declares, "There are sheep in Scotland; we have already detected and confirmed white ones." The mathematician says, "In Scotland there is at least one sheep, at least one side of which is hite."
No, I didn't pull that from Singh (it's there, though). It's an old mathematician's joke but it's true. The most anal Rainman you've ever seen is incredibly chaotic compared to mathematicians (at least when they're working on a proof or theorem).
woof.
"No ma. You don't have to worry about Code Red. Yes, I know CNN told you that you do. Ma, do you run a Web server? No, Netscape is a browser, not a server. Yes, there's a difference. You don't want to know. No, and the Internet didn't die last week, either..." -- my side of a phone call two nights ago.
And the Scotsman says, "You lads keep yer filthy, stinkin' eyes off me wife!!!!"
(Sorry, couldn't resist...)
Of course, finding the appropriate sequence would be the challenge...
Pi is an integer in base pi...
Ok. The standard solution for solving e^(i*A*X) was useful for solving second-order differential equations (ordinary form: P(x)y" + Q(x)y' + R(x)y = G(x) ) . My notes are sparse and my brain is tired, so I'd rather not try to remember the whole series of steps required to get to the part where this is actually useful. If you want the long version of the proof, ok, I'll post it tomorrow if you say so.
I'm not even sure if this was derived in class or just one of those few equations that were given to us "just trust me" sorta things. I had the bad habit of rarely writing down proofs, so I'd probably have to hunt this one down online or in my of my Calc books.
But, magically, here in my notes it says:
e^(i x B[beta] * x) = cos (Bx) + isin Bx .
My apologies for not desiring to hunt down the appropriate symbols.
So, with B = Pi and x = 1, you get:
e^(i*Pi) = cos Pi + isin Pi .
The cosine of Pi is -1 and the sine of Pi is 0, so it becomes = -1 + (i)*0 = -1 . Notice that e^(-i*Pi) also gives -1.
So in summary it really wasn't much of a feat for me to reproduce the final parts of the proof, only a matter of remembering what that standard equation was (shortcuts are wonderful things). Let me know if you really want me to derive the top part though.
"The universe seems neither benign nor hostile, merely indifferent." --Carl Sagan
Thanks. I always knew that not writing down the derivations in class would catch me at some point =).
Well, in one case they already did. In my Calc III class my professor proved a relatively simple theorem and promptly put that theorem on the first page of the test. I wrote the theorem down, but I never really committed it to memory. That was one of the few tests I got a B on...oh well.
"The universe seems neither benign nor hostile, merely indifferent." --Carl Sagan
Trying to imagine why every n digit number shows up the exact same amount of times is hard to imagine at first. But then, once you think about it, on an infinite scale, it would seem to attest to Pi's true randomness.
On a side note, I had a Calc II professor awhile back that wrote on the board:
e^(i*Pi) = -1 (of course, using the real symbols).
Then, he proved it. I have the proof written down in a notebook and I even managed to work through the final parts of the proof (it uses a standard solution for finding e^(i*A*X) without using it. If anyone is really interested in seeing it, I can post it (in rough ascii math =) For those of you with TI-92s that don't believe me, type it in. That magical machine can do more than I give it credit for sometimes.
Anyway, I just thought it was absolutely incredible that you could mix the two most popular transcendental numbers with the imaginary number (square root of -1) and spit out plain old -1.
"The universe seems neither benign nor hostile, merely indifferent." --Carl Sagan
Before his untimely death, Alan Turing did a lot of work on a new theoretical machine capable of transcending the limits of conventional Turing machines. The new machine (the so-called O-machine), would be a Turing machine connected to an "oracle", which would store some irrational quantity that would be able to do things like solve the Halting Problem, since it would contain an infinite amount of information (including overy possible program that could be created). At least, that's as much as I can remember (no links, sorry).
Who knows, maybe pi would suffice for such an oracle?
Sounds hairy. *cough*
--
Not being a troll, but I still don't see the big deal about one irrational number.
