On Thu, Nov 19, 2015 at 9:27 PM, Paul Rubin <no.email@nospam.invalid> wrote:
> 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)=
>
Where do I find out more about the how to calculate information per digit?
Lots of nice little tricks you used below. Thanks for sharing.
> 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()
>
Vincent Davis
720-301-3003
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