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  • 【OpenCV】SIFT原理与源码分析:关键点搜索与定位

    《SIFT原理与源码分析》系列文章索引:http://www.cnblogs.com/tianyalu/p/5467813.html

    由前一步《DoG尺度空间构造》,我们得到了DoG高斯差分金字塔:

    如上图的金字塔,高斯尺度空间金字塔中每组有五层不同尺度图像,相邻两层相减得到四层DoG结果。关键点搜索就在这四层DoG图像上寻找局部极值点。

    DoG局部极值点

    寻找DoG极值点时,每一个像素点和它所有的相邻点比较,当其大于(或小于)它的图像域和尺度域的所有相邻点时,即为极值点。如下图所示,比较的范围是个3×3的立方体:中间的检测点和它同尺度的8个相邻点,以及和上下相邻尺度对应的9×2个点——共26个点比较,以确保在尺度空间和二维图像空间都检测到极值点。 

    在一组中,搜索从每组的第二层开始,以第二层为当前层,第一层和第三层分别作为立方体的的上下层;搜索完成后再以第三层为当前层做同样的搜索。所以每层的点搜索两次。通常我们将组Octaves索引以-1开始,则在比较时牺牲了-1组的第0层和第N组的最高层

    高斯金字塔,DoG图像及极值计算的相互关系如上图所示。

    关键点精确定位

    以上极值点的搜索是在离散空间进行搜索的,由下图可以看到,在离散空间找到的极值点不一定是真正意义上的极值点。可以通过对尺度空间DoG函数进行曲线拟合寻找极值点来减小这种误差。
    利用DoG函数在尺度空间的Taylor展开式:
    则极值点为:
    程序中还除去了极值小于0.04的点。如下所示:
    // Detects features at extrema in DoG scale space.  Bad features are discarded
    // based on contrast and ratio of principal curvatures.
    // 在DoG尺度空间寻特征点(极值点)
    void SIFT::findScaleSpaceExtrema( const vector<Mat>& gauss_pyr, const vector<Mat>& dog_pyr,
                                      vector<KeyPoint>& keypoints ) const
    {
        int nOctaves = (int)gauss_pyr.size()/(nOctaveLayers + 3);
        
        // The contrast threshold used to filter out weak features in semi-uniform
        // (low-contrast) regions. The larger the threshold, the less features are produced by the detector.
        // 过滤掉弱特征的阈值 contrastThreshold默认为0.04
        int threshold = cvFloor(0.5 * contrastThreshold / nOctaveLayers * 255 * SIFT_FIXPT_SCALE);
        const int n = SIFT_ORI_HIST_BINS; //36
        float hist[n];
        KeyPoint kpt;
    
        keypoints.clear();
    
        for( int o = 0; o < nOctaves; o++ )
            for( int i = 1; i <= nOctaveLayers; i++ )
            {
                int idx = o*(nOctaveLayers+2)+i;
                const Mat& img = dog_pyr[idx];
                const Mat& prev = dog_pyr[idx-1];
                const Mat& next = dog_pyr[idx+1];
                int step = (int)img.step1();
                int rows = img.rows, cols = img.cols;
    
                for( int r = SIFT_IMG_BORDER; r < rows-SIFT_IMG_BORDER; r++)
                {
                    const short* currptr = img.ptr<short>(r);
                    const short* prevptr = prev.ptr<short>(r);
                    const short* nextptr = next.ptr<short>(r);
    
                    for( int c = SIFT_IMG_BORDER; c < cols-SIFT_IMG_BORDER; c++)
                    {
                        int val = currptr[c];
    
                        // find local extrema with pixel accuracy
                        // 寻找局部极值点,DoG中每个点与其所在的立方体周围的26个点比较
                        // if (val比所有都大 或者 val比所有都小)
                        if( std::abs(val) > threshold &&
                           ((val > 0 && val >= currptr[c-1] && val >= currptr[c+1] &&
                             val >= currptr[c-step-1] && val >= currptr[c-step] && 
                             val >= currptr[c-step+1] && val >= currptr[c+step-1] && 
                             val >= currptr[c+step] && val >= currptr[c+step+1] &&
                             val >= nextptr[c] && val >= nextptr[c-1] && 
                             val >= nextptr[c+1] && val >= nextptr[c-step-1] && 
                             val >= nextptr[c-step] && val >= nextptr[c-step+1] && 
                             val >= nextptr[c+step-1] && val >= nextptr[c+step] && 
                             val >= nextptr[c+step+1] && val >= prevptr[c] && 
                             val >= prevptr[c-1] && val >= prevptr[c+1] &&
                             val >= prevptr[c-step-1] && val >= prevptr[c-step] && 
                             val >= prevptr[c-step+1] && val >= prevptr[c+step-1] && 
                             val >= prevptr[c+step] && val >= prevptr[c+step+1]) ||
                            (val < 0 && val <= currptr[c-1] && val <= currptr[c+1] &&
                             val <= currptr[c-step-1] && val <= currptr[c-step] && 
                             val <= currptr[c-step+1] && val <= currptr[c+step-1] && 
                             val <= currptr[c+step] && val <= currptr[c+step+1] &&
                             val <= nextptr[c] && val <= nextptr[c-1] && 
                             val <= nextptr[c+1] && val <= nextptr[c-step-1] && 
                             val <= nextptr[c-step] && val <= nextptr[c-step+1] && 
                             val <= nextptr[c+step-1] && val <= nextptr[c+step] && 
                             val <= nextptr[c+step+1] && val <= prevptr[c] && 
                             val <= prevptr[c-1] && val <= prevptr[c+1] &&
                             val <= prevptr[c-step-1] && val <= prevptr[c-step] && 
                             val <= prevptr[c-step+1] && val <= prevptr[c+step-1] && 
                             val <= prevptr[c+step] && val <= prevptr[c+step+1])))
                        {
                            int r1 = r, c1 = c, layer = i;
                            
