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Re: [mg5137] simplify problem
Posted:
Nov 9, 1996 2:06 AM


On Wed, 6 Nov 1996, Tomasz Bieruta wrote:
> I have following problem: > > In[1] u=(a+b)^2; v=(c+d)^2; > In[2] uv=Expand[u+v] > Out[2] a^2+2ab+b^2+c^2+2cd+d^2 > > How can I revers the above Operation, in this example from > > a^2+2ab+b^2+c^2+2cd+d^2 > > I want to have (a+b)^2 + (c+d)^2. > Can someone help me, please ? > > Tomasz B. > >  > email: bieruta@chemie.unihalle.de > >
Consider a function called fullFactor that factors individuals terms of any expression of sum defined as follow:
(* Edmundo's contribution 1996 *) Clear[fullFactor,nextFactor] fullFactor[Plus[args__]]:= Module[{best=Factor[Plus[args]],next,tabu}, tabu={best}; While[Head[next=nextFactor[Plus[args],tabu]]=!=nextFactor, AppendTo[tabu,next]; If[ByteCount[best]>ByteCount[next],best=next] ]; best ] nextFactor[Plus[a__,b__],tabu_List] := Module[{x=Factor[Plus[a]]+Factor[Plus[b]]}, x /; Not[MemberQ[tabu,x]] ] /; Length[{a}]<=Length[{b}]
Then, here is an example of his working:
Expand[(a+b)^2+(c+d)^2] fullFactor[%]
which returns:
2 2 2 2 a + 2 a b + b + c + 2 c d + d
2 2 (a + b) + (c + d)
This solution can be expanded easily for general simplifying of expressions of sum by replacing the word "Factor" by "Simplify" along the definitions of functions above. The new function called fullSimplify is slightly slowest than fullFactor.
I hope this solve your problem.
With kindest regards,
Edmundo [8)  Universidade Federal do Ceara Department of Computing edmundo@lia.ufc.br 



