Mathematica Asked by Ghersic on March 22, 2021
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
.
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.
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] *)
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]} *)
Correct answer by Michael E2 on March 22, 2021
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