[imjustsaying]

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.

[aidos]

Surely you could put as many of them in there as you wanted? So you have an alphabet of 5 characters to make an infinite number of words from.

[imjustsaying]

Good point. 0-9 work the same way. Glad I asked :)

[Tuna-Fish]

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.

[ncza]

55555 if you limit to a 5 character word!

[aidos]

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

[ncza]

Ah, I missed the 4-1 length ones. And yes those were *s