6
$\begingroup$

On this site I found, for example here, some very interesting questions and answers on the use of delayed rules with conditions. I would very much like to understand these answers, but so far I failed on a very basic level.

The documentation states that Condition is an expression with two arguments: a pattern and a test. Such an expression is itself a pattern, which is matched only by expressions that match the first argument and moreover for which the evaluation of test gives True.

Since the lefthand side of a rule is a pattern, the following works:

lhs /. (lhs/;True) :> rhs
(* rhs *)

My problem in understanding has to do with Condition in the right hand side of a rule. The documentation states that lhs :> rhs /; test represents a rule which applies only if the evaluation of test yields True. Such a rule is in fact a normal RuleDelayed, with the Condition expression as the second argument.

A Condition expression evaluates to itself, even when the second argument is True:

Condition[rhs,True]
(* rhs/;True *)

When I use a RuleDelayed, my understanding is that the unevaluated right hand side is substituted for any subexpression that matches the pattern of the left hand side, and then a further evaluation of the expression takes place. Therefore, I expected the result of the following two commands to be respectively Condition[rhs, False] and Condition[rhs, True].

lhs /. lhs:>(rhs /; False) (* lhs *)
lhs /. lhs:>(rhs /; True) (* rhs *)

In the first example the rule is not used at all, in agreement with the documentation for Condition, but not in accordance with my understanding of RuleDelayed. In the second example, the rule is used, but moreover at some point the right hand side is evaluated to its first argument.

So my impression is that when we have a RuleDelayed with a Condition in the right hand side, the evaluation rules seem to be different from those without the Condition. That is something I cannot believe. So can someone point me to what I do not properly understand?

$\endgroup$
  • $\begingroup$ You may find this discussion relevant. $\endgroup$ – Leonid Shifrin Jan 23 '16 at 20:46
  • 3
    $\begingroup$ [1/2] Since, as you mentioned, Condition can be viewed as a pattern, or, pattern-buiding block, it does not surprise me that it behaves in the special way in the context of the pattern-matching. But I think that the source of this behavior is not in Condition, but in the pattern-matcher, into which Condition must be wired pretty deeply. When the pattern-matcher sees a rule with Condition, it evaluates and matches that rule in a special way. The fact that Condition does keep the test code unevaluated, does not mean that it is actually the function that eventually evaluates it - ... $\endgroup$ – Leonid Shifrin Jan 23 '16 at 21:03
  • 3
    $\begingroup$ [2/2] ... rather, it may be a top-level container, that passes that code to some lower-level functionality that actually does the checks. Some analysis in my answer I linked above seems to suggest that this is indeed the case, and the actual heads used for checks are RuleCondition and $ConditionHold. The evaluations in Condition are induced by the pattern-matcher, and are sub-evaluations from the point of view of the main evaluation process for the original expression. Condition taken standalone may behave completely differently, and that does not violate the evaluation rules. $\endgroup$ – Leonid Shifrin Jan 23 '16 at 21:06
  • $\begingroup$ @Leonid. Many thanks for your further reference and your comments. It will give me a lot to study the next days ... Glad to hear that this behaviour is not simple to explain. $\endgroup$ – Fred Simons Jan 24 '16 at 12:13
  • $\begingroup$ Glad it you found it helpful. This is certainly one of the things that I wish to have better understood myself. $\endgroup$ – Leonid Shifrin Jan 24 '16 at 12:19
4
$\begingroup$

This was somewhat too long for a comment. Not sure it answers the question but it might give a way of thinking about the elusive handling of Condition that proves helpful.

The behavior in question is fairly well documented (I think people know this, I just wanted to state it for the record). The general idea to RuleDelayed is that it fires on a match and delivers its rhs. The devil in the details can be broken into two questions. (1) What comprises a match? (2) What is the evaluation of the rhs?

When Condition is present on the rhs we do get into some murky semantics issues. The gist is that it is part of what determines whether there is a match, which is to say that the lhs alone does not suffice (I believe this is the only exception to the general rule that the lhs determines a match). As for what is the evaluated rhs, if the Condition is True then the rhs is what preceded that condition. If it is not True then the issue is moot because the delayed rule is not going to fire. Which is the point of having this construct to begin with.

$\endgroup$
  • $\begingroup$ Thanks for this illuminating and helpful comment. A consequence of your explanation is that a match for Rule is not automatically a match for RuleDelayed, and that the second argument of Rule and of RuleDelayed may be evaluated differently. This surprises me, but it answers my question. $\endgroup$ – Fred Simons Jan 25 '16 at 7:13
0
$\begingroup$

Actually, both

FullForm@Hold[lhs /. lhs :> (rhs /; test)]

and

FullForm@Hold[lhs /. lhs :> rhs /; test]

yield the same:

Hold[Replace[lhs,RuleDelayed[lhs, Condition[rhs,test]]]]

and this Replace evaluates as the documentation says it does. I think you misunderstand Condition as an If construct that evaluates to something on its own. But it seems to me that Condition is closely tied to ReplaceAll as you can make reference to pattern names appearing only in the left hand side of some (delayed) rule. Consider, for example,

lhs[1, 2, 3] /. x_ :> (rhs /; (x > 2))]

(lhs[1, 2, rhs]) and its FullForm.

On the other hand, you can "switch-off and on" the condition by typing

Condition@@(lhs/.lhs:>NoCondition[rhs,True])

Im not sure if that answers your question, though. (Are you asking out of curiosity or do you have some application in mind, where you would need the ReplaceAll to return expressions with head Condition?)

$\endgroup$
  • 1
    $\begingroup$ Yes, as you observed as in my question, in both situations we have a RuleDelayed with a right hand side Condition[rhs, test]. But this ReplaceAll does NOT evaluate as the documentation says it does. When it did, the result had to be Condition[rhs, test] (Replace e.g. Condition with condition to see the difference.) It seems that when we have a Condition in the second argument of RuleDelayed (not Rule!), this rule is used in a non standard way, and that is what my question is about. Despite the fact that I do not understand it, I have no problems in using it. $\endgroup$ – Fred Simons Jan 24 '16 at 7:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.