Hi, can anyone help me in base62 encoding?? if i Encode "ilovemyworld" i should get a code with format[A-Za-z0-9]. If i decode the format, i should get back "ilovemyworld" For Example, encode "ilovemyworld" # eDr4e3S decode "eDr4e3S" # ilovemyworld Thanks, Srikanth.J http://www.srikanthjeeva.blogspot.com

on 2009-05-05 22:09

on 2009-05-06 00:58

On Tue, May 5, 2009 at 1:10 PM, Srikanth J. <removed_email_address@domain.invalid> wrote: > Hi, > can anyone help me in base62 encoding?? > > if i Encode "ilovemyworld" i should get a code with format[A-Za-z0-9]. > If i decode the format, i should get back "ilovemyworld" > > For Example, > encode "ilovemyworld" # eDr4e3S > decode "eDr4e3S" # ilovemyworld Would make for a good ruby quiz. I don't think it's possible, though, unless you change your intermittent base. 64 is there because "111111".to_i(2) is 63. In other words the information you store by bits (0 and 1) will have to cover the gamut of 64 characters and no less. Now, you can limit the number of characters to 32 (using 5 bits), if you're willing to have the encoding severely larger. If you used something other than "bits", like a different base, I thought it would be possible, but 62 is only divisible by 2 and 31. I suppose the pragmatic approach would be to take your undesirable characters and alter them; replacing "=" with "555", for example. I never actually knew anything about base64, so thanks for getting me to learn something :-) Todd

on 2009-05-06 20:32

On Wed, May 6, 2009 at 2:26 AM, Todd B. <removed_email_address@domain.invalid> wrote: > > I don't think it's possible, though, unless you change your > intermittent base. 64 is there because "111111".to_i(2) is 63. In it's definitely possible, just inefficient. first you set up your lookup table: > digits = (0..9).to_a.map(&:to_s) + ("a".."z").to_a + ("A".."Z").to_a > then you repeatedly find the lowest digit > out = "" > while n > 0 > rest, units = n.divmod(62) > out = digits[units] + out > n = rest > end martin

on 2009-05-06 22:13

On Wed, May 6, 2009 at 11:32 AM, Martin DeMello <removed_email_address@domain.invalid> wrote: >> rest, units = n.divmod(62) >> out = digits[units] + out >> n = rest >> end > > martin > > Isn't that just a simple cipher (i.e. map)? I must be missing something. According to what I've read so far, base64 is not, and base62 is, except for that paper written in a scientific journal that I don't have access to (but, for the summary, of course). I suppose that is what the OP wanted anyway. Todd

on 2009-05-07 09:45

On Wed, May 6, 2009 at 11:42 PM, Todd B. <removed_email_address@domain.invalid> wrote: > something. According to what I've read so far, base64 is not, and > base62 is, except for that paper written in a scientific journal that > I don't have access to (but, for the summary, of course). I suppose > that is what the OP wanted anyway. No, it's a number base transformation. Here's an example using base 7 (as being easier to work with than 62 :)): letting n = 1250, and using # as a divmod operator: 1250 # 7 = 178, 4 178 # 7 = 25, 3 25 # 7 = 3, 4 3 # 7 = 0, 3 <-- we have reached n=0, so the loop terminates so 1250[base 10] = 3434 [base 7] If you think about it, base 10 works the same way: 1250 # 10 = 125, 0 125 # 10 = 12, 5 12 # 10 = 1, 2 1 # 10 = 0, 1 so 1250[base 10] = 1250[base 10] To go the other way, you repeatedly add the least significant digit and multiply by the base so 3434[7] start with 0, and read the digits in forward order (0 * 7) + 3 = 3 3 * 7 + 4 = 25 24 * 7 + 3 = 178 178 * 7 + 4 = 1250 <--- et voila! martin martin