Copy Link
Add to Bookmark
Report
NULL mag Issue 03 20 Draw lines in math
this is a very good tutor for making circles and lines in code. its written
by denthor of asphyxia. at the end you can find two procedures i use for
circles and lines in the blockart editor :)
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
ÕÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ͸
³ W E L C O M E ³
³ To the VGA Trainer Program ³ ³
³ By ³ ³
³ DENTHOR of ASPHYXIA ³ ³ ³
³ (updated by Snowman) ³ ³ ³
ÔÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ; ³ ³
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ ³
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ
--==[ PART 3 ]==--
[Note: things in brackets have been added by Snowman. The original text
has remained mostly unaltered except for the inclusion of C++ material]
þ Introduction
Greetings! This is the third part of the VGA Trainer series! Sorry it
took so long to get out, but I had a running battle with the traffic
department for three days to get my car registered, and then the MailBox
went down. Ahh, well, life stinks. Anyway, today will do some things
vital to most programs : Lines and circles.
Watch out for next week's part : Virtual screens. The easy way to
eliminate flicker, "doubled sprites", and subjecting the user to watch
you building your screen. Almost every ASPHYXIA demo has used a virtual
screen (with the exception of the SilkyDemo), so this is one to watch out
for. I will also show you how to put all of these loose procedures into
units.
If you would like to contact me, or the team, there are many ways you
can do it : 1) Write a message to Grant Smith in private mail here on
the Mailbox BBS.
2) Write a message here in the Programming conference here
on the Mailbox (Preferred if you have a general
programming query or problem others would benefit from)
3) Write to ASPHYXIA on the ASPHYXIA BBS.
4) Write to Denthor, Eze or Livewire on Connectix.
5) Write to : Grant Smith
P.O.Box 270 Kloof
3640
6) Call me (Grant Smith) at 73 2129 (leave a message if you
call during varsity)
NB : If you are a representative of a company or BBS, and want ASPHYXIA
to do you a demo, leave mail to me; we can discuss it.
NNB : If you have done/attempted a demo, SEND IT TO ME! We are feeling
quite lonely and want to meet/help out/exchange code with other demo
groups. What do you have to lose? Leave a message here and we can work
out how to transfer it. We really want to hear from you!
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
þ Circle Algorithim
You all know what a circle looks like. But how do you draw one on the
computer?
You probably know circles drawn with the degrees at these points :
0
ÜÛ³ÛÜ
ÛÛÛ³ÛÛÛ
270 ----+---- 90
ÛÛÛ³ÛÛÛ
ßÛ³Ûß
180
Sorry about my ASCII ;-) ... anyway, Pascal and C++ don't work that way ...
they work with radians instead of degrees. (You can convert radians to
degrees, but I'm not going to go into that now. Note though that in Pascal
and C++, the circle goes like this :
270
ÜÛ³ÛÜ
ÛÛÛ³ÛÛÛ
180 ----+---- 0
ÛÛÛ³ÛÛÛ
ßÛ³Ûß
90
Even so, we can still use the famous equations to draw our circle ...
(You derive the following by using the theorem of our good friend
Pythagoras)
Sin (deg) = Y/R
Cos (deg) = X/R
(This is standard 8(?) maths ... if you haven't reached that level yet,
take this to your dad, or if you get stuck leave me a message and I'll
do a bit of basic Trig with you. I aim to please ;-))
Where Y = your Y-coord
X = your X-coord
R = your radius (the size of your circle)
deg = the degree
To simplify matters, we rewrite the equation to get our X and Y values :
Y = R*Sin(deg)
X = R*Cos(deg)
This obviousy is perfect for us, because it gives us our X and Y co-ords to
put into our Putpixel routine (see Part 1). Because the Sin and Cos
functions return a Real [long] value, we use a round function to transform
it into an Integer.
[Pascal]
Procedure Circle (oX,oY,rad:integer;Col:Byte);
VAR deg:real;
X,Y:integer;
BEGIN
deg:=0;
repeat
X:=round(rad*COS (deg));
Y:=round(rad*sin (deg));
putpixel (x+ox,y+oy,Col);
deg:=deg+0.005;
until (deg>6.4);
END;
[C++]
void Circle(int X, int Y, int rad, int col) {
float deg = 0;
do {
X = round(rad * cos(deg));
Y = round(rad * sin(deg));
Putpixel (X+160, Y+100, col);
deg += 0.005;
}
while (deg <= 6.4);
}
In the above example, the smaller the amount that deg is increased by,
the closer the pixels in the circle will be, but the slower the procedure.
