It is for a scaling algorithm, and the conversion will be processed several hundreds of times for a single full screen picture. I need to save every millisecond here.
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I almost made the scaling algorithm! In fact I even completely implemented pixel replication, but it is awfully slow.
I had an idea though: If I have a list of lines to duplicate, I can simply shift the screen, without shifting anything in, starting at those lines. This will duplicate and render the whole line automatically!
Unfortunately I don't have such an assembly routine, that would do the shifting.
I could either shift the whole screen with xLib and then mask off the modification with may area selection algorithm, which will still require 9 commands + big cleanup.
Or, if some kind assembly programmer could write a routine which can shift any part of the screen in any direction, I could do it with a single library call which will probably be almost instant.
Also, a new idea I had:
When saving a file, you will be able to select fullscreen or not fullscreen (I mean the bottom screen row, which is unavailable in BASIC normally). Just for simplicity, you will always be able to draw on that line, the fullscreen or not setting will only determine whether it is saved.
Or I might just save that line by default and only ask for fullscreen or not when exporting files.
I had an idea though: If I have a list of lines to duplicate, I can simply shift the screen, without shifting anything in, starting at those lines. This will duplicate and render the whole line automatically!
Unfortunately I don't have such an assembly routine, that would do the shifting.
I could either shift the whole screen with xLib and then mask off the modification with may area selection algorithm, which will still require 9 commands + big cleanup.
Or, if some kind assembly programmer could write a routine which can shift any part of the screen in any direction, I could do it with a single library call which will probably be almost instant.
Also, a new idea I had:
When saving a file, you will be able to select fullscreen or not fullscreen (I mean the bottom screen row, which is unavailable in BASIC normally). Just for simplicity, you will always be able to draw on that line, the fullscreen or not setting will only determine whether it is saved.
Or I might just save that line by default and only ask for fullscreen or not when exporting files.
 Nik
 Power User (Posts: 483)
 13 Nov 2015 11:50:54 am
 Last edited by Nik on 13 Nov 2015 05:00:47 pm; edited 1 time in total
The enlarging algorithm is done!
It is relatively fast ()!
I would say currently it takes approximately 2 minutes for a fullscreen picture, at least exactly two minutes at any magnification factor from 2 to 3, which I believe will be the most used.
I am still optimizing it, the runtime could halve I think.
I will post the code today or tomorrow, depending on when I finish optimizing.
It is relatively fast ()!
I would say currently it takes approximately 2 minutes for a fullscreen picture, at least exactly two minutes at any magnification factor from 2 to 3, which I believe will be the most used.
I am still optimizing it, the runtime could halve I think.
I will post the code today or tomorrow, depending on when I finish optimizing.
Great, I look forward to seeing the code, and perhaps offering any optimizations if I can spot any. It might be interesting to start learning a little bit of assembly at some point as well, because (I assume) this would be an ideal small, integerbased, relatively simple algorithm to transfer to ASM for speed reasons. You'd just need to learn a little bitmath.
 Nik
 Power User (Posts: 483)
 13 Nov 2015 05:07:07 pm
 Last edited by Nik on 14 Nov 2015 06:11:00 pm; edited 1 time in total
Here is the code:
Code:
Code:
I managed to get the runtime down to 1:38 minutes when enlarging by 250%. Still some optimization I can do, as mentioned in the comments, so I suppose to get it down to about 1:30.
It currently only supports fullscreen. Shouldn't be hard to change, though.
Edit:
Optimized, runtime reduced to 56 seconds!
Also it is slightly smaller, but only a few bytes.
Code coming in 1 minute.
Code:
prgmENLARGE
ClrDraw
RecallPic 0
StorePic 1
// This just gets my testing image (a chessboard pattern)
Disp "SCALE BY:
Input "",Z
// No validation, should be a real number, n=>1
seq(int(2A),A,0,48,2Z^1>L1
seq(int(2A),A,0,32,2Z^1>L2
// Get the rows which should be dupicated, vertical to L1, horizontal to L2
For(A,1,dim(L1
{L1(A),0
prgmSHIFT
End
// Call the shifting routine for all vertical lines
For(A,1,dim(L2
{L2(A),1
prgmSHIFT
End
// Call the shifting routine for all horizontal lines
Code:
prgmSHIFT
ClrDraw
If not(Ans(2
real(12,6,0,0,Ans(1),62,1
// If the argument list says it was vertical, draw vetical mask for masking routine
If Ans(2
real(12,6,0,0,94,Ans(1),1
// If the argument list says it was vertical, draw horizontal mask for masking routine
real(12,8,0,0,94,62,1
// I know I can save that line by optimizing the above part. Just too late at me right now to work. I will generally optimize tomorrow, probably I will put that whole part into prgmENLARGE.
