Consider the following Catalan function in JavaScript.
class Pair {
constructor(fst, snd) {
this.fst = fst;
this.snd = snd;
}
}
const catalan = (x, xs) => {
if (xs.length === 0) return [x];
const result = [];
for (let i = 0; i < xs.length; i ) {
const ys = catalan(x, xs.slice(0, i));
const zs = catalan(xs[i], xs.slice(i 1));
for (const y of ys) for (const z of zs) result.push(new Pair(y, z));
}
return result;
};
const show = (x) => x instanceof Pair
? `(${show(x.fst)} <> ${show(x.snd)})`
: JSON.stringify(x);
const log = (x) => console.log(x);
catalan(1, []).map(show).forEach(log);
catalan(1, [2]).map(show).forEach(log);
catalan(1, [2, 3]).map(show).forEach(log);
catalan(1, [2, 3, 4]).map(show).forEach(log);It returns all the possible ways of associating n applications of a binary operator, where n = xs.length.
I want to do something similar, but with types in TypeScript. However, I don't know how to implement the “else” branch.
class Pair<A, B> {
constructor(public fst: A, public snd: B) {}
}
type Catalan<X, XS extends unknown[]> = XS extends []
? X
: /* how to define this “else” branch? */;
type C0 = Catalan<1, []>; // 1
type C1 = Catalan<1, [2]>; // Pair<1, 2>
type C2 = Catalan<1, [2, 3]>; // Pair<1, Pair<2, 3>> | Pair<Pair<1, 2>, 3>
type C3 = Catalan<1, [2, 3, 4]>; /* Pair<1, Pair<2, Pair<3, 4>>> |
* Pair<1, Pair<Pair<2, 3>, 4>> |
* Pair<Pair<1, 2>, Pair<3, 4>> |
* Pair<Pair<1, Pair<2, 3>>, 4> |
* Pair<Pair<Pair<1, 2>, 3>, 4>
* /
Any help will be greatly appreciated. By the way, I want to use this Catalan type to define the following function.
declare const flatten: <X, XS extends unknown[]>(
x: Catalan<X, XS>
) => [X, ...XS];
Here's how the flatten function is implemented in JavaScript.
class Pair {
constructor(fst, snd) {
this.fst = fst;
this.snd = snd;
}
}
const catalan = (x, xs) => {
if (xs.length === 0) return [x];
const result = [];
for (let i = 0; i < xs.length; i ) {
const ys = catalan(x, xs.slice(0, i));
const zs = catalan(xs[i], xs.slice(i 1));
for (const y of ys) for (const z of zs) result.push(new Pair(y, z));
}
return result;
};
const flatten = (x) => x instanceof Pair
? [...flatten(x.fst), ...flatten(x.snd)]
: [x];
const log = (x) => console.log(JSON.stringify(x));
catalan(1, []).map(flatten).forEach(log);
catalan(1, [2]).map(flatten).forEach(log);
catalan(1, [2, 3]).map(flatten).forEach(log);
catalan(1, [2, 3, 4]).map(flatten).forEach(log);Edit: If it helps, here's an implementation of the value-level catalan function in Haskell.
import Data.List (inits, tails)
data Catalan a = Catalan a :<>: Catalan a | Lift a deriving Show
split :: [a] -> [([a], [a])]
split = init . (zipWith (,) <$> inits <*> tails)
catalan :: a -> [a] -> [Catalan a]
catalan x [] = [Lift x]
catalan x xs = do
(ys, z:zs) <- split xs
y <- catalan x ys
z <- catalan z zs
return $ y :<>: z
main :: IO ()
main = do
mapM_ print $ catalan 1 []
mapM_ print $ catalan 1 [2]
mapM_ print $ catalan 1 [2, 3]
mapM_ print $ catalan 1 [2, 3, 4]
Here's the output of the above Haskell program.