0.005 seem to be best for the 320x200 screen. NOTE : ASPHYXIA does not use
this particular circle algorithm, ours is in assembly language, but this
one should be fast enough for most. If it isn't, give us the stuff you are
using it for and we'll give you ours.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
þ Line algorithms
There are many ways to draw a line on the computer. I will describe one
and give you two. (The second one you can figure out for yourselves; it
is based on the first one but is faster)
The first thing you need to do is pass what you want the line to look
like to your line procedure. What I have done is said that x1,y1 is the
first point on the screen, and x2,y2 is the second point. We also pass the
color to the procedure. (Remember the screens top left hand corner is (0,0);
see Part 1)
Ie. o (X1,Y1)
ooooooooo
ooooooooo
oooooooo (X2,Y2)
Again, sorry about my drawings ;-)
To find the length of the line, we say the following :
[Pascal]
XLength = ABS (x1-x2)
YLength = ABS (y1-y2)
[C++]
xlength = abs(x1-x2);
ylength = abs(y1-y2);
The ABS function means that whatever the result, it will give you an
absolute, or posotive, answer. At this stage I set a variable stating
wheter the difference between the two x's are negative, zero or posotive.
(I do the same for the y's) If the difference is zero, I just use a loop
keeping the two with the zero difference posotive, then exit.
If neither the x's or y's have a zero difference, I calculate the X and Y
slopes, using the following two equations :
[Pascal]
Xslope = Xlength / Ylength
Yslope = Ylength / Xlength
[C++]
[Note: C++ is significantly different here. I had to add special divide
by zero checking. C++ doesn't like x/0. I also added in type-casting
which might not have been necessary. :)]
if ((xlength != 0) && (ylength != 0)) {
xslope = (float)xlength/(float)ylength;
yslope = (float)ylength/(float)xlength;
}
else {
xslope = 0.0;
yslope = 0.0;
}
As you can see, the slopes are real numbers.
NOTE : XSlope = 1 / YSlope
Now, there are two ways of drawing the lines :
X = XSlope * Y
Y = YSlope * X
The question is, which one to use? if you use the wrong one, your line
will look like this :
o
o
o
Instead of this :
ooo
ooo
ooo
Well, the solution is as follows :
*\``³``/*
***\³/***
----+----
***/³\***
*/``³``\*
If the slope angle is in the area of the stars (*) then use the first
equation, if it is in the other section (`) then use the second one.
What you do is you calculate the variable on the left hand side by
putting the variable on the right hand side in a loop and solving. Below
is our finished line routine :
[Pascal]
Procedure Line (x1,y1,x2,y2:integer;col:byte);
VAR x,y,xlength,ylength,dx,dy:integer;
xslope,yslope:real;
BEGIN
xlength:=abs (x1-x2);
if (x1-x2)<0 then dx:=-1;
if (x1-x2)=0 then dx:=0;
if (x1-x2)>0 then dx:=+1;
ylength:=abs (y1-y2);
if (y1-y2)<0 then dy:=-1;
if (y1-y2)=0 then dy:=0;
if (y1-y2)>0 then dy:=+1;
if (dy=0) then BEGIN
if dx<0 then for x:=x1 to x2 do
putpixel (x,y1,col);
if dx>0 then for x:=x2 to x1 do
putpixel (x,y1,col);
exit;
END;
