Conference rusure::math

Following the form-feed is a solution to the first problem.  This is
a freebie, just to ensure that the problem is understood.

    1.  t = a + b*c


Let f(a,t) = (t-a)*a.  Note that this must be expressed without b's or c's.
Then f = ((a+b*c)-a)*a = a*b*c, which is symmetric in a, b, and c, since
f(a,b,c) = f(a,c,b) = f(b,a,c) = f(c,b,a).

    For t = 1/(b+c), an f is a+1/t.
    
    For t = (1+a^2)/(b+c)^2, an f is {a+sqrt([1+a^2]/t)}^2 -- the last
    square is to make the signs work.
    
    
    				-- edp 

     Re .-1
    
>>    For t = (1+a^2)/(b+c)^2, an f is {a+sqrt([1+a^2]/t)}^2 -- the last
>>    square is to make the signs work.

     With the proposed f(a,t) you get

          f(a,t) = {a + sqrt((1 + a^2)/t)}^2
                 = {a + sqrt((b+c)^2)}^2
                 = (a + |b + c|)^2

     which is not symmetrical:

          f(a=3, b=-1, c=-1) = 25            (This replaces an earlier
          f(a=-1, b=3, c=-1) =  1             reply where I had these
                                              wrong)

     Dan

    Items 3 and 4 are still unsolved.  Given t as a function of a, b, and c
    that is symmetric in b and c, find a non-trivial function f(a,t) such
    that f(a,t) is symmetric in a, b, and c.
    
    	3.  t = (1+a^2)/(b+c)^2       
    
    	4.  t = b + c + 1/(b*c)
    
    
    				-- edp

3.  t = (1+a^2)/(b+c)^2

This one is not too hard.  Suppose / let ...

	f1(a,t) = (1+a^2)/t = (b+c)^2

So far so good.  Let's guess that the desired f is:

	f(a,t) = (a+b+c)^2

In other words, we'll try f = f1 + 2ab + 2ac + a^2.
Now we massage those last terms: f = f1 + 2a(b+c) + a^2.
This works, since we have b+c = sqrt(f1)!

Solution:

	f(a,t) = (1+a^2)/t + 2 a sqrt((1+a^2)/t) + a^2

	       = (a+b+c)^2

    Re .6:
    
    The problem described in .4 still applies -- sqrt((1+a^2)/t) =
    sqrt((b+c)^2) = |b+c|, not b+c.
    
    
    				-- edp

>>    Re .6:
>>    
>>    The problem described in .4 still applies -- sqrt((1+a^2)/t) =
>>    sqrt((b+c)^2) = |b+c|, not b+c.

	Maybe we just have to restrict b+c to be nonnegative in
	order to solve #3.

	Dan

    3.	t = (1+a^2)/(b+c)^2

We can easily (and reversibly) change this problem to:

    3a.	t = (b+c)^2

We want F (t,a) = a symmetric function of a, b, and c.
         1

Let's suppose that F (t,a) takes the form of a polynomial in t, where each
                    1
coefficient in the polynomial is itself a polynomial in a.  We will below

that this cannot yield a symmetric function.


Suppose
	F(b+c,a) = P (a) + P (a) (b+c) + P (a) (b+c)^2 + P (a) (b+c)^3 + ...
	            0       1             2               3
where
	P (a) = p   + p  a + p  a^2 + p  a^3 + ...
	 i       i0    i1     i2       i3

Then the symmetry constraint

	F(b+c,a) = F(a+b,c)

greatly constrains the p  , so that F can be expressed as (below, p  = p  ).
                        ij                                         j    j0

	F(b+c,a) =
		(p  + p  a + p  a^2 + p  a^3 + p  a^4 + ...) +
		  0    1      2        3        4

		(p  + 2 p  a + 3 p  a^2 + 4 p  a^3 + 5 p  a^4 + ...) (b+c) +
		  1      2        3          4          5

		(p  + 3 p  a + 6 p  a^2 + 10 p  a^3 + 15 p  a^4 + ...) (b+c)^2 +
		  2      3        4           5           6

		(p  + 4 p  a + 10 p  a^2 + 20 p  a^3 + 35 p  a^4 + ...) (b+c)^3
		  3      4         5           6           7

		+ ...

		          j              i
	    =	Sum  (b+c)  Sum  C(i,j) a
		j>=0        i>=i

where C(i,j) is `i choose j', the binomial coefficient.  For any choice of the
p  values in the above equation for F, we have a symmetric function in a, b, c.
 j


Now for our problem at hand, we want the odd terms (those (b+c) terms with odd
exponent) to be identically zero.  For the first odd (b+c) term, this forces

	p  = p  = p  = ... = 0.
	 1    2    3

So the only polynomial function F ((b+c)^2,a) that is symmetric in a, b, and c
                                 1
is the trivial one -- a constant value.


A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
it's implausible.

        re .9
        
>>	A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
>>	it's implausible.
        
        I agree most rigorously!
        
        Dan

	I haven't worked through the details, but yours seems to be a
very plausible approach.   Tell me, why did you select the four example
polynomials that you did?   Is this a puzzle with a known solution?

	A restatement of the problem which formalizes the definition of the
problem, but which doesn't seem to help to reach any solutions is the following:

	Let K be a field.   For which t lying in K(a,b,c) [the field of
fractions in three indeterminates, a, b & c] is the intersection of
the two fields K(a,t) & K(a+b+c, ab+ac+bc, abc) strictly greater than K?

Regards,
Andrew.

> Tell me, why did you select the four example polynomials that you did?
> Is this a puzzle with a known solution?

I got the problem from Stan Rabinowitz.  I'll ask him.

    I'm not comfortable with the statement that a non-polynomial function
    of (b+c)^2 and a could not be symmetric.
    
    Consider F(t,a) =
    	0 for t= 4,a=1
    	1 for t= 9,a=1 and t=4,a=2
    	2 for t=16,a=1 and t=9,a=2
    	3 for t=16,a=2
    	4 otherwise
    
    Now for f(a,b,c) = F( (b+c)^2, a) =
    	0 when all of a, b, and c are 1.
    	1 when one of a, b, and c is 2 and the others are 1.
    	2 when two of a, b, and c are 2 and the other is 1.
    	3 when all of a, b, and c are 2.
    	4 otherwise.
    
    This is symmetric.  It's also not "trivial", although it's close.  It
    demonstrates symmetry can be achieved without a polynomial.  Certainly
    we could add as many more values to F's range as we please.  Can it be
    done in a more aesthetic manner?
    
    
    				-- edp

	re .9 (I think)

>>	A non-polynomial function of (b+c)^2 and a couldn't be symmetric --
>>	it's implausible.

	re .13

>>    I'm not comfortable with the statement that a non-polynomial function
>>    of (b+c)^2 and a could not be symmetric.
    
	Well, for a, b, and c nonnegative reals, we have already
	seen that F(a, t) = a + sqrt(t) yields F(a, (b+c)^2) = a+b+c.
	So for some domains nonpolynomial functions of a and (b+c)^2
	can be symmetric.  Over the whole real line it just becomes
	harder.

