Re: Baby X is bor nagain

Tim Rentsch <tr.17687@z991.linuxsc.com> writes:

Ben Bacarisse <ben@bsb.me.uk> writes:
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Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
>

Ben Bacarisse <ben@bsb.me.uk> writes:
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Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
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Ben Bacarisse <ben@bsb.me.uk> writes:
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Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
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Ben Bacarisse <ben@bsb.me.uk> writes:
[...]
>

On a C language point, I don't think the standard says anything
about sorting with non-order functions like the one above.  Is
an implementation of qsort permitted to misbehave (for example
by not terminating) when the comparison function does not
implement a proper order relation?

>
N1570 7.22.5p4 (applies to bsearch and qsort):
"""
When the same objects (consisting of size bytes, irrespective of
their current positions in the array) are passed more than once
to the comparison function, the results shall be consistent with
one another.  That is, for qsort they shall define a total
ordering on the array, and for bsearch the same object shall
always compare the same way with the key.
"""
>
That's a "shall" outside a constraint, so violating it results in
undefined behavior.

>
I think it should be clearer.  What the "that is" phrase seems to
clarify in no way implies a total order, merely that the repeated
comparisons or the same elements are consistent with one another.
That the comparison function defines a total order on the elements
is, to me, a major extra constraint that should not be written as
an apparent clarification to something the does not imply it:
repeated calls should be consistent with one another and, in
addition, a total order should be imposed on the elements present.

>
I think you're misreading the first sentence.

>
Let's hope so.  That's why I said it should be clearer, not that it
was wrong.
>

Suppose we are in
court listening to an ongoing murder trial.  Witness one comes in
and testifies that Alice left the house before Bob.  Witness two
comes in (after witness one has gone) and testifies that Bob left
the house before Cathy.  Witness three comes in (after the first
two have gone) and testifies that Cathy left the house before
Alice.  None of the witnesses have contradicted either of the
other witnesses, but the testimonies of the three witnesses are
not consistent with one another.

>
My (apparently incorrect) reading of the first sentence is that
the consistency is only required between the results of multiple
calls between each pair.  In other words, if the witnesses are
repeatedly asked, again and again, if Alice left before Bob and/or
if Bob left before Alice the results would always be consistent
(with, of course, the same required of repeatedly asking about the
other pairs of people).

>
Let me paraphrase that.  When the same pair of objects is passed
more than once to individual calls of the comparison function, the
results of those different calls shall each be consistent with
every other one of the results.

>
No, only with the results of the other calls that get passed the same
pair. [...]

>
Sorry, my oversight.  That's is what I meant.  "When the same pair
of objects is passed more than once to individual calls of the
comparison function, the results of those different calls shall
each be consistent with every other one of THOSE results."  The
consistency is meant to be only between results of comparisons
of the same pair.  (This mistake illustrates how hard it is to
write good specifications in the C standard.)
>

To paraphrase my reading, when some set of "same" objects is each
passed more than once to individual calls of the comparison
function, the results of all of those calls taken together shall
not imply an ordering contradiction.
>
Are the last two paragraphs fair restatements of our respective
readings?

>
I don't think so.  The first does not seem to be what I meant, and the
second begs a question:  what is an ordering contradiction?

>
A conclusion that violates the usual mathematical rules of the
relations less than, equal to, greater than:  A<B and B<C implies
A<C, A<B implies A!=B, A=B implies not A<B, A<B implies B>A, etc.
>

Maybe I could work out what you mean by that if I thought about it
some more, but this discussion has reminded me why I swore not to
discuss wording and interpretation on Usenet.  You found the wording
adequate.  I didn't.  I won't mind if no one ever knows exactly why
I didn't.  C has managed fine with this wording for decades so there
is no practical problem.  I think enough time has been spent on this
discussion already, but I can sense more is likely to spent.

>
A small correction:  I found the wording understandable.  If the
question is about adequacy, I certainly can't give the current
wording 10 out of 10.  I would like to see the specification for
qsort stated more plainly.  Although, as you can see, I'm having
trouble figuring out how to do that.
>

Is the second paragraph plain enough so that you
would not misconstrue it if read in isolation?  Or if not, can
you suggest a better phrasing?

