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Using CP-SAT of google or-tools I'm trying to write this constraint:
q >= (50x + 100y + 150z + 200k + 250p + 300v) / (x + y + z + k + p + v)
Where q is a simple integer.
The thing is I need to round the right side of the equation (let's call it expression) as follows:
if(expression < 75) {
expression = 50;
} else if(expression < 125) {
expression = 100;
} else if(expression < 175) {
expression = 150;
} else if(expression < 225) {
expression = 200;
} else if(expression < 275) {
expression = 250;
} else {
expression = 300;
}
So I need to round the expression
(50x + 100y + 150z + 200k + 250p + 300v) / (x + y + z + k + p + v)
So that it gets one of the following values:
{50, 100, 150, 200, 250, 300}
Let's review 2 cases:
Case 1
q = 180 and expression = 176.
Although the condition 180 >= 176 is true, after rounding up 176 to 200 the tested condition should be 180 >= 200 which is false.
So for q = 180 and expression = 176 I would like the condition to return false.
Case 2
q = 210 and expression = 218.
Although the condition 210 >= 218 is false, after rounding down 218 to 200 the tested condition should be 210 >= 200 which is true.
So for q = 210 and expression = 218 I would like the condition to return true.
I got a great answer here for resolving this challenge over a linear expression but now I need to resolve it for a non-linear expression.
Any suggestions?
Let's rephrase
you have an integer variable e with a value between 0 and 300.
You want to round it to the nearest multiple of 50
if you do:
(e div 50) * 50
you will get the max multiple of 50 less or equal to e
(70 / 50) * 50 -> 50
(99 / 50) * 50 -> 50
(102 / 50) * 50 -> 100
To get a round to nearest, you need to add 25 to e before the division
((e + 25) div 50) * 50
Which will do the correct rounding
((70 + 25) / 50) * 50 -> 50
((99 + 25) / 50) * 50 -> 100
((102 + 25) / 50) * 50 -> 100
with the correct or-tools CP-SAT python code:
numerator = model.NewIntVar(...)
model.Add(numerator == 50x + 100y + 150z + 200k + 250p + 300v)
denom = model.NewIntVar(...)
model.Add(denom == 50x + 100y + 150z + 200k + 250p + 300v)
e = model.NewIntVar(0, 300, 'e')
model.AddDivisionEquality(e, numerator, denom)
shifted_e = model.NewIntVar(25, 325, 'shifted_e')
model.Add(shifted_e == e + 25)
multiple_of_fifty = model.NewIntVar(0, 6, 'multiple_of_fifty')
model.AddDivisionEquality(multiple_of_fifty, shifted_e, 50)
result = model.NewIntVar(0, 300, 'result')
model.Add(result = multiple_of_fifty * 50)
if a and b are positive then
a div b >= q
is equivalent to
a >= q * b
now, your example does not specify how to round (nearest or down)
if you want to round down
q * (x + y + z + k + p + v) <= (50x + 100y + 150z + 200k + 250p + 300v)
If you want to round to nearest, you need to add q / 2 in the right place
q * (x + y + z + k + p + v) <= (50x + 100y + 150z + 200k + 250p + 300v + q / 2)
Now, if you want the other direction
a div b <= q
is equivalent to
a <= q * b + q - 1
The rest of the transformation is the same.
I wanted to find out in which order the VBA evaluates the terms inside the expression.
I have read the document about operator precedence and thought it will simply evaluate from left to right if the terms are all operated with "+".
However I get these strange results.
Option Explicit
Dim x
Sub test()
Debug.Print CreateObject("Scripting.FileSystemObject") _
.GetFileVersion(Application.Path & "\Excel.exe") ' 16.0.12624.20086
x = 0: Debug.Print inc_x + x '(a) 2
x = 0: debug_print inc_x + x '(b) 2
x = 0: Debug.Print x + inc_x '(c) 2 <=== ?
x = 0: debug_print x + inc_x '(d) 2 <=== ?
x = 0: Debug.Print inc_x + 0 + x '(e) 2
x = 0: debug_print inc_x + 0 + x '(f) 2
x = 0: Debug.Print 0 + x + inc_x '(g) 2 <=== ?
x = 0: debug_print 0 + x + inc_x '(h) 1 <=== ???
