[Python-Dev] Comparison of recursive objects
Tim Peters
tim.one@home.com
Sat, 20 Jan 2001 04:28:18 -0500
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[Guido's checkin msg]
> ...
> In a discussion with Tim, where we discovered that our intuition
> on when a<=b should be true was failing, we decided to outlaw
> ordering comparisons on recursive objects. (Once we have fixed our
> intuition and designed a matching algorithm that's practical and
> reasonable to implement, we can allow such orderings again.)
[Jeremy]
> Sounds sensible to me! I was quite puzzled about what <= should
> return for recursive objects.
That's easy: x <= y for recursive objects should return true if and only if
x < y or x == y return true <0.9 wink>.
x == y isn't a problem, although Python gives a remarkable answer:
recursive objects in Python are instances of rooted, ordered, directed,
finite, node-labeled graphs, and "x == y" in Python answers whether their
graphs are isomorphic.
Viewed that way (which is the correct way <0.5 wink>), the *natural* meaning
for "x <= y" is "y contains a subgraph isomorphic to x". And that has
*almost* all the nice properties we like:
x <= x is true
(x <= y and y <= z) implies x <= z
(x <= y and y <= x) if and only if x == y
However,
1. That's much harder to compute.
2. It implies, e.g., [2] <= [1, 2], and that's not what we *want*
non-recursive sequence comparison to mean.
3. It's a partial ordering: given arbitrary x and y, it may be that
neither contains an isomorphic image of the other.
4. We've again given up on avoiding surprises in *simple* comparisons
among builtin types, like (under current CVS):
>>> 1 < [1] < 0L < 1
1
>>> 1 < 1
0
>>>
so it's hard to see why we should do any work at all to avoid
violating "intuition" when comparing recursive objects: we're
already scrubbing the face of intuition with steel wool,
setting it on fire, then putting it out with an axe <wink>.
Now let's look at Guido's example (or one of them, anyway):
>>> a = []
>>> a.append(a)
>>> a.append("x")
>>> b = []
>>> b.append(b)
>>> b.append("y")
>>> a
[[...], 'x']
>>> b
[[...], 'y']
>>>
I think it's a trick of *typography* that caused my first thought to be
"well, clearly, a < b". That is, the *display* shows me two 2-element
lists, each with the same "blob" as the first element, and where a[1] is
obviously less than b[1]. Since "the blobs" are the same, the second
elements control the outcome.
But those "blobs" aren't really the same: a[0] is a, and b[0] is b, so
asking whether a < b by looking first at their first elements just leads
back to the original question: asking whether a[0] < b[0] is again asking
whether a < b, and that makes no progress. Saying that a is less than b by
fiat is *consistent* with the rules for lexicographic ordering, but so is
insisting that a is greater than b. There's no basis for picking one over
the other, and so no clear hope of coming up with a generally consistent
scheme. Well, one clear hope: if recursive comparison says "not equal", it
could resolve the dilemma by comparing object id instead. That would be
consistent (I mostly think at the moment ...), but if you run the program
above multiple times it may say a < b on some runs and b < a on others.
WRT "the right way", it should be clear from the attached picture that
neither a nor b contains an isomorphic image of the other, so from that POV
they're not comparable (a != b, but neither a <= b nor b <= a holds).
So this is what Guido made Python do:
>>> a == b # still cool: they're not isomorphic and Python knows it
0
>>> a < b
Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: can't order recursive values
>>> a <= b
Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: can't order recursive values
In light of that, I still find these mildly surprising:
>>> a < a
0
>>> a <= a
1
>>>
I guess some recursive values are more orderable than others <wink -- but
that's true! the ones Python can prove are equal are indeed "more
orderable">.
>>> import copy
>>> c = copy.deepcopy(a)
>>> c
[[...], 'x']
>>> a == c
1
>>> a <= c
1
>>> a < c
0
>>>
BTW, this kind of construction appears to give equality-testing that's at
best(!) exponential-time in the size of the dicts:
def timeeq(x, y):
from time import clock
import sys
s = clock()
result = x == y
f = clock()
print x, result, round(f-s, 1), "seconds"
sys.stdout.flush()
d = {}
e = {}
timeeq(d, e)
d[0] = d
e[0] = e
timeeq(d, e)
d[1] = d
e[1] = e
timeeq(d, e)
d[2] = d
e[2] = e
timeeq(d, e)
Output:
{} 1 0.0 seconds
{0: {...}} 1 0.0 seconds
{1: {...}, 0: {...}} 1 6.5 seconds
After more than 15 minutes, the 3-element dict comparison still hasn't
completed (yikes!).
ackerman's-function-eat-your-heart-out-ly y'rs - tim
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