StorePic 2
// Save above work to Pic2 (Mask)
ClrDraw
RecallPic 1
real(4,32Ans(2),1,0
StorePic 3
// Shift Pic1 (Original) and save as Pic3 (Modified). For the Ans: Remember the argument list is still in Ans!
real(3,2,1,0
real(3,1,2,0
StorePic 1
real(3,2,0,0
real(12,8,0,0,94,62,0
real(3,3,2,0
real(3,1,1,0
StorePic 1
// This chunk is the masking routine. Check out the first page of this thread for detailed commments.
I managed to get the runtime down to 1:38 minutes when enlarging by 250%. Still some optimization I can do, as mentioned in the comments, so I suppose to get it down to about 1:30.
It currently only supports fullscreen. Shouldn't be hard to change, though.
Edit:
Optimized, runtime reduced to 56 seconds!
Also it is slightly smaller, but only a few bytes.
Code coming in 1 minute.
I created a probably faster algorithm for scaling up images 200%. This is my code:
Code:
How it works:
Code:
You could modify it for another zoom.
That will look like this:
Code:
For(A,0,96,2
StorePic 0
real(12,7,A+1,0,96,64,1
StorePic 1
real(3,0,0,1,1
If A
real(12,7,0,0,A1,64,1
real(4,3,1,1
real(3,1,2,1,1
End
For(A,0,64,2
StorePic 0
real(12,7,0,A+1,96,64,1
StorePic 1
real(3,0,0,1,1
If A
real(12,7,0,0,94,A1,1
real(4,1,1,1
real(3,1,2,1,1
End
How it works:
Code:
For(A,0,96,2
Save whole picture
Erase the right part, without the line which will be doubled
Save the left part+'cursor'
Recall whole picture
If the cursor is not the left line, erase the left part of the screen, again without the cursor
Shift the screen to right = cursor+right part
Add the left part. There you are
End
You could modify it for another zoom.
That will look like this:
Awesome work PT_!
I made it work for any magnification factor and it runs 17 seconds for a fullscreen picture.
Code:
But I just noticed this only works fullscreen. I will put the algorithm in the filters section though, it deserves it!
Edit: PT_ also put some code together which made the algorithm work in both directions (shrinking and enlarging). There were some minor problems, though. As soon as I fix everything I will post the code once more!
Also, I think it is possible to use this algorithm for a specific area on the screen. I just need to mask out the scaled part and here we go!
Overall, PT_ did a very good job here, thank you very much for the awesome idea!
Edit2:
I still couldn't solve everything. I noticed the algorithms can't properly scale to every factor.
However, I made a quick draft for a logo:
What do you think?
I made it work for any magnification factor and it runs 17 seconds for a fullscreen picture.
Code:
ClrDraw
RecallPic 0
Disp "Magnify by:
Input "",Z
For(A,0,48,2Z^1
int(2A
StorePic 2
real(12,7,Ans+1,0,96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,Ans1,64,1
real(4,3,1,1
real(3,1,2,1,1
End
For(A,0,32,2Z^1
int(2A
StorePic 2
real(12,7,0,Ans+1,96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,94,Ans1,1
real(4,1,1,1
real(3,1,2,1,1
End
But I just noticed this only works fullscreen. I will put the algorithm in the filters section though, it deserves it!
Edit: PT_ also put some code together which made the algorithm work in both directions (shrinking and enlarging). There were some minor problems, though. As soon as I fix everything I will post the code once more!
Also, I think it is possible to use this algorithm for a specific area on the screen. I just need to mask out the scaled part and here we go!
Overall, PT_ did a very good job here, thank you very much for the awesome idea!
Edit2:
I still couldn't solve everything. I noticed the algorithms can't properly scale to every factor.
However, I made a quick draft for a logo:
What do you think?
