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Re: precedence for ReplaceAll?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg110546] Re: precedence for ReplaceAll?
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Fri, 25 Jun 2010 07:26:01 -0400 (EDT)
On 6/24/10 at 4:27 AM, mfripp at gmail.com (Matthias Fripp) wrote:
>I am having trouble using ReplaceAll to replace symbols that already
>have a delayed assignment.
>e.g., this input:
>In[287]:=
>a := A c
>b := B c
>a /. {a -> x, b -> y}
>b /. {a -> x, b -> y}
>a + b /. {a -> x, b -> y}
>a * b /. {a -> x, b -> y}
>gives this output:
>Out[289]= x
>Out[290]= y
>Out[291]= x + y
>Out[292]= A B c^2
>All of this works as expected except for the final term. I would
>have expected to get the result "x y". Is there any way to force
>Mathematica to produce that result?
In[10]:= a := A c
b := B c
In[12]:= a*b /. {A -> x/c, B -> y/c}
Out[12]= x y
The reason this needs to be different than the others is due to
the way Mathematica's evaluator works. In everyone of your
examples a,b are replaced by A c and B c when Mathematica
encounters them as expected by using SetDelayed. After the
replacement occurs, then evaluation of the specified operations
on the left hand side of /. occurs. In each case except for the
last, this leaves an expression with in terms of A c and B c.
So, in those cases when a,b on the right hand side get
evaluated, they match something on the left hand side and the
replacement occurs as you want.
But in the last case, the multiplication operation yields the
result (in FullForm) of
Times[A, B, Power[c,2]]
Since nothing in this matches either Times[A, c] or Times[B, c],
the replacement you are looking for does not occur. My work
around does what you want since there is a match for A and B.
>If on the other hand the original assignment is a := A + c and b :=
>B + c, I get an unexpected output for the sum, but the expected
>output (x y) for the product. If I insert d instead of one of the
>c's, I get various other (unpredictable) kinds of result.
This fails for the same reason a * b /. {a -> x, b -> y} failed.
That is, with these assignments,
a + b will evaluate to (in FullForm)
Plus[A, B, Times[2, c]]
Again, there will be nothing to match either A+c or B+c and the
replacement fails.
>My first guess is that Mathematica is doing a sort of "double
>ReplaceAll", where it first tries the pattern given in the delayed
>assignment, and any symbols matched by that are not tested against
>the explicit ReplaceAll. But that doesn't explain why the sum works
>and not the product. Am I thinking about this the wrong way?
Yes, your thinking here is incorrect. Note you can verify what
occurs by using Trace. For example,
In[18]:= a + b /. {a -> x, b -> y} // Trace
Out[18]= {{{a,A+c},{b,B+c},(A+c)+(B+c),A+c+B+c,A+B+c+c,A+B+2
c},{{{a,A+c},A+c->x,A+c->x},{{b,B+c},B+c->y,B+c->y},{A+c->x,B+c->y}},A+B+2=
c/.{A+c->x,B+c->y},A+B+2 c}
Here you can see the evaluation of a+b to A+B+2c occurs before
the replacement rule evaluates as I described above. The
consequence is, there is no match for either A+c or B+c.
A key point about replacement rules is these are implemented as
literal replacements. There is no mathematic operation being by
the replacement rules. Any mathematics is done on the left hand
side before the replacement. Further mathematics may be done
after the replacement depending on the nature of the
replacement. But the replacement itself does not do mathematic operations.
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