From: rick.c.hodgin@nospicedham.gmail.com   
      
   On 10/15/20 2:40 AM, Terje Mathisen wrote:   
   > The idea is that we need to compare the left edge of rectangle one   
   > against both the left and right edge of rectangle2, and the same for the   
   > right edge, so this is 8 compares for the vertical lines. Add the same   
   > for the horizontal and you need 16 compares which would fit in a much   
   > more common AVX register.   
   >   
   > R2 totally inside R1 becomes   
   >   
   >    (R2.l >= R1.l) AND (R2.r <= R1.r) AND   
   >    (R2.b >= R1.b) AND (R2.a <= R1.a)   
   >   
   > i.e. relatively simple, while zero overlap is much harder, and related   
   > to the clipping question from last week: It requires all of R2 to be   
   > either to the left, right, above or below, so there are 4 different   
   > masks that give the same result and you have to OR those together:   
   >   
   >   (R2.r < R1.l) OR (R2.a < R1.b) OR (R2.l >= R1.r) OR (R2.b >= R1.a)   
   >   
   > All that remains is to setup all those values in the correct spot so   
   > that these 8 compares can be done (some of them needs to be inverted   
   > after the fact obviously!) with a single compare_greater_or_equal()   
   > instruction. Seems doable by duplicating the R1 values so that each   
   > occur twice, then use a permute on the R2 values to put them in the   
   > right spots, do the compare, flip half the results and then the slowest   
   > part: Horizontally combine them with and AND for the inside result, with   
   > OR for outside, and then all partial overlaps fall out by being neither   
   > of these.   
      
   I'm thinking there's a relationship between the various values, such   
   that for an inside test we will know:   
      
            t   
           ___   
        l |   | r     // left, right, top, bottom   
          |___|   
      
            b   
      
        r2.l is between r1.l and r1.r   
        r2.r is between r1.l and r1.r   
      
   And the same will be true for r2.top and r2.bottom:   
      
        r2.t is between r1.t and r1.b   
        r2.b is between r1.t and r1.b   
      
   And if we assert that each l is less than r, and each t is less than b,   
   is there some mathematical formula we can conclude?   
      
        if     r2.r - r1.l - r2.l > x   
           and r2.r - r1.r - r2.l > x   
           and r2.b - r1.t - r2.t > y   
           and r2.t - r1.t - r2.t > y   
      
   then we know the relationship is true?  It would be determining x and y,   
   which surely are proportional to the rectangles and/or their   
   relationships.  And once determined, we could possibly simplify the   
   expressions to their minimums.   
      
   Something like that?  I don't know.  My mind is hazy on all this.   
      
   --   
   Rick C. Hodgin   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)