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algorithmlanguage-agnosticgeometry2dangle

check if two segments on the same circle overlap / intersect


Given two circle segments of the same circle: A=[a1, a2] and B=[b1, b2], with:

  • a1, a2, b1, b2 values in degree between -inf and +inf
  • a1 <= a2 ; b1 <= b2
  • a2-a1<=360; b2-b1<=360

How can I find out if these two circle segments overlap? (i.E. if they intersect or touch in at least one point)

Examples:

A=[  -45°,    45°]; B=[   10°,   20°] ==> overlap
A=[  -45°,    45°]; B=[   90°,  180°] ==> no overlap
A=[  -45°,    45°]; B=[  180°,  360°] ==> overlap
A=[ -405°,  -315°]; B=[  180°,  360°] ==> overlap
A=[-3600°, -3601°]; B=[ 3601°, 3602°] ==> overlap (touching counts as overlap)
A=[ 3600°,  3601°]; B=[-3601°,-3602°] ==> overlap (touching counts as overlap)
A=[    -1°,    1°]; B=[ 3602°, 3603°] ==> no overlap 

This looks like a deceptively simple problem but I cannot wrap my head around it. I currently have a basic idea for a solution which involves splitting each segment into two if it crosses 0°, but I am not sure if that covers all cases, and I was wondering if there is an elegant formula.


Solution

  • As @admaoldak mentioned, normalize the degrees first:

    a1_norm = a1 % 360
    a2_norm = a2 % 360
    b1_norm = b1 % 360
    b2_norm = b2 % 360
    

    Now to check if b1 is within (a1,a2),

    def intersect(b, as, ae
        Intersect = False
        If as > ae:
            if b >= as or b <= ae:
                return True
        Else:
            if b>=as and b<=ae:
                return True
        return False
    

    Final answer is:

    intersect(b1_norm,a1_norm,a2_norm)||intersect(b2_norm,a1_norm,a2_norm)||
    intersect(a1_norm,b1_norm,b2_norm)||intersect(a2_norm,b1_norm,b2_norm)