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inst-tester: split _pick_choice into two

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Signed-off-by: Niels Thykier <niels@thykier.net>
debian
Niels Thykier 9 years ago
parent 176fe04c98
commit 97663ebaab
1 changed files with 85 additions and 73 deletions
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158
installability/tester.py
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@ -328,11 +328,11 @@ class InstallabilityTester(object):
#
# * A package is installable if never and musts are disjointed
# and both check and choices are empty.
# - exception: _pick_choice may determine the installability
# - exception: resolve_choice may determine the installability
# of t via recursion (calls _check_inst). In this case
# check and choices are not (always) empty.
def _pick_choice(rebuild, set=set, len=len):
def _prune_choices(rebuild, set=set, len=len):
"""Picks a choice from choices and updates rebuild.
Prunes the choices and updates "rebuild" to reflect the
@ -382,75 +382,14 @@ class InstallabilityTester(object):
# all alternatives would violate the conflicts or are uninstallable
# => package is not installable
stats.choice_presolved += 1
return None
return False
# The choice is still deferred
rebuild.add(frozenset(remain))
if check or not rebuild:
return False
choice = iter(rebuild.pop())
last = next(choice) # pick one to go last
for p in choice:
musts_copy = musts.copy()
never_tmp = set()
choices_tmp = set()
check_tmp = set([p])
if not self._check_loop(universe, testing, eqv_table,
stats, musts_copy, never_tmp,
cbroken, choices_tmp,
check_tmp):
# p cannot be chosen/is broken (unlikely, but ...)
continue
# Test if we can pick p without any consequences.
# - when we can, we avoid a backtrack point.
if never_tmp <= never and choices_tmp <= rebuild:
# we can pick p without picking up new conflicts
# or unresolved choices. Therefore we commit to
# using p.
#
# NB: Optimally, we would go to the start of this
# routine, but to conserve stack-space, we return
# and expect to be called again later.
musts.update(musts_copy)
stats.choice_resolved_without_restore_point += 1
return False
if not musts.isdisjoint(never_tmp):
# If we pick p, we will definitely end up making
# t uninstallable, so p is a no-go.
continue
return True
stats.backtrace_restore_point_created += 1
# We are not sure that p is safe, setup a backtrack
# point and recurse.
never_tmp |= never
choices_tmp |= rebuild
if self._check_inst(p, musts_copy, never_tmp,
choices_tmp):
# Success, p was a valid choice and made it all
# installable
return True
# If we get here, we failed to find something that
# would satisfy choice (without breaking the
# installability of t). This means p cannot be used
# to satisfy the dependencies, so pretend to conflict
# with it - hopefully it will reduce future choices.
never.add(p)
stats.backtrace_restore_point_used += 1
# Optimization for the last case; avoid the recursive call
# and just assume the last will lead to a solution. If it
# doesn't there is no solution and if it does, we don't
# have to back-track anyway.
check.add(last)
musts.add(last)
stats.backtrace_last_option += 1
return False
# END _pick_choice
# END _prune_choices
while check:
if not check_loop(choices, check):
@ -459,15 +398,20 @@ class InstallabilityTester(object):
if choices:
rebuild = set()
# We have to "guess" now, which is always fun, but not cheap
r = _pick_choice(rebuild)
if r is None:
if not _prune_choices(rebuild):
verdict = False
break
if r:
# The recursive call have already updated the
# cache so there is not point in doing it again.
return True
while not check and rebuild:
# We have to "guess" now, which is always fun, but not cheap. We
# stop guessing:
# - once we run out of choices to make (obviously), OR
# - if one of the choices exhaust all but one option
if self.resolve_choice(check, musts, never, rebuild):
# The recursive call have already updated the
# cache so there is not point in doing it again.
return True
choices = rebuild
if verdict:
@ -479,6 +423,74 @@ class InstallabilityTester(object):
return verdict
def resolve_choice(self, check, musts, never, choices):
universe = self._universe
testing = self._testing
eqv_table = self._eqv_table
stats = self._stats
cbroken = self._cache_broken
choice = iter(choices.pop())
last = next(choice) # pick one to go last
for p in choice:
musts_copy = musts.copy()
never_tmp = set()
choices_tmp = set()
check_tmp = set([p])
if not self._check_loop(universe, testing, eqv_table,
stats, musts_copy, never_tmp,
cbroken, choices_tmp,
check_tmp):
# p cannot be chosen/is broken (unlikely, but ...)
continue
# Test if we can pick p without any consequences.
# - when we can, we avoid a backtrack point.
if never_tmp <= never and choices_tmp <= choices:
# we can pick p without picking up new conflicts
# or unresolved choices. Therefore we commit to
# using p.
#
# NB: Optimally, we would go to the start of this
# routine, but to conserve stack-space, we return
# and expect to be called again later.
musts.update(musts_copy)
stats.choice_resolved_without_restore_point += 1
return False
if not musts.isdisjoint(never_tmp):
# If we pick p, we will definitely end up making
# t uninstallable, so p is a no-go.
continue
stats.backtrace_restore_point_created += 1
# We are not sure that p is safe, setup a backtrack
# point and recurse.
never_tmp |= never
choices_tmp |= choices
if self._check_inst(p, musts_copy, never_tmp,
choices_tmp):
# Success, p was a valid choice and made it all
# installable
return True
# If we get here, we failed to find something that
# would satisfy choice (without breaking the
# installability of t). This means p cannot be used
# to satisfy the dependencies, so pretend to conflict
# with it - hopefully it will reduce future choices.
never.add(p)
stats.backtrace_restore_point_used += 1
# Optimization for the last case; avoid the recursive call
# and just assume the last will lead to a solution. If it
# doesn't there is no solution and if it does, we don't
# have to back-track anyway.
check.add(last)
musts.add(last)
stats.backtrace_last_option += 1
return False
def _check_loop(self, universe, testing, eqv_table, stats, musts, never,
cbroken, choices, check, len=len,
frozenset=frozenset):

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