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I am a newbie in the field so please not angry. I am reading the tutorial below and having a problem. Link: How to Create and Use Rules

From the link:
If you give two rules for the same variable, the Wolfram Language will use only the first rule:

In[4]:= 2 x + y /. {x -> 3, x -> 4}
Out[4]= 6 + y

Could you explain how Mathematica do the substitution here? Is this the order of implementation:

  1. Substitute x-> 3 to the expression
  2. Substitute x-> 4 to the expression gotten from No. 1
Sascha
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emnha
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    As far as I understand it, once a subexpression gets replaced, it can't get replaced again. Try this: a*b /. {a -> a, a -> 1, b -> c}. The first replacement just replaces a with a, but since that expression has already been changed, it is ignored after that, so the Rule a - > 1 is only applied to other parts of the expression, and there are no more a's, but b gets replaced with c. So you're kind of right and kind of not. – march Jan 29 '16 at 18:51
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    Use a List of rule lists. `In[265]:= 2 x + y /. {{x -> 3}, {x -> 4}}

    Out[265]= {6 + y, 8 + y}`

     – Daniel Lichtblau Jan 29 '16 at 18:56
  • @march is correct that ReplaceAll (/.) doesn't apply several rules to the same part of a given expression. For that case you want to use ReplaceRepeated (//.). In your case however this won't chance the output since x -> 3 changes x to be 3 so there is no more x for the second rule to match against – Sascha Jan 29 '16 at 18:56
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    I can't help but feel that this is a dupe (although it's a great question). Does anyone have a link? I'll go check the Pitfalls thread. If it's not there, we might consider adding a post about how exactly ReplaceAll works. – march Jan 29 '16 at 19:01
  • This is also simply bad programming style. You should be able to eliminate duplicate substitution rules and retain only the one you want. – David G. Stork Jan 29 '16 at 19:06
  • @march I agree that this should be added in the common pitfalls thread, especially since this behavior of ReplaceAll is somewhat hidden in the "Details" section of the documentation – Sascha Jan 29 '16 at 19:06
  • Thank you all for these answers! – emnha Jan 29 '16 at 19:09

1 Answers1

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As far as I understand it, once a subexpression gets replaced, it can't get replaced again. Try this:

a*b /. {a -> a, a -> 1, b -> c}

resulting in

a*c

The first replacement just replaces a with a, but since that expression has already been changed, it is ignored after that. So the Rule a -> 1 is only applied to other parts of the expression (in this case the b), and those other parts don't depend on a, so nothing changes. Finally, b gets replaced with c. (By the way, in somewhat-advanced Mathematica programming, this fact can be taken advantage of in many clever ways. Here is an example by Mr. Wizard, provided in the comments.)

So you're kind of right and kind of not. The Rules are applied sequentially, but the rules are not applied to the new expressions resulting from previous replacements.

Here's another interesting example:

a b c /. {a -> b, b -> c, c -> d}
% /. {a -> b, b -> c, c -> d}
% /. {a -> b, b -> c, c -> d}
(* b c d *)
(* c d^2 *)
(* d^3 *)

In the first case, the a instance is replaced by b, but it is not in turn replaced by c. However, the original b is replaced by c. And so on. To do all of them at once, use ReplaceRepeated:

a b c //. {a -> b, b -> c, c -> d}
(* d^3 *)

(Be careful of this one, because it can run into infinite recursions.) For completeness, note that ReplaceRepeated does not act the same as a sequence of ReplaceAlls if the list of Rules are different:

a b c //. {a -> b, c -> d, b -> c}
a b c /. {a -> b} /. {c -> d} /. {b -> c}
(* d^3 *)
(* c^2 d*)

Finally, as noted by Daniel Lichtblau in a comment, if you do want to apply the two rules to get different expressions, do this:

2 x + y /. {{x -> 3}, {x -> 4}}
(* {6 + y, 8 + y} *)

Alternatively, do

2 x + y /. # &/@ {x -> 3, x -> 4}

or

2 x + y /. List/@{x -> 3, x -> 4}
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march
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