End Sub
Function inc_x()
x = x + 1
inc_x = x
End Function
Function debug_print(msg)
Debug.Print msg
End Function
I thought (c), (d), and (g) should print 1 but they print 2.
And it gets even weirder when you look at (h) which is the only one printing 1.
Can someone explain what is happening ?
added more patterns
I added (i),(j),(k),(l),(m), and (n) which also shows interesting patterns.
#SJR also taught me that specifying the type of x as Long will change the result of (d).
Option Explicit
Dim x 'As Long 'specifying the type will change the result of (d)
Sub test()
Debug.Print CreateObject("Scripting.FileSystemObject") _
.GetFileVersion(Application.Path & "\Excel.exe") ' 16.0.12624.20086
x = 0: Debug.Print inc_x + x '(a) 2
x = 0: debug_print inc_x + x '(b) 2
x = 0: Debug.Print x + inc_x '(c) 2 <=== ?
x = 0: debug_print x + inc_x '(d) 2 <=== ? (This becomes 1 when x is Long)
x = 0: Debug.Print inc_x + 0 + x '(e) 2
x = 0: debug_print inc_x + 0 + x '(f) 2
x = 0: Debug.Print 0 + x + inc_x '(g) 2 <=== ?
x = 0: debug_print 0 + x + inc_x '(h) 1 <=== ???
x = 0: Debug.Print 100 * inc_x + 10 * x + 1 * inc_x '(i) 122
x = 0: debug_print 100 * inc_x + 10 * x + 1 * inc_x '(j) 112
x = 0: Debug.Print 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(k) 2122
x = 0: debug_print 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(l) 112
x = 0: Debug.Print 10000 * inc_x + 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(m) 13233
x = 0: debug_print 10000 * inc_x + 1000 * x + 100 * inc_x + 10 * x + 1 * inc_x '(n) 11223
End Sub
Function inc_x()
x = x + 1
inc_x = x
End Function
Function debug_print(msg)
Debug.Print msg
End Function
So I have a "main" function (SolveSixODES) that calls a secondary function (AllODEs). And when it does this, the x value in the main function gets modified. I don't understand how this can be possible, seeing as it is not a global variable.
Here is the code, my inputs I used are as follows:
x=0, xmax=3, y=0-6, h=0.1, error=0.1
Public Function SolveSixODE(x As Double, xmax As Double, Y As Range, h As Double, error As Double) 'Weird bug: You must leave the first y4 value blank
Dim i As Integer, k(7, 7) As Double, j As Integer, m As Integer 'k(Order #, equation #)
Dim Y5(7) As Double, Y4(7) As Double, Y4Old(7) As Double
Dim delta0(7) As Double, delta1(7) As Double, delRatio(7) As Double, Rmin As Double
For i = 1 To 6 'Moving the input data so it can acutally be used
Y4(i) = Y(i)
Next i
While x < xmax
If x + h < xmax Then
x = x + h
Else
h = xmax - x
x = xmax
End If
For j = 1 To 6 'j is the order i is equation number
For i = 1 To 6 'Calculating all of the k(1) values for eq 1 to 6
k(j, i) = AllODES(x, Y4, i, j, k, h) '!!!!!SOME HOW THIS LOOP MAKES X negative...!!!!!!!
