Re: Simplifying inequalities

*To*: mathgroup at smc.vnet.net*Subject*: [mg36809] Re: [mg36790] Simplifying inequalities*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Thu, 26 Sep 2002 04:57:20 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

The modification to FullSimplify that I sent earlier works correctly only for assumptions of the form Or[a,b] (and even then not is not always what one would like). For what it's worth here is a better (but slow) version: In[1]:= Unprotect[FullSimplify]; In[2]:= FullSimplify[expr_, x_ || y__] := FullSimplify[ FullSimplify[expr, x] || FullSimplify[expr, Or[y]]]; In[3]:= Protect[FullSimplify]; For example: In[4]:= FullSimplify[Sqrt[(x - 1)^2] + Sqrt[(x - 2)^2] + Sqrt[(x - 3)^2], x > 1 || x > 2 || x > 3] Out[4]= -1 + x + Abs[-3 + x] + Abs[-2 + x] || -3 + 2*x + Abs[-3 + x] || 3*(-2 + x) Andrzej Kozlowski Toyama International University JAPAN On Thursday, September 26, 2002, at 11:14 AM, Andrzej Kozlowski wrote: > The reason why InequalitySolve returns it's answer in what sometimes > turns out to be unnecessarily complicated form is that the underlying > algorithm, Cylindrical Agebraic Decomposition (CAD) returns its > answers in this form. Unfortunately it seems to me unlikely that a > simplification of the kind you need can be can be accomplished in any > general way. To see why observe the following. First of all: > > In[1]:= > FullSimplify[x > 0 || x == 0] > > Out[1]= > x >= 0 > > This is fine. However: > > In[2]:= > FullSimplify[x > 0 && x < 2 || x == 0 && x < 2] > > Out[2]= > x == 0 || 0 < x < 2 > > Of course what you would like is simply 0 <= x < 2. One reason why you > can't get it is that while Mathematica can perform a "LogicalExpand", > as in: > In[3]:= > LogicalExpand[(x > 0 || x == 0) && x < 2] > > Out[3]= > x == 0 && x < 2 || x > 0 && x < 2 > > There i no "LogicalFactor" or anything similar that would reverse what > LogicalExpand does. if there was then you could perform the sort of > simplifications you need for: > > In[4]:= > FullSimplify[(x > 0 || x == 0) && x < 2] > > Out[4]= > 0 <= x < 2 > > However, it does not seem to me very likely that such "logical > factoring" can be performed by a general enough algorithm (though I am > no expert in this field). In any case, certainly Mathematica can't do > this. > > I also noticed that Mathematica seems unable to show that the answer > it returns to your problem is actually equivalent to your simpler one. > In fact this looks like a possible bug in Mathematica. Let's first try > the function ImpliesQ from the Experimental context: > > << Experimental` > > Now Mathematica correctly gives: > > In[6]:= > ImpliesQ[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > > Out[6]= > True > > However: > > In[7]:= > ImpliesQ[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > > Out[7]= > False > > That simply means that ImpliesQ cannot show the implication, not that > it does not hold. ImpliesQ relies on CAD, as does FullSimplify. > Switching to FullSimplify we see that: > > > > In[8]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > > Out[8]= > True > > while > > In[9]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > > Out[9]= > y4 >= -1 && y6 <= y5 <= 1 + y4 + y6 > > On the other hand, taking just the individual summands of Or as > hypotheses; > In[10]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6] > > Out[10]= > True > > In[11]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 ] > > Out[11]= > True > > In fact FullSimplify is unable to use Or in assumptions, which can be > demonstrated on an abstract example: > > > In[12]:= > FullSimplify[C,(A||B)&&(C)] > > Out[12]= > True > > In[13]:= > FullSimplify[C,LogicalExpand[(A||B)&&(C)]] > > Out[13]= > C > > This could be fixed by modifying FullSimplify: > > In[14]:= > Unprotect[FullSimplify]; > > In[14]:= > FullSimplify[expr_,Or[x_,y__]]:=Or[FullSimplify[expr,x],FullSimplify[ex > pr,y]]; > > In[15]:= > Protect[FullSimplify]; > > Now at least we get as before: > > In[16]:= > FullSimplify[y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && > y6 <= y5 <= 1 + y4 + y6, y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + > y4 + y6] > > Out[16]= > True > > but also: > > In[17]:= > FullSimplify[y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6, > y4 == -1 && y6 >= -1 && y5 == y6 || y4 > -1 && y6 >= -1 && y6 <= y5 > <= 1 + y4 + y6] > > Out[17]= > True > > This seems to me a possible worthwhile improvement in FullSimplify, > though of course not really helpful for your problem. > > > Andrzej Kozlowski > Toyama International University > JAPAN > > > On Wednesday, September 25, 2002, at 02:51 PM, Vincent Bouchard wrote: > >> I have a set of inequalities that I solve with InequalitySolve. But >> then >> it gives a complete set of solutions, but not in the way I would like >> it >> to be! :-) For example, the simple following calculation will give: >> >> In[1]:= ineq = {y4 >= -1, y5 >= -1, y6 + y4 >= y5 - 1, y5 >= y6, y6 >> >= -1}; >> InequalitySolve[ineq,{y4,y6,y5}] >> >> Out[1]:= y4 == -1 && y6 >= -1 && y5 == y6 || >> y4 > -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 >> >> the result is good, but I would like it to be in the simpler but >> equivalent form >> >> y4 >= -1 && y6 >= -1 && y6 <= y5 <= 1 + y4 + y6 >> >> How can I tell InequalitySolve to do that? It is simple for this >> example, >> but for a large set of simple inequalities InequalitySolve gives >> lines and >> lines of results instead of a simple result. >> >> Thanks, >> >> Vincent Bouchard >> DPHil student in theoretical physics in University of Oxford >> >> >> >> > >