ReplaceAll behavour [Mathematica]

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From: magma on 12 Oct 2007 03:13

I do not understand what Mathematica 6 is doing here:

a=5
a/.a->3

gives 3

But...

a+2/. a->3

gives 7

Why is that?

Thank you for your help

From: Szabolcs Horvát on 13 Oct 2007 03:50

magma wrote:
> I do not understand what Mathematica 6 is doing here:
>
> a=5
> a/.a->3
>
> gives 3
>
> But...
>
> a+2/. a->3
>
> gives 7
>
> Why is that?
>
> Thank you for your help
>

Hi,

Use Trance[a /. a->3] or On[] and Off[] to find out how Mathematica
evaluates an expression. The following output should explain what happens:

In[1]:= On[]
During evaluation of In[1]:= On::trace: On[] --> Null. >>

In[2]:= a=5
During evaluation of In[2]:= Set::trace: a=5 --> 5. >>

Out[2]= 5
In[3]:= a/.a->3
During evaluation of In[3]:= a::trace: a --> 5. >>
During evaluation of In[3]:= a::trace: a --> 5. >>
During evaluation of In[3]:= Rule::trace: a->3 --> 5->3. >>
During evaluation of In[3]:= Rule::trace: 5->3 --> 5->3. >>
During evaluation of In[3]:= ReplaceAll::trace: a/.a->3 --> 5/.5->3. >>
During evaluation of In[3]:= ReplaceAll::trace: 5/.5->3 --> 3. >>
Out[3]= 3

In[4]:= (a+2)/.a->3
During evaluation of In[4]:= a::trace: a --> 5. >>
During evaluation of In[4]:= Plus::trace: a+2 --> 5+2. >>
During evaluation of In[4]:= Plus::trace: 5+2 --> 7. >>
During evaluation of In[4]:= a::trace: a --> 5. >>
During evaluation of In[4]:= Rule::trace: a->3 --> 5->3. >>
During evaluation of In[4]:= Rule::trace: 5->3 --> 5->3. >>
During evaluation of In[4]:= ReplaceAll::trace: a+2/.a->3 --> 7/.5->3. >>
During evaluation of In[4]:= ReplaceAll::trace: 7/.5->3 --> 7. >>
Out[4]= 7

In[5]:= Off[]

Szabolcs

P.S.

This is why it is a bad idea to do things like

sqrt[a_?NumericQ] := x /. FindRoot[x^2 == a, {x, 1}]

(Even though this function works correctly almost all the time -- one
must come up with a really tricky definition for 'x' to make it fail,
like x = Sequence[1])

So use Module[] to localise 'x':

sqrt[a_?NumericQ] := Module[{x}, x /. FindRoot[x^2 == a, {x, 1}]]

From: adriano.pascoletti on 13 Oct 2007 04:08

> I do not understand what Mathematica 6 is doing here:
>
> a=5
> a/.a->3
>
> gives 3
>
> But...
>
> a+2/. a->3
>
> gives 7
>
> Why is that?
>
> Thank you for your help
>
Being a=5 the statement a+2/. a->3 becomes 7/.5->3.
You can check it with Trace:

In[5]:= Trace[a + 2 /.a -> 3]

Out[5]={{{a,5},5+2,7},{{a,5},5->3,5->3},7/.5->3,7}

Adriano Pascoletti

From: Bill Rowe on 13 Oct 2007 04:17

On 10/12/07 at 2:59 AM, maderri2(a)gmail.com (magma) wrote:

>I do not understand what Mathematica 6 is doing here:

>a=5
>a/.a->3

>gives 3

>But...

>a+2/. a->3

>gives 7

>Why is that?

What is happening in each case is evaluation occurs before replacement.

=46or a/.a-> evaluation causes this to become

5/.5->3 then the replacement occurs to give the result 3.

=46or a+2/.a->3 evaluation gives

5+2/.5->3

Arithmetic operators such as + or - have higher precedence than
patter replacement. So, this becomes
7/.5->3. And finally since there is no 5 to replace, the result
is 7.

You can easily verify this is the way things work by looking at
the output from

Trace[a/.a->3] and
Trace[a+2/.a->3]
--
To reply via email subtract one hundred and four

From: Szabolcs Horvát on 14 Oct 2007 04:18

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