Next i
Next j
For i = 1 To 6
Y4Old(i) = Y4(i) 'Saving old y4 value to calc delta0
Y4(i) = Y4(i) + h * (k(1, i) * (37 / 378) + k(3, i) * (250 / 621) + k(4, i) * (125 / 594) + k(6, i) * (512 / 1771))
Y5(i) = Y4(i) + h * (k(1, i) * (2825 / 27648) + k(3, i) * (18575 / 48384) + k(4, i) * (13525 / 55296) + k(5, i) * (277 / 14336) + k(6, i) * (0.25))
delta0(i) = error * (Abs(Y4Old(i)) + Abs(h * AllODES(x, Y4Old, i, 1, k, h))) 'First order because we don't want to use the k vals
delta1(i) = Abs(Y5(i) - Y4(i))
delRatio(i) = Abs(delta0(i) / delta1(i)) 'Ratio of errors
Next i
Rmin = delRatio(1)
For i = 2 To 6
If delRatio(i) < Rmin Then
Rmin = delRatio(i) 'Determine the smallest error ratio
End If
Next i
If Rmin < 1 Then 'If this is true then the step size was too big must repeat step
x = x - h 'Set x and y's back to previous values
For i = 1 To 6
Y4(i) = Y4Old(i)
Next i
h = 0.9 * h * Rmin ^ 0.25 'adjust h value; 0.9 is a safety factor
Else
h = 0.9 * h * Rmin ^ 0.2 'Otherwise, we march on
End If
m = m + 1
Wend
SolveSixODE = Y4
End Function
Public Function AllODES(x As Double, Y() As Double, EqNumber As Integer, order As Integer, k() As Double, h As Double) As Double
Dim conc(7) As Double, i As Integer, j As Integer
If order = 1 Then
x = x - h
For i = 1 To 6 'Movin the data so I can use it
conc(i) = Y(i) 'also adjusting the x and y values for RK4 (Cash Karp values)
Next i
ElseIf order = 2 Then
x = x - h + h * 0.2
For i = 1 To 6
conc(i) = Y(i) + h * k(1, i) * 0.2
Next i
ElseIf order = 3 Then
x = x - h + 0.3 * h
For i = 1 To 6
conc(i) = Y(i) + h * (0.075 * k(1, i) + 0.225 * k(2, i))
Next i
ElseIf order = 4 Then
x = x - h + 0.6 * h
For i = 1 To 6
conc(i) = Y(i) + h * (0.3 * k(1, i) - 0.9 * k(2, i) + 1.2 * k(3, i))
Next i
ElseIf order = 5 Then
x = x - h + h
For i = 1 To 6
conc(i) = Y(i) + h * ((-11 / 54) * k(1, i) + 2.5 * k(2, i) - (70 / 27) * k(3, i) + (35 / 27) * k(4, i))
Next i
ElseIf order = 6 Then
x = x - h + 0.875 * h
For i = 1 To 6
conc(i) = Y(i) + h * ((1631 / 55296) * k(1, i) + (175 / 512) * k(2, i) + (575 / 13824) * k(3, i) + (44275 / (110592) * k(4, i) + (253 / 4096) * k(5, i)))
Next i
Else
MsgBox ("error")
End If
If EqNumber = 1 Then 'These are the actual equations
AllODES = x + Y(1)
ElseIf EqNumber = 2 Then
AllODES = x
ElseIf EqNumber = 3 Then
AllODES = Y(3)
ElseIf EqNumber = 4 Then
AllODES = 2 * x
ElseIf EqNumber = 5 Then
AllODES = 2 * Y(2)
ElseIf EqNumber = 6 Then
AllODES = 3 * x
Else
MsgBox ("You entered an Eq Number that was dumb")
End If
End Function
It's possible that it is something really trivial that I missed but this seems to contradict my knowledge of how variables work. So if you understand how the function is able to manipulate a variable from another function in this case, I would appreciate any advice and/or explanation!
Thanks in advance!
the x value in the main function gets modified. I don't understand how this can be possible, seeing as it is not a global variable
This is normal because you are passing x by reference to the function AllODES and you do change it there. When the keyword ByVal is not explicitly specified in the function/sub prototype, the default passing mechanism is ByRef, that is, by reference.
Public Function AllODES(x As Double, ...
means
Public Function AllODES(ByRef x As Double, ....
