public OverlappingModel(Bitmap bitmap, int N, int width, int height, bool periodicInput, bool periodicOutput, int symmetry, int ground)
{
this.N = N;
FMX = width;
FMY = height;
periodic = periodicOutput;
int SMX = bitmap.Width, SMY = bitmap.Height;
byte[,] sample = new byte[SMX, SMY];
colors = new List <Color>();
for (int y = 0; y < SMY; y++)
{
for (int x = 0; x < SMX; x++)
{
Color color = bitmap.GetPixel(x, y);
int i = 0;
foreach (var c in colors)
{
if (c == color)
{
break;
}
i++;
}
if (i == colors.Count)
{
colors.Add(color);
}
sample[x, y] = (byte)i;
}
}
int C = colors.Count;
long W = Stuff.Power(C, N * N);
Func <Func <int, int, byte>, byte[]> pattern = f => {
byte[] result = new byte[N * N];
for (int y = 0; y < N; y++)
{
for (int x = 0; x < N; x++)
{
result[x + y * N] = f(x, y);
}
}
return(result);
};
Func <int, int, byte[]> patternFromSample = (x, y) => pattern((dx, dy) => sample[(x + dx) % SMX, (y + dy) % SMY]);
Func <byte[], byte[]> rotate = p => pattern((x, y) => p[N - 1 - y + x * N]);
Func <byte[], byte[]> reflect = p => pattern((x, y) => p[N - 1 - x + y * N]);
Func <byte[], long> index = p => {
long result = 0, power = 1;
for (int i = 0; i < p.Length; i++)
{
result += p[p.Length - 1 - i] * power;
power *= C;
}
return(result);
};
Func <long, byte[]> patternFromIndex = ind => {
long residue = ind, power = W;
byte[] result = new byte[N * N];
for (int i = 0; i < result.Length; i++)
{
power /= C;
int count = 0;
while (residue >= power)
{
residue -= power;
count++;
}
result[i] = (byte)count;
}
return(result);
};
Dictionary <long, int> weights = new Dictionary <long, int>();
List <long> ordering = new List <long>();
for (int y = 0; y < (periodicInput ? SMY : SMY - N + 1); y++)
{
for (int x = 0; x < (periodicInput ? SMX : SMX - N + 1); x++)
{
byte[][] ps = new byte[8][];
ps[0] = patternFromSample(x, y);
ps[1] = reflect(ps[0]);
ps[2] = rotate(ps[0]);
ps[3] = reflect(ps[2]);
ps[4] = rotate(ps[2]);
ps[5] = reflect(ps[4]);
ps[6] = rotate(ps[4]);
ps[7] = reflect(ps[6]);
for (int k = 0; k < symmetry; k++)
{
long ind = index(ps[k]);
if (weights.ContainsKey(ind))
{
weights[ind]++;
}
else
{
weights.Add(ind, 1);
ordering.Add(ind);
}
}
}
}
T = weights.Count;
this.ground = (ground + T) % T;
patterns = new byte[T][];
stationary = new double[T];
propagator = new int[2 * N - 1][][][];
int counter = 0;
foreach (long w in ordering)
{
patterns[counter] = patternFromIndex(w);
stationary[counter] = weights[w];
counter++;
}
wave = new bool[FMX][][];
changes = new bool[FMX][];
for (int x = 0; x < FMX; x++)
{
wave[x] = new bool[FMY][];
changes[x] = new bool[FMY];
for (int y = 0; y < FMY; y++)
{
wave[x][y] = new bool[T];
changes[x][y] = false;
for (int t = 0; t < T; t++)
{
wave[x][y][t] = true;
}
}
}
Func <byte[], byte[], int, int, bool> agrees = (p1, p2, dx, dy) => {
int xmin = dx < 0 ? 0 : dx, xmax = dx < 0 ? dx + N : N, ymin = dy < 0 ? 0 : dy, ymax = dy < 0 ? dy + N : N;
for (int y = ymin; y < ymax; y++)
{
for (int x = xmin; x < xmax; x++)
{
if (p1[x + N * y] != p2[x - dx + N * (y - dy)])
{
return(false);
}
}
}
return(true);
};
for (int x = 0; x < 2 * N - 1; x++)
{
propagator[x] = new int[2 * N - 1][][];
for (int y = 0; y < 2 * N - 1; y++)
{
propagator[x][y] = new int[T][];
for (int t = 0; t < T; t++)
{
List <int> list = new List <int>();
for (int t2 = 0; t2 < T; t2++)
{
if (agrees(patterns[t], patterns[t2], x - N + 1, y - N + 1))
{
list.Add(t2);
}
}
propagator[x][y][t] = new int[list.Count];
for (int c = 0; c < list.Count; c++)
{
propagator[x][y][t][c] = list[c];
}
}
}
}
}
public OverlappingModel(string name, int N, int width, int height, bool periodicInput, bool periodicOutput, int symmetry, int ground)
: base(width, height)
{
this.N = N;
periodic = periodicOutput;
