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MathGroup Archive 2006

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Re: ? about Rule (rewritten)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67204] Re: ? about Rule (rewritten)
  • From: Bill Rowe <readnewsciv at earthlink.net>
  • Date: Sun, 11 Jun 2006 23:08:31 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 6/11/06 at 2:17 AM, jackgoldberg at comcast.net wrote:

>I want to replace an inequality of the type  1 <=  x <= 2 which
>appears in a list some of whose entries are y==3 or z >= 1 and so
>forth.  (This arose in using Reduce on a set of linear
>inequalities.)  My life is simpler if  1 <= x <=2  is replace by an
>interior point of the interval, say  x ==1.5.   It has frustrated me
>for about a week since I thought I knew the ins and outs of Rule and
>ReplaceAll.  So, 2 questions.  The first and most important is how
>to do it (keep in mind that the list I am speaking of is constructed
>by programming and I do not want to examine it and make the
>replacement that way).  The second question is, why doesn't the
>following technique work?

>(1)       {y==z,  t==7, 1 <=x<=2, u>=-4}/. (a_<=w_<=b_)->(a+b)/2

Here, you asked Mathematica to replace 1 <=x<=2 with the average of the end points, i.e. 3/2. Instead of getting just the average of the end points, did you want a replacement rule, i.e., the result to be x->3/2? If so, it seems

In[4]:=
{y == z, t == 7, 1 <= x <= 2, u >= -4} /. 
  (a_) <= (w_) <= (b_) -> w -> (a + b)/2

Out[4]=
{y == z, t == 7, x -> 3/2, u >= -4}

does what you want
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