Yes, I did too. It's not hard. But that's not the point I'm trying to make

Basically he wants to only compare the LSB, as far as I can tell.

If the foci are all at the same y-coordinate, they will never intersect no matter what the value of d is. As long as their y-coordinates are not all the same, it is solvable.

I've been thinking about this all day and I can't figure it out...

Say you have three known points p₀, p₁, and p₂. Each of these points serves as the focus of a parabola, and all three parabolas share a single line at y=d as their directrix. Assume that d is such that all three parabolas open in the same direction, and assume that the three foci do NOT all lie on a single line parallel to the directrix.

So the idea is to solve for value of d that makes all three parabolas intersect at a single point. There should only be one value that makes this true.

Basically, so far I've got quite a bit conceptually figured out in terms of what needs to happen...but I don't really know how to make it happen.

Here's kinda what I have:



The intersection point(s) of two parabolas (z_n)can be calculated fairly easily:
z₀=a₀x²+b₀x+c₀
z₁=a₁x²+b₁x+c₁
Find the quadratic equation that is the difference of the two, and set it equal to zero:
0=(a₀-a₁)x²+(b₀-b₁)x+(c₀-c₁)
Use the quadratic formula to find the zeros:
x=(-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁)



A parabola can also be written as (x-h)²=4p(y-k), where (h,k) is the vertex of the parabola, (h,k+p) is the focus, and the directrix lies at y=k-p
With a focal point F and a directrix at y=d, all other necessary variables can be calculated as follows:
p=(F_y-d)/2
h=F_x
k=F_y-p
a=1/4p
b=-h/2p
c=h²/4p+k

Through this, all variables can be simplified down to only use F and d:
a=1/2(F_y-d)
b=-F_x/(F_y-d)
c=F_x²/2(F_y-d)+F_y-(F_y-d)/2[/tt]

So you could technically write out the equation of a parabola as
y = x²/2(F_y-d)   +   -F_xx/(F_y-d)   +   F_x²/2(F_y-d)+F_y-(F_y-d)/2



As you can see, it gets pretty hectic pretty quickly. For example, the quadratic formula would then be:
x=(F_x/(F_y-d)±√((-F_x/(F_y-d))²-(2/(F_y-d))(F_x²/2(F_y-d)+F_y-(F_y-d)/2)))/(1/(F_y-d))


Remember, this needs to be solving for d...

Really, I don't know where to go from here. Anyone have any insight?


---EDIT:
Okay, so I'm slowly figuring things out.

The equation is crazy complex, but you have to solve for the intercepts of two parabolas, then plug that quadratic formula calculation in for X in one of those equations, and set it equal to the equation for the third parabola with the quadratic formula calculation plugged in for X as well.

a₂((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))²+b₂((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))+c₂=a₀((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))²+b₀((-(b₀-b₁)±√((b₀-b₁)²-4(a₀-a₁)(c₀-c₁)))/2(a₀-a₁))+c₀

Looks like fun, right? Now substitute the As, Bs, and Cs with their F/d equivalents that I mentioned earlier, accounting for the fact that the Fs have to be F_nx and F_ny for each variable of the corresponding n value.

Here's one I did a while ago.

Everything's done with only red, blue, yellow, and black. I do use both a dark blue and a light blue, however.



If for some reason you want to see it, here's a larger version.

Well what about the AI? It is certainly more complex than just constantly moving toward the player. Sure, some enemies do that, but there are certainly more complex actions as well.

What enemies are going to be included in the game? That will help to figure out what needs to be done for the AI.

Triangulate, then remove unnecessary vertices. That's honestly the best way to do it with any simplicity. Anything more and you'll be shooting yourself in the head.

Reminds me of this: