The second half of redo-stamp: out-of-order building.

If a depends on b depends on c, and c is dirty but b uses redo-stamp
checksums, then 'redo-ifchange a' is indeterminate: we won't know if we need
to run a.do unless we first build b, but the script that *normally* runs
'redo-ifchange b' is a.do, and we don't want to run that yet, because we
don't know for sure if b is dirty, and we shouldn't build a unless one of
its dependencies is dirty.  Eek!

Luckily, there's a safe solution.  If we *know* a is dirty - eg. because
a.do or one of its children has definitely changed - then we can just run
a.do immediately and there's no problem, even if b is indeterminate, because
we were going to run a.do anyhow.

If a's dependencies are *not* definitely dirty, and all we have is
indeterminate ones like b, then that means a's build process *hasn't
changed*, which means its tree of dependencies still includes b, which means
we can deduce that if we *did* run a.do, it would end up running b.do.

Since we know that anyhow, we can safely just run b.do, which will either
b.set_checked() or b.set_changed().  Once that's done, we can re-parse a's
dependencies and this time conclusively tell if it needs to be redone or
not.  Even if it does, b is already up-to-date, so the 'redo-ifchange b'
line in a.do will be fast.

...now take all the above and do it recursively to handle nested
dependencies, etc, and you're done.

vars.py

@@ -22,3 +22,6 @@ RUNID = atoi.atoi(os.environ.get('REDO_RUNID')) or None
 BASE = os.environ['REDO_BASE']
 while BASE and BASE.endswith('/'):
     BASE = BASE[:-1]
+
+UNLOCKED = os.environ.get('REDO_UNLOCKED', '') and 1 or 0
+os.environ['REDO_UNLOCKED'] = ''  # not inheritable by subprocesses