11 ms·

Your birthdate in the decimal expansion of Pi

48 points by carljoseph 12y ago

kibibu 12y ago
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!
- Aldo_MX 12y ago
  Haha, I laugh at you! I got a 56k modem.
  - BuildTheRobots 12y ago
    Yeah, but does yours do v92? If you don't get the two bongs during connect then you're missing out ;)
    - programmer_dude 12y ago
      Do the bongs come with some grass?
      - BuildTheRobots 12y ago
        Well I did always find dial up far more bearable when I was drugged out of my head... though I'm kinda upset you don't remember the v92 connect noise... this is HN not reddit ;)
- kubiiii 12y ago
  Hopefully some modems firmware can be upgraded to reach 33.6k :)
- chrismcb 12y ago
  If I kick off a download, it will be done when I wake up. Better check the "hang up after download is finished"
- mathattack 12y ago
  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 12y ago
    "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 12y ago
      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.
zak_mc_kracken 12y ago
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 ?
- myrmidon 12y ago
  (buf[i]&0xf0) >> 8 would be zero (always). One hex digit (0x0f) is 4 bits wide, not 8 (one byte <=> two hex digits).
- gjm11 12y ago
  No, it's extracting the upper 4 bits of the byte, not the upper 8 bits of a 16-bit quantity.
  - zak_mc_kracken 12y ago
    Doh, of course.
  - fhars 12y ago
    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.
Aardwolf 12y ago
> 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? :)
chid 12y ago
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 12y ago
  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 12y ago
    Even if it weren't a normal number it would still be true.
- devNoise 12y ago
  I found my birthday if I used MMDDYYY, but it didn't find it for YYYYMMDD. I tend to use the latter format when I'm date stamping files.
  - chid 12y ago
    It's there. You just have to look harder for it :)
nirai 12y ago
at what index does the ascii encoding of the entire works of Shakespeare begin?
- mnw21cam 12y ago
  I could write the index number here, but it would probably be about as long as the entire works of Shakespeare.
- dmoo 12y ago
  Have they found this bit from Contact yet "When Ellie looks at what the computer has found, she sees a circle rasterized from 0s and 1s that appear after 10^20 places in the base 11 representation of π. This gives her a way to convince the world of something greater – that intelligence is built into the universe itself."
worldsayshi 12y ago
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 12y ago
  (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 12y ago
    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.
- deleted 12y ago
  [deleted]
- BuildTheRobots 12y ago
  It's in no way efficient (and is obviously a joke -though one with compile-able and usable code), but piFS [1] kinda does what you want... [1] https://github.com/philipl/pifs https://github.com/philipl/pifs
- andyjohnson0 12y ago
  I don't know about (1), but (2) could be achieved by scanning through the first n digits and building a lookup tree. If we restrict ourselves to Pi then (3) could be achieved by either just keeping all the digits in memory or using the Bailey–Borwein–Plouffe formula [1], which allows the nth digit to be determined without needing to calculate the preceeding digits. Your're essentially describing a compression mechanism. As you might expect, this isn't a new idea [2] and it generally fails because (as you said) the index quickly gets large as the size of the block of data increases - basically destroying any advantage for non-trivial block sizes. There is some discussion of the space trade-offs in [3], and I wouldn't be surprised if the speed of this approach is poor when compared to conventional techniques. Another problem is that it is still not known whether number like Pi actually contains every possible finite sequence of digits [3], so not all input blocks can be compressed. In practice, this shouldn't be a problem for small blocks though. [1] http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula [2] https://www.google.co.uk/search?q=pi+compression https://www.google.co.uk/search?q=pi+compression [3] http://penduin.blogspot.co.uk/2006/10/pi-compression.html http://penduin.blogspot.co.uk/2006/10/pi-compression.html [4] http://www.askamathematician.com/2009/11/since-pi-is-infinite-can-i-draw-any-random-number-sequence-and-be-certain-that-it-exists-somewhere-in-the-digits-of-pi/ http://www.askamathematician.com/2009/11/since-pi-is-infinit...
ccozan 12y ago
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/
rookonaut 12y ago
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.
plg 12y ago
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 12y ago
  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.