We observe that x is manipulated in this function, so the change will appear in the caller. If you want that the change of x does not report back in the caller's scope, pass x by value:
Public Function AllODES(ByVal x As Double, ....
' ^^^^^
Only in this case the x of the caller and the x of the callee will be two different variables.
Not sure how to solve this. "a" was meant to start from 1 to NumData, but I have deliberately change the start from 44200 to check the ELSEIF. For NumData = 117,350,
I would expect the 3rd ELSEIF to be activated. Instead throughout the whole run, it only step-into the first ELSEIF even though the "a" value does not meet the conditions.
What should I do?
For a = 44200 To NumData 'Int1
If a > 1 Then
If UCase(Trim(Range1(a, 3))) = UCase(Trim(Range1(a - 1, 3))) Then
GoTo Line1 'Next count loop if next Platform name the same
End If
End If
For b = 1 To NumData
lat1 = Range1(a, 5)
lat2 = Range1(b, 5)
long1 = Range1(a, 6)
long2 = Range1(b, 6)
CompRad = Dist(lat1, lat2, long1, long2)
If (CompRad <= Radius And CompRad >= 0) Then
z = CLng(NumData / 8)
If a <= CLng(NumData / 8) Then
For c = 1 To 6
Range2(d, c) = Range1(b, c)
Next c
Acc_Sum2 = Acc_Sum2 + Range2(d, 4)
d = d + 1
ElseIf CLng(NumData / 8) < a <= 2 * CLng(NumData / 8) Then
z = 2 * CLng(NumData / 8)
For c = 1 To 6
Range3(e, c) = Range1(b, c)
Next c
Acc_Sum3 = Acc_Sum3 + Range3(e, 4)
e = e + 1
ElseIf 2 * CLng(NumData / 8) < a <= 3 * CLng(NumData / 8) Then
For c = 1 To 6
Range4(f, c) = Range1(b, c)
Next c
Acc_Sum4 = Acc_Sum4 + Range4(f, 4)
f = f + 1
ElseIf 3 * CLng(NumData / 8) < a <= 4 * CLng(NumData / 8) Then
z = 3 * CLng(NumData / 8)
For c = 1 To 6
Range5(g, c) = Range1(b, c)
Next c
Acc_Sum5 = Acc_Sum5 + Range5(g, 4)
g = g + 1
ElseIf 4 * CLng(NumData / 8) < a <= 5 * CLng(NumData / 8) Then
For c = 1 To 6
Range6(h, c) = Range1(b, c)
Next c
Acc_Sum6 = Acc_Sum6 + Range6(h, 4)
h = h + 1
ElseIf 5 * CLng(NumData / 8) < a <= 6 * CLng(NumData / 8) Then
For c = 1 To 6
Range7(i, c) = Range1(b, c)
Next c
Acc_Sum7 = Acc_Sum7 + Range7(i, 4)
i = i + 1
ElseIf 6 * CLng(NumData / 8) < a <= 7 * CLng(NumData / 8) Then
For c = 1 To 6
Range8(j, c) = Range1(b, c)
Next c
Acc_Sum8 = Acc_Sum8 + Range8(j, 4)
j = j + 1
ElseIf 7 * CLng(NumData / 8) < a <= NumData Then
For c = 1 To 6
Range9(k, c) = Range1(b, c)
Next c
Acc_Sum9 = Acc_Sum9 + Range9(k, 4)
k = k + 1
End If
End If
Next b
Line1:
Next a
Your conditions like:
1 < a <= 10
are always true. First part (1 < a) evaluates to True or False and then it is converted to integer (True = 1, False = 0). Both values are <=10.
You should change thes conditions to:
(1 < a) And (a <= 10)
Brackets are optional, comparison operators have higher precedence.
user3964075 nailed the core issue. I think making the code a little bit more readable will help to filter out other possible trouble areas. Why do you set the value of 'z' and then not use it?
z = CLng(NumData / 8)
Where are the variables d, e, f...; Range2, Range3, Range4...; Acc_Sum2, Acc_Sum3... assigned, and what are they doing? Can each group be replaced by a single variable?
