java - 如何将 2d 阵列旋转小于 90°，以获得最佳近似值？

Question

假设我有一个以 0° 旋转存储的数组:

0 0 1 0 0
0 0 1 0 0 
1 1 1 0 0 
0 0 0 0 0
0 0 0 0 0 

如果我通过，我希望它以一个很好的近似值返回，例如 30° 作为参数，它会是这样的:

0 0 0 1 0
1 1 0 1 0
0 0 1 0 0 
0 0 0 0 0 
0 0 0 0 0 

45°会是

1 0 0 0 1
0 1 0 1 0
0 0 1 0 0 
0 0 0 0 0 
0 0 0 0 0

I am aware of the solutions posted for 90° rotations.但我认为这对我没有帮助？

我没有任何示例代码，因为我目前甚至不知道从哪里开始寻找。如果有任何关键字我可以在谷歌上指出我可以适应的一些公式的方向，那也很好。

Spectre的C#代码解决方案:

    class Rotation
{

    public Rotation() {
        A = new int[xs,ys]{
{0,0,0,9,0,0,0},
{0,0,0,9,0,0,0},
{0,0,0,9,0,0,0},
{9,9,9,9,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
{0,0,0,0,0,0,0},
};
        B = new int[xs, ys];

        deg = (float)(Math.PI / 180.0);
    }

    public const int xs = 7; // matrix size
    public const int ys = 7;
    const int x0 = 3; // rotation center cell
    const int y0 = 3;
    readonly float deg; 
    public int[,] A;
    public int[,] B;

    //---------------------------------------------------------------------------

    public void rotcv(float ang) {
        rotcw(Rotation.x0, Rotation.y0, ang);
    }
    private void rotcw(int x0, int y0, float ang) // rotate A -> B by angle ang around (x0,y0) CW if ang>0
    {
        int x, y, ix0, iy0, ix1, iy1, q;
        double xx, yy, fx, fy, c, s;
        // circle kernel
        c = Math.Cos(-ang); s = Math.Sin(-ang);
        // rotate
        for (y = 0; y < ys; y++)
            for (x = 0; x < xs; x++)
            {
                // offset so (0,0) is center of rotation
                xx = x - x0;
                yy = y - y0;
                // rotate (fx,fy) by ang
                fx = ((xx * c) - (yy * s));
                fy = ((xx * s) + (yy * c));
                // offset back and convert to ints and weights
                fx += x0; ix0 = (int)Math.Floor(fx); fx -= ix0; ix1 = ix0 + 1; if (ix1 >= xs) ix1 = ix0;
                fy += y0; iy0 = (int)Math.Floor (fy); fy -= iy0; iy1 = iy0 + 1; if (iy1 >= ys) iy1 = iy0;
                // bilinear interpolation A[fx][fy] -> B[x][y]
                if ((ix0 >= 0) && (ix0 < xs) && (iy0 >= 0) && (iy0 < ys))
                {
                    xx = (A[ix0,iy0]) + ((A[ix1,iy0] - A[ix0,iy0]) * fx);
                    yy = (A[ix0,iy0]) + ((A[ix1,iy0] - A[ix0,iy0]) * fx);
                    xx = xx + ((yy - xx) * fy); q =(int) xx;
                }
                else q = 0;
                B[x,y] = q;
            }
    }
}

测试:

 static void Main(string[] args)
    {
        Rotation rot = new Rotation();

        for (int x = 0; x < Rotation.xs; x++) {
            for (int y = 0; y < Rotation.xs; y++) {
                Console.Write(rot.A[x,y] + " ");
            }
            Console.WriteLine();
        }
        Console.WriteLine();
        float rotAngle = 0;
        while (true)
        {
            rotAngle += (float)(Math.PI/180f)*90;
            rot.rotcv(rotAngle);
            for (int x = 0; x < Rotation.xs; x++)
            {
                for (int y = 0; y < Rotation.xs; y++)
                {
                    Console.Write(rot.B[x, y] + " ");
                }
                Console.WriteLine();
            }
            Console.WriteLine();
            Console.ReadLine();
        }

    }

Answer 1

好的，这是我们 promise 的。第一个 C++ 代码:

//---------------------------------------------------------------------------
#include <math.h>
//---------------------------------------------------------------------------
const int xs=7; // matrix size
const int ys=7;
const int x0=3; // rotation center cell
const int y0=3;
const float deg=M_PI/180.0;
int A[xs][ys]=
    {
    {0,0,0,9,0,0,0},
    {0,0,0,9,0,0,0},
    {0,0,0,9,0,0,0},
    {9,9,9,9,0,0,0},
    {0,0,0,0,0,0,0},
    {0,0,0,0,0,0,0},
    {0,0,0,0,0,0,0},
    };
int B[xs][ys];
//---------------------------------------------------------------------------
void rotcw(int x0,int y0,float ang) // rotate A -> B by angle ang around (x0,y0) CW if ang>0
    {
    int x,y,ix0,iy0,ix1,iy1,q;
    float xx,yy,fx,fy,c,s;
    // circle kernel
    c=cos(-ang); s=sin(-ang);
    // rotate
    for (y=0;y<ys;y++)
     for (x=0;x<xs;x++)
        {
        // offset so (0,0) is center of rotation
        xx=x-x0;
        yy=y-y0;
        // rotate (fx,fy) by ang
        fx=float((xx*c)-(yy*s));
        fy=float((xx*s)+(yy*c));
        // offset back and convert to ints and weights
        fx+=x0; ix0=floor(fx); fx-=ix0; ix1=ix0+1; if (ix1>=xs) ix1=ix0;
        fy+=y0; iy0=floor(fy); fy-=iy0; iy1=iy0+1; if (iy1>=ys) iy1=iy0;
        // bilinear interpolation A[ix0+fx][iy0+fy] -> B[x][y]
        if ((ix0>=0)&&(ix0<xs)&&(iy0>=0)&&(iy0<ys))
            {
            xx=float(A[ix0][iy0])+(float(A[ix1][iy0]-A[ix0][iy0])*fx);
            yy=float(A[ix0][iy0])+(float(A[ix1][iy0]-A[ix0][iy0])*fx);
            xx=xx+((yy-xx)*fy); q=xx;
            } else q=0;
        B[x][y]=q;
        }
    }
//---------------------------------------------------------------------------

此处 7x7 预览 15 度步长:

可能需要将中心稍微调整一半的单元格或其他东西(我喜欢中心流血太多)

矩阵 A 是源，B 是目标 ...

您还可以添加阈值...例如:

if (q>=5) q=9; else q=0;