Lift 1
Lift 1 :<>: Lift 2
Lift 1 :<>: (Lift 2 :<>: Lift 3)
(Lift 1 :<>: Lift 2) :<>: Lift 3
Lift 1 :<>: (Lift 2 :<>: (Lift 3 :<>: Lift 4))
Lift 1 :<>: ((Lift 2 :<>: Lift 3) :<>: Lift 4)
(Lift 1 :<>: Lift 2) :<>: (Lift 3 :<>: Lift 4)
(Lift 1 :<>: (Lift 2 :<>: Lift 3)) :<>: Lift 4
((Lift 1 :<>: Lift 2) :<>: Lift 3) :<>: Lift 4
CodePudding user response：
So here is my attempt:
First of all, I am not sure if I understood the Catalan algorithm correctly. I created this type purely by looking at the examples you gave. You need to test if this works for bigger tuples as well.
Explanation
I used some utilities to solve this problem. I needed a type to splice arrays, so I used the type Splice from here.
type Absolute<T extends string | number | bigint> = T extends string
? T extends `-${infer R}`
? R
: T
: Absolute<`${T}`>;
type isNegative<T extends number> = `${T}` extends `-${infer _}` ? true : false;
type SliceLeft<
Arr extends any[],
Index extends number,
Tail extends any[] = []
> = Tail["length"] extends Index
? [Tail, Arr]
: Arr extends [infer Head, ...infer Rest]
? SliceLeft<Rest, Index, [...Tail, Head]>
: [Tail, []];
type SliceRight<
Arr extends any[],
Index extends string,
Tail extends any[] = []
> = `${Tail["length"]}` extends Index
? [Arr, Tail]
: unknown extends Arr[0]
? [[], Tail]
: Arr extends [...infer Rest, infer Last]
? SliceRight<Rest, Index, [Last, ...Tail]>
: [[], Tail];
type SliceIndex<
Arr extends any[],
Index extends number
> = isNegative<Index> extends false
? SliceLeft<Arr, Index>
: SliceRight<Arr, Absolute<Index>>;
type Slice<
Arr extends any[],
Start extends number = 0,
End extends number = Arr["length"]
> = SliceIndex<SliceIndex<Arr, End>[0], SliceIndex<Arr, Start>[0]["length"]>[1];
I also needed a way to convert the indizes of the array produzed by keyof X (which are strings) back to numbers. I used this helper type:
type Indizes = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Now to the algorithm itself.
It was not clear to me why X and XS needed to be seperate things. For the first step I took X and XS and merged them into a single array.
type Catalan<X, XS extends unknown[]> = _Catalan<[X, ...XS]>
To evaluate the result I created the recursive type _Catalan which takes a tuple. The first two steps are simple. If the tuple is of length 1, return the element inside the array. If it is of length 2 ,return a Pair of the first two elements.
X["length"] extends 1
? X[0]
: X["length"] extends 2
? Pair<X[0], X[1]>
: /* ... */
The rest is a bit more complicated. My approach was to "cut" the array once at every possible index and recursively call _Catalan with both resulting sub-arrays.
[ 1 , 2 , 3 , 4 ]
| <- first cut here
[ 1 ] [ 2 , 3 , 4 ]
| <- next cut here
[ 1 ] [ 2 ] [ 3 , 4 ]
=> Pair<1, Pair<2, Pair<3,4>>
So on the highest level I iterate through each element and convert all iterations to a union with [keyof X & '${bigint}'].
{
[I in keyof X & `${bigint}`]: /* ... */
}[keyof X & `${bigint}`]
I then convert the index to the corresponding number and also skip the first iteration since we only need to cut the array n - 1 times.
I extends keyof Indizes
? Indizes[I] extends 0
? never
: /* ... */
: never
And last of all I create both sub-arrays with Slice and create a Pair of both sub-arrays by calling _Catalan with them.