>
Since I don't know what an ordering contradiction is, I can't suggest
an alternative.

>
Now that I have explained that phrase, I hope you will have a go
at finding a better wording.

I would not introduce your new term, "an ordering contradiction", since
it still leaves exactly what kind of order vague.  You interpret
"consistent" as "consistent with a total order" so I'd use that phrase:

  "when some set of 'same' objects is each passed more than once to
  individual calls of the comparison function, the results of all of
  those calls taken together shall be consistent with a total order"

Presumably you came to interpret "consistent with one another" as
implying a total order rather because of the sentence that follows
("That is, for qsort they shall define a total ordering on the array").

I could not do that because I was interpreting the text about multiple
calls differently.

... The important point is the "consistent with" is something of an
idiomatic phrase, and it doesn't mean "equivalent to" or "the same
as".  Maybe you already knew that, but I didn't, and learning it
helped me see what the quoted passage is getting at.

...

If you care to be less cryptic, maybe you will say what it was
about the meaning of "consistent with" that helped you see what
the text in question was getting at.

>
I think the key thing is that "consistent with" doesn't mean the
same.  If we're comparing the same pair of objects over and over,
the results are either the same or they are different.  It would
be odd to use "consistent with one another" if all that mattered
is whether they are all the same.

I never thought they were the same.  The trouble is that (a) different
results imply the same order (e.g. -1 and -34 all mean <) and (b) the
(old) wording does not say that the objects are passed in the same order
and the result of cmp(a, b) can't be the same as cmp(b, a) but they can
be consistent.  This makes "consistent with one another" a perfectly
reasonable thing to say even in my limited view of what results are
being talked about.

...

I have a second objection that promoted that remark.  If I take the
(apparently) intended meaning of the first sentence, I think that
"consistent" is too weak to imply even a partial order.  In dog club
tonight, because of how they get on, I will ensure that Enzo is
walking behind George, that George is walking behind Benji, Benji
behind Gibson, Gibson behind Pepper and Pepper behind Enzo.  In what
sense is this "ordering" not consistent?  All the calls to the
comparison function are consistent with each other.

>
I understand the objection, and this is the point I was trying to
make in the paragraph about children in the Jones family.  The
phrase "one another" in "the results shall be consistent with one
another" is meant to be read as saying "all the results taken
together".  It is not enough that results not be contradictory taken
two at a time;  considering all the results at once must not lead to
an ordering contradiction.

...

All the results of the dog-order comparison function, taken together,
are consistent with the circular order, which is obviously not a total
order.

>
If A<B, B<C, C<D, D<E, and E<A, we can infer from the transitivity
of the "less than" relation that A<A.  But A<A can never be true, so
this set of comparison results is no good.

[Technical aside.  The relation should be seen as <=. not <.  You can't
conclude that I intended A < A from the informal presentation -- no dog
can be behind itself.  However, this does not alter your argument in any
significant way.]

So I guess what we have
discovered is that "consistent with one another" is intended to mean
"obeys the usual mathematical rules for ordering relations".

I would say this is backwards.  You are assuming the usual rules where I
gave an order that is not at all usual with the purpose of showing that
some sets of comparisons between pairs can be "consistent with one
another" when the ordering is very peculiar.

On a more mathematical note, imagine that the text was describing a
topological sort function.  Is there anything in your reading of the
first sentence that would make it inappropriate?  If not, that
"consistent with one another" can't imply a total order.

...

It occurs to me now to say that "consistent with" is meant to
include logical inference.

Sure.

That distinction is a key difference
between "consistent" and "consistent with" (at least as the two
terms might be understood).  The combination of:  one, the results
of the comparison function are seen as corresponding to an ordering
relation;

But, according to you, only some ordering relations.

and two, that "consistent with one another" includes
logical inferences considering all of the results together;  is what
allows us to conclude that the results define a total order.

Could the sentence in question be used in the description of a
topological sort based (rather unusually) on a partial order?

I'm sorry if any of the above sounds like it's just stating the
obvious.  I'm strugging trying to find a way to explain what to
me seems straightforward.

--
Ben.