You may also want to create a simple Between function
to clean up a lot of your conditional statements.
Public Function Between(x As Integer, min As Integer, max As Integer) As Boolean
Between = x <= max And x >= min
End Function
Thanks, user3964075 & Carl for your prompt response. It's worked! Carl, the z variable was slotted in when I was trying to figure out the problem. I thought it would be too much to post the whole code. It does require a lot of cleaning up. Appreciate your tips.
I'm trying to make a function in Lua or VB based code to
draw / plot an ellipse and also a filled ellipse.
I don't have much knowledge about this math and I can use some help.
I googled everything there is to google about drawing ellipses with code but I can't find a good simple working example that i can code into my Lua / VB code.
here are a few websites i visited but couldn't make the code work or couldn't convert the code to Lua or VB properly...
https://sites.google.com/site/ruslancray/lab/projects/bresenhamscircleellipsedrawingalgorithm/bresenham-s-circle-ellipse-drawing-algorithm
http://groups.csail.mit.edu/graphics/classes/6.837/F98/Lecture6/circle.html
http://www.blitzbasic.com/codearcs/codearcs.php?code=2817
http://hackipedia.org/Algorithms/Graphics/pdf/A%20Fast%20Bresenham%20Type%20Algorithm%20For%20Drawing%20Ellipses%20by%20John%20Kennedy.pdf
https://scratch.mit.edu/projects/49873666/
http://www.sourcecodesworld.com/source/show.asp?ScriptID=112
How do I draw an ellipse with arbitrary orientation pixel by pixel?
Can anyone help me make code that can draw an ellipse and a filled ellipse?
here is some code I tried to convert to Lua from here:
https://gist.github.com/Wollw/3291916
this code has some problems (missing pixels) and I think it's not converted properly but I don't know how to do it otherwise.
function plotEllipseRect(x0, y0, x1, y1)
-- values of diameter
a = math.abs(x1-x0)
b = math.abs(y1-y0)
b1 = 2.5
-- error increment
dx = 4*(1-a)*b*b
dy = 4*(b1+1)*a*a
-- error of 1.step
err = dx+dy+b1*a*a
-- e2 = 0
if (x0 > x1) then -- if called with swapped points
x0 = x1
x1 = x1 + a
end
if (y0 > y1) then -- .. exchange them
y0 = y1
end
-- starting pixel
y0 = y0 + (b+1)/2
y1 = y0-b1
a = a * 8*a
b1 = 8*b*b
repeat
dot(x1, y0) -- I. Quadrant
dot(x0, y0) -- II. Quadrant
dot(x0, y1) -- III. Quadrant
dot(x1, y1) -- IV. Quadrant
e2 = 2*err
if (e2 <= dy) then -- y step
y0 = y0 + 1
y1 = y1 - 1
dy = dy + a
err = err + dy
end
if (e2 >= dx or 2*err > dy) then -- x step
x0 = x0 + 1
x1 = x1 - 1
dx = dx + b1
err = err + dx
end
until (x0 >= x1)
while (y0-y1 < b) do -- too early stop of flat ellipses a=1
dot(x0-1, y0) -- -> finish tip of ellipse
y0 = y0 + 1
dot(x1+1, y0)
dot(x0-1, y1)
y1 = y1 - 1
dot(x1+1, y1)
end
end
[EDIT:]
I almost got it for the filled one!
see the comments in this code below to know what the problem is...
