How many addresses can one given gate dial?

[Pharaoh Hamenthotep]

39 symbols with 1 always being the point of origin leaves 38 symbols...

38 x 37 36 x35 x 34 x 33 = 1,987,690,320 possible combinations of six symbol addresses (not including the point of origin)

But for every stargate address made up of six symbols (not including the PoO) there are 720 possible combinations of those symbols

6x5x4x3x2x1= 720 possible combinations of six symbols for one valid stargate address

So for every 720 combinations there is only one address so:

720/720=1 address for every 720 combinations of any given set of six symbols

So to find the number of possible stargates in the galaxy you have to divide the number of combinations of symbols by 720 to give the total number of valid stargate addresses.

1,987,690,320/720 = 2,760,681 possible stargate addresses in the Milky Way galaxy.. all of which can be dialled from a single DHD



36 symbols on the Pegasus gate with one always used as a PoO, which gives a total number of combinations of gate symbols of:

35x34x33x32x31x30 = 1,168,675,200 possible combinations

1,168,675,200/720 gives a maximum of 1,623,160 stargate addresses in the Pegasus Galaxy.

[=SGA=Jason_Binn]

wow.....well i greened you for that one PH

[stargater1990]

39 symbols with 1 always being the point of origin leaves 38 symbols...

38 x 37 36 x35 x 34 x 33 = 1,987,690,320 possible combinations of six symbol addresses (not including the point of origin)

But for every stargate address made up of six symbols (not including the PoO) there are 720 possible combinations of those symbols

6x5x4x3x2x1= 720 possible combinations of six symbols for one valid stargate address

So for every 720 combinations there is only one address so:

720/720=1 address for every 720 combinations of any given set of six symbols

So to find the number of possible stargates in the galaxy you have to divide the number of combinations of symbols by 720 to give the total number of valid stargate addresses.

1,987,690,320/720 = 2,760,681 possible stargate addresses in the Milky Way galaxy.. all of which can be dialled from a single DHD



36 symbols on the Pegasus gate with one always used as a PoO, which gives a total number of combinations of gate symbols of:

35x34x33x32x31x30 = 1,168,675,200 possible combinations

1,168,675,200/720 gives a maximum of 1,623,160 stargate addresses in the Pegasus Galaxy.

thats not exactly true. for all you know the same set of 7 symbols could yeild about a dozen valid gate adresses or just 1. its impossible to know for sure but to assume that for every set of 7 symbols theres just 1 valid adress is illogical.

[Pharaoh Hamenthotep]

Not illogical.. going on what has been established in the show for every set of six symbols (not including the PoO because it's not part of the address for whatever planet you're dialling to) there is only one valid address:

Atlantis: Rising part 2:
SHEPPARD: When can you tell me where the Wraith took Colonel Sumner and the others?
McKAY: Even with the six symbols Lieutenant Ford provided there are still hundreds of permutations ...
SHEPPARD: Seven hundred and twenty.
McKAY: Yes. I knew that of course. I’m just surprised you did.
SHEPPARD: Take away the co-ordinates you can’t get a lock on, and that’s your one.

There might be other examples from SG-1.. this was just the first one I thought of.

[Pharaoh Hamenthotep]

A stargate address is a set of co-ordinates to find a point in space A, B, C, D, E, F with G being the point of origin. If you imagine the six symbols that make up the destination as the centre point on each side of a cube with A and B on opposite sides (left and right for example) C and D top and bottom, then E, F front and back then the lines between them would cross over at the central point where the stargate would be located.

Each set of two symbols forms a line from one side of the cube to the other X(A,B) Y(C,D) Z(E,F) The stargate address is a set of 3-Dimensional co-ordinates X, Y and Z. Changing the order of these coordintes would not change the position of the points at either end of each line, meaning the lines would also not change... but as co-ordinates are always given as X, Y (and Z when dealing with 3-Dimensional space) and not Y, Z, X it makes sense that only one address works.