if (dx=0) then BEGIN
if dy<0 then for y:=y1 to y2 do
putpixel (x1,y,col);
if dy>0 then for y:=y2 to y1 do
putpixel (x1,y,col);
exit;
END;
xslope:=xlength/ylength;
yslope:=ylength/xlength;
if (yslope/xslope<1) and (yslope/xslope>-1) then BEGIN
if dx<0 then for x:=x1 to x2 do BEGIN
y:= round (yslope*x);
putpixel (x,y,col);
END;
if dx>0 then for x:=x2 to x1 do BEGIN
y:= round (yslope*x);
putpixel (x,y,col);
END;
END
ELSE
BEGIN
if dy<0 then for y:=y1 to y2 do BEGIN
x:= round (xslope*y);
putpixel (x,y,col);
END;
if dy>0 then for y:=y2 to y1 do BEGIN
x:= round (xslope*y);
putpixel (x,y,col);
END;
END;
END;
[C++]
void Line2(int x1, int y1, int x2, int y2, int col) {
int x, y, xlength, ylength, dx, dy;
float xslope, yslope;
xlength = abs(x1-x2);
if ((x1-x2) < 0) dx = -1;
if ((x1-x2) == 0) dx = 0;
if ((x1-x2) > 0) dx = +1;
ylength = abs(y1-y2);
if ((y1-y2) < 0) dy = -1;
if ((y1-y2) == 0) dy = 0;
if ((y1-y2) > 0) dy = +1;
if (dy == 0) {
if (dx < 0)
for (x=x1; x<x2+1; x++)
Putpixel (x,y1,col);
if (dx > 0)
for (x=x2; x<x1+1; x++)
Putpixel (x,y1,col);
}
if (dx == 0) {
if (dy < 0)
for (y=y1; y<y2+1; y++)
Putpixel (x1,y,col);
if (dy > 0)
for (y=y2; y<y1+1; y++)
Putpixel (x1,y,col);
}
if ((xlength != 0) && (ylength != 0)) {
xslope = (float)xlength/(float)ylength;
yslope = (float)ylength/(float)xlength;
}
else {
xslope = 0.0;
yslope = 0.0;
}
if ((xslope != 0) && (yslope != 0) &&
(yslope/xslope < 1) && (yslope/xslope > -1)) {
if (dx < 0)
for (x=x1; x<x2+1; x++) {
y = round (yslope*x);
Putpixel (x,y,col);
}
if (dx > 0)
for (x=x2; x<x1+1; x++) {
y = round (yslope*x);
Putpixel (x,y,col);
}
}
else {
if (dy < 0)
for (y=x1; y<x2+1; y++) {
x = round (xslope*y);
Putpixel (x,y,col);
}
if (dy > 0)
for (y=x2; y<x1+1; y++) {
x = round (xslope*y);
Putpixel (x,y,col);
}
}
}
Quite big, isn't it? Here is a much shorter way of doing much the same
thing :
[Pascal]
function sgn(a:real):integer;
begin
if a>0 then sgn:=+1;
if a<0 then sgn:=-1;
if a=0 then sgn:=0;
end;
procedure line(a,b,c,d,col:integer);
var u,s,v,d1x,d1y,d2x,d2y,m,n:real;
i:integer;
begin
u:= c - a;
v:= d - b;
d1x:= SGN(u);
d1y:= SGN(v);
d2x:= SGN(u);
d2y:= 0;
m:= ABS(u);
n := ABS(v);
IF NOT (M>N) then
BEGIN
d2x := 0 ;
d2y := SGN(v);
m := ABS(v);
n := ABS(u);
END;
s := INT(m / 2);
FOR i := 0 TO round(m) DO
BEGIN
putpixel(a,b,col);
s := s + n;
IF not (s<m) THEN
BEGIN
s := s - m;
a:= a +round(d1x);
b := b + round(d1y);
END
ELSE
BEGIN
a := a + round(d2x);
b := b + round(d2y);
END;
end;
END;
[C++]
int sgn (long a) {
if (a > 0) return +1;
else if (a < 0) return -1;
else return 0;
}
void Line(int a, int b, int c, int d, int col) {
long u,s,v,d1x,d1y,d2x,d2y,m,n;
int i;
u = c-a;
v = d-b;
d1x = sgn(u);
d1y = sgn(v);
d2x = sgn(u);
d2y = 0;
m = abs(u);
n = abs(v);
if (m<=n) {
d2x = 0;
d2y = sgn(v);
m = abs(v);
n = abs(u);
}
s = (int)(m / 2);
for (i=0;i<round(m);i++) {
Putpixel(a,b,col);
s += n;
if (s >= m) {
s -= m;
a += d1x;
b += d1y;
}
else {
a += d2x;
b += d2y;
}
}
}
This routine is very fast, and should meet almost all of your requirements
(ASPHYXIA used it for quite a while before we made our new one.)
