Anyone have a suggestion of how this would be used in practice for uniquely identifying something?
So there are a certain number of non-width space characters. As far as I can find in the links in the OP, U+FEFF, U+180E, U+200B, U+200C, U+200D would make 5.
So we have at least 5 values to work with, which would make... 120 combinations if they're ordered differently? Surely we would need more if we want to uniquely identify something such as a referral, or are there more non-width spaces?
I'm kind of dumb and bad at probability, also. You're encouraged to correct my thoughts on this and show me the errors of my thinking.
120 are the combinations if you only use 5 characters and just modify the order. However, if you allow the amount of characters to vary, and allow duplicates, you can represent any number you want. As it's a 5-bit number system, the count of distinct possibilities is 5^n for amount of numbers, or with just 10 digits, you can fit any 32-bit integer there. Plenty good enough for tracking.
You have 5 choices for each character. If you have limit to 5 character words you can assume that all 5 char, 4 char, 3, 2, 1 and 0 char words are included too.
I get:
5^0+5^1+5^2+5^3+5^4+5^5 = 3,906
But I may have totally missed something.
Edit: looking a little closer I see some italics in your 55555 so I guess you have
5 * 5 * 5 * 5 * 5
So there are a certain number of non-width space characters. As far as I can find in the links in the OP, U+FEFF, U+180E, U+200B, U+200C, U+200D would make 5.
So we have at least 5 values to work with, which would make... 120 combinations if they're ordered differently? Surely we would need more if we want to uniquely identify something such as a referral, or are there more non-width spaces?
I'm kind of dumb and bad at probability, also. You're encouraged to correct my thoughts on this and show me the errors of my thinking.