Changing the order of the symbols would mean the lines no longer cross in the middle of the cube, A would go to C, B to F and D to E, making an invalid address.

[Net effects unknown]

The best approximation is 41,410,215 possible gate addresses can be dialled from a gate.

Someone put it quite well here and I checked the math. There is no reason to believe that there are 720 unique combinations that lead to the same gate even accounting for the quote from above. 720 is merely the total number of permutations of 7 glyphs.



We can name the three destination lines x, y and z, made up of the points x1, x2, y1, y2, z1 and z2.

A ?7 address could thus be understood in a notation format like this:

[?1 ?2 ?3 ?4 ?5 ?6 ?7] = [(x1, x2);(y1, y2);(z1, z2);PoO]

Assumption 1
No glyph is used twice in any given address.
38 choose 6 addresses = 1,987,690,320

Assumption 2
In any destination line (x, y z are destination lines) the points are interchangeable.
I.e. [(x1, x2);(y1, y2);(z1, z2);PoO] = [(x2, x1);(y1, y2);(z1, z2);PoO]
Or (x1,x2) = (x2, x1), (y1,y2) = (y2, y1)
This means that (at least) two addresses lead to the same gate. But there are three destination lines, we can do this three times:(2!)^3 = 8At least eight addresses lead to the same gate

Assumption 3
For any destination, the lines may be entered in random order. After all, in the end we will have three lines that intersect in the destination. It should not matter which line is first or which is second and which is last.(x, y, z) = (y, x, z) = (x, z, y), …
3! = 6

Combining assumptions 2 and 3
The interchangeability of coordinates within a line combined with the interchangeability of lines within an address:(2!)^3 * 3! = 48

One gate can dial 1,987,690,320 addresses but 48 of them will lead to the same destination.

(38 choose 6) / ((2!)^3 * 3!) = 1,987,690,320 / 48 = 41,410,215

[jfarrell327]

I believe Assumption 1 to be correct. I do believe it said somewhere in Stargate Atlantis in the first season that no glyph can be dialed twice, but also it makes sense due to the dialing device, we always see 7 with the 7th always being the point of origin. Therefore its only possible to have 6 different digits with the 7th always being the POI. However MWG gates have 39 glyphs not 38 so its, 39x38x37x36x35x34 plus the POI which equals 2,349,088,560.

[Net effects unknown]

I believe Assumption 1 to be correct. I do believe it said somewhere in Stargate Atlantis in the first season that no glyph can be dialed twice, but also it makes sense due to the dialing device, we always see 7 with the 7th always being the point of origin.

I didn't know that they say that, haven't seen Stargate Atlantis. The thing with using a glyph twice (or more) is:
It doesn't work if we assume that the Stargate needs six unique points in space to calculate the destination.
If it didn't, I could think of a combination that allows a glyph to be used trice:
[(x1, x2);(y1, y2);(z1, z2);PoO] whereas x2 = y2 = z2 = the symbol 'd' that is used three times
[(x1, d);(y1, d);(z1, d);PoO] now all the lines intersect in point d, thus d must be the destination itself. Possibly the Stargate could dial using only one glyph as destination, the other five chevrons would be mere fillers because the 7 is hard-wired in the dial home device. However I still assume that no glyph is used more than once.

http://vignette3.wikia.nocookie.net/...20070112001028
(found here)

However MWG gates have 39 glyphs not 38 so its, 39x38x37x36x35x34 plus the POI which equals 2,349,088,560.

The Stargate has 39 glyphs but one of them is the PoO which is always chevron 7. It makes it easier to take chevron 7 out of the equation right away and use a subset of 6 out of 38 glyphs.
By means of using a subset of 6 out of 39 you run into a problem later on.
39x38x37x36x35x34 would assume 39 valid symbols for chevron 1 but one of them is invalid because it is the PoO. Therefore only 38/39 of the addresses you calculate this way will be valid but 1 in 39 addresses will have the PoO as chevron 1 and be invalid, this goes on for chevron 2 etc. In the end you have a bunch of 'poisoned' addresses that have the PoO in one of chevrons 1-6 and you need to filter them out.

