带有三次插值的正弦图像失真

Posted 2023-02-22

【中文标题】带有三次插值的正弦图像失真

【英文标题】：Image distortion with sine with cubic interpolation

【发布时间】：2015-03-30 11:44:32

【问题描述】：

我正在尝试使用下一个代码的三次插值在 y 轴上使用正弦/余弦进行失真，我将新图像的大小与旧图像保持相同。而且结果看起来不太好。我只在 y 轴上进行插值。我根据Math 进行了计算

Mat cubic(Mat old, int freq)
int height = old.size().height;
int width = old.size().width;
int p0,p1,p2,p3;
int Bp0,Bp1,Bp2,Bp3;
int Gp0,Gp1,Gp2,Gp3;
int Rp0,Rp1,Rp2,Rp3;
unsigned char B,G,R;

double Ba,Bb,Bc,Bd;
double Ga,Gb,Gc,Gd;
double Ra,Rb,Rc,Rd;
double fi,fj;
Mat res(height,width,CV_8UC3,cv::Scalar(100));
for (int i = 0; i < width; i++) 
    for (int j = 0; j < height; j++) 
        fi = i;
        fj = (j - (((height/30)*(sin((M_PI*i)/(freq/2)))))) + sqrt(height) ;
        //Get the indexes of the for neighbors 
         if(fj > 20 &  fj < height)
        p1 = floor(fj);
        p2 = ceil(fj);
        p0 = p1-1;
        p3 = p2+1;
        //get the color values
        Vec3b p0Color = old.at<Vec3b>(Point(floor(fi),p0));
        Vec3b p1Color = old.at<Vec3b>(Point(floor(fi),p1)); 
        Vec3b p2Color = old.at<Vec3b>(Point(floor(fi),p2)); 
        Vec3b p3Color = old.at<Vec3b>(Point(floor(fi),p3));
        //cout <<" fj:" <<fj<< " p0: " << p0 <<" p1: "<< p1 << " p2: "<<p2 <<" p3:" <<p3;

        Bp0 = p0Color.val[0]; Bp1 = p1Color.val[0]; Bp2 = p2Color.val[0]; Bp3 = p3Color.val[0]; 
        Ba = -0.5*Bp0 + 1.5*Bp1 - 1.5*Bp2 + 0.5*Bp3;
        Bb = Bp0 - 2.5*Bp1 + 2*Bp2 -0.5*Bp3;
        Bc = -0.5*Bp0 + 0.5*Bp2;
        Bd = Bp1;

        Gp0 = p0Color.val[1]; Gp1 = p1Color.val[1]; Gp2 = p2Color.val[1]; Gp3 = p3Color.val[1]; 
        Ga = -0.5*Gp0 + 1.5*Gp1 - 1.5*Gp2 + 0.5*Gp3;
        Gb = Gp0 - 2.5*Gp1 + 2*Gp2 -0.5*Gp3;
        Gc = -0.5*Gp0 + 0.5*Gp2;
        Gd = Gp1;

        Rp0 = p0Color.val[2]; Rp1 = p1Color.val[2]; Rp2 = p2Color.val[2]; Rp3 = p3Color.val[2]; 
        Ra = -0.5*Rp0 + 1.5*Rp1 - 1.5*Rp2 + 0.5*Rp3;
        Rb = Rp0 - 2.5*Rp1 + 2*Rp2 -0.5*Rp3;
        Rc = -0.5*Rp0 + 0.5*Rp2;
        Rd = Rp1;

        B = floor(Ba*pow(fj,3) + Bb*pow(fj,2) + Bc*fj + Bd);
        G = floor(Ga*pow(fj,3) + Gb*pow(fj,2) + Gc*fj + Gd);
        R = floor(Ra*pow(fj,3) + Rb*pow(fj,2) + Rc*fj + Rd);
        if(B==0)
            cout <<"list:" << Ba <<" " << Bb <<" "<< Bc <<" " << Bd;
        
        Vec3b finalColor(B,G,R);
        res.at<Vec3b>(j,i) = finalColor;
       
    

namedWindow( "cubic", CV_WINDOW_AUTOSIZE );
imshow( "cubic", res); 
return res;

出处照片

双线性插值

三次插值

谢谢

【问题讨论】：

可能是整数溢出？

确实在段的边缘存在整数溢出。不知道怎么解决。

也许可以通过某种因素改变量化范围？您可以在计算之前将每个像素除以 2（或某个常数），然后再将它们全部乘以 2。

【参考方案1】：

问题解决了。

Mat cubic(Mat old, int freq)
int height = old.size().height;
int width = old.size().width;
int p0,p1,p2,p3;
int Bp0,Bp1,Bp2,Bp3;
int Gp0,Gp1,Gp2,Gp3;
int Rp0,Rp1,Rp2,Rp3;
double B,G,R;

double fi,fj;
Mat res(height,width,CV_8UC3,cv::Scalar(100));
for (int i = 0; i < width; i++) 
    for (int j = 0; j < height; j++) 
        fi = i;
        fj = (j - (((height/30)*(sin((M_PI*i)/(freq/2)))))) + sqrt(height) ;
        //Get the indexes of the for neighbors 
         if(fj > 20 &  fj < height)
        p1 = floor(fj);
        p2 = ceil(fj);
        p0 = p1-1;
        p3 = p2+1;
        //get the color values
        Vec3b p0Color = old.at<Vec3b>(Point(floor(fi),p0));
        Vec3b p1Color = old.at<Vec3b>(Point(floor(fi),p1)); 
        Vec3b p2Color = old.at<Vec3b>(Point(floor(fi),p2)); 
        Vec3b p3Color = old.at<Vec3b>(Point(floor(fi),p3));
        if(i == 258 && j==59)
        
        Bp0 = p0Color.val[0]; Bp1 = p1Color.val[0]; Bp2 = p2Color.val[0]; Bp3 = p3Color.val[0]; 
        Gp0 = p0Color.val[1]; Gp1 = p1Color.val[1]; Gp2 = p2Color.val[1]; Gp3 = p3Color.val[1]; 
        Rp0 = p0Color.val[2]; Rp1 = p1Color.val[2]; Rp2 = p2Color.val[2]; Rp3 = p3Color.val[2];

       //Normalize the fj
       fj = (fj - floor(fj));

        B= Bp1 + 0.5 * fj*(Bp2 - Bp0 + fj*(2.0*Bp0 - 5.0*Bp1 + 4.0*Bp2 - Bp3 + fj*(3.0*(Bp1 - Bp2) + Bp3 - Bp0)));
        G= Gp1 + 0.5 * fj*(Gp2 - Gp0 + fj*(2.0*Gp0 - 5.0*Gp1 + 4.0*Gp2 - Gp3 + fj*(3.0*(Gp1 - Gp2) + Gp3 - Gp0)));   
        R= Rp1 + 0.5 * fj*(Rp2 - Rp0 + fj*(2.0*Rp0 - 5.0*Rp1 + 4.0*Rp2 - Rp3 + fj*(3.0*(Rp1 - Rp2) + Rp3 - Rp0)));

        if(B > 255)
            B=255;
        if(G>255)
            G=255;
        if(R>255)
            R=255;
        if(B < 0)
            B=0;
        if(G<0)
            G=0;
        if(R<0)
            R=0;
        Vec3b finalColor(B,G,R);
        res.at<Vec3b>(j,i) = finalColor;
       
    

namedWindow( "cubic", CV_WINDOW_AUTOSIZE );
imshow( "cubic", res); 
return res;

我忘记在将 fj 放入最终三次函数之前对其进行归一化

新结果：