Structural comparison of expressions

[rwst]

In many cases it is not necessary to prove equality of expressions but just check structural identity. An undocumented trick for doing this is (ex1-ex2).is_trivial_zero(). This ticket implements a well visible member function for this task.

Component: symbolics

Author: Ralf Stephan

Branch/Commit: 5955d56

Reviewer: Eric Gourgoulhon

Issue created by migration from https://trac.sagemath.org/ticket/24236

[rwst]

Branch: u/rws/structural_comparison_of_expressions

[rwst]

Author: Ralf Stephan

[rwst]

Commit: 92f95ce

[sagetrac-git]

Branch pushed to git repo; I updated commit sha1. New commits:

e53fd2b

24236: Structural comparison of expressions

[sagetrac-git]

Changed commit from 92f95ce to e53fd2b

[egourgoulhon]

comment:4

Thanks for implementing this!

A quick question: is this strictly equivalent to (ex1-ex2).is_trivial_zero()? I mean, can we have expressions ex1 and ex2 for which (ex1-ex2).is_trivial_zero() and ex1.is_trivially_equal(ex2) do not give the same result?

[rwst]

comment:5

One case is:

sage: inf = SR(oo)
sage: inf-inf
...
RuntimeError: indeterminate expression: infinity - infinity encountered.
sage: inf.is_trivially_equal(inf)
True

When I tried to construct another I found a segfault:

sage: half = SR(QQbar(1/2))
sage: (half-1/2).is_trivial_zero()
True
sage: half.is_trivially_equal(1/2)
BOOM

which is fixed in the next commit. Nice question! It turns out this gives True as well.

[sagetrac-git]

Branch pushed to git repo; I updated commit sha1. New commits:

5955d56

24236: fix

[sagetrac-git]

Changed commit from e53fd2b to 5955d56

[rwst]

comment:7

So to answer your question, no, except for objects that can't be subtracted the result is the same, and for those objects the answer is now correct.

[egourgoulhon]

comment:8

LGTM. Thanks!

One last question: ex1.is_trivially_equal(ex2) is always faster than (ex1-ex2).is_trivial_zero(), isn't it?

[egourgoulhon]

Reviewer: Eric Gourgoulhon

[rwst]

comment:9

Replying to @egourgoulhon:

One last question: ex1.is_trivially_equal(ex2) is always faster than (ex1-ex2).is_trivial_zero(), isn't it?

Yes, it's a tree walk in C++ that returns as soon as a difference in the tree is seen.

[vbraun]

Changed branch from u/rws/structural_comparison_of_expressions to 5955d56