Hi. I just wanted to leave that here. I need it from time to time, when someone wants to reminisce and talks about Stargeate. In case you watched Stargate back then, you might remember that scene where Daniel explains addresses.
“To find a destination within any three-dimensional space, you need six points.”
No. Sorry. You don’t. You need minimum 1 point, if it’s the one you want already. Otherwise things get complicated. So Daniel draws three straight lines, each of which is defined by two of those six points. Of course, you can draw a line through any two points. The problem is that arbitrary lines in space do not intersect at all. While in a plane two lines are either parallels or intersect, in 3D they can also be skewed. Just pick up two pencils and imagine them being infinite straight lines. You can hold them touching one another making an intersect, you can hold them exactly parallel, or you can hold them any other way passing by one another. That’s skew. And adding a third line doesn’t make it better. In fact you might now get two intersections or even a triangle. It’s a wondrous thing that the air force officers in the scene do not figure that out.
Furthermore, the 7th symbol as point of origin is idiotic. If the gate didn’t know where it currently is, how does it work in the first place?
So let’s think again. We have some dials that fit constellations. First we need to figure out what that means. A constellation is not a point but an extended area in the night sky, creating an infinte 3D slice out into space. We could say that we are talking about the brightest star in the depicted constellation as seen from earth. That means every Stargate needs custom labels for the planet where it’s located, referring to the sky there. At its latest that becomes problematic, when they start driving around gates via starships in the series, but for the movie at least it’s workable. Let’s stick to the movie.
So we have 38 symbols referring to stars from which we choose up to 8 (a stargate has 9 chevrons, it’s just that two are are not usually visible because of the runway). How could we make a somewhat sensible addressing system? So first of all, we might mean one of our 38 star systems itself, call it A. What would be sensible address for A itself? A+ENTER. Dial that very symbol, then “origin” to end your sequence. So we would want sequences ranging in length from 1 to 8 signs. The ORIGIN terminates the sequence, but is otherwise meaningless.
How do longer sequences work? We want the point that has the shortest total distance to all the symbols we reference; the “center” of our chosen reference points. Using this approach there is no reason to prohibit repitition of symbols. We could do A+A+B+ORIGIN, demarcating the point that is on the line from A to B, but twice as close to A as it is to B.
We could even allow up to 9 target glyphs. Since ORIGIN is just an ENTER, the sequence must stop when all 9 places are used up. The reason they have to dial ORIGIN in the movie is that they have only a 6 symbol address.
Finally, we could make a difference between A+B+ENTER and B+A+ENTER, by giving the first point a little more weight, making its pull on the “center point” a little bit stronger. How many addresses does that give us? We have 38 addresses with a single glyph, 38 squared for two glyphs and so on up to the power of 9. But we must substract the options that all places are the same each time. A+A+…+A+ENTER ist just A+Enter.
38 + 38^2-38 + 38^3-38 +...+ 38^9-38 = 38 + 38^2 + 38^3 +...+ 38^9 - 38*8 = 169,681,401,296,978 - 304 = 169,681,401,296,674
Note the number for Daniel’s method is smaller than you might expect, even if it worked. Since it uses three pairs to define three straight lines, certain addresses are eqivalent like
(A,B) + (C,D) + (E,F) = (B,A) + (D,C) + (F,E) || switching the fix points within pairs = (E,F) + (A,B) + (C,D) || moving pairs around = ... || any combination thereof