How Convexity Defect is calculated in OpenCV?

2020-04-10 23:35发布

What is the algorithm used in OpenCV function convexityDefects() to calculate the convexity defects of a contour?

Please, describe and illustrate the high-level operation of the algorithm, along with its inputs and outputs.

1条回答
【Aperson】
2楼-- · 2020-04-11 00:21

Based on the documentation, the input are two lists of coordinates:

  • contour defining the original contour (red on the image below)
  • convexhull defining the convex hull corresponding to that contour (blue on the image below)

Example contour and convex hull

The algorithm works in the following manner:

If the contour or the hull contain 3 or less points, then the contour is always convex, and no more processing is needed. The algorithm assures that both the contour and the hull are accessed in the same orientation.

N.B.: In further explanation I assume they are in the same orientation, and ignore the details regarding representation of the floating point depth as an integer.

Then for each pair of adjacent hull points (H[i], H[i+1]), defining one edge of the convex hull, calculate the distance from the edge for each point on the contour C[n] that lies between H[i] and H[i+1] (excluding C[n] == H[i+1]). If the distance is greater than zero, then a defect is present. When a defect is present, record i, i+1, the maximum distance and the index (n) of the contour point where the maximum located.

Distance is calculated in the following manner:

dx0 = H[i+1].x - H[i].x
dy0 = H[i+1].y - H[i].y

if (dx0 is 0) and (dy0 is 0) then
    scale = 0
else
    scale = 1 / sqrt(dx0 * dx0 + dy0 * dy0)

dx = C[n].x - H[i].x
dy = C[n].y - H[i].y

distance = abs(-dy0 * dx + dx0 * dy) * scale

It may be easier to visualize in terms of vectors:

  • C: defect vector from H[i] to C[n]
  • H: hull edge vector from H[i] to H[i+1]
  • H_rot: hull edge vector H rotated 90 degrees
  • U_rot: unit vector in direction of H_rot

H components are [dx0, dy0], so rotating 90 degrees gives [-dy0, dx0].

scale is used to find U_rot from H_rot, but because divisions are more computationally expensive than multiplications, the inverse is used as an optimization. It's also pre-calculated before the loop over C[n] to avoid recomputing each iteration.

|H| = sqrt(dx0 * dx0 + dy0 * dy0)

U_rot = H_rot / |H| = H_rot * scale

Then, a dot product between C and U_rot gives the perpendicular distance from the defect point to the hull edge, and abs() is used to get a positive magnitude in any orientation.

distance = abs(U_rot.C) = abs(-dy0 * dx + dx0 * dy) * scale


In the scenario depicted on the above image, in first iteration, the edge is defined by H[0] and H[1]. The contour points tho examine for this edge are C[0], C[1], and C[2] (since C[3] == H[1]).

First edge

There are defects at C[1] and C[2]. The defect at C[1] is the deepest, so the algorithm will record (0, 1, 1, 50).

The next edge is defined by H[1] and H[2], and corresponding contour point C[3]. No defect is present, so nothing is recorded.

The next edge is defined by H[2] and H[3], and corresponding contour point C[4]. No defect is present, so nothing is recorded.

Since C[5] == H[3], the last contour point can be ignored -- there can't be a defect there.

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