> [Aside: if I remember right, you wrote the tacit version there at rosetta
code?]
That is right.
> After sleeping on this thread for a few days, I've finally realized
> that I don't have any idea what you mean by "ad hoc encodings".
>
> If you have time for some questions: What does "ad hoc encodings"
> mean? Is that a good thing or a bad thing? Why?
As far as I can see, when the argument of Y, the higher-order function, for
which the product of Y is a fixed-point, is represented by means of
standard code associated with recursions in J then one can produce
relatively simple versions of Y which are, as much as possible (given the
lack of direct support of higher-order functions as arguments by the
current official interpreters), in compliance with the specifications of
the Rosetta Code (RC) task.
(Beware of line-wrapping)
As an illustration, this is a comparison between the slightly revised
version of my non-tacit version of the Y combinator for monadic recursions,
X=. 1 :'<(<,'':''),<(<1;~":0),<(":0);,(''u u`:6('',(5!:5<''u''),'')`:6
y'')'(1 :'u u`:6')
renamed as X to be able to distinguish it from the explicit version of Y in
the RC entry,
Explicit alternate implementation
https://rosettacode.org/wiki/Y_combinator#Explicit_alternate_implementation
(which I suppose you wrote).
Using recursive computations of factorials as an example,
M=. 1 :'1:`(* u@:<:)@.*'
M=. (5!:1)<'M'
Due to J`s aforementioned limitation, it is unavoidable the use of a
representation of the adverb (i.e., the higher-order function), the AR of
the adverb (M) is chosen as the argument for X,
M X ("0) i.11
1 1 2 6 24 120 720 5040 40320 362880 3628800
This argument is closely related to usual ways to define recursive verbs,
( $:M`:6) ("0) i.11
1 1 2 6 24 120 720 5040 40320 362880 3628800
(f=. f M`:6) ("0) i.11
1 1 2 6 24 120 720 5040 40320 362880 3628800
or, more generaly,
(f=. 3 :'f M`:6 y')("0) i.11
1 1 2 6 24 120 720 5040 40320 362880 3628800
M=. 1 :'if. * y do. y * u <: y else. 1 end.'
M=. (5!:1)<'M'
(f=. 3 :'f M`:6 y')("0) i.11
1 1 2 6 24 120 720 5040 40320 362880 3628800
In contrast, Y (as per the explicit RC J entry) is a noun (and a gerund)
and by itself does not produce anything,
wrap=. _66 [\ (5!:5)@:<
wrap'Y'
,<(<,':'),<(<(,'0');3),<(,'0');3 21$'g=.y recur=
. sivelY`:6 g recur`:6 recur '
Not surprisingly, the argument for Y Ev (Ev is a name referring to `:6) is
also a noun (and a gerund) and by itself does not produce anything either,
wrap'almost_factorial'
,<(<(<,':'),<(<(,'0');2),<(,'0');5 26$' if. (_1 {:: m) <: #m do.
v |. y;_1 }. m else. (y;m) Defer
v`'''' end. '),<(<(<,'0'),<,<2),<(<,':'
),<(<(,'0');3),<(,'0');3 25$'''f n''=.y if. 0 >:
n do. 1 else. n * f`:6 n-1 end.'
and, it seems, its particular purpose is to be used as an argument for (Y
Ev).
On the one hand, regarding the recursive verbs produced by each of the
combinators,
v=. (Y Ev almost_factorial)Ev
is a verb,
wrap'v'
((<,<(<(<,':'),<(<(,'0');2),<(,'0');5 26$' if. (_1 {:: m) <: #m d
o. v |. y;_1 }. m else. (y;m) De
fer v`'''' end. '),<(<(<,'0'),<(<,<(<(<,
':'),<(<(,'0');2),<(,'0');5 26$' if. (_1 {:: m) <: #m do. v |.
y;_1 }. m else. (y;m) Defer v`''''
end. '),<(<(<,'0'),<,<2),<(<,':'),<(<(,
'0');3),<(,'0');3 25$'''f n''=.y if. 0 >: n do.