	Dan

>                       -< Symmetric function -- f(a,t) >-
>
>Let t be a function of a, b, and c that is symmetric in b and c.  [I.e.,
>t(a,b,c) = t(a,c,b)].  For the t given below, the problem is to find a
>non-trivial function f(a,t) that is symmetric in a, b, and c.

>    (3)  t = (1+a^2)/(b+c)^2

	This is all much easier than I had been thinking.   Let's make
the critical restriction that we want f to 'work' for all real values of
a,b & c.

	Equivalently to (3), take t = (b+c)^2.

	Now suppose that we have some symmetric f.   Then for any real x & y:

f(x,y^2) = 
f(x, (-x +(x�y))^2) =	(by symmetry, with a = x, b = -x, c = x�y)
f(x�y, (-x +x)^2) =
f(x�y, 0)

	ie:
f(x+y,0) = f(x-y,0)
	for any real x,y.

	From this it's trivial to see that
f(z,0) = f(0,0) = k, say, for any real z (take x = y = z/2).

	So
f(x,y^2) = k
	for all real x & y.

	So 
f(x,w) = k 
	for all real x and all non-negative w.   Now for w negative, we can
specify that f is anything we choose, and certainly it is certainly symmetric 
on a, b & c, since it maps straight to a constant for any non-negative w.
Whether we want to count such functions as trivial is a matter of taste, but
it is sure that these are the only kind of functions which will do the job.

Regards,
Andrew.

re .13:

    No.  Actually, given your definition of F(t,a),

    f(a,b,c) = F((b+c)^2, a) =
	0 for |b+c|=2, a=1
	1 for |b+c|=3, a=1, or |b+c|=2, a=2
	2 for |b+c|=4, a=1, or |b+c|=3, a=2
	3 for |b+c|=4, a=2
	4 otherwise

    f(a,b,c) is not symmetric, since f(1,1/2,3/2) = 0,
    but f(1/2,3/2,1) = 4.  Or using positive integers,
    f(1,1,3) = 2, while f(3,1,1) = 4.

.15:

    Thanks.  I was looking for a proof like this, but
    couldn't find it.  Can you find one for (4)?

Using Andrew's reply (.15) as a guidepost, let's attack:

>	(4)  t(b,c) = b + c + 1/(bc)


For any real x & u, we will have u = t(�,�), with � and � chosen later.  Now,

	f(x,u) = f(x,t(�,�))

	= f(�,t(�,x))		! by the symmetry condition

	= f(�,k)		! where k a constant

Thus, for all x and u,

	f(x,u) = f(�,k)

Below, � will turn out to be multi-valued, so we'll have

	f(x,u) = f(� ,k) = f(� ,k)
	            �         �
Then we'll see that we can make �  and �  take any values we like, by choosing
x and u.  So for any �  and � ,  �      �
                      �      �
	f(� ,k) = f(� ,k) = F(k), a function of k alone.
	   �         �

And since we've shown for any x and u (and incidentally k) there's a � such that

	f(x,u) = f(�,k),
we have
	f(x,u) = a constant.


Now we proceed with the aforementioned details.


Let T(z,a) be the inverse t function;

	z = t(a,b)	<=>	T(z,a) = b, and T(z,b) = a

	T(z,a) = �(z-a) � � sqrt((z-a)� - 4/a)

Note that T(z,a) is multi-valued (except, FWIW, when z = a - 2/sqrt(a).


How are � and � chosen?  We require that t(�,x) = k, so

	� = T(k,x) = �(k-x) � � sqrt((k-x)� - 4/x)

And t(�,�) = u, so

	� = T(u,�) = �(u-�) � � sqrt((u-�)� - 4/�)


Now let's choose x and u to get any desired values for �  and � .
	                                                �      �
	� = �(u-�) + � sqrt((u-�)� - 4/�)
	 �
	� = �(u-�) - � sqrt((u-�)� - 4/�)
	 �

	�  + �  = u-�
	 �    �

	(�  - � )� = (u-�)� - 4/� = (�  + � )� - 4/�
	  �    �                      �    �

	4/� = (�  + � )� - (�  - � )� = 4 � � 
	        �    �       �    �        � �

So for any desired �  and � , we can take
                    �      �	
	� = 1/(� � )
	        � �

	u = �  + �  + � = t(� ,� )	(of course)
	     �    �          �  �

	x = T(k,�) = �(k-�) � � sqrt((k-�)� - 4/�)


For what it's worth, note that nowhere did we actually choose or bind k.
It's a free constant.

>           <<< Note 682.17 by TRACE::GILBERT "Ownership Obligates" >>>
>Using Andrew's reply (.15) as a guidepost, let's attack:
						  ^^^^^^
	Aha!   Someone else whose mathematical muse is martial! (ref: 1258.3)

	Your approach seems good, but I think it needs to be tidied up
(as does my .15).   Basically, we can make it clearer to any readers what
things are independent variables, dependent variables, constants or
indeterminates at what stage.   This may seem nit-picking, but I don't believe
it is:  half the difficulty with this problem is the surprising difficulty
in coming up with the right formalism.   More concretely, I think you need
to say what fields you are working over at what stage.   You mention the
reals a lot, but T(x,y) will often be complex.   Do you need to explicitly
assume that f is symemtric when it takes constant arguments?   And what 
happens when b or c = 0 and t is infinite?

	For instance, here is a tidied version of the proof for Example (3).

	Let K be a field.   Let f: K x K -> K be such that 
		f(a,(b+c)�)) = f(b,(a+c)�) for all a,b & c in K.
	If char(K) <> 2, then there exists a constant l such that 
		f(a,(b+c)�) = l, for all a,b & c in K.

Proof: 
Assume char(K) <> 2.
	f(a, (b+c)�)
=	f(a, (a+b+c -a)�)
=	f(a+b+c, (a -a)�)	(by symmetry)
=	f(a+b+c, 0)
=	f(d, 0)			(where d = a+b+c)
=	f(d, (d/2 -d/2)�)
=	f(d/2, (d -d/2)�)	(by symmetry)
=	f(d/2, (d/2)�)
=	f(d/2, (0 -d/2)�)
=	f(0, (d/2 -d/2)�)	(by symmetry)
=	f(0, 0)
=	l, say.

	Note that if char(K) = 2, then (b+c)� = b�+c�.   So f(a,t) = a�+t
is a non-trivial function symmetric in a,b & c.

Regards,
Andrew.

.18>	Your approach seems good, but I think it needs to be tidied up

Yes, I think so, too, and have a dozen pages of `tidyings', and proofs that
many other functions t(a,b) allow no solution.  Allow me a day or two, please.

.17>	u = �  + �  + � = t(� ,� )	(of course)
.17>	     �    �          �  �

This 'of course' is not at all obvious, nor is it true.  However, it's true for
t(b,c) = b + c + 1/(bc) (as in .17), and t(b,c) = bc(b+c), and ...?

Question: I suspect that if the function t(b,c) permits an F(a,t) that is
symmetric in a, b, and c, then t(b,c) take the form:

	t(b,c) = R( P(b) + P(c) )

where P and R are arbitrary functions.  Is this true? (note that the addition above
also accounts for multiplications)  Or are there some non-trivial t(b,c) that permit
a symmetric F(a,t)?