string resPath = $"samples/{name}";
Texture2D texture = Resources.Load <Texture2D>(resPath);
int SMX = texture.width;
int SMY = texture.height;
byte[,] sample = new byte[SMX, SMY];
colors = new List <Color>();
for (int y = 0; y < SMY; y++)
{
for (int x = 0; x < SMX; x++)
{
Color color = texture.GetPixel(x, y);
int i = 0;
foreach (var c in colors)
{
if (c == color)//为减少colors长度的操作
{
break;
}
i++;
}
if (i == colors.Count)//i==colors.Count证明遍历完colors还没有color，所以add
{
colors.Add(color);
}
sample[x, y] = (byte)i;//此xy记录color在colors位置
}
}
int C = colors.Count;
long W = Stuff.Power(C, N * N);
/// <summary>
/// 根据遍历 byte[N*N] 提供的 x，y 参数代入 f 从中获取 byte 填充。
/// </summary>
/// <returns> N*N长度的byte[] </returns>
/// <param name="f"> byte取样函数 </param>
byte[] pattern(Func <int, int, byte> f)
{
byte[] result = new byte[N * N];
for (int y = 0; y < N; y++)
{
for (int x = 0; x < N; x++)
{
result[x + y * N] = f(x, y);
}
}
return(result);
};
/// <summary>
/// 以xy为起点取长宽为N*N的色块，以xy为起点，dx，dy为宽高。
/// </summary>
byte[] patternFromSample(int x, int y) => pattern((dx, dy) => sample[(x + dx) % SMX, (y + dy) % SMY]);
byte[] rotate(byte[] p) => pattern((x, y) => p[N - 1 - y + x * N]); //向右转
byte[] reflect(byte[] p) => pattern((x, y) => p[N - 1 - x + y * N]); //x左右翻转
long index(byte[] p)
{
long result = 0;
long power = 1;
for (int i = 0; i < p.Length; i++)
{
result += p[p.Length - 1 - i] * power;//乘以power次方，防止index有冲突。
power *= C;
}
return(result);
};
byte[] patternFromIndex(long ind)
{
long residue = ind;
long power = W;
byte[] result = new byte[N * N];
for (int i = 0; i < result.Length; i++)
{
power /= C;
int count = 0;
while (residue >= power)
{
residue -= power;
count++;
}
result[i] = (byte)count;
}
return(result);
};
Dictionary <long, int> weights = new Dictionary <long, int>();
List <long> ordering = new List <long>();
//periodicInput含义，如果periodicInput周期性输入为true，那么以xy为起点的N*N色块取值到xy最，到尽头可以从原料图片colors起始（0，0）取。
//反之，不是周期性，那么xy取最大值预留最大为max(xy)-N，因为非周期性N*N色块的xy>SMYSMX时不能从原料图片colors起始（0，0）取。
for (int y = 0; y < (periodicInput ? SMY : SMY - N + 1); y++)
{
for (int x = 0; x < (periodicInput ? SMX : SMX - N + 1); x++)
{
//ps:8个N*N大小的索引自sample色块
byte[][] ps = new byte[8][];
//逐渐顺时针旋转90度的四个方向 0，2，4，6
//逐渐对应的反射四个方向 1，3，5，7
ps[0] = patternFromSample(x, y);
ps[1] = reflect(ps[0]);
ps[2] = rotate(ps[0]);
ps[3] = reflect(ps[2]);
ps[4] = rotate(ps[2]);
ps[5] = reflect(ps[4]);
ps[6] = rotate(ps[4]);
ps[7] = reflect(ps[6]);
//symmetry默认值8
for (int k = 0; k < symmetry; k++)//对称
{
long ind = index(ps[k]);
if (weights.ContainsKey(ind))
{
weights[ind]++;
}
else
{
weights.Add(ind, 1);
ordering.Add(ind);
}
}
}
}
T = weights.Count;
this.ground = (ground + T) % T;
patterns = new byte[T][];
base.weights = new double[T];
int counter = 0;
foreach (long w in ordering)
{
patterns[counter] = patternFromIndex(w);
base.weights[counter] = weights[w];
counter++;
}
/// <summary>
/// dx，dy当前p1周围其他色块坐标，判断p1和p2色块交界处像素是否一致
/// </summary>
bool agrees(byte[] p1, byte[] p2, int dx, int dy)
{
int xmin = dx < 0 ? 0 : dx;
int xmax = dx < 0 ? dx + N : N;
int ymin = dy < 0 ? 0 : dy;
int ymax = dy < 0 ? dy + N : N;
for (int y = ymin; y < ymax; y++)
{
for (int x = xmin; x < xmax; x++)
{
if (p1[x + N * y] != p2[x - dx + N * (y - dy)])
{
return(false);
}
}
}
return(true);
};
propagator = new int[4][][];
for (int d = 0; d < 4; d++)
{
propagator[d] = new int[T][];
for (int t = 0; t < T; t++)
{
List <int> list = new List <int>();
for (int t2 = 0; t2 < T; t2++)
{
if (agrees(patterns[t], patterns[t2], DX[d], DY[d]))//对d(4)个方向进行匹配，如果相邻tile的边边color一(agrees)那么说明t2添加进可匹配集合。
{
list.Add(t2);
}
}
//propagator记录T的4个方向相邻匹配的tile色块集合。
propagator[d][t] = new int[list.Count];
for (int c = 0; c < list.Count; c++)
{
propagator[d][t][c] = list[c];
}
}
}
}