On Wed, Nov 30, 2011 at 9:26 AM, Alec Taylor <alec.tayl...@gmail.com> wrote: > On Thu, Dec 1, 2011 at 3:18 AM, Ian Kelly <ian.g.ke...@gmail.com> wrote: >> On Wed, Nov 30, 2011 at 8:19 AM, Alec Taylor <alec.tayl...@gmail.com> wrote: >>> Good evening, >>> >>> I have defined a new numbering structure for certain mathematical >>> advantages. >>> >>> How do I implement this in Python, or would I be better off writing >>> this in C or C++? >>> >>> Ultra concise definition: http://i42.tinypic.com/af7w4h.png >>> LaTeX source: http://pastebin.tlhiv.org/Kf6jPRkI >>> >>> Thanks for all suggestions, >> >> So if I am understanding your definition correctly your hexavigesimals >> are ordered like this? >> >> 0, 1, ..., 9, A, B, ..., P, >> >> 0A, 0B, ..., 0P, >> 1A, 1B, ..., 1P, >> ..., >> 9A, 9B, ..., 9P, >> >> A0, A1, ..., A9, >> B0, B1, ..., B9, >> ..., >> P0, P1, ..., P9 >> >> And that's it, since your constraints preclude anything with more than 2 >> digits? > > To put it simply: > > I want a hexavigesimal of length n where each element contains at > least one letter {A, B, ..., P} and one number {0, 1, 2, ... }. > (unless n is less than 2, in which case only one of those constraints > need to be met) > > and I want addition only for incrementing the element. > > Why do I want all this I hear you ask? - For use as a > unique-identifier. I want a unique-ID of size 3. (3 positions, i.e. > P0A) > > Any suggestions on how to implement this in python would be appreciated.
Okay, that's much clearer. If you don't actually care about the ordinal value of each element, only about incrementing them, then I suggest implementing increment as follows: 1. Convert the string to an int using the regular base-26 conversion with the int() built-in. 2. In a loop, add 1 to the value. 3. Convert the value back to a string using the same base-26 conversion. I don't know of a built-in that does this, so you'll need to write your own, which isn't too difficult. 4. Check whether the new string meets all your constraints. If it does, return it. If not, go back to 2. and try again with the next integer. At most it will take 17 iterations to find a valid value, but most of the time the first value tried will be valid. I assume you're already aware that with 3 digits you'll only have 12480 possible unique IDs using this scheme? It just doesn't seem like very many to me. Cheers, Ian -- http://mail.python.org/mailman/listinfo/python-list
