if min(ay) > min(by)
  temp = ay;
  temp2 = ax;
  ay = by;
  ax = bx;
  by = temp;
  bx = temp;
  %swapping rectangle coordinates. too lazy to explain why
end

if max(ax) =max(bx)
  overlap = 0;
elseif max(ay) >min(by)
  overlap = 1;
else
  overlap = 0;
end

That did not seem to work, may have gotten garbled a bit since the first if statement has an assignment where it should have a ==.

Doug

PS. You are welcome for the puzzles!

http://www.blinkdagger.com/blog/mathematical-body-art

if min(ay) > min(by)
  temp = ay;
  temp2 = ax;
  ay = by;
  ax = bx;
  by = temp;
  bx = temp;
  %swapping rectangle coordinates. too lazy to explain why
end

if max(ax) = max(bx)
  overlap = 0;
elseif max(ay) >min(by)
  overlap = 1;
else
  overlap = 0;
end

Man, that was a lot of work! Do I win a prize if my submission passes the stress test? My old MATLAB t-shirt is getting worn down from wearing it so much. Need new one!

I did not know about that function, that is easy and clean. I like seeing how many different solutions there are for these very simple problems. It helps us all run into new techniques and functions.

Thanks all!
Doug

Doug never specifically said whether the coordinates are sorted. Note that most of the posted solutions will fail if they aren’t.

http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=11837&objectType=file

For each pair of line segments, the linear algebra to determine their intersection point is of course easy put expensive, so the code first checks to see if the smallest rectangles enclosing them overlap. See lines 99-109 of the code.

 

% define rectangle 1
rect1 = [ax(1) ay(1) diff(ax) diff(ay)];

% define rectangle 2
rect2 = [bx(1) by(1) diff(bx) diff(by)];

% the lazy solution
overlap = (rectint(rect1,rect2)~=0)

 

 

voverlap =  (ax(1) < bx(2) & bx(1) < ax(2)) & (ay(1) < by(2) & by(1) < ay(2));