Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!wuarchive!mit-eddie!bloom-beacon!ZIA.AOC.NRAO.EDU!cflatter
From: cflatter@ZIA.AOC.NRAO.EDU (Chris Flatters)
Newsgroups: comp.windows.x
Subject: Re:  GXinvert & planes
Message-ID: <9012112324.AA03770@zia.aoc.nrao.edu>
Date: 11 Dec 90 23:24:01 GMT
Sender: daemon@athena.mit.edu (Mr Background)
Organization: The Internet
Lines: 47

Brian Powell writes:
>     I was trying to invert some text (drawn with Xlib) to show a selection.
> At first, I made a GC with a function of GXinvert, and a plane_mask of the
> default AllPlanes.  This worked fine under OpenWindows, but failed under
> the Sun MIT server (it drew all black).
>     I went looking for some sample client code, and found a couple of
> different examples where the plane_mask was set to the exclusive-or of the
> foreground and background colors.  I tried this, and it worked.  But I didn't
> understand why.  Perhaps it's because I'm not a color guru.

Suppose that the bits in the index of the background colour are b(0) ... b(depth-1) and the bits in the index of the foreground colour are f(0) ...
f(depth-1).  Applying GXinvert to a colour with index c (with bits c(0) ...
c(depth-1) will produce a colour with bits not(c(0)) ... not(c(depth-1).
In order for this to reverse the foreground and background colours we
require

        f(i) = not(b(i)) for all i: 0..depth-1

This is clearly not true in all cases.  Take for example the case where
the foreground has index 1 (0000 0001 in binary) and the background has
index 0 (0000 0000): the inverted bit patterns are 1111 1110 and 1111 1111.
This case is interesting since black appears both at index 1 and 254 in
the initial X11/NeWS colormap on 8-bit colour displays and white appears
at indices 0 and 255:  thus a straight invert of the display will appear
to work.

Suppose we now add a plane mask with bits m(i).  If the initial colour
at a given pixel is c then the output bits are

       (m(i) and not(c(i))) or (not(m(i)) and c(i))

(recall that if the i-th bit of the mask is 0 then the source pixel is
unchanged).  If we set the mask to the exclusive or of f and b the
the output bits are

	o(c(i)) = ((f(i) xor b(i)) and not(c(i))) or
                  (not (f(i) xor b(i)) and c(i))

You may verify that 

        o(f(i)) = b(i)
 and
        o(b(i)) = f(i)

(as required) by Boolean algebra or by constructing truth tables.

			Chris Flatters