It’s APL…but not as we know it!

by the Dyalog Duck
As the party behind Dyalog’s Twitter account I often search our feed for #APL or just APL to see if anyone is talking about us or APL in general and I am frequently surprised by what pops up. We have often had a giggle in the office over some of the things we find and it was suggested to me that others might find these interesting too hence this rather random blog post.

Probably the most well known alternative APL is the container shipping and ocean freight company American President Lines (@APLShipping). The Dyalog staff often stand at the railway crossing on our way to the village bakery for lunch and chuckle at the APL containers whizzing past on the train … we are easily amused!

My personal favourite other APL has to be Athletic Propulsion Labs (@APLrunning) with their ‘oh so gorgeous running shoes’ … their trainers are things that I desire on a daily basis!! For you non-shoe lovers out there, they are also the official footwear supplier for the Renault Sport Formula One team.

The APL alter-ego that comes up most frequently in my Twitter search is a musical one … it seems we also share #APL with a 4 member boyband from South Korea. This APL (@APL_world) is extremely popular with teenage girls from Japan and Korea who frequently tweet their undying love for the boys … it makes you wonder who thought it was a good idea to name a teenage pop sensation after a programming language!

Still in the musical world we also have an uber cool APL namesake in rapper apl.de.ap (@apldeap) who is also one of the founding members of the Black Eyed Peas.

Another APL that we are proud to share #APL with is the John Hopkins University Applied Physics Laboratory (@JHUAPL) in Maryland USA. NASA’s Juno spacecraft successfully entered Jupiter’s orbit in July carrying an instrument suite that includes the Jupiter Energetic Particle Detector Instrument (JEDI) built by the Johns Hopkins Applied Physics Laboratory. APL in space … how cool is that!

twitterapl_6The Association of Professional Landscapers (@The_APL) are the APL that fill my screen with greenness, tweeting some amazing photos of their garden and landscaping projects.

twitterapl_7Finally, if you are having a bad day and just need cute puppies and kittens to coo over, the Animal Protective League (@APLSpringfield) in Springfield USA is the perfect APL for you. As well as providing cuteness overload they also do a great job re-homing animals in need.

So there you go folks … a few of my favourite alternative APLs for your amusement. Of course there are lots of other APLs out there in Twitterland but frankly most them are just not that interesting … and a few, to spare your blushes, are just not the sort of thing we want to share!

A Memo Operator

Abstract

Some functions are simply stated and easily understood as multiply-recursive functions. For example, the Fibonacci numbers are {1≥⍵:⍵ ⋄ (∇ ⍵-2)+∇ ⍵-1}. A drawback of multiple recursion is abysmal performance even for moderate-sized arguments, consequently motivating resorts to circumlocutions. The memo operator addresses the performance problem and restores multiple recursion to the toolkit of thought. We present a memo operator modelled as a 6-line d-operator.

⎕io←0 throughout.

The Man Who Knew Infinity

Jay sent out an e-mail asking who would be interested to go see the film The Man Who Knew Infinity. Fiona’s response: “Ah, Mr. 1729; I’m in.” John was game, and so was I.

The Number of Partitions
Nick found an essay by Stephen Wolfram on Ramanujan written on occasion of the film, wherein Wolfram said:

A notable paper [Ramanujan] wrote with Hardy concerns the partition function (PartitionsP in the Wolfram Language) that counts the number of ways an integer can be written as a sum of positive integers. …

In Ramanujan’s day, computing the exact value of PartitionsP[200] was a big deal — and the climax of his paper. But now, thanks to Ramanujan’s method, it’s instantaneous:

In[1]:= PartitionsP[200]
Out[1]= 3 972 999 029 388

A partition of a non-negative integer n is a vector v of positive integers such that n = +/v, where the order in v is not significant. For example, the partitions of the first seven non-negative integers are:

┌┐
││
└┘
┌─┐
│1│
└─┘
┌─┬───┐
│2│1 1│
└─┴───┘
┌─┬───┬─────┐
│3│2 1│1 1 1│
└─┴───┴─────┘
┌─┬───┬───┬─────┬───────┐
│4│3 1│2 2│2 1 1│1 1 1 1│
└─┴───┴───┴─────┴───────┘
┌─┬───┬───┬─────┬─────┬───────┬─────────┐
│5│4 1│3 2│3 1 1│2 2 1│2 1 1 1│1 1 1 1 1│
└─┴───┴───┴─────┴─────┴───────┴─────────┘
┌─┬───┬───┬─────┬───┬─────┬───────┬─────┬───────┬─────────┬───────────┐
│6│5 1│4 2│4 1 1│3 3│3 2 1│3 1 1 1│2 2 2│2 2 1 1│2 1 1 1 1│1 1 1 1 1 1│
└─┴───┴───┴─────┴───┴─────┴───────┴─────┴───────┴─────────┴───────────┘

Can we compute the partition counting function “instantaneously” in APL?

