Cheers. Just been playing it :)

Can't wait for the game! :)

Agree with eliskan on the double-quote issue.

I can't reproduce error. (Ipad3, 6.1.3 ios)

The (pseudo)code snippet there:
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)

Return true if line segments AB and CD intersect

def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)

Looks pretty interesting(and pretty concise!) I will try that.

Thanks will look into it. I have a Marware carbon fibre case which is really good and was £20, so I am a bit loath to change it!
(P.s. it's not REAL carbon fibre;)

Okay Coder123, that's slightly easier! :)

This sorta works...
<pre>
#line1...
xd= line1[2]-line1[0]
yd= line1[3]-line1[1]
g1=yd/xd
c1=line1[1]-(g1*line1[0])
#formula is y= g1 * x + c1

#line2...
xd2= line2[2]-line2[0]
yd2= line2[3]-line2[1]
g2=yd2/xd2
c2 = line2[1] - (g2 * line2[0])
#formula is y= g2 * x + c2

both formulas will be equal at the intersection,

i.e. one minus the other equals zero

(g1 * x) + c1 - (g2 * x) - c2 = 0

#solve for x...
xinter = (c2-c1) / (g1-g2)

but is xinter somewhere in the correct range?

eg the range of one of the lines x bounds

note: needs work as needs to limit to overlapping x range

if xinter in range(line1[0],line1[2]):
return True
else:
return False
</pre>

I just sketched this out using linear equations and it worked, and it seems like others on the web do the same - find the point where both linear equations equal each other ( or if they don't, then your lines don't intersect)
Nice example here :
http://keisan.casio.com/has10/SpecExec.cgi?id=system/2006/1223519249
But wonder if something more elegant using the "in" function for rectangles might work...hmmm

Thanks, this looks very interesting. You get an A* for working this out! :)