symbol x("x"); symbol y("y"); symbol a("a"); symbol z("z"); ex e1=a-y+x+z; ex e2=a+x-y+z; cout <<"e1="<< e1 << endl; // e1=a-y+z+x cout <<"e2="<< e2 << endl; // e2=a-y+z+x What if I want to get the same order of the variables which was in expression on creation. How can I do this? And how can I even get ex e3=x+x; cout <<"e3="<< e3 << endl; // e3=x+x ? best regards, Dmitri Gorbenko
Hi, On Tue, 4 Dec 2001 gorb@krasu.ru wrote: [...]
ex e1=a-y+x+z; ex e2=a+x-y+z; cout <<"e1="<< e1 << endl; // e1=a-y+z+x cout <<"e2="<< e2 << endl; // e2=a-y+z+x
What if I want to get the same order of the variables which was in expression on creation. How can I do this?
You can't. This is intentional, in order to establish a canonical ordering so that ex e3 = e1-e2 immediately simplifies to 0 by syntactic comparison.
And how can I even get ex e3=x+x; cout <<"e3="<< e3 << endl; // e3=x+x
Again, you can't. This is intentional for the same reason. This anonymous evaluation mechanism drastically reduces the trees representing the expression. In the same level of a (polynomial) tree, no two children are equal. If this would be allowed, we would also allow three, four or more identical subexpressions which increses the complexity of any tree traversing algorithm substantially and would soon become horribly inefficient. By the way, all major algebraic engines do this. Regards -richy. -- Richard B. Kreckel <Richard.Kreckel@Uni-Mainz.DE> <http://wwwthep.physik.uni-mainz.de/~kreckel/>
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gorb＠krasu.ru
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Richard B. Kreckel