Problem:
	Given t(b,c) symmetric in b and c, find F(a,t(b,c)) symmetric in a, b, and c.
	Or prove that no such F exists. 

Symmetry means:

	t(b,c) = t(c,b)

	F(a,t(b,c)) = F(b,t(c,a)) = F(c,t(a,b)) = F(a,t(c,b)) = F(b,t(a,c)) = F(c,t(b,a))

Method of showing non-existence:

	Suppose for every b and c we can:

	    (step 1)	Solve t(u,b) = t(u,c) for u.

	Now u = u(b,c).  (the u needn't be unique, nor need it be explicit; we're pretty lax).
	We have:

	    F(c,t(u,a)) = F(a,t(u,c)) = F(a,t(u,b)) = F(b,t(u,a))

		F(c,t(u,a)) = F(b,t(u,a))	-- for all b and c.

	Now if we can:

	    (step 2)	Solve k = t(u,a) for a.

	Then for any k, we can choose a so that k = t(u,a).  And then

		F(c,k) = F(b,k)			-- for all b and c.

	Thus F(a,t) is independent of a, and (by symmetry) independent of b and c.

For example:

	Given: t(b,c) = bc(b+c).

	t(u,b) = t(u,c) = ub(u+b) = uc(u+c)	-- (step 1)

		=>  u (b-c) ( u + b + c ) = 0

	So take u = -(b+c).  (We could take u = 0, but that makes (step 2) difficult/impossible.)

	For step 2, k = t(u,a) = ua(u+a).  Ayup, that's a quadratic.  We could solve for `a', but
	we're lazy -- all we care about is the existence of such an `a' for any k and u.

	And that's it.

Another example:

	t(b,c) = (b+c)�.

	t(u,b) = t(u,c)  =>  (b-c)( 2u + b+c ) = 0  =>  u = -(b+c)/2  (ignore the b=c solution).

	k = t(u,a)  =>  a = sqrt(k) - u, so an inverse t-function exists.

	Done.

	Knowing that it is a quadratic doesn't imply real solutions.

	Dan

.21> Knowing that it is a quadratic doesn't imply real solutions.

True, but it doesn't matter -- only existence matters.

        But if the only roots that exist are complex, then it
        says nothing about whether there is a nontrivial F(a,t)
        over the reals that satisfies the symmetry constraint.
        
        Dan

	I have a new result, which indicates the sort of directions my
thoughts have been taking:

	Let M be a finite set of cardinality m.   Let t be a symmetric 
function from M x M *onto* a set N, of cardinality n.   If n > m�/3, then there 
exists a non-trivial symmetric function, �: M x M x M to L, and a function
F: M x N to L such that � = Ft.

	Idea: if we have a finite underlying set on which t acts, and
if t is not too 'destructive of information', then the kind of F we want will
always exist.

	I would like to abstract further away from sets, to categories, since
I think that is the natural lair of this problem, but I am not familiar
enough with categorical talk.

	Btw, I agree with Dan.   If you permit u & a to be complex, then you
must extend t & F to deal with complex arguments.   Then you prove that
such an extended F must be constant.   But why can't an *inextensible* F
exist, which is not constant?   This is not nit-picking, it is a fundamental
logical flaw in your argument.

Cheers,
Andrew.

1. Counter-example to Peter's algorithm.

	I think I've found a function t(b,c) which satisfies Peter's
step(1) & step(2) as he posed them, and yet admits a non-constant
symmetric F.   I was looking for such a function in order to bolster my
rather unsupported arguments that Peter's treatment of complex numbers was
inappropriate.

	t(b,c) = b� - bc + c�

	First, step 1.

	t(u,c) = t(u,b)
=>	-uc + c� = -ub + b�
=>	u = b+c

	No problems there.   Now for step 2.

	k = t(u,a)
=>	k = u� - ua + a�
=>	a = �u � �sqrt(4k - 3u�)

	According to Peter's approach, this is OK, even though a will be
complex for some k.   Therefore, t does not admit a non-constant function F.

	Aha, but it does, sneakily.   Consider:

F(a,t) = 1 iff a = t = 0
	 0 otherwise	

	The key thing is that t(b,c) = 0 *iff* b = c = 0.   End of story.

	Why is this a natural thing to consider?

Lemma:	Let M is a set, and t a symmetric function from M x M onto a set n.
Further, say there is an x in M such that for no y, z in M does
		t(x,x) = t(y,z)
Then there exists a non-constant symmetric function, �: M x M x M to L, and
a function F: M x N to L, such that � = Ft.


2. Contined exploration of the finite case.

>	Let M be a finite set of cardinality m.   Let t be a symmetric 
>function from M x M *onto* a set N, of cardinality n.   If n > m�/3, then 
>there exists a non-trivial symmetric function, �: M x M x M to L, and a 
>function F: M x N to L, such that � = Ft.

	Puzzle: assume that m > 0 (& finite as before).   Can we sharpen the
result above to include the case where n = m�/3?   This is instructive.

3. Hint as to the approach for 1. & 2.

	Let M be a set.   Let M�, M� be the set of unordered pairs, triplets
of elements in M.

	We can view t as a function from M� -> N.   Our question is, can we
find  P,F,w, such that the following diagram commutes (Ft' = wu):

			M x M� -----> M�
			  |     u     |
			  | t'        | w  
			  v           v
			M x N ------> P
				F

where: t' is simply an 'enhanced' t mapping M x [a,b] -> M x t(a,b)
       u is the 'obvious' map, mapping {a} x [b,c] -> [a,b,c]
       P is the pullback: (ie any other P',F',w' satisfying the commuting
property are such that a map z: P -> P' can be defined such that zF = F',
zw = w'.)

	You can perhaps see how the finite case in 2. above, under this
model, translates into a simple graph theory problem.   The lemma also becomes 
a trivial special case.  More interesting will be (when I have time) to 
examine how a repaired version of Peter's algorithm can be interpreted in this 
categorical space, and how the earlier arguments for t = (b+c)� reach the same 
conclusions without demanding the quadratic closure of M.   Unless some
person cares to attain the destination before me...

Regards,
Andrew.

	In addressing the n = m�/3 question, the following t is useful:
Let N = M = {stones, scissors, paper}, and let t(b,c) = "the better of b & c".
ie:	t(stones,scissors) = t(scissors, stones) = t(stones,stones) = stones.

	If we represent N as a partition of M�, then I think that the following
is a complete list of the exceptions to the n = m�/3 conjecture:

{ab, bb}	{ac, aa}	{bc, cc}	[the example given above]
{ab, bb}	{ac, cc}	{bc, aa}	[N = Z_3, t(b,b) = b
						otherwise t(b,c) = b+c.]
{ab, bb, cc}	{ac, aa}	{bc}
{ab, bb, cc}	{ac}		{bc, aa}
{ab, ac, bb}	{bc}		{aa, cc}

	I don't think that one should look to find any special patterns in
this lot, apart from the obvious one that MxN is given a tree structure where
the edges are defined by M�.   Oops, what a giveaway... :-).

Regards,
Andrew.

