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?



enter image description here










share|improve this question





























    1















    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?



    enter image description here










    share|improve this question

























      1












      1








      1








      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?



      enter image description here










      share|improve this question














      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?



      enter image description here







      c++ opencv






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 4 at 15:00









      HanzDieterHanzDieter

      327




      327
























          1 Answer
          1






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          1














          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)


          enter image description here






          share|improve this answer





















          • 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












          Your Answer






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          1 Answer
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          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          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)


          enter image description here






          share|improve this answer





















          • 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














          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)


          enter image description here






          share|improve this answer





















          • 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








          1







          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)


          enter image description here






          share|improve this answer















          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)


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          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














          • 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




















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