In Carl Sagan's book, Contact, there is an interesting revelation made to Ellie by the alien she visits light years away. It tells her that buried deep in Pi is an important message.
(Here's where my memory gets a little iffy.)
So when she returns home, she writes a program that searches for non-random data in Pi, in multiple bases, and sure enough she finds a message in base-11 composed of all 1's and 0's.
When laid out in rows of equal columns, a perfect circle is formed out of 1's, with 0's as the background.
So according to Contact, embedded into the digits of Pi is the picture of a perfect symbol. If this were true, it would be proof that the universe was created by intelligent life.
Or at least a real funny joke.
"And like that
Well, as others have rightly noted, this solution wouldn't work. It takes N digits to represent a number of N digits, quantum mathematics aside, as long as those digits are more or less random. Compression programs like ZIP only work at all because certain strings of numbers are more common than others in computer files (if I understand the technology correctly).
However, this idea could go the way of all complex (yet failed) compression algorythms: encryption! Imagine trying to decode the resulting index, with no idea that Pi was even involved. Not gonna happen.
I can't say the idea didn't intrigue me for a few seconds, though; adding infinity to any equation always makes for the most fascinating possibilities.
---
I remember watching Northern Exposure when I was about 13 and there was this episode where Chris Stevens dates this mathematician chic and she talks about a string of eight 8's. Years later when I read about a Pi search engine I tried it and was actually surprised to see it worked.
/. = useless posts.
alcohol +
:o)
BOSTON SUCKS!
Every mathematical "definition" is arbitrary. The logic in mathematics lies in the derivations from the definitions, not in the definition themselves. You may have useless definitions (which are in contrast with "usual" maths), fruitful definitions (which let people find a lot of new things), or boring definitions, which do not give a lot of interesting results.
In the case of factorization, 1 is called "a unity", that is something separated from the prime numbers, because the most important result is the unique factorization ( up to unities!!!) in the group of integer numbers with the usual "multiplication" operation.
Notice that there are also a lot of theorems which have to distinguish the case of 2 and of another prime: the reason is not that 2 is the only even prime number, but rather than in the field generated from the first 2 numbers, "1" and "-1" are the same thing. But those results are not so important to deserve a modification of the definition of prime numbers to cast out 2.
A last comment: it is not difficult to write down a number which is normal in base 10. It suffices to digit 0.1234567891011121314151617181920... But if someone ever shows that \pi (or e, or a "common" irrational number) is normal in a certain base, it would be a really important result - for mathematicians, that is. The practical importance would be zero.
ciao,If a five-hundred digit sequence of numbers can be represented instead by, say, a twenty-five digit offset value, that seems like quite good compression -- until you remember that the only way it would work is if either: the (de)compressor carries around a datafile of "all" (well, a huge chunk of) for reference, or it has to recalculate up to the position of that offset on each run. And there's where it all breaks down.
Liberty in your lifetime
0123456789 : from 17,387,594,880-th of pi
0123456789 : from 26,852,899,245-th of pi
0123456789 : from 30,243,957,439-th of pi
0123456789 : from 34,549,153,953-th of pi
0123456789 : from 41,952,536,161-th of pi
0123456789 : from 43,289,964,000-th of pi
Liberty in your lifetime
It's also at position 242424 not counting the "3." since the sequence starting at 242422 (242424 counting the "3.") is "42424242."
Liberty in your lifetime
Liberty in your lifetime
Are you sure you know what you're looking for/at? Is the number 42424242 one that you would search for first, or a pattern one found after the fact? These are VERY different things. If it was a pattern that was interesting, one must look at the space of all interesting patterns and look at the probability that one might be found. This is the difference between math and numerology, or astrology and astronomy. Coincidence has a role, why it should even be expected; for christs sake it is even predictable. Perhaps some of the people reading this are aware of the fact that if you get about thirty people or so, together two of them are likely to share a birthday. This does not mean that if you pick any person that someone else is likely to share their birthday. Rather of the group there is a good chance two of the individuals, whomever they may be, will likely share the birthday. If you pick 42424242 before hand, that makes it unlikely, if you don't it makes it coincidence.