This was a hard piece of code. Still doesnt work for a factor of x4:
Code:
Code:
ClrHome
Input "Scale by: ",Z
If Z=1 or not(Z
Then
If not(Z
ClrDraw
Return
End
2(Z>1)1>X
Z^~Ans>Y
For(A,0,Y96,1+Y
int(AZ^Y
StorePic 2
real(12,7,Ans+1,0,95,63,1
StorePic 1
real(3,2,0,1
If Ans or X=~1
real(12,7,0,0,Ans(X=1),63,1
real(12,2,Ans,0,Ans,63,1
real(12,1,0,95,63,95,1
real(12,2,Ans,0,Ans,63,1
real(4,2+(X=1),1,1
real(3,1,2,1,1
End
real(12,1,0,95,63,95,1
For(A,0,Y64,2Y
int(A2^(X=1
StorePic 2
real(12,7,0,Ans+1,95,63,1
StorePic 1
real(3,2,0,1,1
If Ans or X=~1
real(12,7,0,0,95,AnsX,1
real(12,2,0,Ans,95,Ans,1
real(12,1,0,63,95,63,1
real(12,2,0,Ans,95,Ans,1
real(4,(X=1),1,1
real(3,1,2,1,1
End
real(12,1,0,63,95,63,1
I solved the enlarging bug with this code:
Code:
Let's start for shrinking...
EDIT: here's the code for shrinking:
Code:
Note that Z > 1 for shrinking, you could change it to 1/Z though.
Now let's combine it into one section.
EDIT2:
Code:
Code:
Input Z
For(B,0,96/Z
int(ZB
For(A,2,Z
StorePic 0
real(12,7,Ans+1,0,96,64,1
StorePic 1
real(3,0,0,1,1
If Ans or X=~1
real(12,7,0,0,Ans(X=1),64,1
real(4,2+(X=1),1,1
real(3,1,2,1,1
End
End
EDIT: here's the code for shrinking:
Code:
Input Z
For(B,0,96/Z
B+1
For(A,2,Z
StorePic 0
real(12,7,Ans,0,96,64,1
StorePic 1
real(3,0,0,1,1
If Ans
real(12,7,0,0,Ans1,64,1
real(4,2,1,1
real(3,1,2,1,1
End
End
Now let's combine it into one section.
EDIT2:
Code:
Input Z
Z(Z>1)+Z^^1(Z<1>U
For(B,0,96/U
ZB
If Z<1
B+1
For(A,2,U
StorePic 0
real(12,7,Ans+(Z>1),0,96,64,1
StorePic 1
real(3,0,0,1,1
If Ans
real(12,7,0,0,Ans1,64,1
real(4,3(Z<1),1,1
real(3,1,2,1,1
End
End
Here is what I got so far:
I erase the bottom and the left rows, so when shrinking, white pixels are shifted in.
However, a weird thing does not work.
Code:
This line doesn't work for some reason, I tried with real(12,0,... and real(12,2,... both gave no result. Seems like something on the screen gets overwritten?
I erase the bottom and the left rows, so when shrinking, white pixels are shifted in.
However, a weird thing does not work.
Code:
Input Z
Z(Z>1)+Z^^1(Z<1>U
For(B,0,96/U
ZB
If Z<1
B+1
For(A,2,U
real(12,1,95,0,95,63,1
StorePic 2
real(12,7,Ans+(Z>1),0,96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,Ans1,64,1
real(4,3(Z<1),1,1
real(3,1,2,1,1
End
End
For(B,0,64/U
ZB
If Z<1
B+1
For(A,2,U
real(12,1,0,63,95,63,1 // This is the line that thinks it is an exception.
StorePic 2
real(12,7,0,Ans+(Z>1),96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,96,Ans1,1
real(4,(Z>1),1,1
real(3,1,2,1,1
End
End
This line doesn't work for some reason, I tried with real(12,0,... and real(12,2,... both gave no result. Seems like something on the screen gets overwritten?
Code:
For(B,0,64/U
ZB
If Z<1
B+1
For(A,2,U
StorePic 2
real(12,7,0,Ans+(Z>1),96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,96,Ans1,1
real(12,1,0,63,95,63,1
real(4,(Z>1),1,1
real(3,1,2,1,1
End
End
real(12,1,0,63,95,63,1
I made an algorithm to render Bézier Curves!