I use EGSL to test this Lua code:
http://www.egsl.retrogamecoding.org//pages/downloads.php
function DrawEllipse(xc,yc,w,h)
local w2 = w * w
local h2 = h * h
local fw2 = 4 * w2
local fh2 = 4 * h2
xc = xc + w
yc = yc + h
local x = 0
local y = h
local s = 2 * h2 + w2 * (1 - h)
while h2 * x <= w2 * y do
dot(xc + x, yc + y)
dot(xc - x, yc + y)
dot(xc + x, yc - y)
dot(xc - x, yc - y)
redraw()
inkey()
color(int(rnd()*255),int(rnd()*255),int(rnd()*255)) --random color to see changes
if s >= 0 then
s = s + fw2 * (1 - y)
y = y - 1
color(255,0,255)
line(xc + x, yc + y, xc - x, yc + y)
line(xc + x, yc - y, xc - x, yc - y)
end
s = s + h2 * ((4 * x) + 6)
x = x + 1
end
x = w
y = 0
s = 2 * w2 + h2 * (1 - w)
line(xc + x, yc + y, xc - x, yc + y) --to prevent the first line to be drawn twice
redraw()
inkey()
s = s + w2 * ((4 * y) + 6)
y = y + 1
while w2 * y < h2 * (x-2) do
line(xc + x, yc + y, xc - x, yc + y)
redraw()
inkey()
color(int(rnd()*255),int(rnd()*255),int(rnd()*255))
line(xc + x, yc - y, xc - x, yc - y)
redraw()
inkey()
color(int(rnd()*255),int(rnd()*255),int(rnd()*255))
if s >= 0 then
s = s + fh2 * (1 - x)
x = x - 1
end
s = s + w2 * ((4 * y) + 6)
y = y + 1
end
dot(xc + x, yc + y)
dot(xc - x, yc + y)
redraw()
inkey()
color(int(rnd()*255),int(rnd()*255),int(rnd()*255))
dot(xc + x, yc - y)
dot(xc - x, yc - y)
redraw()
inkey()
end
openwindow (70,70,32,"Resize Window")
color(255,255,0)
DrawEllipse(10,10,20,20) --works perfect!
inkey()
cls()
DrawEllipse(10,10,10,20) --problems with last 2 horizontal lines between the pixels!
inkey()
cls()
DrawEllipse(10,10,20,10) --works perfect to!
closewindow()
The following VB works for me, based on the first link provided; the only difference between mine here and the code at your link is I move xc and yc over, since you cannot can't have negative x or y values for the pixels in a bitmap.
Public Shared Function DrawEllipse(ByVal xc As Integer, ByVal yc As Integer, ByVal w As Integer, ByVal h As Integer, ByVal doFill As Boolean) As Drawing.Bitmap
Dim w2 As Integer = w * w
Dim h2 As Integer = h * h
Dim fw2 As Integer = 4 * w2
Dim fh2 As Integer = 4 * h2
// cheat by moving xc and yc so that we can handle quadrants
xc = w
yc = h
Dim bm As New Drawing.Bitmap(w2, h2)
// first half
Dim x As Integer = 0
Dim y As Integer = h
Dim s As Integer = 2 * h2 + w2 * (1 - h)
While h2 * x <= w2 * y
If doFill Then
For i As Integer = -y To y
bm.SetPixel(xc + x, yc + i, Drawing.Color.Red)
bm.SetPixel(xc - x, yc + i, Drawing.Color.Red)
Next
Else
bm.SetPixel(xc + x, yc + y, Drawing.Color.Red)
bm.SetPixel(xc - x, yc + y, Drawing.Color.Red)
bm.SetPixel(xc + x, yc - y, Drawing.Color.Red)
bm.SetPixel(xc - x, yc - y, Drawing.Color.Red)
End If
If s >= 0 Then
s += fw2 * (1 - y)
y -= 1
End If
s += h2 * ((4 * x) + 6)
x += 1
End While
// second half
x = w
y = 0
s = 2 * w2 + h2 * (1 - w)
While w2 * y <= h2 * x
If doFill Then
For i As Integer = -x To x
bm.SetPixel(xc + i, yc + y, Drawing.Color.Red)
bm.SetPixel(xc + i, yc - y, Drawing.Color.Red)
Next
Else
bm.SetPixel(xc + x, yc + y, Drawing.Color.Red)
bm.SetPixel(xc - x, yc + y, Drawing.Color.Red)
bm.SetPixel(xc + x, yc - y, Drawing.Color.Red)
bm.SetPixel(xc - x, yc - y, Drawing.Color.Red)
End If
If s >= 0 Then
s += fh2 * (1 - x)
x -= 1
End If
s += w2 * ((4 * y) + 6)
y += 1
End While
Return bm
End Function
(Aside: I used // instead of ' for the comments... just for readability here. If you copy to Visual Studio you'll have to fix that)
Ok, I managed to find a solution for the filled ellipse by checking
if the pixel from the second half is gonna be drawn in the x-range of the first half of the ellipse.