1. Insight before foresight

	A teeny error in my treatment of the finite case earlier, in the case
where m = 1.   To summarize, F can always be picked to be non-constant, 
symmetric in the way we want if:

		(i) n > m�/3  &  m <> 1
or		(ii) n = m�/3  & m <> 3

	Eg: suppose m = n = 2.   Then n = 2 > 4/3 = m�/3.   t can be, wlog:
{aa}		{ab, bb}	(permits F profiting from the singleton aa)
{aa, bb}	{ab}		(t is + over Z_2, so permits F)


2. step(1) & step(2)

	Neither of these are necessary for "all symmetric F are constant"
to hold.   step(1) is falsified by the counter-example of Z_3 given previously.
		t(u,1) = t(u,2)		has no solutions
As for step(2), when m=3, the *only* symmetric t which satisfies step(2) is +
over Z_3.


Regards,
Andrew.

I should've mentioned in .20 that some of the steps may require that the range
of b & c be reduced.  Unfortunately, that doesn't apply to the argument in .25.

.25>	The key thing is that t(b,c) = 0 *iff* b = c = 0.   End of story.

Eh?  I find b = c(1�sqrt(-3))/2 also gives t(b,c) = 0.  What's the value of
your t function for complex numbers?  I was assuming a complex number domain,
even though I hadn't so stated.

.23>    But if the only roots that exist are complex, then it
.23>    says nothing about whether there is a nontrivial F(a,t)
.23>    over the reals that satisfies the symmetry constraint.

You mean that if we extend the problem domain to complex numbers, we can prove
there are no solutions, but if we restrict ourselves to the reals, then there's
still a question.  This is very curious.

>Eh?  I find b = c(1�sqrt(-3))/2 also gives t(b,c) = 0.  What's the value of
>your t function for complex numbers?  I was assuming a complex number domain,
>even though I hadn't so stated.

	In this example, I am taking M to be the reals.   I assumed from your
remarks that you were doing the same, since your remarks about 'laziness'
didn't make sense if you were working over the complex field from the outset.   

>You mean that if we extend the problem domain to complex numbers, we can prove
>there are no solutions, but if we restrict ourselves to the reals, then there's
>still a question.  This is very curious.

	Exactly!   (Assuming that our function t is algebraic.   There are
plenty of strange functions which are not algebraic that would give us trouble 
over the complex field.
eg:
	  t(b,b) = b
otherwise t(b,c) = 0

	This is a perfectly good function, whatever set (containing zero) we 
work over.   And cutely F(x,y) = t(x,y) is non-constant symmetric.)

	Note that if t(x,y) is any symmetric function then we can create a
pathological but symmetric mutant of it given any function z of M.   Just 
define:
	  t*(b,b) = z(b)
otherwise t*(b,c) = t(b,c)

	Remember that N is defined by the fact that t is *onto*.   So we
must remember that range(t*) needs to be calculated carefully.

	Incidentally, t = b+c+1/bc for M = the reals is still an open
question.

(A) NEW THEORY

	A minor modification of Peter's algorithm seems to do the trick.

	Given t, to show that all symmetric f are constant, it is clearly 
sufficient to show that for all b,c in M and all x in N:

		f(b,x) = f(c,x)

	Peter's idea relied on two conditions:

		(1) for all b,c in M, we can find u in M with:
			t(b,u) = t(c,u)

		(2) for all k in N, and u in M, we can find a in M with: 
			t(u,a) = k


	Given k in N, suppose that I can define a binary relation ~ on M,
and a subset M' of M, such that:

		(1) for all b,c in M with b~c, we can find u in M' with:
			t(b,u) = t(c,u)

		(2) for all u in M', we can find a in M with:
			t(u,a) = k

	Now examine the partition of M under the transitive closure of ~.
(This is valid since ~ is symmetric).   If there is exactly one equivalence
class (ie ~ 'connects' M), then all symmetric f are constant.


(B) EXAMPLES

	This is much easier to see in the context of an example.

	Example1: t = b + c + 1/bc,	M = R \ {0}.

	In this example, ~ and M' can be chosen independently of k, so:
	We define b~c iff bc < 0
	We define M' = R- (the negative reals)

	step (1) b + u + 1/ub = c + u + 1/uc =>
		 u = 1/bc.
		b~c => u lies in M'.

	step (2) a + u + 1/ua = k =>
		 ua� + (u-k)ua + 1 = 0 =>
		 u = �(k-u) � �sqrt( (u-k)� - 4/u )

	If u lies in M', the discriminant is evidently positive and a is real.
There may be some positive values of u which make the discriminant positive,
but they aren't important.   Also, u <> 0.

	Now ~ connects M.   In fact, abusing notation, the graph (M,~) has 
diameter 2.


	Example2: t = bc(b+c), M = R.

	Here, we cannot pick ~ & M' independently of k.
	Suppose that k <> 0.
	We define b~c iff k/(b+c) < 0
	We define M' = R+ if k > 0
		     = R- if k < 0

	step (1) bu(b+u) = cu(c+u) has a solution
		 u = -b-c
		b~c => k/(b+c) < 0 => k/u > 0 => u lies in M'

	     (2) ua(u+a) = k =>
		 a� + ua -k/u = 0 =>
		 a = -�u � �sqrt( u� + 4k/u )
		u lying in M' => discriminant is +ve, so a is real.

	Now, given k, does ~ connect M?   Well, again the graph (M,~) has
diameter 2.   If k < 0, then can find d > max{|b|,|c|} such that k/(b+d) < 0
& k(c+d) < 0.   So b~d~c.   Similarly if k > 0.

	It remains to deal with the case k = 0.
	Define b~c iff b+c <> 0.
	Define M' = R \ {0}.
	And it all works out even easier than the cases k <> 0.

	
	Example3: t = (b+c)�,		M = R.   
Note that N = R+ & {0}.   This means that the discriminant, k, in step(2) is
always non-negative, and so a is real-valued.   Thus Peter's algorithm is quite
adequate to handle this case without getting upset.   

	In terms of the new algorithm, we can select b~c for all b,c & M' = M, 
whatever value of k we want.


	Example4: t = b� - bc + c�,	M = R \ {0}   Note, N = R+.
	Given k > 0,
	Define b~c iff |b+c| < k
	Define M' = (-k, k)

	step (1) yields u = b+c
	step (2) yields a = -�u � �sqrt( 4k - 3u�)
The discriminant is always > 0, so a is real-valued.   a is also non-zero.

	Returning to (M,~), we find it is connected, with no finite diameter.
Between any two values b & c, there is a path of length ceil(|b+c|/k).


(C) WHAT NEXT?

	So the next questions this suggests to me are:

	(NQ1)	How close is this new result to being *necessary*?
	(NQ2)	t = b + c + 1/bc,	M = R+.


Regards,
Andrew.

>	Example4: t = b� - bc + c�,	M = R \ {0}   Note, N = R+.
>	Given k > 0,
>	Define b~c iff |b+c| < k
>	Define M' = (-k, k)
>
>	step (1) yields u = b+c
>	step (2) yields a = -�u � �sqrt( 4k - 3u�)
>The discriminant is always > 0, so a is real-valued.   a is also non-zero.
>
>	Returning to (M,~), we find it is connected, with no finite diameter.
>Between any two values b & c, there is a path of length ceil(|b+c|/k).

