> Russell McMahon wrote:
>> E = [(A^2-B^2)/(C^2+ 2.C.D + D^2)] x [(C+D)/(A-B)] x [ (C+D)/(A+B)] 
>> x
>> F
>>
>> E will always equal F :-)

> But that could just as well have been implemented by simply checking 
> A-D
> according to some rules. ...


Indeed. I think the object is to include a parameter (in my case 4 
parameters) in the algorithm but for it/them to have no effect on the 
answer. I added the security part as a guess as to reason.


        RM

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