Conference rusure::math

    There's a hole in your argument at 'b'. Two areas inside a third
    are governed by shape as well as size, and any probabilities are
    affected by the probability of their overlapping. For many areas
    it becomes more and more complicated; as the number increases, you
    have to alternately add and subtract the areas where 2,3,... subareas
    overlap. It's a mess.
    
    I suspect you will have to find a good Measure Theory book to attack
    the problem adequately. Good luck.
    

    Are you talking about finding the union of the smaller polygon areas?
    I think I can assume I have an algorithm to do that.  
    For what I'd like to define/prove, the assumption is that the polygon
    shape is not explicitly known, rather that its area is known and
    that the areas of the other shapes are known.  Given just areas,
    can you derive probabilities of their intersection?
    I have an idea that you'd start with proving it for a line
    segment which is part of a larger line segment (countably infinite)
    and then extend to deal with planes and solids as per topology...
    
    Also, I remember reading something a long time ago which dealt
    with the question of land mass distribution on the earth -
    why there is more "up top"   - the explanation showed that
    the distribution was well within what would be expected for 
    randomly adding polygons to a planar shape, which reasoning was
    workable for sphere surfaces as well....
    
    mp

Given two smaller *squares* inside a larger square, what is the
probability that they intersect?

We'll assume that the larger square has side 1, and that the two smaller squares
have sides a and b, and that the lower left corners of these smaller squares may
be at (0..1-a,0..1-a) and (0..1-b,0..1-b), with uniform distribution.

The intersection of the squares is based on two *independent* events: that
the x-c�ordinates overlap, and the y-c�ordinates overlap. 

Now the probability that the x-c�ordinates overlap can be determined by drawing
a square, labelling the edges with xa and xb (the x-c�ordinates of the left
sides of the small squares), and shading the region where the values imply
there's overlap (I've assumed that a+2b < 1 and a < b), and figuring only the
region bounded by 1-a and 1-b (since xa may not exceed 1-a).  This gives the
probability that the x-c�ordinates overlap as:

	b [a+(a+b)]/2 + (1-2a-b) [b+a] + a [(a+b)+b)]/2
	-----------------------------------------------
			(1-a) (1-b)
Or:
	2 - 3 a� - b�
	------------- - 1
	2 (1-a) (1-b)

Now, because the x- and y-c�ordinates are independent, we have for the
probability that the squares overlap:

	  2 - 3 a� - b�      2
	[ ------------- - 1 ]
	  2 (1-a) (1-b)

This seems reasonable; if a = b, and a << 1, this reduces to about 4a�,
which is what we'd get if we first placed one square (in the middle),
and considered the probability that the second square would intersect it.

    If you're going to get anywhere I think you'll have to assume these
    are convex polygons; i.e., any two points interior to the polygon
    have the line joining them entirely interior to the polygon.
    
    Otherwise you can use a Cantor-set approach to construct a fractal-
    like object with arbitrarily small area X but which covers the entire
    enclosing rectangle so much that you can't put a square of area X
    in the rectangle without crossing the first polygon.
    
      John

    Some tractable forms of the 'intersecting regions in a square' problem
    may use aligned squares of a given size, or intersecting line segments.
    
    Like John, I wondered what a random polygon *looks* like.

Peter:

I like your method in .3, but...

Aren't the corners of the shaded region:

	(xa,xb) = { (0,0), (0,a), (1-a-b,1-b), (1-a,1-b), (1-a,1-a-b),
			(b,0), (0,0) }

and isn't the area of the shaded region:

	(1-a)(1-b) - (1-a-b)^2

and isn't the probability of overlap:

	       (1-a-b)^2
	[ 1 - ------------ ] ^ 2
	       (1-a)(1-b)

which still approximates, for  a = b << 1:

	4a^2

John

    re: .4
    
    Good point - it's definitely reasonable to assume convexity for
    purposes of the paper.  I can mention that concavities would be
    an exception and pass on your suggestion for dealing with them.
    
    re: .3, .6            \
    
    thanks - I'll have to take a closer look at what you're saying.
    It looks like the kind of expressions I'd been fooling with;
    but how do you do the proof?  
    
    I guess what I'd ideally like to show is a direct correlation 
    between area and probability.  It may also be helpful to deal
    with it in terms of uncertainty - If you can quantitiate 
    the uncertainty by measuring  the presence of  factors that will
    throw you off (e.g. complex polygon shape, variability of "filling"
    density, number of objects in bounded area) it may be easier to
    show the basic correlation between area and probability - working towards
    the case where, for an uncertainty of 0, the relationships are directly
    provable...                                   
    
    What you said about Cantor sets sounds to me like "what you
    derive/prove is true, but only within an uncertainty limit of
    this area X"....?
    
    MP
                                                        
    

The conditions being given here "Put a RANDOM polygon (shape?) at a RANDOM
location inside some other shape" are very badly undefined.  The meaning of
"random" and "probability" are completely intertwined; just because the WORDS
sound like they mean something you understand doesn't mean they really DO mean
what you seem to think.

There is a classic conundrum that illustrates the problem:  Suppose a stick
is broken at random into three pieces.  What's the probability that the three
pieces form a triangle.

There are (at least) two reasonable interpretations of "breaking a stick into
three pieces at random".  In one, you choose two breakpoints independently.
In the other, you choose one breakpoint, then proceed to break the larger of
the remaining pieces.

If you work it out - not too complicated; this was in Martin Gardner's column
many years ago - you get two different answers - I think 1/2 one way, 1/3 the
other.
							-- Jerry

re .6:
	Yes (that's what I get for missing the easy way to calculate
	the area).

re .-1
	Lacking any other definition, I'd say the interpretation of
	'random' should mean that each possibility is equally likely.
	Of course, this is *not* what's intended in .0, but perhaps
	it leads to an interesting problem, nonetheless:

	Roughly how many different polygons are there in a bounded
	convex region?  There may be different answers for convex vs.
	arbitrary polygons, and also depending on whether the polygon
	may intersect itself (that last one seems the easiest).  The
	answer is infinity, but infinity of *what* order?

>	Roughly how many different polygons are there in a bounded
>	convex region?
    
    It would seem that there are no more than alef-null sides to a polygon,
    and each side is associated with one endpoint (the other endpoint
    belonging to the next side).  Each endpoint has c * c possibilities,
    so an upper bound is seen to be alef-null * c * c = c.  Also, it
    is easy to see that c is a lower bound (take one polygon and translate
    or rotate it a bit).
    
    The above probably assumes that a polygon must be a SCROC.  If you
    really want to include fractals, then the question remains open.
    
      John