                            // 关键点精确定位
                            if( !adjustLocalExtrema(dog_pyr, kpt, o, layer, r1, c1,
                                                    nOctaveLayers, (float)contrastThreshold,
                                                    (float)edgeThreshold, (float)sigma) )
                                continue;
                            
                            float scl_octv = kpt.size*0.5f/(1 << o);
                            // 计算梯度直方图
                            float omax = calcOrientationHist(
                                gauss_pyr[o*(nOctaveLayers+3) + layer],
                                Point(c1, r1),
                                cvRound(SIFT_ORI_RADIUS * scl_octv),
                                SIFT_ORI_SIG_FCTR * scl_octv,
                                hist, n);
                            float mag_thr = (float)(omax * SIFT_ORI_PEAK_RATIO);
                            for( int j = 0; j < n; j++ )
                            {
                                int l = j > 0 ? j - 1 : n - 1;
                                int r2 = j < n-1 ? j + 1 : 0;
    
                                if( hist[j] > hist[l]  &&  hist[j] > hist[r2]  &&  hist[j] >= mag_thr )
                                {
                                    float bin = j + 0.5f * (hist[l]-hist[r2]) / 
                                    (hist[l] - 2*hist[j] + hist[r2]);
                                    bin = bin < 0 ? n + bin : bin >= n ? bin - n : bin;
                                    kpt.angle = (float)((360.f/n) * bin);
                                    keypoints.push_back(kpt);
                                }
                            }
                        }
                    }
                }
            }
    }

    删除边缘效应

    除了DoG响应较低的点,还有一些响应较强的点也不是稳定的特征点。DoG对图像中的边缘有较强的响应值,所以落在图像边缘的点也不是稳定的特征点。
    一个平坦的DoG响应峰值在横跨边缘的地方有较大的主曲率,而在垂直边缘的地方有较小的主曲率。主曲率可以通过2×2的Hessian矩阵H求出:
    D值可以通过求临近点差分得到。H的特征值与D的主曲率成正比,具体可参见Harris角点检测算法。
    为了避免求具体的值,我们可以通过H将特征值的比例表示出来。令为最大特征值,为最小特征值,那么:
    Tr(H)表示矩阵H的迹,Det(H)表示H的行列式。
    表示最大特征值与最小特征值的比值,则有:
    上式与两个特征值的比例有关。随着主曲率比值的增加,也会增加。我们只需要去掉比率大于一定值的特征点。Lowe论文中去掉r=10的点。
    // Interpolates a scale-space extremum's location and scale to subpixel
    // accuracy to form an image feature.  Rejects features with low contrast.
    // Based on Section 4 of Lowe's paper.
    // 特征点精确定位
    static bool adjustLocalExtrema( const vector<Mat>& dog_pyr, KeyPoint& kpt, int octv,
                                    int& layer, int& r, int& c, int nOctaveLayers,
                                    float contrastThreshold, float edgeThreshold, float sigma )
    {
        const float img_scale = 1.f/(255*SIFT_FIXPT_SCALE);
        const float deriv_scale = img_scale*0.5f;
        const float second_deriv_scale = img_scale;
        const float cross_deriv_scale = img_scale*0.25f;
    
        float xi=0, xr=0, xc=0, contr;
        int i = 0;
    
        //三维子像元插值
        for( ; i < SIFT_MAX_INTERP_STEPS; i++ )
        {
            int idx = octv*(nOctaveLayers+2) + layer;
            const Mat& img = dog_pyr[idx];
            const Mat& prev = dog_pyr[idx-1];
            const Mat& next = dog_pyr[idx+1];
    
