Re: Simplifying inequalities

*To*: mathgroup at smc.vnet.net*Subject*: [mg36807] Re: [mg36790] Simplifying inequalities*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Thu, 26 Sep 2002 04:57:14 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

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[exp r,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 > > > >

**a numerical integration**

**Re: Simplifying inequalities**

**Simplifying inequalities**

**Re: Simplifying inequalities**