Re: [Bug-apl] Tree Traversal in gnu-apl

[Dr . Jürgen Sauermann]

Hi Rowan, not sure if it helps, but the term accumulator makes me think of the reduction operator rather than the rank operator:       {⍺,1+⍵} / 1 2 3  1 3 5 Not sure, though what you want to compute. The syntax of the rank operator in the ISO standard is IMHO broken. The syntax: Z←f ⍤ y B has a rather ambiguous right argument j B with no rules as to where j end and B begins. For example: Z←f⍤1 2 3 4 5 could mean: j = 1 and B = 2 3 4 5   or: j = 1 2 and B = 3 4 5   or even: j = 1 2 3 and B = 4 5 GNU APL uses some crude rules to resolve such cases, but in order to get a predictable result, you need to properly use parentheses like this: Z←(f⍤1) 2 3 4 5 or, as a cleaner though non-standard alternative syntax, make j an axis argument: Z←f⍤[1] 2 3 4 5 It is very difficult to get all that right, therefore I never use ⍤. Best Regards, Jürgen Sauermann On 10/9/19 9:44 PM, Rowan Cannaday wrote: this seems to be behaving much better:       foo ← {⍺,{((1+1↑⍵) foo 1↓⍵)}⍣(~0=⍴⍵)⊢⍵}       ⍬ foo 1 2 3 2 3 4       5 4 3 foo 1 2 3 5 4 3 2 3 4 On Wed, Oct 9, 2019 at 3:40 PM Rowan Cannaday <cannad...@gmail.com> wrote: somebody pointed out that the previous fn was failing because i wasn't calling 'foo' w/ dyadic arguments in the inner dfn On Wed, Oct 9, 2019 at 2:16 PM Rowan Cannaday <cannad...@gmail.com> wrote: interestingly enough this is ok:       foo ← {⍬{⍺,(1+1↑⍵),(foo 1↓⍵)}⍣(~0=⍴⍵)⍵}       foo 1 2 3 2 3 4 but this isnt:       foo ← {⍺{⍺,(1+1↑⍵),(foo 1↓⍵)}⍣(~0=⍴⍵)⍵}       ⍬ foo 1 2 3 VALUE ERROR foo[1]  λ←⍺ λ1⍣(∼0=⍴⍵)⍵           ^       ⎕CR 'foo' λ←⍺ λ1 ⍵                         λ←⍺{⍺,(1+1↑⍵),(foo 1↓⍵)}⍣(~0=⍴⍵)⍵ On Wed, Oct 9, 2019 at 1:29 PM Rowan Cannaday <cannad...@gmail.com> wrote: making progress...       foo ← {{(1+1↑⍵),(foo 1↓⍵)}⍣(~0=⍴⍵)⍵}       foo 1 2 3 2 3 4 On Wed, Oct 9, 2019 at 12:47 PM Rowan Cannaday <cannad...@gmail.com> wrote: this is a slightly better way of writing my (still broken) accumulator: acc←{⍺{⍺ acc 1↑⍵}⍣(0=⍴⍵)⊢⍵} On Wed, Oct 9, 2019 at 12:40 PM Rowan Cannaday <cannad...@gmail.com> wrote: Given a recursive factorial definition: fact←{{⍵ × fact ⍵-1}⍣(⍵>2)⊢1⌈⍵} [written by Kacper Gutowski in the 'Recursive Lambda' thread] I am attempting to write a basic accumulator. This should take an empty vector as the left value argument, and a rank 1 array as the right value argument. Every iteration it should drop a value from the right value, and append it to the left, until the right value is an empty vector. I thought I'd be able to do something like the following: acc←{⍺,{acc 1↑⍵}⍣(0=⍴⍵)⊢⍵} ⍬ acc 1 2 3 But modifying this to say, add a 1 to every number, still returns the input vector ⍵. Thoughts? On Fri, Sep 27, 2019 at 3:44 PM Rowan Cannaday <cannad...@gmail.com> wrote: Hello y'all. I have been attempting to learn function composition & higher-order functions in gnu-apl, and how to use it to perform tree traversal. https://en.wikipedia.org/wiki/Function_composition_(computer_science)#APL https://en.wikipedia.org/wiki/Higher-order_function#APL https://rosettacode.org/wiki/Tree_traversal#APL Unfortunately a lot of the syntax used is dyalog & dfn specific, so working out some of the examples is a bit tricky for myself. (the main inconsistencies are '∇' as a recursive function definition, ⍺⍺ & ⍵⍵ to refer to left and right operands, '@' as the 'at' operator, '⍣' operator differences, as well as possibly others). Has anybody done 'idiomatic' tree traversal in gnu-apl? Does anybody use primitive composition functions in their code? Trying to figure out what works and feels natural in the language. Any examples or guidance would be appreciated. Examples: Higher order fns in gnu-apl: ∇Z ← (L twice) B     Z ← L L B ∇ ∇Z ← plusthree B     Z ← B + 3 ∇ ∇Z ← g B     Z ← plusthree twice B ∇