

Your birthdate in the decimal expansion of Pi - carljoseph
http://www.angio.net/pi/piquery.html

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kibibu
I love this sort of thing:

> Be warned that 50 million digits of pi takes up 50 megabytes. This can take
> up to 4 hours to download with a 28.8k modem!

~~~
mathattack
The website doesn't look like it's been updated in a dozen or so years, but
it's still up. It's a good style marker for how the web used to look. I guess
they still make enough in ad sales and Pi t-shirts to stay running.

~~~
leephillips
"It's a good style marker for how the web used to look."

Loads fast and easy to read. I hope our web designer friends are taking note.

~~~
mathattack
This was probably very slow back in the day. The irony is that network speed
is being used to cram more stuff across the web, rather than speed up existing
material. The price of change.

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worldsayshi
An idea I've had, the answer is most probably no, but I found it interesting
to think about, others may find it trivial:

Is there a number that (1) contains most subsequences and where (2) finding a
certain subsequence is computationally efficient? Also, (3) finding a
subsequence given a starting and end position should be efficient. If so, we
have a efficient mean to transfer data. Just send the index positions. And we
can store any data with just those two index positions. Then again, those
index numbers are likely to be very very large. Perhaps we can in turn
transform the index positions to smaller index positions by finding their
positions in the sequence. Then we need a third number to signify the number
of recursive uses of the storage.

~~~
fhars
(1) Yes, 0.1234567891011121314151617181920212223242526....

(2) I'm too lazy to do the calculation right now, but it should not be too
difficult to calculate the sum of the lengths of all numbers smaller than n,
and then you've got the index of n.

(3) In any feasible encoding, the expected size of the start index of n will
be much larger than the size of n. And every recursive step will just blow up
the sizes even more.

~~~
danbruc
If you slightly modify it to 0. 0 1 2 3 4 5 6 7 8 9 00 01 02 03 04 05... then
the number n padded with zeros to k digits can be found starting at position
10 * floor(10^(k - 1) / 9) + k * n.

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rookonaut
Maybe of interest for some of you:
[http://baby.pirthday.com/](http://baby.pirthday.com/) 8 days until the
perfect creation date for your ultimate pi-baby (born 3/14/15).

Other essential services are: your age in pi
([http://pirthday.com/](http://pirthday.com/)) or seeing who has his pirthday
today ([http://happy.pirthday.com/](http://happy.pirthday.com/)). Disclaimer:
Logo has been created by a friend (@turboele), other stuff by me.

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ccozan
Interesting idea. Actually, just thinking, could be used for a pseudo-crypto
app. Just send a series of origin numbers [n1,n2,n3...] and this translates to
[pos1,pos2,pos3...] series of numbers. You can use also instead of Pi any
other transcendental/irrational number [0].

[0] [http://www.numberworld.org/digits/](http://www.numberworld.org/digits/)

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plg
So if I take the latest Lady Gaga mp3 and look at the stream of bytes, and if
I can find that identical stream of digits in the digits of Pi, does Lady Gaga
(or her record company) "own" that portion of Pi? If yes, then WTF!?!? If no,
why not, and what does this mean for the idea of copyright, anyway?

~~~
dragonwriter
Copyright restricts copying, it doesn't restrict similar, or even identical,
expressions that are independently generated (a jury may see similarity as
evidence of copying, so merely the theoretical possibility that a similar work
was independently arrived at doesn't get you a free pass from liability, but
similarity itself isn't restricted by copyright.)

So, no, the fact that a sequence of bytes occurs in a particular digital
recording doesn't mean that the copyright owner of that recording "owns" that
series of bytes. It means they own a specified set of rights _in that
particular recording_.

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Aardwolf
> If, on the other hand, you act like a computer geek and use zero based
> indexing, then you get these numbers: 6, 27, 13598, 43611, 24643510

Heh, it says "act like a computer geek" and the numbers are not even in
binary? :)

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zak_mc_kracken
I wonder if there is not a bug in the decompress program:

    
    
      for (i = 0; i < FILE_SIZE; i++) {
        printf("%d%d", (buf[i]&0xf0)>>4, buf[i] & 0xf);
      }
    

Shouldn't this be >>8 ?

~~~
gjm11
No, it's extracting the upper 4 bits of the byte, not the upper 8 bits of a
16-bit quantity.

~~~
fhars
This is actually a quite standard number format. Real computers like the 6502
[http://www.6502.org/tutorials/decimal_mode.html](http://www.6502.org/tutorials/decimal_mode.html)
and the IBM zSeries have harware support for that encoding
[http://en.wikipedia.org/wiki/Binary-
coded_decimal#Packed_BCD](http://en.wikipedia.org/wiki/Binary-
coded_decimal#Packed_BCD) and there are decoder chips like the 74LS48 that
directly translate these nibbles to 7-segment display numbers.

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nirai
at what index does the ascii encoding of the entire works of Shakespeare
begin?

~~~
mnw21cam
I could write the index number here, but it would probably be about as long as
the entire works of Shakespeare.

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chid
Well mine isn't within the first 50 million. In any case, what's the relevance
of this? Everything (with the correct encoding scheme) can be expressed
somewhere in the decimal expansion of Pi.

~~~
danbruc
Without "(with the correct encoding scheme)" this is not a proven fact, i.e.
it is not known whether Pi is a normal number [1].

[1]
[http://en.wikipedia.org/wiki/Normal_number](http://en.wikipedia.org/wiki/Normal_number)

~~~
chid
Even if it weren't a normal number it would still be true.