We will use a recurrence relation derived from the pentagonal number theorem, proved by Euler more than 160 years before Hardy and Ramanujan:

equation (11) in http://mathworld.wolfram.com/PartitionFunctionP.html. In APL:

      pn  ← {0≥⍵:0=⍵ ⋄ -/+⌿∇¨rec ⍵}
      rec ← {⍵ - (÷∘2 (×⍤1) ¯1 1 ∘.+ 3∘×) 1+⍳⌈0.5*⍨⍵×2÷3}

      pn¨0 1 2 3 4 5 6
1 1 2 3 5 7 11

      pn 30
5604

Warning: don’t try pn 200 because that would take a while! Why? pn 200 would engender pn being applied to each element of rec 200 and:

   rec 200
199 195 188 178 165 149 130 108 83 55 24 ¯10
198 193 185 174 160 143 123 100 74 45 13 ¯22

Each of the non-negative values would itself engender further recursive applications.

A Memo Operator

I recalled from J the memo adverb (monadic operator) M. Paraphrasing the J dictionary: f M is the same as f but may keep a record of the arguments and results for reuse. It is commonly used for multiply-recursive functions. Sounds like just the ticket!

The functionistas had previously implemented a (dyadic) memo operator. The version here is more restrictive but simpler. The restrictions are as follows:

  • The cache is constructed “on the fly” and is discarded once the operand finishes execution.
  • The operand is a dfn {condition:basis ⋄ expression} where none of condition, base and expression contain a and recursion is denoted by in expression
  • The arguments are scalar integers.
  • The operand function does not have ¯1 as a result.
  • A recursive call is on a smaller integer.

With these restrictions, our memo operator can be defined as follows:

M←{
 f←⍺⍺
 i←2+'⋄'⍳⍨t←3↓,⎕CR'f'
 0=⎕NC'⍺':⍎' {T←(1+  ⍵)⍴¯1 ⋄  {',(i↑t),'¯1≢T[⍵]  :⊃T[⍵]   ⋄ ⊃T[⍵]  ←⊂',(i↓t),'⍵}⍵'
          ⍎'⍺{T←(1+⍺,⍵)⍴¯1 ⋄ ⍺{',(i↑t),'¯1≢T[⍺;⍵]:⊃T[⍺;⍵] ⋄ ⊃T[⍺;⍵]←⊂',(i↓t),'⍵}⍵'
}

      pn M 10
42

      pn 10
42

Obviously, the lines in M are crucial. When the operand is pn, the expression which is executed is as follows; the characters inserted by M are shown in red.

{T←(1+⍵)⍴¯1 ⋄ {0≥⍵:0=⍵ ⋄ ¯1≢T[⍵]:⊃T[⍵] ⋄ ⊃T[⍵]←⊂-/+⌿∇¨rec ⍵}⍵}

The machinations produce a worthwhile performance improvement:

      pn M 30
5604
      cmpx 'pn M 30'
3.94E¯4
      cmpx 'pn 30'
4.49E1 
      4.49e1 ÷ 3.94e¯4
113959

That is, pn M 30 is faster than pn 30 by a factor of 114 thousand.

Can APL compute pn M 200 “instantaneously”? Yes it can, for a suitable definition of “instantaneous”.

      cmpx 'm←pn M 200'
4.47E¯3
      m
3.973E12
      0⍕m
 3972999029388

M also works on operands which are anonymous, dyadic, or have non-scalar results.

