To actually solve the problem, I'd suggest reading the wiki page and adapting the old routine to fit this task. Shouldn't be too difficult (though it will be a few more bytes).

EDIT: There is a problem with this line though, I get a Argument error:

randIntNoRep(1+(D=2),4-(D=3),4-(D=3)-(D=2→L₁

0→Xmin:264→Xmax 0→Ymin:164→Ymax AxesOff For(Z,0,Ymax+2,2 Horizontal Z End Pt-Off(Xmax,2 Pt-Off(0,162 2→A:2→B:­3→E:2→F:1→I ClrList L₁ Repeat K=45 getKey→K If I=1 randIntNoRep(1+(D=2),4-(D=3),4-(D=3)-(D=2→L₁ \\Random seed; speeds up generation by not worrying about already checked directions L₁(I→D Pt-Off(B,Ymax-A If D=1 and A≠2:Then pxl-Test(A-5,B→P \\Tests for a lack of pixel in desired direction A-5Ans→A \\Moves the drawer Pt-Off(B,Ymax-Ans-2P \\Draws End If D=2 and A≠162:Then pxl-Test(A+5,B→P A+5Ans→A Pt-Off(B,Ymax-Ans+2P End If D=3 and B≠2:Then pxl-Test(A,B-5→P B-5Ans→B Pt-Off(Ans+2P,Ymax-A End If D=4 and B≠262:Then pxl-Test(A,B+5→P B+5Ans→B Pt-Off(Ans-2P,Ymax-A End 1+Inot(P→I \\Count of how many directions the drawer has tried Pt-Off(B,Ymax-A If I>dim(L₁:Then Repeat Qnot(P E+5→E \\Moves the "main path" point down F+5(Ans=167→F \\Moves the "main path" point right if the column is exhausted If F=267:Goto 1 \\Exit condition E≠167:EAns+2not(Ans→E pxl-Test(Ans,F→P pxl-Test(E-5(E≠2),F)+pxl-Test(E+5(E≠162),F)+pxl-Test(E,F-5(F≠2))+pxl-Test(E,F+5(F≠262→Q \\Keep moving the "main path" point until it arrives somewhere a path could spawn from End E→A:F→B:1→I End:End

One important thing about this generator is that it does not use a stack or queue or other list structure to keep track of its progression. Once a branch of the maze boxes itself in, the "drawer" returns to last point it left the "main path", and starts again. The "main path" point moves across the screen as the maze is generated, and can fill the entire screen in about ten minutes.
I went through heavy optimization for this, so if any bit looks weird or hard to understand I'll do my best to explain it.

0→P If D=1 Then ⁻2→L ⁻1→R ⁻1→U 1→D End If D=2 Then ⁻1→L 1→R ⁻2→U ⁻1→D End If D=3 Then 1→L 2→R ⁻1→U 1→D End If D=4 Then ⁻1→L 1→R 1→U 2→D End For(θ,W+L,W+R For(E,V+U,V+D P+Pxl-Test(E,θ)→P //note that Pxl-Test is in Y,X not in X,Y End End //P will contain 0 if no pixels were on, or it will contain the number of pixels that were on

I recommend looking into how nested For( loops work so you can better undersand what I am trying to do. The best way to visualize them may be to use the code below. Try changing some of the numbers to see how it effects the loop.

For(θ,10,40 For(E,10,25 Pxl-Change(E,θ End End

im not a good developer so i will let u guys do something out of this ;)
Without further, here my idea and how i think its going to work.

- First we will set a PXL at a random position on the screen.

- Then we will set a random direction 1-4.

- Before a new PXL is placed in the random direction, we will test as followed. ▓=Pixel, ▒=Testet Area, ´=Just a placeholder

▒▒
▒▒▓ :if D=1
▒▒

▒▒▒
▒▒▒ :if D=2
´ ▓

´ ▒▒
▓▒▒ :if D=3
´ ▒▒

´ ▓
▒▒▒ :if D=4
▒▒▒

- In this case there will always be a gap between placed PXL.

- If there is no Pixel, only then there will be placeds a new Pixel.

___

I tried it with for loops, with repeats etc. and so on. But it does not work at all.
But im not a good developer.

I hope u could find a way how to solve.