Re: Question on Unevaluated
- To: mathgroup at smc.vnet.net
- Subject: [mg117406] Re: Question on Unevaluated
- From: "Alexey Popkov" <lehin.p at gmail.com>
- Date: Thu, 17 Mar 2011 06:33:48 -0500 (EST)
Leonid,
This must have been a design decision. This also has to do with the semantics of both Block
(dynamic scoping) and local variables shared between the body and the condition. My feeling is that
this is a unique language feature, in the sense that it can not be easily reproduced or imitated
by other means.
It seems that the reason is that the evaluator just does not leave Block scope at all:
In[1]:= Clear[x];
x := Block[{tried = True}, x + 1] /; ! TrueQ[tried]
x /; ! TrueQ[tried] := x + 1
x
x /; TrueQ[tried] := x + 1
x
Out[4]= 1 + x
During evaluation of In[1]:= $RecursionLimit::reclim: Recursion depth of 256 exceeded. >>
Out[6]= 254 +
Hold[RuleCondition[$ConditionHold[$ConditionHold[
Block[{tried = True}, x + 1]]], ! TrueQ[tried]]]
This means that the evaluation is completed inside Block. It is unexpected for me. For what aims have the developers created such a peculiarity? In which cases it can be useful? Is Block unique in this sense?
From: Leonid Shifrin
To: Alexey Popkov ; mathgroup at smc.vnet.net
Sent: Thursday, March 17, 2011 4:02 AM
Subject: Re: [mg117372] Re: Question on Unevaluated
Alexey,
It seems that DownValues can help to ensure that new definition will beon top:
DownValues[f] = Prepend[DownValues[f], < new pattern >];
Yes, sure, direct manipulations with the DownValues are always an option. I just prefer to not do this
when it is not necessary, and in our context I believe it is not.
Very interesting! I do not understand why this happens at all. How does it work?
This must have been a design decision. This also has to do with the semantics of both Block
(dynamic scoping) and local variables shared between the body and the condition. My feeling is that
this is a unique language feature, in the sense that it can not be easily reproduced or imitated
by other means.
Is there a way to generalize this for preventing infinite recursion in arbitrary case, say classical
x=x+1
?
Well, this particular one is easy, and you know the trick already:
In[19]:= Clear[x];
Module[{tried}, x := Block[{tried = True}, x + 1] /; ! TrueQ[tried]]
In[21]:= x
Out[21]= 1 + x
Cheers,
Leonid
Alexey
----- Original Message -----
From: Leonid Shifrin
To: Alexey Popkov ; mathgroup at smc.vnet.net
Sent: Wednesday, March 16, 2011 2:49 AM
Subject: Re: [mg117264] Question on Unevaluated
Alexey,
I have two quesions regarding your function makeHoldN:
1) It works well in simplest cases even if makeHoldN is called before defining a function:
In[13]:= ClearAll[ff];
makeHoldN[ff,3];
ff[args___]:=Hold[args];
ff[1^2,2^2,3^2]
Out[16]= Hold[1,4,3^2]
Probably conflicts are only possible when the target function has definitions for which Mathematica's rule ordering
system cannot make conclusions about their generality?
Yes. Those may end up sitting on top of that added by makeHoldN. The problem for them will be not only that
calls matching their patterns will not be intercepted, but also that they will still be evaluated with HoldAll
attribute, which may not be desirable/intended (makeHoldN effectively releases all arguments except n-th).
Does calling makeHoldN after making all definitions always guarantee that the new definition will be on top?
Actually, I made a mistake in that statement - the call must be the first one, not the last one - sorry about that.
In fact, this is good, since it is always better to set Attributes before we make definitions, especially for Hold-
attributes. Otherwise, quite often some definitions partially evaluateduring the time they are entered (because Set and SetDelayed partially evaluate the l.h.s., I think contrary to some statements in the docs), which may result in very subtle bugs. Regarding the guarantees - I can not give them, but as far as I know, yes, the new definition will be on top, except possibly for definitions containing neither patterns nor conditions, such as memoized definitions like HoldPattern[f[2]]:>4 - they will be higher in the list. There is a way to force these to also go below the one added by make HoldN - for example, when you create them, add a dummy condition like HoldPattern[f[2]]:>4 /;(someSymbol;True). This will however negatively impact the performance - without conditions such no-pattern definitions are stored in a hash table and are (much) more efficient. But, anyway, this is a rather special case.