(More about Bézier Curves on Wikipedia)
Screenies:
Code:
Code:
Explanation:
Takes three points and draws a Bézier Curve between them.
The input "Steps" determines how acurate the curve will be drawn  more steps: Longer calculation time, better and less edged curve, fewer steps: Faster, but worse curve and more edges.
Basically this just says of how many straight lines the curve consists. The more, the shorter they are and the smoother the angles.
I took 20 which makes an absolutely good curve at such a short lenght, probably even 5 would be good enough to notice no difference.
(More about Bézier Curves on Wikipedia)
Screenies:
Code:
Code:
Input
PtOn(X,Y,3
{X,~Y>L1
Input
PtOn(X,Y,3
{X,~Y>L2
Input
PtOn(X,Y,3
{X,~Y>L3
Input "Steps:",theta
theta^^1>theta
// Self explanatory
L2L1>L4
// Get the line lenght from P1 to P2, store in L4
L3L2>L5
// Line lenght from P2 to P3, store in L5
L1(1>A
L1(2>B
// Initialize "Last point"
For(C,1,theta^^1
L1+CthetaL4>L6
// P1 + Current * Step * Line lenght P1_P2 > L6
L2+CthetaL5
// P2 + Current * Step * Line lenght P2_P3 > Ans
L6+Ctheta(AnsL6>L6
// In Line L6_Ans we now have the bottom line, on which we define the next rendering point (At Current * Step)
Line(A,~B,Ans(1),~Ans(2
// Draw a line between last point and the point we just defined
Ans(1>A
L6(2>B
// Update last point
End
Explanation:
Takes three points and draws a Bézier Curve between them.
The input "Steps" determines how acurate the curve will be drawn  more steps: Longer calculation time, better and less edged curve, fewer steps: Faster, but worse curve and more edges.
Basically this just says of how many straight lines the curve consists. The more, the shorter they are and the smoother the angles.
I took 20 which makes an absolutely good curve at such a short lenght, probably even 5 would be good enough to notice no difference.
Now there are Cubic Bézier Curves!
These are the same, except that there are 2 "Weight Points" instead of one.
Screenies:
Code:
No comments now, sorry. This code works exactly like the one above, just with a bigger amount of lines and points. Check the Wikipedia page on Bézier Curves for more information.
Also note the rendering time gretly increased  and the step rate too. Just because the curves get longer, 50 was more than enough. However, as you see, 25 gives significant quality loss, so something like 30/35/40 should be good.
Edit:
Optimized the code, now I use complex vars instead of lists and [recursiveN] for the loop variable (Thanks for the suggestions, lirtosiast!). Also I moved a little bit of the calculations out of the loop so it got faster.
The cubic and quadratic curves got 0.33 and 0.177 times faster and a little bit smaller.
These are the same, except that there are 2 "Weight Points" instead of one.
Screenies:
Code:
Input
PtOn(X,Y,3
{X,~Y>L1
Input
PtOn(X,Y,3
{X,~Y>L2
Input
PtOn(X,Y,3
{X,~Y>L3
Input
PtOn(X,Y,3
{X,~Y>L4
Input "Steps:",theta
theta^^1>theta
L2L1>L5
L3L2>L6
L4L3>LA
L1(1>A
L1(2>B
For(C,1,theta^^1
L1+CthetaL5>LB
L2+CthetaL6>LC
L3+CthetaLA>LD
LB+Ctheta(LCLB>LB
LC+Ctheta(LDLC>LC
LB+Ctheta(AnsLB>LB
Line(A,~B,Ans(1),~Ans(2
Ans(1>A
LB(2>B
End
DelVar LADelVar LBDelVar LCDelVar LD
No comments now, sorry. This code works exactly like the one above, just with a bigger amount of lines and points. Check the Wikipedia page on Bézier Curves for more information.
Also note the rendering time gretly increased  and the step rate too. Just because the curves get longer, 50 was more than enough. However, as you see, 25 gives significant quality loss, so something like 30/35/40 should be good.
Edit:
Optimized the code, now I use complex vars instead of lists and [recursiveN] for the loop variable (Thanks for the suggestions, lirtosiast!). Also I moved a little bit of the calculations out of the loop so it got faster.