In the end program, both the new line routine and the circle routine are
tested. A few of the procedures of the first parts are also used.
Line and circle routines may seem like fairly trivial things, but they are
a vital component of many programs, and you may like to look up other
methods of drawing them in books in the library (I know that here at the
varsity they have books for doing this kind of stuff all over the place)
A good line routine to look out for is the Bressenhams line routine ...
there is a Bressenhams circle routine too ... I have documentaiton for them
if anybody is interested, they are by far some of the fastest routines
you will use.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
þ In closing
Varsity has started again, so I am (shock) going to bed before three in
the morning, so my quote this week wasn't written in the same wasted way
my last weeks one was (For last week's one, I had gotten 8 hours sleep in
3 days, and thought up and wrote the quote at 2:23 am before I fell asleep.)
[ "What does it do?" she asks.
"It's a computer," he replies.
"Yes, dear, but what does it do?"
"It ..er.. computes! It's a computer."
"What does it compute?"
"What? Er? Um. Numbers! Yes, numbers!" He smiles
worriedly.
"Why?"
"Why? Well ..um.. why?" He starts to sweat.
"I mean, is it just something to dust around, or does
it actually do something useful?"
"Um...you can call other computers with it!" Hope lights
up his eyes. "So you can get programs from other computers!"
"I see. Tell me, what do these programs do?"
"Do? I don't think I fol..."
"I see. They compute. Numbers. For no particular reason." He
withers under her gaze.
"Yes, but..."
She smiles, and he trails off, defeated. She takes another look
at the thing. "Although," she says, with a strange look in
her eyes. He looks up, an insane look of hope on his
face. "Does it come in pink?" she asks.
]
- Grant Smith
Tue 27 July, 1993
9:35 pm.
See you next time,
- Denthor
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
circle and line procedures by xqtr
Procedure Circle1(x1, y1, x2, y2 : integer; color : byte; Var Image: TScreenImage);
Var
r,y,x,d : Integer;
Begin
r := (x2 - x1) Div 2;
y := r;
d := -r;
Image[y1][x1+r].Char:='*';
For x := 10 to trunc((r / Sqrt(2))*10) Do Begin
d := d + (2*x)-1;
if ( d >= 0 ) Then Begin
y := y - 1;
d := d - 2* y;
End;
Image[y1+y][x1+x]Char:='*';
End;
End;
procedure Line(x1, y1, x2, y2 : integer; color : byte; Var Image: TScreenImage; C:Char);
var i, deltax, deltay, numpixels,
d, dinc1, dinc2,
x, xinc1, xinc2,
y, yinc1, yinc2 : integer;
begin
{ Calculate deltax and deltay for initialisation }
deltax := abs(x2 - x1);
deltay := abs(y2 - y1);
{ Initialize all vars based on which is the independent variable }
if deltax >= deltay then
begin
{ x is independent variable }
numpixels := deltax + 1;
d := (2 * deltay) - deltax;
dinc1 := deltay Shl 1;
dinc2 := (deltay - deltax) shl 1;
xinc1 := 1;
xinc2 := 1;
yinc1 := 0;
yinc2 := 1;
end
else
begin
{ y is independent variable }
numpixels := deltay + 1;
d := (2 * deltax) - deltay;
dinc1 := deltax Shl 1;
dinc2 := (deltax - deltay) shl 1;
xinc1 := 0;
xinc2 := 1;
yinc1 := 1;
yinc2 := 1;
end;
{ Make sure x and y move in the right directions }
if x1 > x2 then
begin
xinc1 := - xinc1;
xinc2 := - xinc2;
end;
if y1 > y2 then
begin
yinc1 := - yinc1;
yinc2 := - yinc2;
end;
{ Start drawing at }
x := x1;
y := y1;
{ Draw the pixels }
for i := 1 to numpixels do
begin
Image[y][x].Char:=C;
Image[y][x].Attr:=Color;
if d < 0 then
begin
d := d + dinc1;
x := x + xinc1;
y := y + yinc1;
end
else
begin
d := d + dinc2;
x := x + xinc2;
y := y + yinc2;
end;
end;
end;