Fibonacci Numbers

   fib←{1≥⍵:⍵ ⋄ (∇ ⍵-2)+∇ ⍵-1}

   fib¨ ⍳15
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377
   fib M¨ ⍳15
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

   cmpx 'fib 30'
1.45E0 
   cmpx 'fib M 30'
1.08E¯4

Anonymous function:

      {1≥⍵:⍵ ⋄ (∇ ⍵-2)+∇ ⍵-1}M¨ ⍳15
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

Combinations
⍺ comb ⍵ produces a matrix of all the size- combinations of ⍳⍵. The algorithm is from the venerable APL: An Interactive Approach by Gilman and Rose.

      comb←{(⍺≥⍵)∨0=⍺:((⍺≤⍵),⍺)⍴⍳⍺ ⋄ (0,1+(⍺-1)∇ ⍵-1)⍪1+⍺ ∇ ⍵-1}

      3 comb 5
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4

      3 (comb ≡ comb M) 5
1

      cmpx '8 comb M 16' '8 comb 16'
  8 comb M 16 → 1.00E¯3 |     0% ⎕                             
  8 comb 16   → 3.63E¯2 | +3526% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

A faster comb is possible without a memo operator, but it’s not easy. (See for example section 4.1 of my APL87 paper.) A memo operator addresses the performance problem with multiple recursion and restores it to the toolkit of thought.

Presented at the BAA seminar at the Royal Society of Arts on 2016-05-20

I-Beam Mnemonics

I-Beam () is an operator that takes as its operand a numeric code and derives a function which isn’t really considered to be part of the APL language – for example: something which could be experimental, which might provide access to parts of the interpreter that should only be accessed with care, or may set up specific conditions within the interpreter to help in testing. Principally, I-Beam functions exist for internal use within Dyalog but as of version 15.0 there are 55 I-Beam functions that are published and available for general use. How on earth are you supposed to remember all the different codes?

To an extent, you are not. It is perhaps no bad thing that they are a little difficult to remember: I-Beam functions are experimental, liable to be changed or even removed and should be used with care. The codes appear to be somewhat random partly because they are somewhat random – mostly selected on the whim of the developer who implemented the function. However, some codes were chosen less randomly than others – quite a few are grouped so that I-Beams that provide related functionality appear consecutively or at least close together but the most interesting ones are “named” – or, at least, given a numeric code that is derived from a name or otherwise memorable value. So here are some of those unofficial mnemonics that may help you better remember them too.

A favourite trick for deriving a code from a meaningful name is to devise a name containing the letters I, V, X, L, C, D and M, and then convert that into a number as if it were a Roman numeral. Thus:

  • “Inverted table IndeX of” is abbreviated to IIX, which is I-Beam 8
  • “Syntax Colouring” rather awkwardly becomes “Cyntax Colouring”, thence CC and I-Beam 200
  • “Called Monadically” is CM, or 900
  • “Memory Manager” is MM – I-Beam 2000
  • “Line Count” is LC, or rather L,C – 50100

Another favourite is to use numbers that look or sound like something else:

  • The compression I-Beam is 219, which looks a bit like “ZIP”
  • The case folding I-Beam is 819, or “BIg”
  • When support for function trains was in development an I-Beam was used to switch between different implementations – 1060, or “lOCO[motive]” (this I-Beam is no longer in use)
  • The number of parallel threads I-Beam is 1111 – four parallel lines
  • The fork I-Beam is 4000: 4000 is 4K; “Four K” said very quickly might sound like “Fork”

For others the function is associated, or can be associated, with something already numeric:

  • The I-Beam to update function timestamps is 1159 – a memorable time
  • The I-Beam to delete the content in unused pockets is 127 – the same as the ASCII code for the delete character
  • The draft JSON standard is RFC 7159 and support for it is implemented as 7159⌶
  • The now-deprecated I-Beam to switch between Random Number Generator algorithms is 16807 – the default value used to derive the random seed in a clear workspace

Tips for the use of I-Beam functions

  • Do not use I-Beam functions directly in application code – always encapsulate their use in cover-functions that can quickly be modified to protect your application in the event that Dyalog should change the behaviour of, or withdraw, an I-Beam function.
  • Do not use I-Beam functions that are not documented by Dyalog in the Language Reference or elsewhere – those designed to test the interpreter may have undesirable side effects.
  • Do let Dyalog know if you find an I-Beam function particularly useful and/or have suggestions for its development. Some new features are initially implemented as I-Beam functions to allow such feedback to shape their final design.

