# Final Waypoint Name

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I know this has been talked about before, but I couldn't find any topics where it was.

Some people have been saying they think the final waypoint name (GC****) will be given out around the end of October, but I don't think that's true.

When I posted this, there were 322,449 active caches. Assuming I did my math correctly, we can still put out 1,357,167 caches before the last one is given out.

My Math

36x36x36x36 (36 possible digits [letters A-Z and numbers 0-9] in each of four spots)

=1,679,616 possible caches.

1,679,616 - the 322,449 active caches (when this was posted)

=1,357,167 more caches before we run out of waypoints.

_____________________________________________________________________

I'm sure I did something wrong here, so please tell me.

You'll probably find this thread to be interesting.

...4-bows - wrong - it is a base 31 system changed from an original hex system - there are considerably less than you think.

How is it a base-31 system?

To avoid "naughty" words.

Also, it's a bit less than a Base 31 system, because it was a simple hex system through GCFFFF. For example, there is no GCBZZZZ -- after GCBFFF came GCC000. So there is an offset. Fizzymagic has a calculator for this.

We'll run out of four-digit waypoints sometime soon. See the predictions in the thread linked to above.

Edited by The Leprechauns

I'm confused, but okay.

The naming convention uses the 31 alphanumeric characters 0123456789ABCDEFGHJKMNPQRTVWXYZ (the letters ILOSU are omitted).

The system used to use a simple hexadecimal conversion. That means there's a computational gap of 411,120 between GCFFFF (old system) and GCG000 (new system).

Hmmm...if my college algebra is correct, one must use a "combination" to determine how many possible sets of 6 digits are available. To explain it simply, out of a possible 31 usable letters and numbers, and only 6 can be used for the waypoint, one has to calculate it using factorials, i.e., 31! over 6!. There are ways to eliminate duplicates, etc., but the end result is very different from what one might think is possible.

We were taught "permutations" at the same time. A little different, but just as fun.

ETA: I'm no brainiac, and I will welcome correction if needed.

Edited by sparqui66

Hmmm...if my college algebra is correct, one must use a "combination" to determine how many possible sets of 6 digits are available. To explain it simply, out of a possible 31 usable letters and numbers, and only 6 can be used for the waypoint, one has to calculate it using factorials, i.e., 31! over 6!. There are ways to eliminate duplicates, etc., but the end result is very different from what one might think is possible.

We were taught "permutations" at the same time. A little different, but just as fun.

ETA: I'm no brainiac, and I will welcome correction if needed.

It would be 31! over 4!, because the GC is permanent. GC(****)= 4 digits.

Hmmm...if my college algebra is correct, one must use a "combination" to determine how many possible sets of 6 digits are available. To explain it simply, out of a possible 31 usable letters and numbers, and only 6 can be used for the waypoint, one has to calculate it using factorials, i.e., 31! over 6!. There are ways to eliminate duplicates, etc., but the end result is very different from what one might think is possible.

We were taught "permutations" at the same time. A little different, but just as fun.

ETA: I'm no brainiac, and I will welcome correction if needed.

It would be 31! over 4!, because the GC is permanent. GC(****)= 4 digits.

Thanks icefall5! I'd forgotten that the first two digits were locked.

S.

Well, actually, combinations are used to determine how many sets of things can be picked from a larger pool without replacement. The equation goes like this:

n!/((n-k)!*k!)

where n is the total number of things (here, 31), and k is the number of selections (here, 4). It's basically a formal way of writing 31*30*29*... for as much as you need.

It doesn't apply here, because you "replace" characters back into the larger pool - you can have GCTTFF, for example. For this you just do 31^4, or 31*31*31*31. Minus, of course, any naughty combinations that might come up. ;-)

Combinations also are only useful for when order doesn't matter - that is, if picking A, B, C, and D can be in any order. We would be more interested in permutations, which gets rid of the k! term in the denominator.

Edited by Shorelander

Are we a bunch of nerds or what?

Thanks for dusting off my brain.

S.

Me? Former Math Team captain? Naaaah, I'm not a nerd.

Going to college for math in 8th grade? I don't think I'm a nerd either.

I didn't actually "endorse" the 31! over 4!, I just corrected the 6! and 4! part. I didn't bother to actually do the math.

Edited by icefall5

You also ignored all the archived caches. Those numbers are not re-used. So a lot more numbers have been used that what was cited.

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