Pair<_Catalan<Slice<X, 0, Indizes[I]>>, _Catalan<Slice<X, Indizes[I], X["length"]>>>
Result
The end result looks like this:
type _Catalan<X extends unknown[]> = X["length"] extends 1
? X[0]
: X["length"] extends 2
? Pair<X[0], X[1]>
: {
[I in keyof X & `${bigint}`]: I extends keyof Indizes
? Indizes[I] extends 0
? never
: Indizes[I] extends number
? Pair<_Catalan<Slice<X, 0, Indizes[I]>>, _Catalan<Slice<X, Indizes[I], X["length"]>>>
: never
: never
}[keyof X & `${bigint}`]
type Catalan<X, XS extends unknown[]> = _Catalan<[X, ...XS]>
Usage:
class Pair<A, B> {
constructor(public fst: A, public snd: B) {}
}
type _Catalan<X extends unknown[]> = X["length"] extends 1
? X[0]
: X["length"] extends 2
? Pair<X[0], X[1]>
: {
[I in keyof X & `${bigint}`]: I extends keyof Indizes
? Indizes[I] extends 0
? never
: Indizes[I] extends number
? Pair<_Catalan<Slice<X, 0, Indizes[I]>>, _Catalan<Slice<X, Indizes[I], X["length"]>>>
: never
: never
}[keyof X & `${bigint}`]
type Catalan<X, XS extends unknown[]> = _Catalan<[X, ...XS]>
type C0 = Catalan<1, []>; // 1
type C1 = Catalan<1, [2]>; // Pair<1, 2>
type C2 = Catalan<1, [2, 3]>; // Pair<1, Pair<2, 3>> | Pair<Pair<1, 2>, 3>
type C3 = Catalan<1, [2, 3, 4]>; /* Pair<1, Pair<2, Pair<3, 4>>> |
* Pair<1, Pair<Pair<2, 3>, 4>> |
* Pair<Pair<1, 2>, Pair<3, 4>> |
* Pair<Pair<1, Pair<2, 3>>, 4> |
* Pair<Pair<Pair<1, 2>, 3>, 4>
*/
If you hover over the type C3, you will notice that it looks different from what you specified. There are only 3 top-level unions and each member will have unions inside the Pairs.
Pair<1, Pair<2, Pair<3, 4>> | Pair<Pair<2, 3>, 4>>
//instead of
Pair<1, Pair<2, Pair<3, 4>>> | Pair<1, Pair<Pair<2, 3>, 4>>
But this is purely a visual thing and they should functionally not be different.
Some tests to validate the result:
type Test1 = Pair<1, Pair<2, Pair<3, 4>>> extends C3 ? true : false
type Test2 = Pair<1, Pair<Pair<2, 3>, 4>> extends C3 ? true : false
type Test3 = Pair<Pair<1, 2>, Pair<3, 4>> extends C3 ? true : false
type Test4 = Pair<Pair<1, Pair<2, 3>>, 4> extends C3 ? true : false
type Test5 = Pair<Pair<Pair<1, 2>, 3>, 4> extends C3 ? true : false
CodePudding user response：
When encountering type-level problems like these, it's best to look at the original code and look for patterns, or anything that the type system can do for you.
So let's begin:
const catalan = (x, xs) => {
if (xs.length === 0) return [x];
const result = [];
for (let i = 0; i < xs.length; i ) {
const ys = catalan(x, xs.slice(0, i));
const zs = catalan(xs[i], xs.slice(i 1));
for (const y of ys) for (const z of zs) result.push(new Pair(y, z));
}
return result;
};
First we notice that if xs is empty, then we directly return x. We make a mental note to use XS["length"] extends 0 ? X : ... later.
Then we see that this:
const ys = catalan(x, xs.slice(0, i));
const zs = catalan(xs[i], xs.slice(i 1));
is really just partitioning xs in such a way that:
partition [1, 2, 3, 4, 5] at 3 => [1, 2, 3] [5]
In other words, we split the tuple at index 3 and return the two halves. This will be much faster than slicing the tuple two times individually and can be implemented without much trouble.
Finally we notice this nested loop:
for (const y of ys) for (const z of zs) result.push(new Pair(y, z));
No need for this in the type system, we can simply do:
Pair<YS, ZS>
and have it generate all the possible pairs for us from the unions.
Alright, time to get cracking away at the solution.