Shuffle Faster Please!

Andy reported that in the shuffle QA some functions take a long time:

m9249   “4½ days so far”
rankop  21.5 hours
m12466  26.3 hours
points  7.2 hours

Background: Shuffle QA is an intensification of the normal QA. The suite of over 1800 QA functions is run under shuffle, whereby every getspace (memory allocation) is followed by every APL array in the workspace being shifted (“shuffled”) and by a simple workspace integrity check. The purpose is to uncover memory reads/writes which are rendered unsafe by events which in normal operations are rare.

m9249

m9249 tests +\[a], ×\[a], ⌈\[a], and ⌊\[a]. It ran in under 3 seconds in normal operations; for it to take more than 4½ days under shuffle, getspace is clearly the dominant resource and hereafter we focus on that.

m9249 used expressions like:

assert (+⍀x) ≡ {⍺+⍵}⍀x

Since {⍺+⍵} is not recognized by the scan operator and therefore not supported by special code, {⍺+⍵}⍀x is done by the general O(n*2) algorithm for scans, AND on each scalar the operand {⍺+⍵} is invoked and a getspace is done. The improved, faster QA does this:

F ← {1≥≢⍵:⍵ ⋄ x ⍪ (¯1↑⍵) ⍺⍺⍨ ¯1↑x←⍺⍺ ∇∇ ¯1↓⍵}
assert (+⍀x) ≡ +F x

With such a change, the faster m9249 runs in 22 minutes under shuffle instead of over 4½ days, without any reduction in coverage.

The number of getspace are as follows: +⍀x takes O(1). The second expression {⍺+⍵}⍀x does ×\1↓⍴x separate scans, and each one takes O((≢x)*2) getspace; the total is therefore O(((≢x)*2)××/1↓⍴x) ←→ O((≢x)××/⍴x). The third expression +F is linear in ≢x and is therefore O(≢x) in the number of getspace. In m9249 the largest x has size 33 5 7. The following table shows the orders of complexity and what they imply for this largest x:

+⍀x O(1) 1
{⍺+⍵}⍀x O((≢x)××/⍴x) 38115 = 33 × ×/ 33 5 7
+F x O(≢x) 33

The ratio between 38115 and 33 goes a long way towards explaining why the time to run m9249 under shuffle was over 4½ days and is now 22 minutes. (Why does m9249 still take 22 minutes? It tests for the four functions + × ⌈ ⌊, for various axes, for different datatypes, and for different axis lengths.)

rankop

Another shuffle sluggard rankop took 47 hours. rankop tests expressions of the form f⍤r⊢yand x f⍤r⊢y for various functions f. The techniques used for reducing the number of getspace differ from function to function. We look at x↑⍤r⊢y as an illustration.

assert (4↑⍤1⊢y) ≡ 4{⍺↑⍵}⍤1⊢y

⍴y was 2 7 1 8 2 8. The expression took 7.6e¯5 seconds normally and 7.5 seconds under shuffle. The left hand side requires O(1) getspace, the right hand side O(×/¯1↓⍴x). The expression was changed to:

assert (4↑⍤1⊢y) ≡ 2 7 1 8 2 4↑y

Feels like cheating, doesn’t it? :-) The new expression takes 0.115 seconds under shuffle.

m12466

m12466 tests n⌈/[a]x (and n⌊/[a]x). Previously, it did this by comparing n⌈/[a]x against n{⍺⌈⍵}/[a]x for x with shape 7 17 13 11, for each of the four axes, for two different values of n, and for the 1-, 2-, 4-, and 8-byte numeric datatypes. It required 5238426 getspace and 26.3 hours to run under shuffle.

F←{p⍉⍺⍺⌿(⊂(⎕io-⍨⍳|⍺)∘.+⍳(1-|⍺)+(⍴⍵)[⍵⍵])⌷(⍋p←⍒⍵⍵=⍳⍴⍴⍵)⍉⍵}

F is a dyadic operator. The left operand ⍺⍺ is or ; the right operand ⍵⍵ is the axis; and n(⌈ F a)x is the same as n⌈/[a]x. n(⌈ F a)x mainly does n⌈⌿x; other axes are handled by transposing the axis of interest to be the first, then doing n⌈⌿x, then transposing the axes back into the required order. n(⌈ F a)x is much more complicated than n{⍺⌈⍵}/[a]x but has the advantage of using only O(1) getspace.

