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I'd like to assign an "UpUpValue" in a way generalized to any nested head surrounding the value for which the UpUpValue would be defined. That is, if a function h[x] is called and it is nested within another two functions f[g[h[x]]], I'd like it to have a specific behavior generalizable to any head g.

I had thought this would work:

h /: f[g_[h[x_]]] := (f[x] + g[x] + h[x])

However, TagSetDelayed is limited to 2nd level specification (such that it returns that "TagSetDelayed::tagpos : "Tag h in f[g_[h[x_]]] is too deep for an assigned rule to be found."). I then tried bypassing this by defining it manually using:

UpValues[g] = {HoldPattern[f[h_[g[x]]]] :> HoldPattern[f[x] + h[x] + g[x]]}

However, it seems this does not fire successfully.

The following using UpSetDelayed also doesn't work:

f[g_[h[x_]]] ^:= (f[x] + g[x] + h[x])

As this seeks to apply the rule to specific heads only (not general g that can be used on the RHS).

Can anyone conceive of a way to accomplish this in a way that preserves generality in the head of g? For any single function g, I could simply define an UpValue or DownValue, but I would like to do so in a general way such that it is applied to any function g when it is fed the head h.

Clarification on SetDelayed:

xzczd pointed out that the following would work in principal:

f[g_[h[x_]]] := (f[x] + g[x] + h[x])

However, this associates a DownValue with the symbol f. DownValues are checked exhaustively upon calling a function, such that making many additions to DownValues of a function f that is called many times can become inefficient when compared to making UpValues (or "UpUpValues") associated with a more rarely used function h.

For example, if you wanted to define special handling for 1000 different functions sitting in h's position, this would define 1000 different DownValues of f that must be checked each time f is called, rather than one "UpUpValue" for each unique function sitting in h's spot.

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    $\begingroup$ 1. f[g_[h[x_]]] ^:= (f[x] + g[x] + h[x]) doesn't work (and it's expected). 2. Why not simply f[g_[h[x_]]] := (f[x] + g[x] + h[x])? $\endgroup$
    – xzczd
    Aug 10, 2020 at 3:10
  • $\begingroup$ Depending on where you need this, you could try to build your own lightweight evaluator that applies arbitrary replacement rules to expressions $\endgroup$
    – Lukas Lang
    Aug 10, 2020 at 14:26
  • $\begingroup$ Another, (slightly overkill) approach could be to use @MrWizard's step function to do one evaluation step at a time, and then apply some additional rules each time before doing the next "normal" step $\endgroup$
    – Lukas Lang
    Aug 10, 2020 at 14:36
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    $\begingroup$ @xzczd This would certainly work. However, for efficiency reasons it is good to have an UpValue (or my aspiring UpUpValue) attached directly to the symbol h such that it does not populate the list of Downvalues for f, which would slow down very large development projects for which f might be called many, many times whereas h is only called rarely. This is an important enough point I will edit the main blog post to avoid confusion, because I too felt my request foolish for a minute after seeing your response. $\endgroup$
    – Ghersic
    Aug 10, 2020 at 20:34

1 Answer 1

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If you can stand a slight tweak:

ClearAll[h];
h /: g_[h[x_]] := h[g, x];
h /: f[h[g_, x_]] := f[x] + g[x] + h[x];
MakeBoxes[h[g_, x_], form_] := MakeBoxes[g[h[x]], form];

f[g[h[x]]]
(*  f[x] + g[x] + h[x]  *)
r[h[y]]
FullForm[%]
f[%%]
(*
  r[h[y]]   <-- Output form 
  h[r, y]   <-- Internal form 
  f[y] + h[y] + r[y]
*)

It's just a bit unclear how this is supposed to work: Just what is h[x]? Does it evaluate to something else or is h inert? Keeping it from evaluating may be hard if g is arbitrary. Consider this simplified example:

ClearAll[hh];
hh /: ff[hh[x_]] := ff[x] + hh[x];
hh[x_] := x^2;

ff[hh[x]]

(*  ff[x^2]  *)

The arguments of ff are evaluated individually before upvalues for hh are searched. The upvalue fails to be applied. However, if ff holds its argument, then the upvalue works:

SetAttributes[ff, HoldAll];
ff[hh[x]]
(*  x^2 + ff[x]  *)

Addendum: Comment on performance

Performance is one on the motivating factors in the OP's desire for an UpUpValue. Let's examine it.

First, make a 1000 symbols to serve as our potential h's.

syms = Table[Unique[], {1000}];
sym0 = syms[[500]]
(*  $591  <-- will vary *)

A comparison of the standard downvalue approach with the upvalue approach above shows that the OP has some justification:

ClearAll[fDown]; ClearAll @@ syms;
(fDown[g_[#[x_]]] := fDown[x] + g[x] + #[x]) & /@ syms;

fDown[Sin[Cos[x]]] // RepeatedTiming
fDown[Sin[sym0[x]]] // RepeatedTiming
(*
  {2.*10^-8, fDown[Sin[Cos[x]]]}
  {0.000068, fDown[x] + Sin[x] + $591[x]}
*)
ClearAll @@ syms;
(# /: g_[#[x_]] := #[g, x];
 # /: fUp[#[g_, x_]] := fUp[x] + g[x] + #[x];) & /@ syms;

fUp[Sin[Cos[x]]] // RepeatedTiming
fUp[Sin[sym0[x]]] // RepeatedTiming
(*
  {3.1*10^-8, fUp[Sin[Cos[x]]]}           <-- same
  {3.1*10^-6, fUp[x] + Sin[x] + $591[x]}  <-- faster
*)

Now, let's consider another downvalue method, which is just as fast as the upvalue method:

ClearAll[fDown2]; ClearAll @@ syms;
SetAttributes[fDown2, HoldAll];
assoc = AssociationThread[syms -> True]; 
fDown2[g_[h_[x_]]] /; Lookup[assoc, h, False] := 
 fDown2[x] + g[x] + h[x];

fDown2[Sin[Cos[x]]] // RepeatedTiming
fDown2[Sin[sym0[x]]] // RepeatedTiming
(*
  {3.1*10^-8, fDown2[Sin[Cos[x]]]}
  {2.2*10^-6, fDown2[x] + Sin[x] + $591[x]}
*)
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  • $\begingroup$ Thanks, Michael. This seems like it would work for general functions. In my case, h is typically one of many overridden system functions, which do generally evaluate to something else; I should've stated this initially. Since your approach will work for general functions, I'll wait awhile to see if anyone else proposes anything, and then select it as an answer if not. Thanks again. $\endgroup$
    – Ghersic
    Aug 12, 2020 at 18:07
  • $\begingroup$ @Ghersic You're welcome. You may be interested in fDown2, which appears at the end of the addendum I just appended. $\endgroup$
    – Michael E2
    Aug 12, 2020 at 22:02
  • $\begingroup$ That's a very interesting point about DownValues not necessarily being slow, if properly implemented. If, say, fDown2 was overloaded to have thousands or tens of thousands of DownValues in this fashion, I wonder if it would remain of comparable efficiency to the UpValues method. $\endgroup$
    – Ghersic
    Aug 27, 2020 at 18:46
  • $\begingroup$ @Ghersic I suppose it's as easy to check it out for 10000 symbols as 1000, though. $\endgroup$
    – Michael E2
    Aug 27, 2020 at 20:20

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