4 ms·

Can you copyright a tweet?

5 points by winanga 17y ago

noonespecial 17y ago
Hmmm. How large would a book need to be to contain all possible tweets?
- buugs 17y ago
  impossibly large
- patio11 17y ago
  140 bytes means there are 2.70768525 × 10^126 (thank you, Google) possible tweets. I hope your book doesn't need a hydrogen atom to list each tweet, because since the universe probably contains less than 10^80 of them, we'd run out before you were finished.
  - ubernostrum 17y ago
    That doesn't seem right. Assuming: 1. A tweet can be anywhere from 0 to 140 bytes. 2. Each byte can have any one of 256 possible values. Then the number of possible tweets is 256^0 + 256^1 + 256^2 + ... + 256^140 (the sum of number of permutations at each length in bytes). If I remember my number theory correctly, the sum of the first n powers of a number, k, is (k^(n+1) - 1)/(k - 1). Or, in this case, (256^141 - 1)/255. Which is a much larger number than the one you just gave.
    - derefr 17y ago
      A tweet is a zero-terminated string. It's 256^140, as the rest of the series can simply be unified with their representations padded with zeros.
      - ubernostrum 17y ago
        Depends on how you want to define a tweet, I guess, but 256^140 is still a larger number than quoted in the post I replied to.
        patio11 17y ago
        I tried calculating 8^140. Eight bits in a byte, you know. This is what happens at the end of a week of crunch. (On the plus side, you guys aren't paying me for what I'm writing today.)
- gojomo 17y ago
  Clearly some unicode is allowed, but I can't find a good reference as to whether all unicode characters are allowed, or whether each counts equally as one character against the 140-character limit. But let's assume 140 bytes is the actual limit. That suggests around 255^140 possible tweets (assuming NULs unused). RFC678 defines a 'logical page' based on the printable area of a letter-sized page as 60 lines of 72 characters each; thus figuring 2 lines per tweet, that's 60 tweets per double-sided page. So we need (255^140)/60 pages. An unopened ream (500 sheets) of 30lb paper I have is 5cm thick. So the stack of all printed tweets would be ((255^140)/60)×(5cm/500). WolframAlpha calculates: http://www86.wolframalpha.com/input/?i=((255^140)%2F60)*(5cm%2F500) http://www86.wolframalpha.com/input/?i=((255^140)%2F60)*(5cm... = 1.451x10^315 ly (light years) ...and they also helpfully offer a "Comparison as diameter"... ≅ 1.6x10^304 x diameter of the observable universe (≅ 93 billion ly ) So, even with an allowance for the smaller print of commercial publishing, there'd be no place in our universe to put this book. A Kindle version, implemented in a line or two of code, seems more practical. ObBorges: http://jubal.westnet.com/hyperdiscordia/library_of_babel.html http://jubal.westnet.com/hyperdiscordia/library_of_babel.htm...
  - eru 17y ago
    I propose not storing/generating every possible tweet, put a careful selection: Take a statistical model of tweet/sms-language, say with Markov-Chains, and generate only the tweets that fit the model with a large-enough significance. That should reduce the number to a more manageable level, although a book may be still off limits.
    - gojomo 17y ago
      But the ones that look like noise are compressed versions of brilliant longer tweets! (If decompression does not yield a suitably brilliant tweet, try another algorithm until the desired level of brilliance and length is achieved. Generation and storage of the algorithms is left as an exercise for the reader; see Kolmogorov.)
- anigbrowl 17y ago
  Depends...how small can I print them?
brl 17y ago
This link from the article is a lot more interesting than the article itself: http://www.canyoucopyrightatweet.com/ http://www.canyoucopyrightatweet.com/

5 points by winanga 17y ago

noonespecial 17y ago

Hmmm. How large would a book need to be to contain all possible tweets?

buugs 17y ago
impossibly large
patio11 17y ago
140 bytes means there are 2.70768525 × 10^126 (thank you, Google) possible tweets. I hope your book doesn't need a hydrogen atom to list each tweet, because since the universe probably contains less than 10^80 of them, we'd run out before you were finished.
- ubernostrum 17y ago
  That doesn't seem right. Assuming: 1. A tweet can be anywhere from 0 to 140 bytes. 2. Each byte can have any one of 256 possible values. Then the number of possible tweets is 256^0 + 256^1 + 256^2 + ... + 256^140 (the sum of number of permutations at each length in bytes). If I remember my number theory correctly, the sum of the first n powers of a number, k, is (k^(n+1) - 1)/(k - 1). Or, in this case, (256^141 - 1)/255. Which is a much larger number than the one you just gave.
  - derefr 17y ago
    A tweet is a zero-terminated string. It's 256^140, as the rest of the series can simply be unified with their representations padded with zeros.
    - ubernostrum 17y ago
      Depends on how you want to define a tweet, I guess, but 256^140 is still a larger number than quoted in the post I replied to.
      - patio11 17y ago
        I tried calculating 8^140. Eight bits in a byte, you know. This is what happens at the end of a week of crunch. (On the plus side, you guys aren't paying me for what I'm writing today.)
gojomo 17y ago
Clearly some unicode is allowed, but I can't find a good reference as to whether all unicode characters are allowed, or whether each counts equally as one character against the 140-character limit. But let's assume 140 bytes is the actual limit. That suggests around 255^140 possible tweets (assuming NULs unused). RFC678 defines a 'logical page' based on the printable area of a letter-sized page as 60 lines of 72 characters each; thus figuring 2 lines per tweet, that's 60 tweets per double-sided page. So we need (255^140)/60 pages. An unopened ream (500 sheets) of 30lb paper I have is 5cm thick. So the stack of all printed tweets would be ((255^140)/60)×(5cm/500). WolframAlpha calculates: http://www86.wolframalpha.com/input/?i=((255^140)%2F60)*(5cm%2F500) http://www86.wolframalpha.com/input/?i=((255^140)%2F60)*(5cm... = 1.451x10^315 ly (light years) ...and they also helpfully offer a "Comparison as diameter"... ≅ 1.6x10^304 x diameter of the observable universe (≅ 93 billion ly ) So, even with an allowance for the smaller print of commercial publishing, there'd be no place in our universe to put this book. A Kindle version, implemented in a line or two of code, seems more practical. ObBorges: http://jubal.westnet.com/hyperdiscordia/library_of_babel.html http://jubal.westnet.com/hyperdiscordia/library_of_babel.htm...
- eru 17y ago
  I propose not storing/generating every possible tweet, put a careful selection: Take a statistical model of tweet/sms-language, say with Markov-Chains, and generate only the tweets that fit the model with a large-enough significance. That should reduce the number to a more manageable level, although a book may be still off limits.
  - gojomo 17y ago
    But the ones that look like noise are compressed versions of brilliant longer tweets! (If decompression does not yield a suitably brilliant tweet, try another algorithm until the desired level of brilliance and length is achieved. Generation and storage of the algorithms is left as an exercise for the reader; see Kolmogorov.)
anigbrowl 17y ago
Depends...how small can I print them?

brl 17y ago

This link from the article is a lot more interesting than the article itself: http://www.canyoucopyrightatweet.com/ http://www.canyoucopyrightatweet.com/