The revamped m12466 function uses 13717 getspace and 2 minutes to run under shuffle.

points

points is a function to test monadic format () for different values of ⎕pp.

points←{⍺←17 ⋄ ⎕pp←⍺
  tt←{(⌽∨\⌽⍵≠⍺)/⍵}                   ⍝ trim trailing ⍺s.    
  ~0∊∘.{                             ⍝ all pairs (⍳⍵)       
    num←'0'tt↑,/⍕¨⍺'.'⍵              ⍝ eg '9.3'             
    num≡⍕⍎num                        ⍝ check round trip     
  }⍨⍳⍵
}

The expression is 14 15 16 17 points¨ 100, meaning that for each of the 100 numbers num as text,

  1.1   1.2   1.3   1.4   1.5     1.100
  2.1   2.2   2.3   2.4   2.5     2.100
  3.1   3.2   3.3   3.4   3.5 ... 3.100
...
 99.1  99.2  99.3  99.4  99.5    99.100
100.1 100.2 100.3 100.4 100.5   100.100

a “round-trip” num≡⍕⍎num is evaluated. The expression requires 1.3 million getspace and 7.2 hours to run under shuffle.

A much more efficient version with respect to getspace is possible. Let x be a vector. ⍕¨x requires O(≢x) getspace; ⍕x requires O(1) getspace and, like ⍕¨x, formats each individual element of x independently.

points1←{
  ⎕io←⎕ml←1                                                 
  tt←{(⌽∨\⌽⍵≠⍺)/⍵}                   ⍝ trim trailing ⍺s     
  x←1↓∊(' ',¨s,¨'.')∘.,'0'tt¨s←⍕¨⍳⍵  ⍝ numbers as text      
  y←⍎x                                                      
  {⎕pp←⍵ ⋄ x≡⍕y}¨⍺                   ⍝ check round trip     
}

14 15 16 17 points1 100 requires 11050 getspace and less than 2 minutes to run under shuffle.

What to do?

Andy says that the shuffle QA takes over 2000 hours (!). The shuffle sluggards will be tackled as done here by using more parsimonious APL expressions to compare against the aspect being QA-ed. Going forward, a new QA function will be run under shuffle as it is being developed. The developers will be unlikely to tolerate a running time of 4½ days.

Loops, Folds and Tuples

For-loops
Given an initial state defined by a number of variables, a for-loop iterates through its argument array modifying the state.

    A←... ⋄ B←... ⋄ C←...       ⍝ initial state
    :For item :In items         ⍝ iterating through array "items"
        A←A ... item            ⍝ new value for A depending on item
        C←C ... A ... item      ⍝ new value for C depending on A and item
        ...                     ⍝ state updated
    :EndFor
    A B C                       ⍝ final state

In the above example the state comprises just three variables. In general, it may be arbitrarily complex, as can the interactions between its various components within the body of the loop.

Dfns
Dfns don’t have for-loops. Instead, we can use reduction (or “fold”) with an accumulating vector “tuple” of values representing the state. Here is the D-equivalent of the above code:

    ⊃{                      ⍝ next item is ⍺
        (A B C)←⍵           ⍝ named items of tuple ⍵
        A∆←A ... ⍺          ⍝ new value for A depending on item ⍺
        C∆←C ... A∆ ... ⍺   ⍝ new value for C depending on A∆ and item ⍺
        ...                 ⍝ ...
        A∆ B C∆             ⍝ "successor" tuple (A and C changed)
    }/(⌽items),⊂A B C       ⍝ for each item and initial state tuple A B C

In this coding, the accumulating tuple arrives as the right argument (⍵) of the operand function, with the next “loop item” on the left (⍺). Notice how the items vector is reversed (⌽items) so that the items arrive in index-order in the right-to-left reduction.

