Total Gate Addresses

[Kralizec]

Okay, so to start out, I've calculated that there are 1,987,690,320 possibleSeven symbol addresses per gate {38*37*36*35*34*33=1987690320}, six symbols and the PoO (Point of Origin). Likewise, there are 63,606,090,240 possible Eight Symbol Addresses {38*37*36*35*34*33*32=63606090240}. An Furthermore, there are a total of 2,035,394,887,680 possible Nine Symbol addresses per gate (maybe we'll actually find out what they're for) {38*37*36*35*34*33*32*31=2035394887680}. All of this meaning, that if there were all of these possible gates, then there would be 1987690320!*63606090240!*2035394887680! possible addresses that you could dial total from all gates! I don't know about you, but that's one hell of a lot. Can anyone tell me what the grand total comes out to? Both my calculater and my laptop can't cruch that many numbers. Think about it.

[Darth Buddha]

Do we know for certain that the same Chevron can't appear twice in the same address? In that case, instead of 38!/32! seven symbol gate addresses, there could instead be 38^7 possible addresses.

[PyRo1896]

Ya i dont think it can repeat itself. And the people in Shi'en mountain tried dieling other gates randomly and nothing came out. So..

[immhotep]

my calulator give the grand total of:
2.57333367408764330783564365824e+32 - have no idea!

[immhotep]

Do we know for certain that the same Chevron can't appear twice in the same address? In that case, instead of 38!/32! seven symbol gate addresses, there could instead be 38^7 possible addresses.

you cant dial twice simply because the gate cant lock the same cheveron in two different places.......and the location wouldnt be locked because you need 6 different to locate, otherwise its just 5 your using the same one again for the same thing......maybe you could use the PoO as a cheveron and the PoO but thats different.

[Kralizec]

you cant dial twice simply because the gate cant lock the same cheveron in two different places.......and the location wouldnt be locked because you need 6 different to locate, otherwise its just 5 your using the same one again for the same thing......maybe you could use the PoO as a cheveron and the PoO but thats different.

Yeah, you can't use the same cheveron twice. It'd be like {a=1, b=2, a=b, so 1=2}? , no can't be done. I didn't add the PoO to my equations because it is always the same for the accompaning gate. But, even so there are a lot of possible gates out there. Question to TPTB, when will we see a nine symbol address, or learn the use(s) of the ninth cheveron?

[immhotep]

its for ascended use only, so well only ever get to see it if danielhurrys up and dies!

[LtColCarter]

its for ascended use only, so well only ever get to see it if danielhurrys up and dies!

Been there...done that!

[immhotep]

yes lets do it again!

[Kralizec]

its for ascended use only, so well only ever get to see it if danielhurrys up and dies!

::deep thought:: More gates, only a death way ::sigh:: .

[NowIWillDestroyAbydos]

Okay, so to start out, I've calculated that there are 1,987,690,320 possibleSeven symbol addresses per gate {38*37*36*35*34*33=1987690320}, six symbols and the PoO (Point of Origin). Likewise, there are 63,606,090,240 possible Eight Symbol Addresses {38*37*36*35*34*33*32=63606090240}. An Furthermore, there are a total of 2,035,394,887,680 possible Nine Symbol addresses per gate (maybe we'll actually find out what they're for) {38*37*36*35*34*33*32*31=2035394887680}. All of this meaning, that if there were all of these possible gates, then there would be 1987690320!*63606090240!*2035394887680! possible addresses that you could dial total from all gates! I don't know about you, but that's one hell of a lot. Can anyone tell me what the grand total comes out to? Both my calculater and my laptop can't cruch that many numbers. Think about it.

Wow , Nice calculations! And that's a lot. But did you calculate the addresses on the Atlantis gate.

[Kralizec]

Wow , Nice calculations! And that's a lot. But did you calculate the addresses on the Atlantis gate.

Yeah, I did. But All my calculations are in my thread starter. There are a maximum of 198769032!*63606090240!*2035394887680! addresses from every possible gate, to every possible gate.

[Kralizec]

my calulator give the grand total of:
2.57333367408764330783564365824e+32 - have no idea!

I know that that's not right, because it should have a much higher exponent. The equation 198769032!*63606090240!*2035394887680!=X, means that you multiply 198769032*198769031*198769030*...*3*2*1*63606090240*63606090239*...*3*2*1*203539 4887680*2035394887679*...*3*2*1=X. So as you can see, it's quite large. My best guess is that it's more than 1x10^100,000,000,000,000,000,000, but it's just a guess.

[immhotep]

i dont know i pasted that right out of the windows scientific calc, not my fault.

[Kralizec]

i dont know i pasted that right out of the windows scientific calc, not my fault.

Im not saying that it's your fault, I'm just saying.