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MathGroup Archive 2006

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Re: Simplifying a Boolean Expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68886] Re: [mg68790] Simplifying a Boolean Expression
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 22 Aug 2006 05:20:22 -0400 (EDT)
  • References: <200608180712.DAA02183@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hebhardt, Jon wrote:
> Can Mathametica be used to simplify the following type of logic typically
> found in programming applications, and if so, how do I go about setting it
> up?
> 
>  
> 
> EXAMPLE
> 
> Variables A and T can have the values "none", "dense", "sparse"
> 
> The variable X will be assigned a value of either "onepass" or "twopass"
> based on the following logic:
> 
>  
> 
> IF
> 
> [(A="none") AND (T="none")] OR [(A="dense") OR (A="sparse") AND (T="none")]
> OR [(A="dense") AND (T="dense")] THEN X = "onepass"
> 
> ELSE IF
> 
> [{A="sparse") AND (T="sparse")] OR [(A="sparse") AND (T="dense")] THEN X =
> "twopass"
> 
>  
> 
> I want to be able to express this type of problem in Mathematica and have it
> return the simplest logic for assigning a value to X.  
> 
>  
> 
> Regards,
> 
> Jon
> 
>  
> 
> Work:  401-752-3821
> 
> Cell:    508-873-9780
> 

Here is an approach that might help. Use integers 1-3 for "none", 
"dense", and "sparse", and likewise for the number of passes (where 
"threepass" is a dummy for the case where neither condition holds). Then 
  use Reduce to solve a system of equations and bounded inequalities, 
removing from consideration the threepass case. It can be done as below.

{none,dense,sparse} = {1,2,3};

{onepass,twopass,threepass} = {1,2,3};

cond1 = (aa==none && tt==none) ||
   ((aa==dense || aa==sparse) && tt==none);

cond2 = (aa==sparse && tt==sparse) ||
   (aa==sparse && tt==dense);

In[51]:= Reduce[{Implies[cond1,xx==onepass] &&
     Implies[!cond1 && cond2, xx==twopass] &&
         Implies[!cond1 && !cond2, xx==threepass],
     Element[{aa,tt,xx},Integers],
     1<=aa<=3, 1<=tt<=3, 1<=xx<=2}, xx]

Out[51]= (aa == 1 && tt == 1 && xx == 1) ||
   (aa == 2 && tt == 1 && xx == 1) ||
   (aa == 3 && tt == 1 && xx == 1) ||
   (aa == 3 && tt == 2 && xx == 2) ||
   (aa == 3 && tt == 3 && xx == 2)

I'm not sure if this is quite what you seek but it might be a start.


Daniel Lichtblau
Wolfram Research




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