How it could be checked?
Well, you can always query the function information with ? and see theorder of the definitions. If your question
is how to ensure the generality of this behavior - I don't know, but I use this kind of tricks routinely and had no
problems with this. An informal proof of this behavior can go along these lines:
1. It is necessarily impossible for the Mathematica's rule-ordering system to determine relative generality of 2 rules where at least one contains a condition attached, with a user-defined symbol. The reason is that the fact of the match of such a rule depends on the global state and can only be determined at run-time.
2. When Mathematica can not determine the relative generality of some rules, it keeps them in the order they were entered - I think this statement can be found in the documentation somewhere.
I do not exclude the possibility for some exceptions, but generally this should work and my experience confirms that.
2) I do not well understand for what
/; Hold[result] =!= Hold[f[args]]]]) /;
FreeQ[DownValues[f], keepOnTop]]
is added. In simple cases makeHoldN works well wihout it.
Indeed, the line Hold[result] =!= Hold[f[args]] can be left out - this is a remnant of my intermediate implementation.
But the reason that we can do this appears quite subtle. Once the function (say f[1^2,2^2,3^2]) computes into
f[1,2^2,9], a new attempt to compute f is of course made. But now, it is made from within the Block scope, and
the definition does not match since keepOnTop is now False. The non-trivial part here is that the new evaluation
is not attempted after we leave the Block scope - this is why there is noinfinite recursion. I think that this, more
subtle part of the trick, I did not realize before (this looks different from the version of the trick with overloading
the built-ins, where the system definitions *are* used even when the function ends up being returned "unevaluated" - say due to unknown number and/or types of arguments).
The line FreeQ[DownValues[f], keepOnTop]] is needed to not add a new definition more than once.
Regards,
Leonid
Alexey
----- Original Message -----
From: Leonid Shifrin
To: Alexey Popkov ; mathgroup at smc.vnet.net
Sent: Monday, March 14, 2011 2:40 PM
Subject: Re: [mg117264] Question on Unevaluated
Alexey,
Doing what you request is generally not possible or at least extremely hard (emulating exact behavior of Unevaluated in all cases), since Unevaluated is one of the "magic symbols" (together with Evaluate and Sequence), wired deep into the system. I discuss this a bit more here:
http://stackoverflow.com/questions/4856177/preventing-evaluation-of-mathematica-expressions/4856721#4856721
Regarding your particular request: it is an interesting exercise in working with held expressions. Assuming
that we can only use Hold attributes, but neither Unevaluated nor Evaluate, the following function will
(hopefully) work as if you had an attribute HoldN (that is, n-th argument held, others not):
joinHeld[a__Hold] :=
Hold @@ Replace[Hold[a], Hold[x___] :> Sequence[x], {1}];
splitHeldSequence[Hold[seq___], f_: Hold] := List @@ Map[f, Hold[seq]];
Clear[makeHoldN];
Module[{keepOnTop},
makeHoldN[f_, n_Integer] :=
(SetAttributes[f, HoldAll];
f[args___] /; (! TrueQ[! keepOnTop]) :=
Block[{keepOnTop = False},
With[{result =
If[n > Length[Hold[args]],
f @@ {args},
Apply[f,
joinHeld @@
Flatten[{Hold[##] & @@@ Take[#, n - 1], #[[n]],
Hold[##] & @@@ Drop[#, n]}] &[
splitHeldSequence@Hold[args]]]]},
result /; Hold[result] =!= Hold[f[args]]]]) /;
FreeQ[DownValues[f], keepOnTop]]
Here are some examples of use:
ClearAll[f]
makeHoldN[f, 2];
f[1^2, 2^2, 3^2]
f[1, 2^2, 9]
In[22]:= ClearAll[ff];
ff[args___] := Hold[args];
makeHoldN[ff, 3];
ff[1^2, 2^2, 3^2]
Out[25]= Hold[1, 4, 3^2]
One can rather easily generalize this to hold an arbitrary subsequence of arguments (specified by
a list of their indices) while evaluating the rest. The implementation employs a number of tricks.
