5

Here is my ungolfed Ruby code for a function I want to try and golf:

```
Iter =-> k {
str = ""
(0..k).map{|m| str += "-> f#{m} {"}
str += "f#{k}.times{f1 = f0[f1]};f1"
(2..k).map{|m| str += "[f#{m}]"}
eval(str + "}" * (k+1))
}
```

The best I can do to golf this is essentially shortening variable names, removing spaces, and rearranging the code to reduce the amount of loops:

```
I=->k{s="f#{k}.times{f1=f0[f1]};f1"
(2..k).map{|m|s+="[f#{m}]"}
(0..k).map{|m|s="->f#{k-m}{"+s+?}}
eval s}
```

This function can be compacted so that it avoids a lot of defining new functions. You can think of it like this:

$$\operatorname{Iter}(k;f_0,f_1,\dots,f_k)=f_0^{f_k}(f_1)(f_2)(f_3)\dots(f_k)$$

where

$$f^n(x)=\underbrace{f(f(\dots f(}_nx)\dots))$$

denotes function iteration. The types of arguments are given as:

$$\operatorname{Iter}(\text{int};\dots,(\text{int}\mapsto\text{int})\mapsto(\text{int}\mapsto\text{int}),\text{int}\mapsto\text{int},\text{int})$$

That is, the last argument is an integer, and each previous argument maps from T to T, where T is the type of the next argument on the right.

It is true that accepting all arguments at once would allow me to golf the above code further:

```
I=->k,*a{a[-1].times{a[1]=a[0][a[1]]};(2..k).map{|m|a[1]=a[1][a[m]]};a[1]}
```

However, the issue is that I need to curry this function. This way, I may treat objects such as

$$\operatorname{Iter}(k)(f_0)$$

as their own object, and thus be able to pass this into Iter(k) to get things such as

$$\operatorname{Iter}(k)(\operatorname{Iter}(k)(f_0))$$

As a specific example of what this function does, we have

\begin{align}\operatorname{Iter}(2)(\operatorname{Iter}(1))(x\mapsto x+1)(2)&=\operatorname{Iter}(1)(\operatorname{Iter}(1)(x\mapsto x+1))(2)\\&=\operatorname{Iter}(1)(x\mapsto x+1)(\operatorname{Iter}(1)(x\mapsto x+1)(2))\\&=\operatorname{Iter}(1)(x\mapsto x+1)(2+1+1)\\&=\operatorname{Iter}(1)(x\mapsto x+1)(4)\\&=4+1+1+1+1\\&=8\end{align}

I'm interested in seeing if this can be golfed down further, with the restriction that the arguments must be curried in the order provided.

Mind if anyone explain the "unclear" close vote? I really did try my best to explain everything, so if there's any part that's unclear, just ask. – Simply Beautiful Art – 2019-06-28T01:37:29.683

1It's a bit unclear because you have tagged this "tips", but it's written almost like a challenge. I'd remove the tips tag and make it more clearly a challenge. Otherwise, perhaps make it clear that you're looking for golfing tips for this specific program. Your last statement makes us unsure of what you want. – mbomb007 – 2019-06-28T02:47:54.280

1

Not sure if this is what you're looking for, but in J

– Jonah – 2019-06-28T06:17:27.260`>:`

is the increment function, and your example`I[2][I[1]][->x{x+1}][2]`

can be written`>:^:2^:3 (2)`

. Try it online!.`^:`

is the power of conjunction, and iteratively applies the verb on its left the number of times specified by the number on its right, with 1 being normal application`f(x)`

, 2 being`f(f(x)`

, etc.@mbomb007 okay, I cleared that part up. – Simply Beautiful Art – 2019-06-28T10:17:20.987

1@lirtosiast Thanks and done – Simply Beautiful Art – 2019-06-28T10:17:54.013

@Jonah interesting syntax, though it seems I'm no longer looking for non-Ruby answers. Am still a bit curious as to how you would write something like

`I[1][I[2][I[1]]][s][2]`

. – Simply Beautiful Art – 2019-06-28T10:25:54.383Apparently with your golfed code (and my solution that improves on your golfed code) hits an infinite loop when you change the last argument to

– Value Ink – 2019-06-29T00:49:46.347`3`

instead of`2`

... This shouldn't happen, right? Or am I missing something? Try it online!1@ValueInk that's not an infinite loop. The issue there is the end result is 3 times 2^402653191 (3<<402653191) – Simply Beautiful Art – 2019-06-29T01:55:08.173

That definitely means I missed something lol... I thought the output was going to be something closer to, I don't know,

`12`

. – Value Ink – 2019-06-29T01:57:57.620@ValueInk Let

`s[n] = n+1`

. See that`I[1][s][n] = 2n`

and`I[1][I[1][s]][n] = n2ⁿ = n<<n`

. From that point, you can directly compute your example with`n = 3; 3.times{n <<= n}`

... – Simply Beautiful Art – 2019-06-29T02:02:22.313Erm, miscalculation, it ought to be

`3 << 402653211`

. – Simply Beautiful Art – 2019-06-29T02:10:05.597