If you prefer your accumulator to be on the left, you can replace the primitive reduction operator (/) with Phil Last’s foldl operator, which also presents its loop items in index-order:

    foldl←{⊃⍺⍺⍨/(⌽⍵),⊂⍺}    ⍝ fold left

then:

    A B C {                 ⍝ final state from initial state (left argument)
        (A B C)←⍺           ⍝ named items of tuple ⍺
        A∆←A ... ⍵          ⍝ new value for A depending on item ⍵
        C∆←C ... A∆ ... ⍵   ⍝ new value for C depending on A∆ and item ⍵
        A∆ B C∆             ⍝ successor tuple (A and C changed)
    } foldl items

If the number of elements in the state tuple is large, it can become unwieldy to name each on entry and exit to the operand function. In this case it simplifies the code to name the indices of the tuple vector, together with an at operator to denote the items of its successor:

    T ← (...) (...) ...         ⍝ intitial state tuple T
    A B C D ... ← ⍳⍴T           ⍝ A B C D ... are tuple item "names"

    T {                         ⍝ final state from initial state (left argument)
        A∆←(A⊃⍺) ... ⍵          ⍝ new value for A depending on item ⍵
        C∆←(C⊃⍺) ... A∆ ... ⍵   ⍝ new value for C depending on A∆ and item ⍵
        A∆ C∆(⊣at A C)⍺         ⍝ successor tuple (A and C changed)
    } foldl items

where:

    at←{A⊣A[⍵⍵]←⍺ ⍺⍺(A←⍵)[⍵⍵]}   ⍝ (⍺ ⍺⍺ ⍵) at ⍵⍵ in ⍵

There is some discussion about providing a primitive operator @ for at in Dyalog V16.

Examples
Function kk illustrates naming tuple elements (S E K Q) at the start of the operand function.
Function scc accesses tuple elements using named indices (C L X x S) and an at operator.

Tuples: a postscript
We might define a “tuple” in APL as a vector in which we think of each item as having a name, rather than an index position.

    bob         ⍝ Tuple: name gender age
┌───┬─┬──┐
│Bob│M│39│
└───┴─┴──┘

    folk        ⍝ Vector of tuples
┌──────────┬────────────┬──────────┬────────────┬─
│┌───┬─┬──┐│┌─────┬─┬──┐│┌───┬─┬──┐│┌─────┬─┬──┐│
││Bob│M│39│││Carol│F│31│││Ted│M│31│││Alice│F│32││ ...
│└───┴─┴──┘│└─────┴─┴──┘│└───┴─┴──┘│└─────┴─┴──┘│
└──────────┴────────────┴──────────┴────────────┴─

    ↓⍉↑ folk    ⍝ Tuple of vectors
┌─────────────────────────┬────────┬───────────────┐
│┌───┬─────┬───┬─────┬─   │MFMF ...│39 31 31 32 ...│
││Bob│Carol│Ted│Alice│ ...│        │               │
│└───┴─────┴───┴─────┴─   │        │               │
└─────────────────────────┴────────┴───────────────┘

    ⍪∘↑¨ ↓⍉↑ folk   ⍝ Tuple of matrices
┌─────┬─┬──┐
│Bob  │M│39│
│Carol│F│31│
│Ted  │M│31│
│Alice│F│32│
 ...   . ..
└─────┴─┴──┘

APLers sometimes talk about “inverted files”. In this sense, a “regular” file is a vector-of-tuples and an inverted file (or more recently: “column store”) is a tuple-of-vectors (or matrices).

The Halting Problem Rendered in APL

The halting problem is the problem of determining, from a description of a program and an input, whether the program will finish running or continue to run forever. It was proved in the negative by Alonzo Church and Alan Turing in 1936, and can be rendered in Dyalog APL as follows:

Suppose there is an operator H such that f H ⍵ is 1 or 0 according to whether f ⍵ halts. Consider:

In dfns

   f←{∇ H ⍵:∇ ⍵ ⋄ ⍬}

For f 0, if ∇ H 0 is 1, then ∇ 0 is invoked, leading to an infinite recursion, therefore ∇ H 0 should have been 0. On the other hand, if ∇ H 0 is 0, then the part is invoked and is the result of f 0, therefore ∇ H 0 should have been 1.

Presented as a table: For f 0

suppose ∇ H 0 is invokes consequence ∇ H 0 should be
1 ∇ 0 infinite recursion 0
0 f 0 results in 1

Therefore, there cannot be such an operator H.

In Tradfns

   ⎕fx 'f x' '→f H x'

For f 0

suppose f H 0 is invokes consequence f H 0 should be
1 →1     infinite loop 0
0 →0 f 0 exits 1