	The fool who wrote this should have written there is a path of
length at most 1 + ceil(||b|-|c||/k).


Regards,
Andrew.

>>	The fool who wrote this should have written there is a path of
>>	length at most 1 + ceil(||b|-|c||/k).

	You should be careful ... not everyone who reads these
	realizes that you are referring to a note that you wrote.
	So they go away thinking that this is a hostile conference.

	Dan

	The much-admired person who wrote .30 inquired as to the fate of
t = b + c + 1/bc when M = R+.   In fact, using a (hopefully familiar) result,
F(1,1) = 1, otherwise F(x,y) = 0, gives Ft as a 3-way symmetric function.

Regards,
Andrew.

.30> Now ~ connects M

	It took me a while to understand this.  In the case of Example2, it
	means that

		F(b,k) = F(c,k), for any bc < 0 and k in M.

	But now we "connect".  For b > 0,

		F(b,k) = F(-1,k) = F(1,k).

	And for b < 0,

		F(b,k) = F(1,k).

    5.  t(b,c) = ( b + c + bc )/( 1 + b + c )

Enjoy.

(1) A solution found.

	t(b,c) = (b + c + bc) / (b + c + 1)

	I haven't looked very thoroughly at this curious function.   In
particular, the handling of cases where t has a zero denominator can no
longer be tackled by restricting M.   What one should have is a value oo
(infinity) in N, which is given to divide-by-zero cases.   The same thing
should happen for fields M which contain roots of the poly x� + x + 1 = 0,
because then there is the possibility of 0/0.

	However, we note that t(b,c) = d => t(b,-d) = -c.   And this inspires
me to propose:

	f(-x,x) = 1	for all x in M-intersect-N
	f(x,y)  = 0	otherwise, for all x in M, y in N.

	f(a,t(b,c)) = 1
<=>	t(b,c) = -a
<=>	t(b,a) = -c
<=>	f(c,t(a,b)) = 1

	That's neat.   What made it occur to you as a function to suggest?
Have you had any feedback from Stan the Man?


(2) two tiny corrections (which make me suspicious that no-one ever reads
anything I write! :-)

	For  n = m�/3, another exception overlooked is for M = N = � (empty).
Since then, the only f from � x � is trivial and hence constant.

	For t = b + c + 1/bc, M = R+, the definition of f should read: 
		f(1,3) = 1, otherwise f(x,y) = 0.


Regards,
Andrew.

.37> That's neat.   What made it occur to you as a function to suggest?

I was looking at functions t(b,c)=N(b,c)/D(b,c) that had a minimum (or maximum).
If M=t(z,z) is an extrema, then F(a,k) = { K1 if a=z and k=M; else K2 }.

Actually, (b+c+bc)/(b+c+1) has a far less trivial solution!

.37> Have you had any feedback from Stan the Man?

Yes, I saw him this weekend.  The problem is his own, and I saw the solutions
given in `Crux' -- these were for t=(b^2+c^2) and t=bc(b+c), and the solutions
were very problem-specific.  The second half of the problem asks for a t(b,c)
that permits a symmetric F(a,t).  It's still open in `Crux'.  I'm still working
on it.

What's Stan doing?  He's still at Avid, and...

Stan is working on a concordance of math problems that have been published in
math magazines over the past 5 (or ten) years.  If you're interested in some
extra cash, by typing some in, or by translating Polish, Rumanian, ..., or if
you know someone who is, then call Stan at (617) 272-1680.

(1) A clarification

.38>	I was looking at functions t(b,c)=N(b,c)/D(b,c) that had a minimum (or 
.38>	maximum).   If M=t(z,z) is an extrema, then 
.38>	F(a,k) = { K1 if a=z and k=M; else K2 }.

	But (b+c+bc)/(b+c+1) is not such a function.   It will have local
extrema at (w,w) & (w�,w�), where w & w� are non-unity roots of x� = 1.
But these will not be *global* extrema.   t(w�,w�) = -w, and
		-w = t(b,-t(b,w)), for almost any b 
by .37.   Consequently, assuming M is a field, even if M contains w, then there
is not a unique solution to t(b,c) = -w, and so the F you propose is not
likely to work.

	Concretely, suppose we proposed:
		F(w�,-w) = K1, else F(a,k) = K2

	K1 
= 	F(w�,-w)
=	F(w�,t(0,-t(0,w))	(taking b = 0 in the expression for -w above)
=	F(0,t(w�,-t(0,w))	(by symmetry)
=	K2			(by definition of F)
	So our F is constant.

	Compare this with t = b� - bc + c�, where (0,0) is a *global* minimum.


(2)	An obvious generalization of .37

	Let ' be an involution on M.   ie: (' is a bijection from M -> M, and
a'' = a for all a in M.)   Suppose that we have some symmetric t such that:
		a = t(b,c) => b' = t(a',c)

	Then defining f(a',a) = 1, otherwise f(a,k) = 0 is ok since:
	f(a,t(b,c)) = 1
<=>	a' = t(b,c)
<=>	b' = t(a,c)
<=>	f(b,t(a,c)) = 1

		What kinds of function t will satisfy this requirement?
Any symmetric t where:
	a' = t(b,t(b,a)')
Now we can regard t(b,x) as a function, B, taking x -> t(b,x).   So the
fundamental requirement on B, is that 'B is an involution.

	For instance:

Example 1:	M = a field, 
		': x -> x 
		t(b,c) = 1/(bc)	
		therefore B: x -> 1/(bx)
	'B'B: x -> 1/(b.(1/bx)) = x.
	Define f(a,t) = 1 iff a=t, otherwise 0 
	If we have g(a,t) = t/a, and h(x) = Kronecker d(x,1) 
then f = hg.   So our f is a very crude non-constant function, compared to
g, but non-constant none the less.

Example 2:	M = a field containing no non-trivial cube root of 1.
		': x -> -x
		t(b,c) = (b+c+bc)/(b+c+1)
		therefore B: x -> (b+x+bx)/(b+x+1)
	'B'B: x-> -(b - (b+1)*(b+x+bx)/(b+x+1) )/( b+1 - (b+x+bx)/(b+x+1) )
	 = -(b*(b+x+1) - (b+1)*(b+x+bx) )/( (b+1)*(b+x+1) - (b+x+bx) )
	 = (b�+b+1)x/(b�+b+1)
	 = x	{as long as b is not a non-trivial cube root of 1}
	 f(a,k) = 1 iff a=-k, otherwise 0

Example 3:	M = the set of bitstrings, length n
		t = reverse (b XOR c)
(fill in the gaps yourself)


(3) More questions:

	(1) How do we go about generating lots of functions 'B?
	(2) When can we get f to be continuous?

Regards,
Andrew.

.39>	But (b+c+bc)/(b+c+1) is not such a function.

    Oops.  I overlooked forest while searching for a certain tree.
    But no matter.  Read on.

>	(1) How do we go about generating lots of functions 'B?
>	(2) When can we get f to be continuous?

    Well, ....  Let T be an inverse of t, T(K,a) = b  =>  t(a,b) = K.

