绘制一个基于椭圆 sqrt 的函数
Draw an ellipse sqrt-based function
我正在尝试在基于 Lua 或 VB 的代码中创建一个函数来
绘制/绘制椭圆和实心椭圆。
我对这个数学知识不多,需要一些帮助。
我 googled google 关于用代码绘制椭圆的所有内容,但我找不到一个很好的简单工作示例,我可以将其编码到我的 Lua / VB代码。
这是我访问过的一些网站,但无法使代码正常工作或无法将代码正确转换为 Lua 或 VB...
http://groups.csail.mit.edu/graphics/classes/6.837/F98/Lecture6/circle.html
http://www.sourcecodesworld.com/source/show.asp?ScriptID=112
How do I draw an ellipse with arbitrary orientation pixel by pixel?
谁能帮我写出可以画椭圆和实心椭圆的代码?
这是我尝试从此处转换为 Lua 的一些代码:
https://gist.github.com/Wollw/3291916
这段代码有一些问题(缺少像素),我认为它没有正确转换,但我不知道该怎么做。
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
[编辑:]
我差一点就得到了!
查看下面这段代码中的注释以了解问题所在...
我使用 EGSL 测试此 Lua 代码:
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()
以下 VB 适合我,基于提供的第一个 link;我这里的代码和你的代码之间的唯一区别 link 是我移动了 xc 和 yc,因为位图中的像素不能有负 x 或 y 值。
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
(另外:我使用 // 而不是 ' 作为注释...只是为了此处的可读性。如果您复制到 Visual Studio,则必须修复它)
在 vb.net 中,您同时拥有 Graphics.DrawEllipse 和 Graphics.DrawArc。在 Lua 中,您也许可以使用 Cairo,我知道它具有 arc 函数。
如果您在 .Net 中的 GraphicsPath 中的何处制作椭圆以及在何处对其存储在内存中的方式进行逆向工程,您会发现它存储为四个贝塞尔曲线。我曾经在 vb.net 中实现了自己的图形库,我就是这样做的。我在 Actionscript 中实现时找到的最好的资源,不幸的是我无法找到我正在谈论的那个图形库。
TLDR;你应该看看贝塞尔曲线。
一个完全不同且非常简单的实现,尽管椭圆看起来不像其他算法 "pretty";这只是使用椭圆的数学定义,循环 x 计算给定 x、w 和 h 的 y 坐标。
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
好的,我设法通过检查找到了填充椭圆的解决方案
如果下半部分的像素将绘制在椭圆前半部分的 x 范围内。
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
我正在尝试在基于 Lua 或 VB 的代码中创建一个函数来 绘制/绘制椭圆和实心椭圆。
我对这个数学知识不多,需要一些帮助。
我 googled google 关于用代码绘制椭圆的所有内容,但我找不到一个很好的简单工作示例,我可以将其编码到我的 Lua / VB代码。
这是我访问过的一些网站,但无法使代码正常工作或无法将代码正确转换为 Lua 或 VB...
http://groups.csail.mit.edu/graphics/classes/6.837/F98/Lecture6/circle.html
http://www.sourcecodesworld.com/source/show.asp?ScriptID=112
How do I draw an ellipse with arbitrary orientation pixel by pixel?
谁能帮我写出可以画椭圆和实心椭圆的代码?
这是我尝试从此处转换为 Lua 的一些代码:
https://gist.github.com/Wollw/3291916
这段代码有一些问题(缺少像素),我认为它没有正确转换,但我不知道该怎么做。
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
[编辑:]
我差一点就得到了! 查看下面这段代码中的注释以了解问题所在...
我使用 EGSL 测试此 Lua 代码: 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()
以下 VB 适合我,基于提供的第一个 link;我这里的代码和你的代码之间的唯一区别 link 是我移动了 xc 和 yc,因为位图中的像素不能有负 x 或 y 值。
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
(另外:我使用 // 而不是 ' 作为注释...只是为了此处的可读性。如果您复制到 Visual Studio,则必须修复它)
在 vb.net 中,您同时拥有 Graphics.DrawEllipse 和 Graphics.DrawArc。在 Lua 中,您也许可以使用 Cairo,我知道它具有 arc 函数。
如果您在 .Net 中的 GraphicsPath 中的何处制作椭圆以及在何处对其存储在内存中的方式进行逆向工程,您会发现它存储为四个贝塞尔曲线。我曾经在 vb.net 中实现了自己的图形库,我就是这样做的。我在 Actionscript 中实现时找到的最好的资源,不幸的是我无法找到我正在谈论的那个图形库。
TLDR;你应该看看贝塞尔曲线。
一个完全不同且非常简单的实现,尽管椭圆看起来不像其他算法 "pretty";这只是使用椭圆的数学定义,循环 x 计算给定 x、w 和 h 的 y 坐标。
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
好的,我设法通过检查找到了填充椭圆的解决方案 如果下半部分的像素将绘制在椭圆前半部分的 x 范围内。
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