New Topic/Question
Reply
Perhaps one could leverage the fact that the first two exchange numbers (the second set of 3 numbers) are never 0 or 1.
16 Replies - 4461 Views - Last Post: 16 January 2013 - 06:44 AM
MultiQuote
#17
Skydiver
- Code herder
-
Reputation: 7915
- Posts: 26,430
- Joined: 05-May 12
Re: Everybody needs a bit of help, here are 32
Posted 16 January 2013 - 06:44 AM
Almost there, but I'm one bit over.
Numbers that comply with the NANPA (http://en.wikipedia.org/wiki/North_American_Numbering_Plan) should fit in 33 bits with my scheme:
NXx-Nxx-xxxx
N can only be 2-9
X can only be 0-8
x can be 0-9
If you take away the two N's, which only require 3 bits each. (eg. N-2 and then convert to bits) That leaves 26 bits of the 32 bits available. If you treat the remaining Xx-xx-xxxx as a single 8 digit number: Xxxxxxxx, then 89999999 is 27 bits.
There's got to be a trick somewhere to save an extra bit because of the special cases where some digits sequences are not allowed.
Numbers that comply with the NANPA (http://en.wikipedia.org/wiki/North_American_Numbering_Plan) should fit in 33 bits with my scheme:
NXx-Nxx-xxxx
N can only be 2-9
X can only be 0-8
x can be 0-9
If you take away the two N's, which only require 3 bits each. (eg. N-2 and then convert to bits) That leaves 26 bits of the 32 bits available. If you treat the remaining Xx-xx-xxxx as a single 8 digit number: Xxxxxxxx, then 89999999 is 27 bits.
There's got to be a trick somewhere to save an extra bit because of the special cases where some digits sequences are not allowed.
Was This Post Helpful?
0
| 