[PATCH] c++: Improve memory usage of subsumption [PR100828]

Jason Merrill jason@redhat.com
Wed Aug 11 13:46:05 GMT 2021


On 8/9/21 5:07 PM, Patrick Palka wrote:
> On Wed, Jul 28, 2021 at 4:42 PM Jason Merrill <jason@redhat.com> wrote:
>>
>> On 7/19/21 6:05 PM, Patrick Palka wrote:
>>> Constraint subsumption is implemented in two steps.  The first step
>>> computes the disjunctive (or conjunctive) normal form of one of the
>>> constraints, and the second step verifies that each clause in the
>>> decomposed form implies the other constraint.   Performing these two
>>> steps separately is problematic because in the first step the
>>> disjunctive normal form can be exponentially larger than the original
>>> constraint, and by computing it ahead of time we'd have to keep all of
>>> it in memory.
>>>
>>> This patch fixes this exponential blowup in memory usage by interleaving
>>> these two steps, so that as soon as we decompose one clause we check
>>> implication for it.  In turn, memory usage during subsumption is now
>>> worst case linear in the size of the constraints rather than
>>> exponential, and so we can safely remove the hard limit of 16 clauses
>>> without introducing runaway memory usage on some inputs.  (Note the
>>> _time_ complexity of subsumption is still exponential in the worst case.)
>>>
>>> In order for this to work we need formula::branch to prepend the copy
>>> of the current clause directly after the current clause rather than
>>> at the end of the list, so that we fully decompose a clause shortly
>>> after creating it.  Otherwise we'd end up accumulating exponentially
>>> many (partially decomposed) clauses in memory anyway.
>>>
>>> Bootstrapped and regtested on x86_64-pc-linux-gnu, and also tested on
>>> range-v3 and cmcstl2.  Does this look OK for trunk and perhaps 11?
>>
>> OK for trunk.
> 
> Thanks a lot, patch committed to trunk as r12-2658.  Since this low
> complexity limit was introduced in GCC 10, what do you think about
> increasing the limit from 16 to say 128 in the 10/11 release branches
> as a relatively safe stopgap?

Now that 11.2 is out, go ahead and apply this patch to the 11 branch.

Won't a limit of 128 in GCC 10 lead to extremely long compile times for 
affected code?  Is that more desirable than an error?