function drawellipse(xc, yc, w, h, dofill)
--trouble with the size, 1 pixel to large on x and y to...
w=w/2 --good solution for making it the right size?
h=h/2 --good solution for making it the right size?
local w2 = w * w
local h2 = h * h
local fw2 = 4 * w2
local fh2 = 4 * h2
-- cheat by moving xc and yc so that we can handle quadrants
xc = xc + w
yc = yc + h
-- first half
local x = 0
local y = h
local s = 2 * h2 + w2 * (1 - h)
while h2 * x <= w2 * y do
if dofill then
for i = -y , y do
color(0,255,0)
dot(xc + x, yc + i)
dot(xc - x, yc + i)
--redraw()inkey()
end
else
color(255,0,255)
dot(xc + x, yc + y)
dot(xc - x, yc + y)
dot(xc + x, yc - y)
dot(xc - x, yc - y)
--redraw()inkey()
end
if s >= 0 then
s =s+ fw2 * (1 - y)
y =y- 1
end
s =s+ h2 * ((4 * x) + 6)
x =x+ 1
end
color(255,0,255)
line(xc + x,0,xc - x,0)
test1 = xc + x
test2 = xc - x
print(test1 .. '/' .. test2)
redraw()inkey()
-- second half
x = w
y = 0
s = 2 * w2 + h2 * (1 - w)
while w2 * y <= h2 * x do
if dofill then
for i = -x , x do
if not(xc + i > test2 and xc + i < test1) then
color(255,255,0)
dot(xc + i, yc + y)
dot(xc + i, yc - y)
redraw()inkey()
end
end
else
color(0,255,255)
dot(xc + x, yc + y)
dot(xc - x, yc + y)
dot(xc + x, yc - y)
dot(xc - x, yc - y)
redraw()inkey()
end
if s >= 0 then
s =s+ fh2 * (1 - x)
x =x- 1
end
s =s+ w2 * ((4 * y) + 6)
y =y+ 1
end
end
In vb.net you have both Graphics.DrawEllipse and Graphics.DrawArc. In Lua you may be able to use Cairo which I know has a arc function.
If you where to make a ellipse in a GraphicsPath in .Net and where to reverse engineer how it is stored in memory, you would find out that it is stored as four bezier curves. I implemented my own graphics library in vb.net once, and that was how I did it. The best resource I found at the time where a implementation in Actionscript, that I unfortunately was unable to locate aswell as that graphics library I was talking about.
TLDR; You should have a look at bezier curves.
A completely different, and very simple take on this, although the ellipse doesn't seem as "pretty" as the other algorithms; this just uses the mathematical definition of an ellipse and, looping over x calculates the y coordinate given x, w, and h.
Public Shared Function DrawEllipse2(ByVal xc As Integer, ByVal yc As Integer, ByVal w As Integer, ByVal h As Integer, ByVal doFill As Boolean) As Drawing.Bitmap
Dim bm As New Drawing.Bitmap(w * w, h * h)
For x As Integer = xc - w To xc + w
Dim y As Integer = CInt((Math.Sqrt(1 - ((x * x) / (w * w)))) * h)
If doFill Then
For j As Integer = -y To y
bm.SetPixel(w + x, h + j, Drawing.Color.Red)
Next
Else
bm.SetPixel(w + x, h + y, Drawing.Color.Red)
bm.SetPixel(w + x, h - y, Drawing.Color.Red)
End If
Next
Return bm
End Function