The cubic and quadratic curves got 0.33 and 0.177 times faster and a little bit smaller.
I think there's an even faster way to compute the Bezier curves that requires <i>no</i> multiplications in the loop. A real*complex multiplication takes 5 ms or so (don't quote me on that though) but an addition can be almost instantaneous. Given that your current method takes about 150 ms per iteration, drawing the Line( cannot be the bottleneck. Therefore, a faster algorithm can speed it up considerably.
There's a method called "forward differencing", which allows you to basically calculate points through iteration of cumSum( on a carefully constructed complex fourelement list, taking the last element to be your new point each time. See this page for the details.
cumSum( has only ~5 ms of listprocessing overhead, and the actual summing will be very fast compared to multiplications (at the machine code level, adding has a rst instruction, while multiplying doesn't).
(Hmm, it looks like I'll need to use backwards differences instead to use the cumSum( optimization. Forward is still faster than the old way though, if it works.)
Here's my implementation (now it works! runtime ~1.52.5s for a 50 point curve):
Code:
There's a method called "forward differencing", which allows you to basically calculate points through iteration of cumSum( on a carefully constructed complex fourelement list, taking the last element to be your new point each time. See this page for the details.
cumSum( has only ~5 ms of listprocessing overhead, and the actual summing will be very fast compared to multiplications (at the machine code level, adding has a rst instruction, while multiplying doesn't).
(Hmm, it looks like I'll need to use backwards differences instead to use the cumSum( optimization. Forward is still faster than the old way though, if it works.)
Here's my implementation (now it works! runtime ~1.52.5s for a 50 point curve):
Code:
For(I,1,4
Input
PtOn(X,Y,3
X+Yi>L1(I
End
//Find coefficients
L1(1)+3L1(2)3L1(3)+L1(4>A
3(L1(1)2L1(2)+L1(3>B
3(L1(2)L1(1>C
L1(1>D
Input "Steps:",S
S1>S
1/S>θ
//differences are U,V,W; point is T
D>T
Aθ^3+Bθ^2+Cθ>U
6Aθ^3+Bθ^2>V
6Aθ^3>W
//old point saved in Z
For(n,1,S)
T>Z
T+U>T
U+V>U
V+W>V
Line(real(Z),imag(Z),real(T),imag(T
End

lirtosiast
 Power User (Posts: 338)
 21 Nov 2015 04:35:40 pm
 Last edited by lirtosiast on 23 Dec 2015 05:33:23 pm; edited 3 times in total
Here's a new, even faster algorithm to generate the list of points for the cubic Bezier curve. Input is four points as complex numbers in L1, and output is the list of Bezier points in L2. Will update with the fivepoint preview version.
Everything is based on backward differences. Call the constructor points P0, P1, P2, and P3. Call the first point of the Bezier curve B₀, and the last B₁, and the Nth point (starting from 0) B_N/(S1) so there are S points in total. Now, there's a formula for the points B:
(From Wikipedia)
(sorry for using B as one of my variables)
However, computing these directly is too expensive! We'd need to raise t and (1t) to powers, multiply them by the P points, and add, for *every single point*. Instead, we can figure out a clever formula. Since this is a cubic polynomial in t evaluted at constant points, it has constant fourth differences (Just read the top two sections). These can be found using a little bit of mathematical trickery from the polynomial coefficients.
The A, B, and C are just the coefficients in the simplified form of the formula above (you can get them just by expanding). So the polynomial, when you expand it, is equal to At^3+Bt^2+Ct+P₀.
Code:
Everything is based on backward differences. Call the constructor points P0, P1, P2, and P3. Call the first point of the Bezier curve B₀, and the last B₁, and the Nth point (starting from 0) B_N/(S1) so there are S points in total. Now, there's a formula for the points B:
(From Wikipedia)
(sorry for using B as one of my variables)
However, computing these directly is too expensive! We'd need to raise t and (1t) to powers, multiply them by the P points, and add, for *every single point*. Instead, we can figure out a clever formula. Since this is a cubic polynomial in t evaluted at constant points, it has constant fourth differences (Just read the top two sections). These can be found using a little bit of mathematical trickery from the polynomial coefficients.