            Vec3f dD((img.at<short>(r, c+1) - img.at<short>(r, c-1))*deriv_scale,
                     (img.at<short>(r+1, c) - img.at<short>(r-1, c))*deriv_scale,
                     (next.at<short>(r, c) - prev.at<short>(r, c))*deriv_scale);
    
            float v2 = (float)img.at<short>(r, c)*2;
            float dxx = (img.at<short>(r, c+1) + 
                    img.at<short>(r, c-1) - v2)*second_deriv_scale;
            float dyy = (img.at<short>(r+1, c) + 
                    img.at<short>(r-1, c) - v2)*second_deriv_scale;
            float dss = (next.at<short>(r, c) + 
                    prev.at<short>(r, c) - v2)*second_deriv_scale;
            float dxy = (img.at<short>(r+1, c+1) - 
                    img.at<short>(r+1, c-1) - img.at<short>(r-1, c+1) + 
                    img.at<short>(r-1, c-1))*cross_deriv_scale;
            float dxs = (next.at<short>(r, c+1) - 
                    next.at<short>(r, c-1) - prev.at<short>(r, c+1) + 
                    prev.at<short>(r, c-1))*cross_deriv_scale;
            float dys = (next.at<short>(r+1, c) - 
                    next.at<short>(r-1, c) - prev.at<short>(r+1, c) + 
                    prev.at<short>(r-1, c))*cross_deriv_scale;
    
            Matx33f H(dxx, dxy, dxs,
                      dxy, dyy, dys,
                      dxs, dys, dss);
    
            Vec3f X = H.solve(dD, DECOMP_LU);
    
            xi = -X[2];
            xr = -X[1];
            xc = -X[0];
    
            if( std::abs( xi ) < 0.5f  &&  std::abs( xr ) < 0.5f  &&  std::abs( xc ) < 0.5f )
                break;
    
            //将找到的极值点对应成像素(整数)
            c += cvRound( xc );
            r += cvRound( xr );
            layer += cvRound( xi );
    
            if( layer < 1 || layer > nOctaveLayers ||
               c < SIFT_IMG_BORDER || c >= img.cols - SIFT_IMG_BORDER  ||
               r < SIFT_IMG_BORDER || r >= img.rows - SIFT_IMG_BORDER )
                return false;
        }
    
        /* ensure convergence of interpolation */
        // SIFT_MAX_INTERP_STEPS:插值最大步数,避免插值不收敛,程序中默认为5
        if( i >= SIFT_MAX_INTERP_STEPS )
            return false;
    
        {
            int idx = octv*(nOctaveLayers+2) + layer;
            const Mat& img = dog_pyr[idx];
            const Mat& prev = dog_pyr[idx-1];
            const Mat& next = dog_pyr[idx+1];
            Matx31f dD((img.at<short>(r, c+1) - img.at<short>(r, c-1))*deriv_scale,
                       (img.at<short>(r+1, c) - img.at<short>(r-1, c))*deriv_scale,
                       (next.at<short>(r, c) - prev.at<short>(r, c))*deriv_scale);
            float t = dD.dot(Matx31f(xc, xr, xi));
    
            contr = img.at<short>(r, c)*img_scale + t * 0.5f;
            if( std::abs( contr ) * nOctaveLayers < contrastThreshold )
                return false;
    
            /* principal curvatures are computed using the trace and det of Hessian */
           //利用Hessian矩阵的迹和行列式计算主曲率的比值
           float v2 = img.at<short>(r, c)*2.f;
            float dxx = (img.at<short>(r, c+1) + 
                    img.at<short>(r, c-1) - v2)*second_deriv_scale;
            float dyy = (img.at<short>(r+1, c) + 
                    img.at<short>(r-1, c) - v2)*second_deriv_scale;
            float dxy = (img.at<short>(r+1, c+1) - 
                    img.at<short>(r+1, c-1) - img.at<short>(r-1, c+1) + 
                    img.at<short>(r-1, c-1)) * cross_deriv_scale;
            float tr = dxx + dyy;
            float det = dxx * dyy - dxy * dxy;
    
            //这里edgeThreshold可以在调用SIFT()时输入;
            //其实代码中定义了 static const float SIFT_CURV_THR = 10.f 可以直接使用
            if( det <= 0 || tr*tr*edgeThreshold >= (edgeThreshold + 1)*(edgeThreshold + 1)*det )
                return false;
        }
    
        kpt.pt.x = (c + xc) * (1 << octv);
        kpt.pt.y = (r + xr) * (1 << octv);
        kpt.octave = octv + (layer << 8) + (cvRound((xi + 0.5)*255) << 16);
        kpt.size = sigma*powf(2.f, (layer + xi) / nOctaveLayers)*(1 << octv)*2;
    
        return true;
    }

    至此,SIFT第二步就完成了。参见《SIFT原理与源码分析

    本文转自:http://blog.csdn.net/xiaowei_cqu/article/details/8087239

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  • 原文地址:https://www.cnblogs.com/tianyalu/p/5467854.html
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