Comparing Two Contours: Rotation invariant?

asked 2017-06-07 13:45:36 -0600

JoeBroesel gravatar image

updated 2017-06-07 14:07:49 -0600

I found one approach for estimate the orientation of two contours here , which rotates one contour and checks the distance to the original. I changed the headers to 

#include <opencv2/core.hpp>
#include <opencv2/shape.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/opencv_modules.hpp>
#include <iostream>
#include <fstream>
#include <string.h>

and the main to:

int main(int argc, char* argv[])

It may be kind of a stupid question, but first of all i don't know, why the transformation of the contours should improve the result of computeDistance. Is the <cv::shapecontextdistanceextractor> not invariant to rotation and translation, because it does an internal fit?

If this would be the case, my results would be coherent, because I always get 0 as distance (but unfortunately no image as well). Also the result from an other program, where i match rotated contours with cv::shapecontextdistanceextractor> as well as the hausdorff metric seems not to be wrong (small distances, but no exact 0).

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Comments

Well, that question specifically asked about orientation, not translation. By removing the translation, it removes a complication and makes the results easier to understand.

Tetragramm gravatar imageTetragramm ( 2017-06-07 16:57:34 -0600 )edit

the cv::shapecontextdistanceextractor IS invariant to rotation and translation, because it does an internal fit.

berak gravatar imageberak ( 2017-06-07 22:51:35 -0600 )edit

so how can this part of the code work if the extractor is invariant?

    for (int i = 0; i < 8; i++)
{
    RotateContour(contours_Trans,contours_Rotated,Angle,ptCCentre);
    TestContour= simpleContour(contours_Rotated);
    float dis = mysc->computeDistance( QueryContour, TestContour);      
    if(Angle>=360)
        Angle=0.0;
    else
        Angle+=45.0;

    if ( dis<bestDis )
    {
        bestMatch  = Angle;
        bestDis = dis;
     }

}

edit: Because if its invariant dist would be independent of the angle - or do I understand something wrong?

JoeBroesel gravatar imageJoeBroesel ( 2017-06-08 00:39:58 -0600 )edit
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