>>>        PR c++/100828
>>>
>>> gcc/cp/ChangeLog:
>>>
>>>        * logic.cc (formula::formula): Use emplace_back.
>>>        (formula::branch): Insert a copy of m_current in front of
>>>        m_current instead of at the end of the list.
>>>        (formula::erase): Define.
>>>        (decompose_formula): Remove.
>>>        (decompose_antecedents): Remove.
>>>        (decompose_consequents): Remove.
>>>        (derive_proofs): Remove.
>>>        (max_problem_size): Remove.
>>>        (diagnose_constraint_size): Remove.
>>>        (subsumes_constraints_nonnull): Rewrite directly in terms of
>>>        decompose_clause and derive_proof, interleaving decomposition
>>>        with implication checking.  Use formula::erase to free the
>>>        current clause before moving on to the next one.
>>> ---
>>>    gcc/cp/logic.cc | 118 ++++++++++++++----------------------------------
>>>    1 file changed, 35 insertions(+), 83 deletions(-)
>>>
>>> diff --git a/gcc/cp/logic.cc b/gcc/cp/logic.cc
>>> index 142457e408a..3f872c11fe2 100644
>>> --- a/gcc/cp/logic.cc
>>> +++ b/gcc/cp/logic.cc
>>> @@ -223,9 +223,7 @@ struct formula
>>>
>>>      formula (tree t)
>>>      {
>>> -    /* This should call emplace_back(). There's an extra copy being
>>> -       invoked by using push_back().  */
>>> -    m_clauses.push_back (t);
>>> +    m_clauses.emplace_back (t);
>>>        m_current = m_clauses.begin ();
>>>      }
>>>
>>> @@ -248,8 +246,7 @@ struct formula
>>>      clause& branch ()
>>>      {
>>>        gcc_assert (!done ());
>>> -    m_clauses.push_back (*m_current);
>>> -    return m_clauses.back ();
>>> +    return *m_clauses.insert (std::next (m_current), *m_current);
>>>      }
>>>
>>>      /* Returns the position of the current clause.  */
>>> @@ -287,6 +284,14 @@ struct formula
>>>        return m_clauses.end ();
>>>      }
>>>
>>> +  /* Remove the specified clause.  */
>>> +
>>> +  void erase (iterator i)
>>> +  {
>>> +    gcc_assert (i != m_current);
>>> +    m_clauses.erase (i);
>>> +  }
>>> +
>>>      std::list<clause> m_clauses; /* The list of clauses.  */
>>>      iterator m_current; /* The current clause.  */
>>>    };
>>> @@ -659,39 +664,6 @@ decompose_clause (formula& f, clause& c, rules r)
>>>      f.advance ();
>>>    }
>>>
>>> -/* Decompose the logical formula F according to the logical
>>> -   rules determined by R.  The result is a formula containing
>>> -   clauses that contain only atomic terms.  */
>>> -
>>> -void
>>> -decompose_formula (formula& f, rules r)
>>> -{
>>> -  while (!f.done ())
>>> -    decompose_clause (f, *f.current (), r);
>>> -}
>>> -
>>> -/* Fully decomposing T into a list of sequents, each comprised of
>>> -   a list of atomic constraints, as if T were an antecedent.  */
>>> -
>>> -static formula
>>> -decompose_antecedents (tree t)
>>> -{
>>> -  formula f (t);
>>> -  decompose_formula (f, left);
>>> -  return f;
>>> -}
>>> -
>>> -/* Fully decomposing T into a list of sequents, each comprised of
>>> -   a list of atomic constraints, as if T were a consequent.  */
>>> -
>>> -static formula
>>> -decompose_consequents (tree t)
>>> -{
>>> -  formula f (t);
>>> -  decompose_formula (f, right);
>>> -  return f;
>>> -}
>>> -
>>>    static bool derive_proof (clause&, tree, rules);
>>>
>>>    /* Derive a proof of both operands of T.  */
>>> @@ -744,28 +716,6 @@ derive_proof (clause& c, tree t, rules r)
>>>      }
>>>    }
>>>
>>> -/* Derive a proof of T from disjunctive clauses in F.  */
>>> -
>>> -static bool
>>> -derive_proofs (formula& f, tree t, rules r)
>>> -{
>>> -  for (formula::iterator i = f.begin(); i != f.end(); ++i)
>>> -    if (!derive_proof (*i, t, r))
>>> -      return false;
>>> -  return true;
>>> -}
>>> -
>>> -/* The largest number of clauses in CNF or DNF we accept as input
>>> -   for subsumption. This an upper bound of 2^16 expressions.  */
>>> -static int max_problem_size = 16;
>>> -
>>> -static inline bool
>>> -diagnose_constraint_size (tree t)
>>> -{
>>> -  error_at (input_location, "%qE exceeds the maximum constraint complexity", t);
>>> -  return false;
>>> -}
>>> -
>>>    /* Key/value pair for caching subsumption results. This associates a pair of
>>>       constraints with a boolean value indicating the result.  */
>>>
>>> @@ -845,31 +795,33 @@ subsumes_constraints_nonnull (tree lhs, tree rhs)
>>>      if (bool *b = lookup_subsumption(lhs, rhs))
>>>        return *b;
>>>
>>> -  int n1 = dnf_size (lhs);
>>> -  int n2 = cnf_size (rhs);
>>> -
>>> -  /* Make sure we haven't exceeded the largest acceptable problem.  */
>>> -  if (std::min (n1, n2) >= max_problem_size)
>>> -    {
>>> -      if (n1 < n2)
>>> -        diagnose_constraint_size (lhs);
>>> -      else
>>> -     diagnose_constraint_size (rhs);
>>> -      return false;
>>> -    }
>>> -
>>> -  /* Decompose the smaller of the two formulas, and recursively
>>> -     check for implication of the larger.  */
>>> -  bool result;
>>> -  if (n1 <= n2)
>>> -    {
>>> -      formula dnf = decompose_antecedents (lhs);
>>> -      result = derive_proofs (dnf, rhs, left);
>>> -    }
>>> +  tree x, y;
>>> +  rules r;
>>> +  if (dnf_size (lhs) <= cnf_size (rhs))
>>> +    /* When LHS looks simpler than RHS, we'll determine subsumption by
>>> +       decomposing LHS into its disjunctive normal form and checking that
>>> +       each (conjunctive) clause implies RHS.  */
>>> +    x = lhs, y = rhs, r = left;
>>>      else
>>> +    /* Otherwise, we'll determine subsumption by decomposing RHS into its
>>> +       conjunctive normal form and checking that each (disjunctive) clause
>>> +       implies LHS.  */
>>> +    x = rhs, y = lhs, r = right;
>>> +
>>> +  /* Decompose X into a list of sequents according to R, and recursively
>>> +     check for implication of Y.  */
>>> +  bool result = true;
>>> +  formula f (x);
>>> +  while (!f.done ())
>>>        {
>>> -      formula cnf = decompose_consequents (rhs);
>>> -      result = derive_proofs (cnf, lhs, right);
>>> +      auto i = f.current ();
>>> +      decompose_clause (f, *i, r);
>>> +      if (!derive_proof (*i, y, r))
>>> +     {
>>> +       result = false;
>>> +       break;
>>> +     }
>>> +      f.erase (i);
>>>        }
>>>
>>>      return save_subsumption (lhs, rhs, result);
>>>
>>
> 



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