12
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Bug introduced in 11.0, persisting through 14.1.

Background info

I enocountered this problem when exploring

Pattern match about Times default value

But this turns out to be a different issue i.e. the behavior of code sample in the link is consistent in version 8 and version 14. So I decide to post this as a separate question.

Consider this example:

Replace[x y, c_. (a : (x y)) :> aaa]

In version 8.0.4, the output is aaa, which looks reasonable for me:

enter image description here

But it's not the case in version 14.1:

enter image description here

I don't have access to version 9 to 13 at the moment so cannot do further test.

My question is:

  1. Is this a bug, or intentional incompatible change (this does happen in the past, see e.g. this and this)?

  2. If it's an intentional change, what's the underlying reason for it?

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    $\begingroup$ To the downvoter, I am interested in what's wrong with my question, would you please elaborate. I'm not trying to complain here, I just want to improve my question if possible. $\endgroup$
    – xzczd ♦
    Commented Oct 2 at 14:26
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    $\begingroup$ It appears to have changed in version 11.0. I do not yet know the cause. $\endgroup$
    – Daniel Lichtblau
    Commented Oct 2 at 17:46
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    $\begingroup$ This returns x y in Version "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)". Note that this only seems to be a problem for multiplication: Both Replace[x + y, c_. (a : (x + y)) :> aaa] and Replace[f[x, y], c_. (a : f[x, y]) :> aaa] yield aaa, so I wonder if this is doing some multiplication simplification first or something like that. $\endgroup$
    – march
    Commented Oct 2 at 19:01
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    $\begingroup$ It's worth noting that the matching is not happening: MatchQ[x y, c_. (a : (x y))] and MatchQ[x + y, c_. (a : (x + y))] yield False and True, respectively. Using Trace doesn't help. $\endgroup$
    – march
    Commented Oct 2 at 19:03
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    $\begingroup$ This has been filed as a bug. $\endgroup$
    – Daniel Lichtblau
    Commented Oct 2 at 19:47

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