to match the left to right processing of 2048, add &.|. adverb to temporarily
reverse list
f/&.|. 8 4 4 4 8 8 4
8 8 4 16 4
though this doesn't quite match the 2048 rules in:
f/&.|. 8 4 4 8
8 16
result should be
16 8
in this case (8 8 8 intermediate merged with the first 2 merged to 16)
On Wednesday, May 22, 2019, 7:58:04 p.m. EDT, Raul Miller
<[email protected]> wrote:
You need to apply your rank operation to the entire operation for the
second case.
This might be easier to see if you eliminated some extra, unnecessary code?
...
Actually ... those rank statements don't help you at all - get rid of them.
Also, your = test should really be (= {.) because your are building a list.
This does what you said you want:
f=: , ` ((+{.),}.@])@.(={.)
And f/"1 is kind of like a 2048 mechanism. Maybe (4 {. f)/"1 is where you
were heading with this?
Anyways... one problem was that f was adding unwanted leading dimensions on
intermediate results, which made your first and rest do something different
from what you intended.
I hope this helps,
--
Raul
On Wed, May 22, 2019 at 6:41 PM Luca Masini <[email protected]> wrote:
> Hi,
>
> I'm trying to come out with a few primitives for a 2048-like-game
> operations.
>
> I did the following
>
> first =: {.
> rest =. }.
> f =: ((dyad def 'x,y')"(0 1)`(dyad def '(x + first y) , rest y'"(0
> 1))@.=)
>
> In words f should do the following
> - if x = first y the reduce it to x+y
> - if x not = first y then give back the sequence x,y
>
>
> The intended use would be, for ex.
>
> f/ 4 4 4
> 4 8
>
> f/ 4 4 4 4
> 8 8
>
> In the first case it works.
>
>
> In the second case the result is the following:
>
> f/ 4 4 4 4
> 8 0
> 4 8
>
> My idea was
>
> f/ x x x x
>
> is equivalent to
> 4 f 4 f 4 f 4
>
> then, working right-to-left, 4 = first 4, so it simplify to
> 4 f 4 f 8
>
> now, 4 is not = first 8, so it should simplify to
> 4 f 4 8
>
> now, 4 is = first 4 8, so it should simplify to
> 8 8
>
>
> What could be a way to implement the behavior that I described?
>
> MfG
>
> Luca.
>
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