1 else. n * f`:6 n-1 end.'),<2),<(<,':'),<(<(,'0');3),<(,'
0');2 29$'''g recur''=.y (recursivelY`:6 g)`:6 r
ecur'),(<,<(<(<,':'),<(<(,'0');2),<(,'0');5 26$' if. (_1 {:: m) <
: #m do. v |. y;_1 }. m else. (y
;m) Defer v`'''' end. '),<(<(<,'0'),<,<2
),<(<,':'),<(<(,'0');3),<(,'0');3 25$'''f n''=.y
if. 0 >: n do. 1 else. n * f`:6 n-1 end.'),<3) (2 : 0) (3
: 0)
'g recur x'=.y
(g`:6 recur`:6 recur)`:6 x
)
if. (_1 {:: m) <: #m do.
v |. y;_1 }. m
else.
(y;m) Defer v`''
end.
)
and this linear representation of (Y Ev almost_factorial)Ev is incomplete;
in other words, the verb produced is not stateless and consequently it is
vulnerable to reassignments,
(Y Ev almost_fibonacci)Ev ("0) i. 11
0 1 1 2 3 5 8 13 21 34 55
Defer=. 1
(Y Ev almost_fibonacci)Ev ("0) i. 11
|syntax error
| (y;m)Defer v`''
Thus, the explicit entry in RC is not an implementation of the Y combinator
complying with the specifications.
On the other hand, the verb produced by M X is stateless and relatively
very simple,
u=. M X
wrap'u'
<(<,':'),<(<(,'0');1),<<;._1 '|0|u u`:6(<(<,'':''),<(<(,''0'');1),
<(,''0'');,:''if. * y do. y * u <: y else. 1 end.'')`:6 y' (1 : 'u
u`:6(<(<,'':''),<(<(,''0'');1),<(,''0'');,:''if. * y do. y * u <:
y else. 1 end.'')`:6 y')
Furthermore, according to the interpreter (Y Ev almost_factorial)Ev seems
to be doing a lot of unnecessary stuff vs M X for the task at hand,
stp=. ] (([ ((<;._1 '|Sentence|Space|Time|Space * Time') , (, */&.:>@:(1
2&{))@:(] ; 7!:2@:] ; 6!:2)&>) (10{a.) -.&a:@:(<;._2@,~) ]) [ (0 0 $
13!:8^:((0 e. ])`(12"_)))@:(2 -:/\ ])@:(".&.>)@:((10{a.) -.&a:@:(<;._2@,~)
]) ::(0 0&$@(1!:2&2)@:('Mismatch!'"_))) ".@:('0( : 0)'"_)
stp 11
(Y Ev almost_factorial)Ev("0) i.11
M X ("0) i.11
)
┌──────────────────────────────────┬───────┬──────────┬────────────┐
│Sentence │Space │Time │Space * Time│
├──────────────────────────────────┼───────┼──────────┼────────────┤
│(Y Ev almost_factorial)Ev("0) i.11│1242368│0.0142416 │17693.3 │
├──────────────────────────────────┼───────┼──────────┼────────────┤
│M X ("0) i.11│250304 │0.00179289│448.768 │
└──────────────────────────────────┴───────┴──────────┴────────────┘
I hope it helps
On Mon, Nov 19, 2018 at 7:11 PM Jose Mario Quintana <
[email protected]> wrote:
> I am on vacation this week. I will have more time (and access to a PC)
> when I get back.
>
> On Monday, November 19, 2018, Raul Miller <[email protected]> wrote:
>
>> [Aside: if I remember right, you wrote the tacit version there at rosetta
>> code?]
>>
>> After sleeping on this thread for a few days, I've finally realized
>> that I don't have any idea what you mean by "ad hoc encodings".
>>
>> If you have time for some questions: What does "ad hoc encodings"
>> mean? Is that a good thing or a bad thing? Why?
>>
>> Thanks,
>>
>> --
>> Raul
>>
>>
>>
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