shorten "compress" long integer to short ascii.

[Paul Rubin]

Vincent Davis <vincent at vincentdavis.net> writes:
> My goal is to shorten a long integer into a shorter set of characters.
> Below is what I have which gets me about a 45-50% reduction. Any suggestion
> on how to improve upon this?

You can't improve much.  A decimal digit carries log(10,2)=3.32 bits
of information.  A reasonable character set for Twitter-style links
might have 80 or so characters (upper/lower alphabetic, digits, and
a dozen or so punctuation characters), or log(80,2)=

> I not limited to ascii but I didn't see how going to utf8 would help.

If you could use Unicode characters like Chinese ideographs, that gives
you a much larger alphabet to work with, so you'd need fewer chars
displayed in the link, but they'd be hard for most people to type.

> l = string.ascii_lowercase + string.ascii_uppercase +
> '!"#$%&\'()*+,-./:;<=>?@[]^_`{|}~'

OK, 83 chars, but it may not be ok to use some of them like #, /, and ?,
since they can have special meanings in urls.

Your algorithm looks basically ok though I didn't examine it closely.
Here is my shortened version:

  import string

  # alphabet here is 83 chars
  alphabet = string.ascii_lowercase + \
       string.ascii_uppercase +'!"#$%&\'()*+,-./:;<=>?@[]^_`{|}~'
  alphabet_size = len(alphabet)

  decoderdict = dict((b,a) for a,b in enumerate(alphabet))

  def encoder(integer):
      a,b = divmod(integer, alphabet_size)
      if a == 0: return alphabet[b]
      return encoder(a) + alphabet[b]

  def decoder(code):
    return reduce(lambda n,d: n*alphabet_size + decoderdict[d], code, 0)

  def test():
      n = 92928729379271
      short = encoder(n)
      backagain = decoder(short)
      nlen = len(str(n))
      print (nlen, len(short), float(len(short))/nlen)
      assert n==backagain, (n,short,b)

  test()