Add rectangles to a trapezoid openCV C++
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I have an image with two areas. I want to add now rectangles with a fixed size randomly into area2. The coordinate origin of the image is in the top-left corner. I have the cordinates of the area2. This are P1, P2, P3(0, y_max) and P4(x_max, y_max). Does anybody knows how to check, if a rectangle lies in this area?
I can try to separate this area into 2 parts, a rectangle (rect_area) and a triangle (trangle_area). For the rect_area I can check with
bool intersects = ((rect_area & rect_random).area() > 0);
if the random rect lies inside the area. For the triangle I have found some complicated stuff like here: How to determine if a point is in a 2D triangle?
Does anybody knows an easier way to do that?
c++ opencv
add a comment |
I have an image with two areas. I want to add now rectangles with a fixed size randomly into area2. The coordinate origin of the image is in the top-left corner. I have the cordinates of the area2. This are P1, P2, P3(0, y_max) and P4(x_max, y_max). Does anybody knows how to check, if a rectangle lies in this area?
I can try to separate this area into 2 parts, a rectangle (rect_area) and a triangle (trangle_area). For the rect_area I can check with
bool intersects = ((rect_area & rect_random).area() > 0);
if the random rect lies inside the area. For the triangle I have found some complicated stuff like here: How to determine if a point is in a 2D triangle?
Does anybody knows an easier way to do that?
c++ opencv
add a comment |
I have an image with two areas. I want to add now rectangles with a fixed size randomly into area2. The coordinate origin of the image is in the top-left corner. I have the cordinates of the area2. This are P1, P2, P3(0, y_max) and P4(x_max, y_max). Does anybody knows how to check, if a rectangle lies in this area?
I can try to separate this area into 2 parts, a rectangle (rect_area) and a triangle (trangle_area). For the rect_area I can check with
bool intersects = ((rect_area & rect_random).area() > 0);
if the random rect lies inside the area. For the triangle I have found some complicated stuff like here: How to determine if a point is in a 2D triangle?
Does anybody knows an easier way to do that?
c++ opencv
I have an image with two areas. I want to add now rectangles with a fixed size randomly into area2. The coordinate origin of the image is in the top-left corner. I have the cordinates of the area2. This are P1, P2, P3(0, y_max) and P4(x_max, y_max). Does anybody knows how to check, if a rectangle lies in this area?
I can try to separate this area into 2 parts, a rectangle (rect_area) and a triangle (trangle_area). For the rect_area I can check with
bool intersects = ((rect_area & rect_random).area() > 0);
if the random rect lies inside the area. For the triangle I have found some complicated stuff like here: How to determine if a point is in a 2D triangle?
Does anybody knows an easier way to do that?
c++ opencv
c++ opencv
asked Jan 4 at 15:00
HanzDieterHanzDieter
327
327
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1 Answer
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Does the graph you draw represent the general case of the problem?
- P1.x == 0
- P2.x == 0
- q1.x == q2.x
- q2.y == q3.y
If the above conditions hold, then you can check
- if q1 is below the line of P1P2 (q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y)
- q2 is abovep2 (q2.y < P2.y)
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
Does the graph you draw represent the general case of the problem?
- P1.x == 0
- P2.x == 0
- q1.x == q2.x
- q2.y == q3.y
If the above conditions hold, then you can check
- if q1 is below the line of P1P2 (q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y)
- q2 is abovep2 (q2.y < P2.y)
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
add a comment |
Does the graph you draw represent the general case of the problem?
- P1.x == 0
- P2.x == 0
- q1.x == q2.x
- q2.y == q3.y
If the above conditions hold, then you can check
- if q1 is below the line of P1P2 (q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y)
- q2 is abovep2 (q2.y < P2.y)
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
add a comment |
Does the graph you draw represent the general case of the problem?
- P1.x == 0
- P2.x == 0
- q1.x == q2.x
- q2.y == q3.y
If the above conditions hold, then you can check
- if q1 is below the line of P1P2 (q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y)
- q2 is abovep2 (q2.y < P2.y)
Does the graph you draw represent the general case of the problem?
- P1.x == 0
- P2.x == 0
- q1.x == q2.x
- q2.y == q3.y
If the above conditions hold, then you can check
- if q1 is below the line of P1P2 (q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y)
- q2 is abovep2 (q2.y < P2.y)
edited Jan 5 at 13:32
answered Jan 4 at 15:57
StoveStove
1164
1164
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
add a comment |
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
1
1
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
P2.x > 0. How can I check, if q1 is above or below Line P1P2?
– HanzDieter
Jan 4 at 16:18
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
you can check q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.x
– Stove
Jan 4 at 16:26
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I don´t see it right now. Can you explain the equation, please?
– HanzDieter
Jan 4 at 16:42
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
I made a typo, it should be q1.y > (q1.x*(p2.y-p1.y)/x_max)+p1.y
– Stove
Jan 4 at 16:47
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
(p2.y-p1.y)/x_max is the slope (dy/dx) of the line, p1.y is the offset. it is y coordinate of line at q.x is given by q.x*(slope of line) + (offset of line)
– Stove
Jan 4 at 16:49
add a comment |
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