--Jimmy has fancy plans; and pants to match.
Then you were taught wrong. In fact I can prove the opposite. If PI reccurs within PI in any number base then PI is a rational number, therefore it cannot reccur.
The reason is that if we have 0.123123123... then we can divide it by 1000 and subtract it from itself to get 0.123, so if x - 0.001 x = 0.123 then x = 0.123 / 0.999 => x = 123/999
The proof can be trivially extended to cover numbers of the form 0.1111123123123123 ...
The flaw in the original logic is that a proof that every finite string occurs in PI does not consititute a proof that every infinite string occurs in PI.
Looking for an Information Security student project suggestion?
Try http://dotcrimeManifesto.com/
Right here in C
If pi has all conceivable messages, pi must contain all of the US military's secrets, DeCSS, kiddie pr0n, violent and explicit sexual films beyond anyone's imagination and much much more. It must therefore be banned. When you get the death penalty for circle possession, don't say I didn't warn you...
Strictly speaking, the property mentioned isn't actually normality, but normality to base 10^n for all n. Normality to base b means that if you write down the base-b expression for the number then every base-b digit occurs with equal frequency. So normality to base 10 means that in the usual decimal expansion, 3 and 7 occur with equal frequency, for instance. Normality to base 100 means that, e.g., in the decimal expansion 34 and 87 occur with equal frequency.
It's known that in a certain precise sense, almost all numbers are normal (i.e. normal to *all* bases). But to this day, not one single specific number has been *proved* to be normal!
A specific string of k bits long will on average occur every 2^k bits of a random bitstream. (For example, it will on average take about 1024 coin tosses to get 10 heads in a row.)[0] The reasoning I have in mind for this relies on the fact that coin tosses are independent events. The digits of pi are certainly not independent random events because we can calculate them. However, because of normality behaves macroscopically much like independent events, k precise bits will occur on average every 2^k bits of pi. I imagine the actual proof is highly nontrivial, but handwavy entropy arguments convince me that this is true.
.320 and they
play 1000 games (made-up numbers).
[0] As an interesting sidenote, many of the "streaks" in sports coincide roughly with the streaks that probability dictates. Baseball hitting streaks aren't necessarily because the hitter is in the zone... they may be because probabalistically, they are due to hit in 30 games in a row if their average is
healyourchurchwebsite.com - WWJB?
(Don't laugh at me! I can still kick your butt in chemistry!)
I recall a method for detecting falsified data that relied on the fact that certain functions (those that followed a log pattern, I think) produce results with the last digit's frequency skewed toward smaller numbers (1, 2, 3,...) and AWAY from larger numbers (9, 8, 7,...).
So my question is, does Normalcy apply to functions (in the way I described), as well as numbers?
Yeah, but what's the point of finding, "First Post!", in pi if it's not first?
As one of the two co-authors of the recent paper on pi, I thought it would be appropriate if I myself made a few comments: 1. Crandall and I did not prove that pi is normal. Ours is a partial proof, reducing this question to a conjecture of chaotic processes. But if nothing else it help explain why the digits of pi appear random -- they are generated by a chaotic sequence generator. 2. Our results deal strictly with base 2 and base 16 -- we can say nothing about decimal (base 10) digits of pi. 3. The "normality" property that we seek to prove is not the same as randomness in the general sense. The digits of pi are generated by a very simple, compact deterministic sequence which is thus not random in the Chaitin sense. Instead, we only claim that Pi is statistically normal -- its binary expansion contains every string of n binary digits, with limiting frequency 2^(-n). 4. The algorithm for computing individual binary or hex digits of pi is very simple. It is completely stated on my web site: http://www.nersc.gov/~dhbailey/pi-alg 5. The latest paper on the normality of pi, and the original paper giving the pi-digit-calcualting algorithm, are also available on my web site: http://www.nersc.gov/~dhbailey Cheers, David H Bailey david@dhbailey.com