Recall that x is returned if xs is empty:
type Catalan<X, XS extends ReadonlyArray<unknown>> =
XS["length"] extends 0 ? X :
And also when XS is only one element then we return that pair. If XS has more than one element, we enter the loop instead:
... : XS["length"] extends 1 ? Pair<X, XS[0]> : CatalanLoop<X, XS>;
Let's see the loop now:
type CatalanLoop<X, XS extends ReadonlyArray<unknown>> = {
[K in keyof XS & `${bigint}`]: ...
}[keyof XS & `${bigint}`];
Now, what's this funny looking thing:
keyof XS & `${bigint}`
keyof XS would give us something in the form of number | "0" | "1" | "2" | "at" | "concat" | "...", but we only want the indices of XS. If we intersect keyof XS with the interpolated bigint, we get the desired "0" | "1" | "2" only.
That means this is just like the loop in the original code! We loop over each index using a mapped type.
Inside the loop body we partition XS at index K:
type CatalanLoop<X, XS extends ReadonlyArray<unknown>> = {
[K in keyof XS & `${bigint}`]:
Partition<XS, K> extends [infer Left, infer Right]
? ...
: ...
}[keyof XS & `${bigint}`];
But we have to assert to TypeScript that our partitioning type will definitely give us tuples like this first:
Partition<XS, K> extends [infer Left, infer Right]
? Left extends ReadonlyArray<unknown>
? Right extends ReadonlyArray<unknown>
Then we call Catalan and make our pairs:
? Catalan<X, Left> extends infer YS
? Catalan<XS[K], Right> extends infer ZS
? Pair<YS, ZS>
This is doing what this original code does:
const ys = catalan(x, xs.slice(0, i));
const zs = catalan(xs[i], xs.slice(i 1));
for (const y of ys) for (const z of zs) result.push(new Pair(y, z));
And let's close off all our ternaries/conditionals with never (because these clauses should never be reached anyways):
: never
: never
: never
: never
: never
Finally, we need to make our partitioning type.
To do that, we need a type to increment a number. This can be done with a tuple like this:
type Increment = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33];
Increment[0] // => 1
Increment[15] // => 16
Increment[32] // => 33
Now that we can increment a number, we define Partition:
type Partition<
XS extends ReadonlyArray<unknown>,
At extends string,
Index extends number = 0,
Left extends ReadonlyArray<unknown> = [],
> = XS extends [infer First, ...infer Rest]
? `${Index}` extends At
? [Left, Rest]
: Partition<Rest, At, Increment[Index], [...Left, First]>
: never
This type loops over XS until it hits At, the index to partition at. It excludes the element at At and stops, giving us [Left, Rest], the two halves. Partition is the type that replaces xs.slice(0, i) and xs.slice(i 1).
Lastly, just for kicks, let's also make a type to mimic the original show function:
type Show<Pairs> = Pairs extends Pair<infer A, infer B> ? `(${Show<A>} <> ${Show<B>})` : `${Pairs & number}`;
And wow! It really does work!
type ShowFifth = Show<Catalan<1, [2, 3, 4, 5]>>;
// =>
// | "(1 <> (2 <> (3 <> (4 <> 5))))"
// | "(1 <> (2 <> ((3 <> 4) <> 5)))"
// | "(1 <> ((2 <> 3) <> (4 <> 5)))"
// | "(1 <> ((2 <> (3 <> 4)) <> 5))"
// | "(1 <> (((2 <> 3) <> 4) <> 5))"
// | "((1 <> 2) <> (3 <> (4 <> 5)))"
// | "((1 <> 2) <> ((3 <> 4) <> 5))"
// | "((1 <> (2 <> 3)) <> (4 <> 5))"
// | "((1 <> (2 <> (3 <> 4))) <> 5)"
// | "((1 <> ((2 <> 3) <> 4)) <> 5)"
// | "(((1 <> 2) <> 3) <> (4 <> 5))"
// | "(((1 <> 2) <> (3 <> 4)) <> 5)"
// | "(((1 <> (2 <> 3)) <> 4) <> 5)"
// | "((((1 <> 2) <> 3) <> 4) <> 5)"
To end off this little adventure, a playground where you can play around with this yourself.