	  F(a ,t(b ,c ))
	= F(a ,   k    )
	= F(a ,t(x1,x2))	where t(x1,x2) = k , or x2 = T(k,x1)
	= F(x1,t(a ,x2))
	= F(x1,t(y1,y2))	where t(y1,y2) = t(a,x2), or y2 = T(t(a,x2),y1)
	= F(y2,t(x1,y1))

    Now x1 and y1 are `free'.  Choose them so the inverses, x2, and y2 exist and
    are easily calculated (i.e., often this means choosing x1=y1=0).
    Now t(x1,y1) is a constant, and we see that

	F(a,k) = F(y2,t(x1,y1)) is a function of the single expresion y2,

	where y2 = T(t(a,T(k,x1)),y1), and x1 and y1 are arbitrary constants.

    This is progress.

    For example, suppose (as in .35,.37) that t(b,c) = (b+c+bc)/(b+c+1).

	T(K,a) = (K - a + K a)/(a - K + 1)

	x2 = T(k,x1) = (k - x1 + k x1)/(x1 - k + 1)

	t(a,x2) = ((a k + k - 1) x1 + k + a)/((k + a) x1 - a k + a + 1)

		(here, we could just set x1 = 0, but we'll leave it in)

	y2 = T(t(a,x2),y1)

		  (k a - a - 1) x1 y1 + (k a + k - 1) (x1 + y1) + (k + a)
		= -------------------------------------------------------
		  (k + a) x1 y1 - (k a - a - 1) (x1 + y1) - (k a + k - 1)

		  (x1 y1 + x1 + y1) (k a - 1) + (x1 + y1 + 1) k - (x1 y1 - 1) a
		= -------------------------------------------------------------
		  (x1 y1 - 1) k + (x1 y1 + x1 + y1) a - (x1 + y1 + 1) (k a - 1)

	(Notice this is symmetric in x1, y1 -- is that a necessary effect of the
	construction?  Also notice other regularities).


	We want F(a,k) = y2 to be a symmetric function in a, b, and c (recall
	that k = t(b,c)).  We need only show F(a,t(b,c)) - F(b,t(a,c)) = 0,
	since t is already symmetric.  Indeed, this verifies, and the function
	is symmetric!


	Of course, the y2 above is not the only solution.  Any function of y2
	is also a solution.  However, if the inverse functions doesn't restrict
	the domain, *any* solution is expressible as a function of y2.  For
	example,

.37>	f(-x,x) = 1	for all x in M-intersect-N
.37>	f(x,y)  = 0	otherwise, for all x in M, y in N.

	This is f(a,k) := (if y2 = -(x1*y1+x1+y1)/(x1+y1+1) then 1 else 0),
	where y2 and the constants x1 and x2 are as described above.  I was
	pleased & relieved to verify that y2 = -(x1*y1+x1+y1)/(x1+y1+1) only
	when (a+k)(x1�+x1+1)(y1�+y1+1) = 0; the a=-k condition appears.


	In general, the function:

		         n0 + n1 (b + c) + n2 b c
		t(b,c) = ------------------------
		         d0 + d1 (b + c) + d2 b c

	permits the above construction of a nontrivial, symmetric F(a,t(b,c)).
	This result is straight-forward, but tedious.  I've tried putting in
	some squared or higher-order terms, but then the problem becomes
	overwhelmingly tedious.  Perhaps summations can be used.

(1) Theory

	I like Peter's new idea, which I tidy slightly below.

	Essentially, this *entire* problem is all about connectivity of a 
certain graph, which is infinite for most of the problem we want to consider.  
Connectivity being what it is, there is unlikely to be any all-encompassing 
necessary & sufficient condition for the graph to be connected.   However, we 
have been successful in identifying some good general *sufficient* conditions.

	We can view the vertices of our graph as an array of elements (a,k), 
where a lies in M, and k runs over N.   We can view the edges of the graph as
lying in two families:
		(i) (a,t(a,b)) <-> (b,t(a,a))			[a <> b]
		(ii) (a,t(b,c)) <-> (b,t(a,c)) <-> (c,t(a,b))   [a<>b<>c<>a]

	The earlier algorithm was interested in showing that any one row was 
connected, (ie: (b,k) = (c,k) for all b,c,k) since from that the connectivity
of the entire graph follows trivially.

	Peter's new algorithm takes a different approach.   He asks, given 
a,b & c can we find x1,x2,y1,y2, such that the following hold true:
		(1) t(x1,x2) = t(b,c)
		(2) t(x2,a) = t(y1,y2)
	If we can, then we know that (a,t(b,c)) is connected to (y2,t(x1,y1)).
Now suppose that:
		(3) y2, & t(y1,x1) are symmetric in a,b,c.   
	Then the two families of edges listed above, can give us no more 
connections than we already have deduced.   So, F(a,k) = (y2, t(x1,y1)) is a 
symmetric function of a & t and as Peter says any other symmetric function of 
a & t(b,c) will be of the form QF, where Q is composed with F.

Notes:
(o)	Yes, F is a mapping from M x N -> M x N (but not nec. surjective!)
(i)	(1), (2) & (3) are not *necessary* for F to exist.
(ii)	x1,x2 & y1 can be non-symmetrical without disrupting the algorithm.
(iii)   (1) & (2) are widely satisfied, they are related to Peter's "step 2".
	(3) is rather tougher to satisfy.

	However, before I go into examples, I want Peter to check his working
on the famous (b+c+bc)/(b+c+1).   I don't believe that the y2 he derives is
symmetric, even for x1=y1=0.

Regards,
Andrew.

.41> We can view the edges of the graph as lying in two families:
.41>		(i) (a,t(a,b)) <-> (b,t(a,a))			[a <> b]
.41>		(ii) (a,t(b,c)) <-> (b,t(a,c)) <-> (c,t(a,b))   [a<>b<>c<>a]
	(also worth mentioning in this reformulation is:)
		(iii) t(b,a) = t(a,b)				[a <> b]


.41> given a,b & c can we find x1,x2,y1,y2, such that the following hold true:
.41>		(1) t(x1,x2) = t(b,c)
.41>		(2) t(x2,a) = t(y1,y2)
.41>	If we can, then we know that (a,t(b,c)) is connected to (y2,t(x1,y1)).

Actually, the x1 and y1 were free variables; but above, they aren't -- it
suggests that *any* x1,x2,y1,y2 satisfying the above constraints suffice.
I think the difference is important, viz:

		(1) t(x1,x2) = t(b,c),		with x2=c, x1=b
		(2) t(x2,a) = t(y1,y2),		with x2=c, y1=a, y2=c

By .41, we now know that (a,t(b,c)) is connected to (c,t(b,a)).  This has made
the problem altogether too easy.


.41> Now suppose that:
.41>		(3) y2, & t(y1,x1) are symmetric in a,b,c.   

	You lost me.  How/why did t(y1,x1) pop in?


I'm still curious about that x1,y1 symmetry.  Returning to it, ...