[thekillman]

The Stargate has 39 glyphs but one of them is the PoO which is always chevron 7. It makes it easier to take chevron 7 out of the equation right away and use a subset of 6 out of 38 glyphs.
By means of using a subset of 6 out of 39 you run into a problem later on.
39x38x37x36x35x34 would assume 39 valid symbols for chevron 1 but one of them is invalid because it is the PoO. Therefore only 38/39 of the addresses you calculate this way will be valid but 1 in 39 addresses will have the PoO as chevron 1 and be invalid, this goes on for chevron 2 etc. In the end you have a bunch of 'poisoned' addresses that have the PoO in one of chevrons 1-6 and you need to filter them out.

I agree. The "other glyph" problem from early SG1 seems to be solved by having 38 static glyphs and 1 unique-per-gate POO. Otherwise, given that the glyphs don't change later on, one could only ever have 39 adresses.

So, a logical conclusion is that pressing the red DHD dome inserts the 7th glyph. It makes sense that the one glyph that changes, would not be used in addresses to prevent confusion.

[jfarrell327]

I didn't know that they say that, haven't seen Stargate Atlantis. The thing with using a glyph twice (or more) is:
It doesn't work if we assume that the Stargate needs six unique points in space to calculate the destination.
If it didn't, I could think of a combination that allows a glyph to be used trice:
[(x1, x2);(y1, y2);(z1, z2);PoO] whereas x2 = y2 = z2 = the symbol 'd' that is used three times
[(x1, d);(y1, d);(z1, d);PoO] now all the lines intersect in point d, thus d must be the destination itself. Possibly the Stargate could dial using only one glyph as destination, the other five chevrons would be mere fillers because the 7 is hard-wired in the dial home device. However I still assume that no glyph is used more than once.

http://vignette3.wikia.nocookie.net/...20070112001028
(found here)


The Stargate has 39 glyphs but one of them is the PoO which is always chevron 7. It makes it easier to take chevron 7 out of the equation right away and use a subset of 6 out of 38 glyphs.
By means of using a subset of 6 out of 39 you run into a problem later on.
39x38x37x36x35x34 would assume 39 valid symbols for chevron 1 but one of them is invalid because it is the PoO. Therefore only 38/39 of the addresses you calculate this way will be valid but 1 in 39 addresses will have the PoO as chevron 1 and be invalid, this goes on for chevron 2 etc. In the end you have a bunch of 'poisoned' addresses that have the PoO in one of chevrons 1-6 and you need to filter them out.

Good point. We also know it can't use 2 of the same symbols because the symbols stay lit up after you press them they don't light up then go off right after. However what I don't know is how they know how to work the gate, the dhd is just a bunch of constellations, obviously its easy to explain in show which they don't really do but if we didn't have the computer I don't think we would have done as much as they've done on the show. A real DHD is hard compared to the computer decoding the addresses.

[Erumaaro]

The points are positions in 3D Space (constellations) and thus, depending on their coordinates, it is possible for the lines not to intersect. It is even possible, that none of the combinations result in a unique point of intersection.

possibilities: (0,1,2,3,infinity)points of intersection/combination.
0: none of the lines intersect
1: all 3 lines intersect in 1 point
2: 2 lines intersect a 3rd
3: all 3 lines are in a single plane
infinit: 2 or more lines are co-linear

It is however possible to dial outside the galaxy with only 6 symbols.

=>possibilities depend entirely on the actual coordinates of the constellations, but there cannot be any more than:
(38 choose 6) / ((2!)^3 * 3!) = 1,987,690,320 / 48 = 41,410,215