New intersection function for rotated rectangles

diff --git a/modules/imgproc/include/opencv2/imgproc.hpp b/modules/imgproc/include/opencv2/imgproc.hpp
index dd849ef..b170e6a 100644 (file)
--- a/modules/imgproc/include/opencv2/imgproc.hpp
+++ b/modules/imgproc/include/opencv2/imgproc.hpp
@@ -1414,6 +1414,9 @@ CV_EXPORTS_W void fitLine( InputArray points, OutputArray line, int distType,
 //! checks if the point is inside the contour. Optionally computes the signed distance from the point to the contour boundary
 CV_EXPORTS_W double pointPolygonTest( InputArray contour, Point2f pt, bool measureDist );
 
+//! computes whether two rotated rectangles intersect and returns the vertices of the intersecting region
+CV_EXPORTS_W bool rotatedRectangleIntersection( const RotatedRect& rect1, const RotatedRect& rect2, OutputArray intersectingRegion  );
+
 CV_EXPORTS Ptr<CLAHE> createCLAHE(double clipLimit = 40.0, Size tileGridSize = Size(8, 8));
 
 } // cv

diff --git a/modules/imgproc/src/intersection.cpp b/modules/imgproc/src/intersection.cpp
new file mode 100644 (file)
index 0000000..6d6f97d
--- /dev/null
+++ b/modules/imgproc/src/intersection.cpp
@@ -0,0 +1,223 @@
+/*M///////////////////////////////////////////////////////////////////////////////////////
+//
+//  IMPORTANT: READ BEFORE DOWNLOADING, COPYING, INSTALLING OR USING.
+//
+//  By downloading, copying, installing or using the software you agree to this license.
+//  If you do not agree to this license, do not download, install,
+//  copy or use the software.
+//
+//
+//                           License Agreement
+//                For Open Source Computer Vision Library
+//
+// Copyright (C) 2000-2008, Intel Corporation, all rights reserved.
+// Copyright (C) 2008-2011, Willow Garage Inc., all rights reserved.
+// Third party copyrights are property of their respective owners.
+//
+// @Authors
+//      Nghia Ho, nghiaho12@yahoo.com
+//
+// Redistribution and use in source and binary forms, with or without modification,
+// are permitted provided that the following conditions are met:
+//
+//   * Redistribution's of source code must retain the above copyright notice,
+//     this list of conditions and the following disclaimer.
+//
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+//     this list of conditions and the following disclaimer in the documentation
+//     and/or other materials provided with the distribution.
+//
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+//     derived from this software without specific prior written permission.
+//
+// This software is provided by the copyright holders and contributors "as is" and
+// any express or implied warranties, including, but not limited to, the implied
+// warranties of merchantability and fitness for a particular purpose are disclaimed.
+// In no event shall the OpenCV Foundation or contributors be liable for any direct,
+// indirect, incidental, special, exemplary, or consequential damages
+// (including, but not limited to, procurement of substitute goods or services;
+// loss of use, data, or profits; or business interruption) however caused
+// and on any theory of liability, whether in contract, strict liability,
+// or tort (including negligence or otherwise) arising in any way out of
+// the use of this software, even if advised of the possibility of such damage.
+//
+//M*/
+#include "precomp.hpp"
+
+namespace cv
+{
+
+bool rotatedRectangleIntersection( const RotatedRect& rect1, const RotatedRect& rect2, OutputArray intersectingRegion )
+{
+    const float samePointEps = 0.00001; // used to test if two points are the same, due to numerical error
+
+    Point2f vec1[4], vec2[4];
+    Point2f pts1[4], pts2[4];
+
+    std::vector <Point2f> intersection;
+
+    rect1.points(pts1);
+    rect2.points(pts2);
+
+    // Line vector
+    // A line from p1 to p2 is: p1 + (p2-p1)*t, t=[0,1]
+    for( int i = 0; i < 4; i++ )
+    {
+        vec1[i].x = pts1[(i+1)%4].x - pts1[i].x;
+        vec1[i].y = pts1[(i+1)%4].y - pts1[i].y;
+
+        vec2[i].x = pts2[(i+1)%4].x - pts2[i].x;
+        vec2[i].y = pts2[(i+1)%4].y - pts2[i].y;
+    }
+