	Define T by  T(k,b) = c  =>  t(b,c) = k

	And the construction:
			k  = t(b,c)
		Define	x2 = T(k,x1)
		Define	k2 = t(a,x2)
		Define	y2 = T(k2,y1)

	When y2 is symmetric in a,b,c, we've seen that it is also symmetric in
	x1,y1.  It's tedious but not difficult to find functions t(p,q) for
	which y2 is not symmetric in a,b,c; in those cases, it's also not
	symmetric in x1,y1.  Is there a pattern?  In general,

		If  t(p,q) = t(q,p)
		and T(t(a,T(t(b,c),x1)),y1) = T(t(b,T(t(c,a),x1)),y1)
					    = T(t(c,T(t(a,b),x1)),y1)
		Does
		    T(t(a,T(t(b,c),x1)),y1) = T(t(a,T(t(b,c),y1)),x1)

	That is, does the a,b,c symmetry imply the x1,y1 symmetry?

	The t operation is commutative (by its definition).  Suppose it is also
	an associative group, then letting p�q = t(p,q), we have T(k,a) = k�a'
	(the ' represents the inverse, or an exponent of -1), and now:

		T(t(a,T(t(b,c),x1)),y1)
			= t(a,T(t(b,c),y1))�x1'
			= a�T(t(b,c),y1)�x1'
			= a�t(b,c)�y1'�x1'
			= a�b�c�y1'�x1'
			= a�b�c�x1'�y1'
			= ...
			= T(t(a,T(t(b,c),y1)),x1)

	But can this be shown when t is not associative?

.42>	(also worth mentioning in this reformulation is:)
.42>		(iii) t(b,a) = t(a,b)				[a <> b]

	t(a,b) = t(b,a) of course, but it doesn't create any links in the
graph over vertex set M x N, so it isn't a "family".

.41> given a,b & c can we find x1,x2,y1,y2, such that the following hold true:
.41>		(1) t(x1,x2) = t(b,c)
.41>		(2) t(x2,a) = t(y1,y2)
.41>	If we can, then we know that (a,t(b,c)) is connected to (y2,t(x1,y1)).

.42>Actually, the x1 and y1 were free variables; but above, they aren't -- it
.42>suggests that *any* x1,x2,y1,y2 satisfying the above constraints suffice.

	Exactly!   As far as generating links is concerned, I am free to
pick whatever x1 & y1 I like, dependent on a,b,c if necessary.   However, I
will only have found a symmetric f if y2 & t(x1,y1) are symmetric.

.42> I think the difference is important, viz:
.42>		(1) t(x1,x2) = t(b,c),		with x2=c, x1=b
.42>		(2) t(x2,a) = t(y1,y2),		with x2=c, y1=a, y2=c
.42> By .41, we now know that (a,t(b,c)) is connected to (c,t(b,a)).  

	True.

.42> This has made the problem altogether too easy.

	Ah, no.   Let's look at what we've actually *achieved* if we take this
approach.   We haven't done anything useful, in terms of identifying the
connectivity of M x N, since the set of points of the form (y2,t(x1,y1)) is
no smaller than M x N itself.   The whole point of Peter's algorithm was that
it identified a subset, S, of M x N (often a column, M x {k}) such that every
vertex in M x N is connected to at least one vertex in S.

	Now, if in addition, every point in S is symmetric in terms of 
a,b & c, we know immediately that no two points in S are connected.   This is
because each edge of M x N belongs to one of the two families above.   It will
have two end-points u & v, and both are already known to be connected to the
*same* point of S (by symmetry).   So no new edge links members of S.   This
gives us our pullback.

.41> Now suppose that:
.41>		(3) y2, & t(y1,x1) are symmetric in a,b,c.   

.42>	You lost me.  How/why did t(y1,x1) pop in?

	I'm generalizing.   You fixed x1, y1 a priori.   I'm saying, there's no
need to do that, we can choose them as we go along, and the only thing we need
to ensure is that our eventual S only contains symmetrical points, so that
we can have the pullback effect discussed above.   Each point of S is of the
form (y2,t(x1,y1)).   So I say that y2 & t(x1,y1) must be symmetrical, and
that's enough.

	And if we end up *not* be able to find a cute continuous f, then the
postponement of definition of x1 & y1 gives much better scope for opportunistic
selection of them, in order to pin down what the connectivity must be.   To
see this in action, I will post a solution of t(b,c) = b/c + c/b using the
modified version of your algorithm.   This is an interesting function, because
for M = R \ {0}, the pullback is non-trivial but *not* continuous.

	So, either way, there are advantages to this modest generalization.

	Note, there is no simple way to repair the main weakness of your
algorithm, which is that it can only work for functions whose graphs over
M x N have radius =< 2 for each connected component.   Do you see what I'm
saying?

.42> I'm still curious about that x1,y1 symmetry.
  
	Yes, me too.   However, I'm loth to invoke associativity (or even
commutativity!) when they're not necessary to prove what you want.

	t [or equivalently �] is a symmetric function from M x M to N.   Let's
work in the assumption that T [we'll write as ^] is well-defined, that there 
is a unique function N x M -> M such that (a�b)^b = a, & (k^a)�a = k.

	If (((b�c)^x1)�a)^y1 = (((b�a)^x1)�c)^y1, then
	((b�c)^x1)�a = ((b�a)^x1)�c = k, say,
=>	b = ((k^a)�x1)^c
&	b = ((k^c)�x1)^a

	So equating these two expressions for b, gives us the result that
you were looking for.

	However, I'm still a bit mystified about the equation:
	(((b�c)^x1)�a)^y1 = (((b�a)^x1)�c)^y1.
Peter seems happy that it's always going to be true for the family of t he
identified.   Algebraically, I can see that it works out, but I would like
to understand *WHY* it all works out so neatly.

	I can see that Peter's family of functions are generalizations of the
M�bius transformations:

	z -> (a + bx)/(c + dz)

	I think that the secret of their behaviour lies there.

Regards,
Andrew.

(1) t(b,c) = (b+c+bc)/(b+c+1)

	Let me admit my slip in verifying Peter's .40 in my .41.

	Once I'd got T the right way round, the 3-way symmetry of
tTt is clear.   Note that we don't have to go all the way to TtTt, since
the symmetry appears at the stage before.   Thus:

	f(a,t) = (s1 + s2) / (s0 + s1 - s3)

where si is the ith symmetric polynomial.   This expression is not y2,
but the earlier-calculated y2�y1, which is also symmetric.



(3) t(b,c) = b/c + c/b,		M = R \ {0}

	Using the modified version of Peter's algorithm, we find there are
four candidates for y2:

	y1 * (a/x1)^(�1) * (b/c)^(�1)

By taking y1 = x1 = pac, we can see that (a,k) connects with (p�abc,2) for all
non-zero reals, p.   So:

	f(a,k) = |ak|/(ak)

is the finest f that exists for this t.

Regards,
Andrew.

(1)	Some more details on the proof of x1<->y1 symmetry...

	We suppose that:
	� is a symmetric function from M x M *onto* N
	^ is a function from N x M *onto* N such that, for m1 & m2 in M, and
n in N, we have:
	(i)	(m1�m2)^m2 = m1	
	(ii)	(n^m1)�m1 = n

	Now, suppose also that mysteriously we know that for all mi in M:
	(((m1�m2)^m3)�m4)^m5 = (((m1�m4)^m3)�m2)^m5

	Now, using (i),
	((m1�m2)^m3)�m4 = ((m1�m4)^m3)�m2 = n in N, say.
=>	((n^m4)�m3)^m2 = m1
      &	((n^m2)�m3)^m4 = m1
=>	((n^m4)�m3)^m2 = ((n^m2)�m3)^m4

	Now since this relation holds for *all* mi in M, and n in N, we are
allowed to substitute n = (b�c), m2 = x1, m3 = a, m4 = y1, and we get the
result that we wanted, that the expression for m1 = y2 is symmetric in x1 & y1.