48 points by carljoseph 12y ago

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!

Aldo_MX 12y ago
Haha, I laugh at you! I got a 56k modem.
- BuildTheRobots 12y ago
  Yeah, but does yours do v92? If you don't get the two bongs during connect then you're missing out ;)
  - programmer_dude 12y ago
    Do the bongs come with some grass?
    - BuildTheRobots 12y ago
      Well I did always find dial up far more bearable when I was drugged out of my head... though I'm kinda upset you don't remember the v92 connect noise... this is HN not reddit ;)
kubiiii 12y ago
Hopefully some modems firmware can be upgraded to reach 33.6k :)
chrismcb 12y ago
If I kick off a download, it will be done when I wake up. Better check the "hang up after download is finished"
mathattack 12y ago
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 12y ago
  "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 12y ago
    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.

zak_mc_kracken 12y ago

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 ?

myrmidon 12y ago
(buf[i]&0xf0) >> 8 would be zero (always). One hex digit (0x0f) is 4 bits wide, not 8 (one byte <=> two hex digits).
gjm11 12y ago
No, it's extracting the upper 4 bits of the byte, not the upper 8 bits of a 16-bit quantity.
- zak_mc_kracken 12y ago
  Doh, of course.
- fhars 12y ago
  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.

Aardwolf 12y ago

> 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? :)

chid 12y ago

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 12y ago
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 12y ago
  Even if it weren't a normal number it would still be true.
devNoise 12y ago
I found my birthday if I used MMDDYYY, but it didn't find it for YYYYMMDD. I tend to use the latter format when I'm date stamping files.
- chid 12y ago
  It's there. You just have to look harder for it :)

nirai 12y ago

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

mnw21cam 12y ago
I could write the index number here, but it would probably be about as long as the entire works of Shakespeare.
dmoo 12y ago
Have they found this bit from Contact yet "When Ellie looks at what the computer has found, she sees a circle rasterized from 0s and 1s that appear after 10^20 places in the base 11 representation of π. This gives her a way to convince the world of something greater – that intelligence is built into the universe itself."

worldsayshi 12y ago

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 12y ago
(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 12y ago
  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.
deleted 12y ago
[deleted]
BuildTheRobots 12y ago
It's in no way efficient (and is obviously a joke -though one with compile-able and usable code), but piFS [1] kinda does what you want... [1] https://github.com/philipl/pifs https://github.com/philipl/pifs
andyjohnson0 12y ago
I don't know about (1), but (2) could be achieved by scanning through the first n digits and building a lookup tree. If we restrict ourselves to Pi then (3) could be achieved by either just keeping all the digits in memory or using the Bailey–Borwein–Plouffe formula [1], which allows the nth digit to be determined without needing to calculate the preceeding digits. Your're essentially describing a compression mechanism. As you might expect, this isn't a new idea [2] and it generally fails because (as you said) the index quickly gets large as the size of the block of data increases - basically destroying any advantage for non-trivial block sizes. There is some discussion of the space trade-offs in [3], and I wouldn't be surprised if the speed of this approach is poor when compared to conventional techniques. Another problem is that it is still not known whether number like Pi actually contains every possible finite sequence of digits [3], so not all input blocks can be compressed. In practice, this shouldn't be a problem for small blocks though. [1] http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula http://en.wikipedia.org/wiki/Bailey-Borwein-Plouffe_formula [2] https://www.google.co.uk/search?q=pi+compression https://www.google.co.uk/search?q=pi+compression [3] http://penduin.blogspot.co.uk/2006/10/pi-compression.html http://penduin.blogspot.co.uk/2006/10/pi-compression.html [4] http://www.askamathematician.com/2009/11/since-pi-is-infinite-can-i-draw-any-random-number-sequence-and-be-certain-that-it-exists-somewhere-in-the-digits-of-pi/ http://www.askamathematician.com/2009/11/since-pi-is-infinit...

ccozan 12y ago

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/

rookonaut 12y ago

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.

plg 12y ago

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 12y ago
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.