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POHough transform returns the collinear and semi collinear points
 

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  1. POHough transform returns the collinear and semi collinear points
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    <p>I have points in an image. I need to detect the most collinear points. The fastest method is to use Hough transform, but I have to modify the opencv method. Actually I need that the semi collinear points to be returned with detected line, for this reason I modified the polar line struct. A tolerance threshold is also needed to detect nearly detected points as shown in the image. Can someone help in how to tune this threshold? I need at least four semi collinear points to detect the line to which they belong. <img src="https://i.stack.imgur.com/Xatig.png" alt="ex 1"> <img src="https://i.stack.imgur.com/eWSto.png" alt="ex 2"> <img src="https://i.stack.imgur.com/cb9wq.png" alt="ex 3"></p> <ol> <li><p>The points of first image were detected by 6 overlapped lines</p></li> <li><p>the point of middle images were detected by nothing</p></li> <li>the third's points were detected by three lines</li> </ol> <p><img src="https://i.stack.imgur.com/YWMTM.jpg" alt="`![enter image description here"> <img src="https://i.stack.imgur.com/b9vJD.png" alt="enter image description here"> <img src="https://i.stack.imgur.com/0KTbm.jpg" alt="enter image description here"></p> <p>Which is the best way to get rid from the overlapped liens?? Or how to tune the tolerance threshold to detect the semi collinear points by only one line?</p> <p>the is my own function call:</p> <pre><code>vector&lt;CvLinePolar2&gt; lines; CvMat c_image = source1; // loaded image HoughLinesStandard(&amp;c_image,1,CV_PI/180,4,&amp;lines,INT_MAX); </code></pre> <hr> <pre><code> typedef struct CvLinePolar2 { float rho; float angle; vector&lt;CvPoint&gt; points; }; void HoughLinesStandard( const CvMat* img, float rho, float theta, int threshold, vector&lt;CvLinePolar2&gt; *lines, int linesMax= INT_MAX ) { cv::AutoBuffer&lt;int&gt; _accum, _sort_buf; cv::AutoBuffer&lt;float&gt; _tabSin, _tabCos; const uchar* image; int step, width, height; int numangle, numrho; int total = 0; int i, j; float irho = 1 / rho; double scale; vector&lt;vector&lt;CvPoint&gt;&gt; lpoints; CV_Assert( CV_IS_MAT(img) &amp;&amp; CV_MAT_TYPE(img-&gt;type) == CV_8UC1 ); image = img-&gt;data.ptr; step = img-&gt;step; width = img-&gt;cols; height = img-&gt;rows; numangle = cvRound(CV_PI / theta); numrho = cvRound(((width + height) * 2 + 1) / rho); _accum.allocate((numangle+2) * (numrho+2)); _sort_buf.allocate(numangle * numrho); _tabSin.allocate(numangle); _tabCos.allocate(numangle); int *accum = _accum, *sort_buf = _sort_buf; float *tabSin = _tabSin, *tabCos = _tabCos; memset( accum, 0, sizeof(accum[0]) * (numangle+2) * (numrho+2) ); //memset( lpoints, 0, sizeof(lpoints) ); lpoints.resize(sizeof(accum[0]) * (numangle+2) * (numrho+2)); float ang = 0; for(int n = 0; n &lt; numangle; ang += theta, n++ ) { tabSin[n] = (float)(sin(ang) * irho); tabCos[n] = (float)(cos(ang) * irho); } // stage 1. fill accumulator for( i = 0; i &lt; height; i++ ) for( j = 0; j &lt; width; j++ ) { if( image[i * step + j] != 0 ) { CvPoint pt; pt.x = j; pt.y = i; for(int n = 0; n &lt; numangle; n++ ) { int r = cvRound( j * tabCos[n] + i * tabSin[n] ); r += (numrho - 1) / 2; int ind = (n+1) * (numrho+2) + r+1; int s = accum[ind]; accum[ind]++; lpoints[ind].push_back(pt); } } } // stage 2. find local maximums for(int r = 0; r &lt; numrho; r++ ) for(int n = 0; n &lt; numangle; n++ ) { int base = (n+1) * (numrho+2) + r+1; if( accum[base] &gt; threshold &amp;&amp; accum[base] &gt; accum[base - 1] &amp;&amp; accum[base] &gt;= accum[base + 1] &amp;&amp; accum[base] &gt; accum[base - numrho - 2] &amp;&amp; accum[base] &gt;= accum[base + numrho + 2] ) sort_buf[total++] = base; } // stage 3. sort the detected lines by accumulator value icvHoughSortDescent32s( sort_buf, total, accum ); // stage 4. store the first min(total,linesMax) lines to the output buffer linesMax = MIN(linesMax, total); scale = 1./(numrho+2); for( i = 0; i &lt; linesMax; i++ ) { CvLinePolar2 line; int idx = sort_buf[i]; int n = cvFloor(idx*scale) - 1; int r = idx - (n+1)*(numrho+2) - 1; line.rho = (r - (numrho - 1)*0.5f) * rho; line.angle = n * theta; line.points = lpoints[idx]; lines-&gt;push_back(line); } </code></pre> <p>}</p>
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