+    // Line test - test all line combos for intersection
+    for( int i = 0; i < 4; i++ )
+    {
+        for( int j = 0; j < 4; j++ )
+        {
+            // Solve for 2x2 Ax=b
+            float x21 = pts2[j].x - pts1[i].x;
+            float y21 = pts2[j].y - pts1[i].y;
+
+            float vx1 = vec1[i].x;
+            float vy1 = vec1[i].y;
+
+            float vx2 = vec2[j].x;
+            float vy2 = vec2[j].y;
+
+            float det = vx2*vy1 - vx1*vy2;
+
+            float t1 = (vx2*y21 - vy2*x21) / det;
+            float t2 = (vx1*y21 - vy1*x21) / det;
+
+            // This takes care of parallel lines
+            if( !std::isnormal(t1) || !std::isnormal(t2) )
+            {
+                continue;
+            }
+
+            if( t1 >= 0.0f && t1 <= 1.0f && t2 >= 0.0f && t2 <= 1.0f )
+            {
+                float xi = pts1[i].x + vec1[i].x*t1;
+                float yi = pts1[i].y + vec1[i].y*t1;
+
+                intersection.push_back(Point2f(xi,yi));
+            }
+        }
+    }
+
+    // Check for vertices from rect1 inside recct2
+    for( int i = 0; i < 4; i++ )
+    {
+        // We do a sign test to see which side the point lies.
+        // If the point all lie on the same sign for all 4 sides of the rect,
+        // then there's an intersection
+        int posSign = 0;
+        int negSign = 0;
+
+        float x = pts1[i].x;
+        float y = pts1[i].y;
+
+        for( int j = 0; j < 4; j++ )
+        {
+            // line equation: Ax + By + C = 0
+            // see which side of the line this point is at
+            float A = -vec2[j].y;
+            float B = vec2[j].x;
+            float C = -(A*pts2[j].x + B*pts2[j].y);
+
+            float s = A*x+ B*y+ C;
+
+            if( s >= 0 )
+            {
+                posSign++;
+            }
+            else
+            {
+                negSign++;
+            }
+        }
+
+        if( posSign == 4 || negSign == 4 )
+        {
+            intersection.push_back(pts1[i]);
+        }
+    }
+
+    // Reverse the check - check for vertices from rect2 inside recct1
+    for( int i = 0; i < 4; i++ )
+    {
+        // We do a sign test to see which side the point lies.
+        // If the point all lie on the same sign for all 4 sides of the rect,
+        // then there's an intersection
+        int posSign = 0;
+        int negSign = 0;
+
+        float x = pts2[i].x;
+        float y = pts2[i].y;
+
+        for( int j = 0; j < 4; j++ )
+        {
+            // line equation: Ax + By + C = 0
+            // see which side of the line this point is at
+            float A = -vec1[j].y;
+            float B = vec1[j].x;
+            float C = -(A*pts1[j].x + B*pts1[j].y);
+
+            float s = A*x + B*y + C;
+
+            if( s >= 0 )
+            {
+                posSign++;
+            }
+            else
+            {
+                negSign++;
+            }
+        }
+
+        if( posSign == 4 || negSign == 4 )
+        {
+            intersection.push_back(pts2[i]);
+        }
+    }
+
+    // Get rid of dupes
+    for( int i = 0; i < (int)intersection.size()-1; i++ )
+    {
+        for( size_t j = i+1; j < intersection.size(); j++ )
+        {
+            float dx = intersection[i].x - intersection[j].x;
+            float dy = intersection[i].y - intersection[j].y;
+            double d2 = dx*dx + dy*dy; // can be a really small number, need double here
+
+            if( d2 < samePointEps*samePointEps )
+            {
+                // Found a dupe, remove it
+                std::swap(intersection[j], intersection.back());
+                intersection.pop_back();
+                i--; // restart check
+            }
+        }
+    }
+
+    if( intersection.empty() )
+    {
+        return false;
+    }
+
+    intersectingRegion.create(intersection.size(), 1, CV_MAKETYPE(intersectingRegion.depth(), 2) );
+
+    Mat m = intersectingRegion.getMat();
+
+    size_t step = !m.isContinuous() ? m.step[0] : sizeof(Point2f);
+
+    for( size_t i = 0; i < intersection.size(); i++ )
+    {
+        *(Point2f*)(m.data + i*step) = intersection[i];
+    }
+
+    return true;
+}
+
+}