(2)	Some more details on the function...

	t = t(b,c) = b/c + c/b
	M = R \ {0}

	t(b,c) = t(x1,x2)
=>	t = x1/x2 + x2/x1
=>	x2� - x1*x2*t + x1� = 0
=>	x2 = �x1*t ��x1*sqrt(t�-4)

	Now t�-4 = (b/c + c/b)� - 4 = (b/c - c/b)�

Thus x2 = �x1*{ (b/c + c/b) � (b/c - c/b) }
	= x1*b/c or x1*c/b

Similarly, we have:
	t(a,x2) = t(y1,y2)
=>	y2 = y1*x2/a or y1*a/x2

& substituting for x2, we have:

	y2 = y1	* (x1/a)^(�1) * (b/c)^(�1)

	Now I want to focus on the solution y2 = y1*(x1/a)*(b/c).   If I take
y1 = x1 = pac, (p real) which I am perfectly free to do, then I get y2 = p�abc.
Note that t(pac,pac) = 2.   In other words:
	f(a,t(b,c)) = f(p�abc,2) for all reals p.

	In particular, let me take p = |abc|^(-�).   Then:  
f(a,t(b,c)) = f(�1,2).   Since the rhs is symmetrical, we have found the
pullback function.

	We write it as f(a,k) = (ak)/|ak|, since k(b,c) has the same sign as bc.

	This fascinating topic, which does not seem to be classifiable
into any standard domain of maths, seems to borrow from Functional Analysis,
Geometry, Galois Theory, Graph Theory and Category Theory.   I suspect that it
is a topic worth much greater study than we are able to devote to it here.   I
want to jot down here some of the directions that occur to me for possible 
future pursuit.


(1) M�bius.   
	Why do the M�bius transformations "work"?
	Is there a "well-behaved" criterion we can give t, such that *only*
the M�bius transformations "work"?
	Can we extend to higher dimensions (ie: t has 3,4,5... parameters,
f has 4,5,6...)?
	Is there a neat geometrical interpretation for what's going on?


(2) Finding suitable f
	How can we extend Peter's algorithm for more general t?
	In particular, can we "walk" the (a,k) graph in a more extended way,
rather than just the two steps currently permitted?
	Are there any simple t which generate very pathological (a,k) graphs
(eg:  connected, but with enormous radius)?
	The (a,k) graph is rather unintuitive in the way the nodes are
connected.   Is there a representation for the pullback f which allows a
more "Cartesian" layout of the edges?
	What can we gain from a more categorical approach still?


(3) Special t
	What if t is composition over M an abelian group or semigroup?
	What if t is distance over M a metric space?
	What if t' is a (non-commutative) binary operator, and t(b,c) is the
unordered pair {t'(b,c), t'(c,b)}?
	What if t is a predicate (ie values T & F only)?
	What if...?
	How do we bring in concepts such as continuity, topology and
differentiability to the topic.


(4) Extremal t [a whole new ballgame]
	Say that t is *extremal* if:
		(i) f must be constant
		(ii) any *finer* symmetric function t' permits a non-constant f
	What interesting things can we say about extremal t in general?
	What extremal t exist for finite M?
	Is the (a,k) graph essentially a tree for an extremal t?

Regards,
Andrew.

I've nailed b+c+1/(bc) for the positive reals.  It wasn't easy or pretty,
but it may illustrate some of the problems with these problems.

We're working with R+ throughout.

	t(b,c) = b + c + 1/(bc)

	t(b,c) = t(1/(bc),c) = t(b,1/(bc))

		(aside: F(a,b,c) = F(a,b,1/(bc)) = F(a,b,c(b/a)) is a nice
			transformation, but I couldn't find what to do with it)

	t(b,c) is minimized when b = c = 1; t(1,1) = 3.

	f(b,t(1/(bc),a))
		= f(a,t(1/(bc),b))
		= f(a,t(1/(bc),c))
		= f(c,t(1/(bc),a))

	So f(b,k) = f(c,k) for any k in { t(1/(bc),a) }.  (a,b,c all in R+)

	k in { t(1/(bc),a) }  <=>  k >= t(1/(bc),sqrt(bc)) = 2sqrt(bc) + 1/(bc)
		(since a=sqrt(bc) minimizes t(1/(bc),a)).

	Is f(b,k)=f(c,k)?  They are equal if k-3 >= (p-1)^2 (2p+1) / p^2, where
	p = sqrt(bc); but that's only a small part of the b-c plane.  But for
	a given k > 3, the graph of this inequality looks roughly like a
	hyperbola with some non-zero `width'.  In a vertical slice through
	(the width of) this curve (i.e., with a constant b), f(b,k)=f(c,k),
	where c varies within the bounds allowed by the inequality.  And
	in a horizontal slice, f(b,k)=f(c,k), with b varying.  Within this
	wide hyperbola, we can zig-zag to connect any two values.

		c3	\--\
			 \ |\
			  \| \
		c2	   \--\
			    \ |\
			     \| \
		c1	      \--\
			       \ |\
			  b1 b2 b3

	To visualize this, see the diagram and see that
		f(c1,k)=f(b,k) for b2 <= b <= b3;
	    and	f(b2,k)=f(c,k) for c1 <= c <= c2;
	    and	f(c2,k)=f(b,k) for b1 <= b <= b2;
	    and	f(b1,k)=f(c,k) for c2 <= c <= c3.
	So f(b3,k)=f(c3,k) even though the point (b3,c3) is not on the
	wide hyperbola.

	The following argument extends the equality between k values:
	Since the value of f(a,k) is independent of a, by symmetry it's also
	independent of b and c, and so f(a,k) must be a constant for any k.
	(We could instead find suitable a,b,c so that f(a,t(b,c)) = f(b,t(a,c))
	connects any/all different k values).


	All that remains is to handle k=3.  It can't be handled as above,
	because when k=3, the hyperbola has no width.  We know that t(x,y)=3
	iff x=y=1.  Now f(a,3)=f(a,t(1,1))=f(1,t(a,1)).  And if a<>1, this
	last expression connects to all the other f(a,k) with k > 3.

	And finally, there is f(1,t(1,1)).  Symmetry constrains it to equal
	itself (which it does), and the only other connection involves finding
	other t(x,y) equal to t(1,1), which we can't.  So f(1,3) remains
	disconnnected, and is permitted its own value.

So we have:

    Problem:
	Let t(b,c) = b + c + 1/(bc) be a function R+ x R+ -> R+.
	Find the most general F(a,k) for which F(a,t(b,c)) is symmetric
	in its parameters.

    Solution:
	Let F(a,k) =
		K1	if k>3,
		K1	if k=3 and a<>1,
		K2	if k=3 and a=1,
	     undefined	if k<3.
	
	Where K1 and K2 are arbitrary constants (i.e., expressions that don't
	involve a or k).