: cos(θ→P : sin(θ→Q : -Ans→R : [[P^2+QR^2,PR,PR+PR^2][PQ+PQR,P^2,QR+RP^2][PQ,Q,P^2→[F]

I've tested it and it seems to work fine, but I wouldn't mind if someone double checked my math just to make sure.

For the second part of your post, I didn't actually know matrix multiplication was associative, so that's a big help, thanks! Also, I tried [B](1,1)/C→[D](A,1) and [B](2,1)/C→[D](A,2) and it made the cube pretty much point-like on the screen. But, when I multiplied [B](1,1) and [B](2,1) by C, it was the normal cube again.

Edit:
To test out some complex number usage, I changed:

: 1/(L-[B](3,1 : [[Ans,0,0][0,Ans,0 : 20Ans[B]→[B] : [B](1,1→[D](A,1 : [B](2,1→[D](A,2

To:

: 20/(L-[B](3,1 : Ans[B](1,1)+iAns(2,1→⸤C(A

And changed the drawing part to:

: For(A,1,4 : For(B,0,4,4 : ⸤C(A+B-(A=4→V : real(Ans→N : imag(V→O : ⸤C(A+B+1+(A=2)-4(A=A→V : real(Ans→S : imag(V : Line(N,O,S,Ans,BLACK,1 : End : ⸤C(A→V : real(Ans→N : imag(V→O : ⸤C(A+4→V : real(Ans→S : imag(V : Line(N,O,S,Ans,BLACK,1 : End

But, when I tested it out, it appears to have made it slightly slower, and I'm guessing that's because of the imag( and real(, but I'm not too sure.

1/(L-[B](3,1 [[Ans,0,0][0,Ans,0 20Ans[B]→[B] [B](1,1→[D](A,1 [B](2,1→[D](A,2

This matrix multiplication is also unnecessary as it just turns a 3x1 matrix into a 2x1 matrix with the values multiplied by C. Just store 20/(L-[B](3,1→C and do [B](1,1)/C→[D](A,1 and [B](2,1)/C→[D](A,2.

On a higher level, matrix multiplication is pretty inefficient. I think eliminating all matrices and using complex numbers could bring an improvement, but converting would be nontrivial.

: [[cos(θ),-sin(θ),0][sin(θ),cos(θ),0][0,0,1→[J] : [[1,0,0][0,cos(θ),-sin(θ)][0,sin(θ),cos(θ→[I] : [[cos(θ),0,-sin(θ)][0,1,0][sin(θ),0,cos(θ→[H]

It could be:

: cos(θ→P : sin(θ→Q : -Ans→R : [[P,R,0][Q,P,0][0,0,1→[J] : [[1,0,0][0,P,R][0,Q,P→[I] : [[P,0,R][0,1,0][Q,0,P→[H]

While I was at it, I changed the following:

: [D](A+B-(A=4),1→N : [D](A+B-(A=4),2→O : [D](A+B+1+(A=2)-4(A=4),1→S : [D](A+B+1+(A=2)-4(A=4),2→T

To:

: A+B-(A=4→V : [D](Ans,1→N : [D](V,2→O : A+B+1+(A=2)-4(A=4→V : [D](Ans,1→S : [D](V,2→T

Both of which save on bytes and run time.

One major thing I want help with is the matrix setup part near the top of the program (it's the part that stores the cube's vertices). Basically I want to say (in code): for any given dimension, make a matrix such that all the spots within itself are a binary representation of coordinates (using positive and negative numbers for 1's and 0's).

Ex:
dimension = 2
matrix = [[1,1][1,-1][-1,1][-1,-1]]

dimension = 3
matrix = [[1,1,1][1,1,-1][1,-1,1][1,-1,-1][-1,1,1][-1,1,-1][-1,-1,1][-1,-1,-1]]
etc.

Another idea to refactor this program to make it run faster is to not use matrices at all, but instead, use the math within them and variables A-Z (and θ) to calculate the rotated point's positions.

Edit #1:
Eventually, I want to expand the drawing capabilities to draw things other than just cubes, so the idea directly above this to use variables instead of matrices would be kind of moot. Plus, I'd also need a way to store edges, probably with lists or something like that. Not to mention that the drawing edges section would need to be redone completely to accommodate for that.

Edit #2:
I just realized I can put the calculations for matrices [H], [I], and [J] right before the for loop on line 38, so it's not calculating them for each point when they're the same for every point.

Hey! I just created an account to ask for some optimizations for my 3D rendering program for a TI-84 Plus CE. I've been coding in TI-BASIC for quite some time, so I have sufficient knowledge of the language. This code was translated from The Coding Train (java) to BASIC, so the original code is not mine, all I did was port it over.

Feel free to drop by and contribute what you may, all help is accepted, including constructive criticism.

Here's my code:

: //Setup : ClrHome : TextColor(BLACK : 2→L : : //Matrix Setup : {8,2→dim([D] : {8,3→dim([A] : Fill(.5,[A] : For(A,2,8,2 : -.5→[A](A,3 : End : For(B,0,4,4 : For(A,3,4 : -.5→[A](A+B,2 : End : End : For(A,5,8 : -.5→[A](A,1 : End : : //Graph Setup : 16.5→Xmax : -Ans→Xmin : 10.25→Ymax : -Ans→Ymin : ClrDraw : GridOff : AxesOff : PlotsOff : FnOff : : //Main Loop : 0→θ : Repeat getKey=45 : //Calcualations : startTmr→W : For(A,1,8 : Matr▶list([A]^^T^^,A,L₁ //"^^T^^" is the superscript transpose token : List▶matr(L₁,[B] : [[cos(θ),-sin(θ),0][sin(θ),cos(θ),0][0,0,1→[J] : [[1,0,0][0,cos(θ),-sin(θ)][0,sin(θ),cos(θ→[I] : [[cos(θ),0,-sin(θ)][0,1,0][sin(θ),0,cos(θ→[H] : [J][B] : [I]Ans : [H]Ans→[B] : 1/(L-[B](3,1 : [[Ans,0,0][0,Ans,0 : 20Ans[B]→[B] : [B](1,1→[D](A,1 : [B](2,1→[D](A,2 : End : : //Drawing Edges : ClrDraw : For(A,1,4 : For(B,0,4,4 : [D](A+B-(A=4),1→N : [D](A+B-(A=4),2→O : [D](A+B+1+(A=2)-4(A=4),1→S : [D](A+B+1+(A=2)-4(A=4),2→T : Line(N,O,S,T,BLACK,1 : End : [D](A,1→N : [D](A,2→O : [D](A+4,1→S : [D](A+4,2→T : Line(N,O,S,T,BLACK,1 : End : : Text(0,0,"SPF: ",checkTmr(W //SPF is seconds per frame : θ+π/16→θ : End