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Final Waypoint Name


icefall5

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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. B)

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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
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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
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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.

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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.

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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
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