我正在为了学习而编写自己的中点位移算法，我决定按照自己的方式实现它，以查看我是否能够理解该算法并且是否可以修改为我所喜欢的方式。
下面是生成分形的代码：
public void Generate(Double rg, int size)
{
Random rand = new Random();
int min = -127;
int max = 128;
// Starting points of the rectangle
MDP_Point s1 = new MDP_Point(0, 0, rand.Next(min, max));
MDP_Point s2 = new MDP_Point(size, 0, rand.Next(min, max));
MDP_Point s3 = new MDP_Point(size, size, rand.Next(min, max));
MDP_Point s4 = new MDP_Point(0, size, rand.Next(min, max));
// Lists containing the rectangles
List<MDP_Rect> newRect = new List<MDP_Rect>(); // Newly created rectangles
List<MDP_Rect> oldRect = new List<MDP_Rect>(); // Rectangles being divided
// Starting rectangle is added to the list
oldRect.Add(new MDP_Rect(s1, s2, s3, s4));
// Distance between 2 points in a rectangle
int h = size;
while (h > 1)
{
foreach (MDP_Rect r in oldRect)
{
// Middle points of rectangle segments
MDP_Point m1 = new MDP_Point();
MDP_Point m2 = new MDP_Point();
MDP_Point m3 = new MDP_Point();
MDP_Point m4 = new MDP_Point();
// Middle point of rectangle
MDP_Point mm = new MDP_Point();
m1.x = (r.C1.x + r.C2.x) / 2;
m1.y = (r.C1.y + r.C2.y) / 2;
m1.z = ((r.C1.z + r.C2.z) / 2) +(rand.Next(min, max) * rg);
m2.x = (r.C2.x + r.C3.x) / 2;
m2.y = (r.C2.y + r.C3.y) / 2;
m2.z = ((r.C2.z + r.C3.z) / 2) +(rand.Next(min, max) * rg);
m3.x = (r.C3.x + r.C4.x) / 2;
m3.y = (r.C3.y + r.C4.y) / 2;
m3.z = ((r.C3.z + r.C4.z) / 2) +(rand.Next(min, max) * rg);
m4.x = (r.C1.x + r.C4.x) / 2;
m4.y = (r.C1.y + r.C4.y) / 2;
m4.z = ((r.C1.z + r.C4.z) / 2) + (rand.Next(min, max) * rg);
mm.x = (r.C1.x + r.C2.x + r.C3.x + r.C4.x) / 4;
mm.y = (r.C1.y + r.C2.y + r.C3.y + r.C4.y) / 4;
mm.z = ((r.C1.z + r.C2.z + r.C3.z + r.C4.z) / 4) + (rand.Next(min, max) * rg);
newRect.Add(new MDP_Rect(r.C1, m1, mm, m4));
newRect.Add(new MDP_Rect(m1, r.C2, m2, mm));
newRect.Add(new MDP_Rect(mm, m2, r.C3, m3));
newRect.Add(new MDP_Rect(m4, mm, m3, r.C4));
}
oldRect.Clear();
oldRect = new List<MDP_Rect>(newRect);
newRect.Clear();
h /= 2;
}
List<MDP_Rect> sorted = new List<MDP_Rect>();
sorted = oldRect.OrderBy(y => y.C1.y).ThenBy(x => x.C1.x).ToList();
List<MDP_Point> mapArray = new List<MDP_Point>();
mapArray.AddRange(CreateArray(sorted));
CreateImage(size, mapArray, rg);
}
MDP_Point仅包含x、y和z值。
MDP_Rectangle包含4个点，创建一个矩形。
CreateArray()方法仅接受有序的矩形列表，并输出正确顺序的点列表以创建图像。
CreateArray():
private List<MDP_Point> CreateArray(List<MDP_Rect> lRect)
{
List<MDP_Point> p = new List<MDP_Point>();
int size = (int)Math.Sqrt(lRect.Count);
int i = 0;
foreach (MDP_Rect r in lRect)
{
p.Add(new MDP_Point((int)r.C1.x, (int)r.C1.y, (int)r.C1.z));
if (i > 0 && i % size == size - 1)
{
p.Add(new MDP_Point((int)r.C2.x, (int)r.C2.y, (int)r.C2.z));
}
i++;
}
for (int a = 0; a < size; a++)
{
p.Add(new MDP_Point((int)lRect[(size * size - size) + a].C4.x,
(int)lRect[(size * size - size) + a].C4.y,
(int)lRect[(size * size - size) + a].C4.z));
if (a > 0 && a % size == size - 1)
{
p.Add(new MDP_Point((int)lRect[(size * size - size) + a].C3.x,
(int)lRect[(size * size - size) + a].C3.y,
(int)lRect[(size * size - size) + a].C3.z));
}
}
return p;
}
这是创建图片的方法：
private void CreateImage(int size, List<MDP_Point> arr, double roughness)
{
Bitmap map = new Bitmap(size, size);
int ver = 0;
for (int i = 0; i < map.Height; i++)
{
for (int n = 0; n < map.Width; n++ )
{
int h = (int)arr[ver].z + 127;
if (h < 0)
{
h = 0;
}
else if (h > 255)
{
h = 255 ;
}
Color c = Color.FromArgb(h, h, h);
//map.SetPixel(n, i, c);
map.SetPixel(i, n, c);
ver++;
}
}
Bitmap m = new Bitmap(map);
bool saved = true;
int num = 0;
while (saved)
{
if (File.Exists("map_S" + size + "_R" + roughness + "_" + num + ".png"))
{
num++;
}
else
{
m.Save("map_S" + size + "_R" + roughness + "_" + num + ".png", System.Drawing.Imaging.ImageFormat.Png);
saved = false;
}
}
map.Dispose();
m.Dispose();
}
任何低于0的值都被设置为0，任何高于255的值都被设置为255，这可能是一个大问题...不确定该怎么办。
以下是代码生成的图像： 大小：1024 粗糙度：0.5
最明显的问题是对角线“脊线”和平铺的外观。此时，我不确定如何修复它以使其看起来更自然。 有什么想法吗？
当我使用时：
rand调用将从均匀分布中抽取（在0到1之间均匀分布）。请参见 https://dev59.com/YHVD5IYBdhLWcg3wI3-L 和 http://stackoverflow.com/questions/8739851/drawing-random-number-from-a-standard-normal-distribution-using-the-standard-ran - sfstewman