The A, B, and C are just the coefficients in the simplified form of the formula above (you can get them just by expanding). So the polynomial, when you expand it, is equal to At^3+Bt^2+Ct+P₀.
Code:
//Get the polynomial coefficients
L1(2)3L1(1)L1(3)3+L1(4>A
3(L1(1)L1(2)2+L1(3>B
3(L1(2)L1(1>C
Input "Steps:",S
S>dim(L2
(Ans1)⁻¹>θ
6Aθ³
Fill(Ans,L2
2Bθ²Ans>L2(1
cumSum(L2>L2
Aθ³Bθ²+Cθ>L2(1
cumSum(L2>L2
L1(1>L2(1
cumSum(L2>L2
Thank you for the code again, it looks really awesome.
Unfortunately, as I use Doors CS, the sum( token is preserved for one of the libraries.
Is there an alternative?
And a note on the project:
It might look like there is not much discussion about this, but actually we are talking a lot in the SAX widget.
If you are interested in reading or participating, you should register at the registration page.
Unfortunately, as I use Doors CS, the sum( token is preserved for one of the libraries.
Is there an alternative?
And a note on the project:
It might look like there is not much discussion about this, but actually we are talking a lot in the SAX widget.
If you are interested in reading or participating, you should register at the registration page.

PT_
 Site Admin (Posts: 2088)
 23 Dec 2015 01:23:21 pm
 Last edited by PT_ on 24 Dec 2015 11:02:02 am; edited 2 times in total
I'm still working on the scaling algorithm, and I'm now working on a new technology; hope it works.
Code:
EDIT: explanation
I've used the code from the website http://techalgorithm.com/articles/nearestneighborimagescaling/ , the second one. It is only for the horizontal enlarging though. It was 95/Z, now it becomes 95. That
Code:
is for defining the count of lines for shifting, which is executed in the second Forloop.
J is at which line you must shift.
Code:
Input Z
95/Z>A
3>B
95>C
63>D
{0>L2
int(A*4096/C+1>P
int(B*4096/D+1>Q
0>J
int(Pseq(X,X,0,94)/4096>L1
For(A,0,92P/4096
sum(L1=A>D
For(B,2,Ans
J
real(12,1,95,0,95,63,1
StorePic 2
real(12,7,Ans+1,0,96,64,1
StorePic 1
real(3,2,0,1,1
If Ans
real(12,7,0,0,Ans1,64,1
real(4,3,1,1
real(3,1,2,1,1
End
J+D>J
End
EDIT: explanation
I've used the code from the website http://techalgorithm.com/articles/nearestneighborimagescaling/ , the second one. It is only for the horizontal enlarging though. It was 95/Z, now it becomes 95. That
Code:
int(Pseq(X,X,0,94)/4096>L1
J is at which line you must shift.
Okay, so a fter (another more) long delay I am back at this project. I am sorry for so many things that came across it...
So, first, lirtosiast, your code seems not to work correctly, and I am sure I typed it correctly. PT_ also confimed this.
Code:
It draws a curve, but a wrong one...
I use libraries, so the real( token is preserved. However this can be worked around with imag([var] * i) so I changed this in the code (But this is not the source of error)...
Thanks for your guys help, I wouldn't have brought it to this point without it!
So, first, lirtosiast, your code seems not to work correctly, and I am sure I typed it correctly. PT_ also confimed this.
lirtosiast wrote:
Code:
For(I,1,4
Input
PtOn(X,Y,3
X+Yi>L1(I
End
//Find coefficients
L1(1)+3L1(2)3L1(3)+L1(4>A
3(L1(1)2L1(2)+L1(3>B
3(L1(2)L1(1>C
L1(1>D
Input "Steps:",S
S1>S
1/S>θ
//differences are U,V,W; point is T
D>T
Aθ^3+Bθ^2+Cθ>U
6Aθ^3+Bθ^2>V
6Aθ^3>W
//old point saved in Z
For(n,1,S)
T>Z
T+U>T
U+V>U
V+W>V
Line(real(Z),imag(Z),real(T),imag(T
End
It draws a curve, but a wrong one...
I use libraries, so the real( token is preserved. However this can be worked around with imag([var] * i) so I changed this in the code (But this is not the source of error)...
Thanks for your guys help, I wouldn't have brought it to this point without it!
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