One that needs a bit of clarification is the f[args___] /; (! TrueQ[!keepOnTop]) line, since it serves
2 purposes. The one related to Block trick is well-known to you. The other is that the presence of
condition involving a user-defined symbol makes it impossible for Mathematica rule ordering
system to make conclusions about the generality of the rule, and therefore the rule does not go
to the bottom of the rule list. This is needed because we want this rule to stay at the top, to intercept
all calls to the function. For the same reason, makeHoldN should be called already after all the
definitions have been given to the function, or the function will not work properly.
As you see, this is sort of possible, but complex and error-prone. In practice, it is best to avoid
this sort of trickery, by changing the design of your functions. In my experience, having HoldFirst,
HoldRest and HoldAll is quite enough. To evaluate any held argument, you can also wrap it in
Evaluate. So, you particular question can be answered quite easily also as
SetAttributes[f,HoldAll];
f[Evaluate[Print[1]], Print[2], Evaluate[Print[3]]]
Note also, that I don't claim to reproduce exactly the behavior of Unevaluated with my function above.
It is just an illustration of a possible poor man's device to accomplish the specific goal of holding
n-th argument without the help of Evaluate and Unevaluated.
HTH
Regards,
Leonid
On Mon, Mar 14, 2011 at 1:06 AM, Alexey Popkov <lehin.p at gmail.com> wrote:
Leonid,
Is it possible to imitate the behavior of Unevaluated by setting At tributes in this case:
f[Print[1], Unevaluated[Print[2]], Print[3]]
?
I am wondering, what attributes are temporarily set when we use Unevaluated and how could I imitate this?
Alexey
----- Original Message -----
From: Leonid Shifrin
To: Alexey ; mathgroup at smc.vnet.net
Sent: Monday, March 14, 2011 1:39 AM
Subject: Re: [mg117264] Question on Unevaluated
Alexey,
You forgot about the CompoundExpression (;). You only attempted to prevent the evaluation of 1+1 inside
(1+1;3), but not the total result for CompoundExpression, which is the value of the last statement
(2+1 in this case). This is what you probably had in mind:
In[9]:= f[Unevaluated[(1 + 1; 2 + 1)]]
Out[9]= f[Unevaluated[1 + 1; 2 + 1]]
What is perhaps less obvious is that you did not prevent the evaluation of 1+1 either. Here is
a simple way to check it:
In[14]:= f[Unevaluated[Print["*"]]; 2 + 1]
During evaluation of In[14]:= *
Out[14]= f[3]
The problem is that Unevaluated is only effective once. To toally prevent something from evaluation,
you have to know the exact number of sub-evaluations (which is generally impossible to know since
it can be data-dependent), and wrap in as many levels of Unevaluated. In this case, the following will do:
In[13]:= f[Unevaluated[Unevaluated[Print["*"]]]; 2 + 1]
Out[13]= f[3]
But as I said, this is not a robust approach, and in such cases you will be better to use Hold or similar for
a persistent holding wrapper, stripping it off later when needed. You may want to check out e.g. this thread
http://groups.google.com/group/comp.soft-sys.math.mathematica/browse_thread/thread/bfd67e9122b1fdec
(my second post there), where I elaborate on these issues.
HTH.
Regards,
Leonid
On Sun, Mar 13, 2011 at 1:26 PM, Alexey <lehin.p at gmail.com> wrote:
Hello,
I am puzzled a bit by the Documentation for Unevaluated. Under "More
information" field we read:
"f[Unevaluated[expr]] effectively works by temporarily setting
attributes so that f holds its argument unevaluated, then evaluating
f[expr].".
After reading this I expect that
f[Unevaluated[1 + 1]; 2 + 1]
will be returned completely unevaluated as it is when I set HoldFirst
attribute to f:
In[2]:= SetAttributes[f, HoldFirst]
f[Unevaluated[1 + 1]; 2 + 1]
Out[3]= f[Unevaluated[1 + 1]; 2 + 1]
But in really we get
In[1]:= f[Unevaluated[1 + 1]; 2 + 1]
Out[1]= f[3]
This leads me to a question: what is implied in documentation? Which
attributes are temporarily set and to which function?