Using the Y Combinator in C#
Asked Answered
H

1

24

I'm trying to figure out how to write recursive functions (e.g. factorial, although my functions are much more complicated) in one line. To do this, I thought of using the Lambda Calculus' Y combinator.

Here's the first definition:

Y = λf.(λx.f(x x))(λx.f(x x))

Here's the reduced definition:

Y g = g(Y g)

I attempted to write them in C# like this:

// Original
Lambda Y = f => (new Lambda(x => f(x(x)))(new Lambda(x => f(x(x)))));
// Reduced
Lambda Y = null; Y = g => g(Y(g));

(Lambda is a Func<dynamic, dynamic>. I first tried to typedef it with using, but that didn't work, so now it's defined with delegate dynamic Lambda(dynamic arg);)

My factorial lambda looks like this (adapted from here):

Lambda factorial = f => new Lambda(n =>  n == 1 ? 1 : n * f(n - 1));

And I call it like this:

int result = (int)(Y(factorial))(5);

However, in both cases (original and reduced forms of the Y combinator), I end up with a stack overflow exception. From what I can surmise from using the reduced form, it seems as if it just ends up calling Y(factorial(Y(factorial(Y(factorial(... and never ends up actually entering the factorial lambda.

Since I don't have much experience debugging C# lambda expressions and I certainly don't understand much of the lambda calculus, I don't really know what's going on or how to fix it.

In case it matters, this question was inspired by trying to write a one-line answer to this question in C#.

My solution is the following:

static IEnumerable<string> AllSubstrings(string input)
{
    return (from i in Enumerable.Range(0, input.Length)
            from j in Enumerable.Range(1, input.Length - i)
            select input.Substring(i, j))
            .SelectMany(substr => getPermutations(substr, substr.Length));
}
static IEnumerable<string> getPermutations(string input, int length)
{
    return length == 1 ? input.Select(ch => ch.ToString()) :
        getPermutations(input, length - 1).SelectMany(
            perm => input.Where(elem => !perm.Contains(elem)),
            (str1, str2) => str1 + str2);
}
// Call like this:
string[] result = AllSubstrings("abcd").ToArray();

My goal is to write getPermutations as a one-line self-recursive lambda so that I can insert it into the SelectMany in AllSubstrings and make a one-liner out of AllSubstrings.

My questions are the following:

  1. Is the Y combinator possible in C#? If so, what am I doing wrong in the implementation?
  2. If the Y combinator is possible in C#, how would I make my solution to the substrings problem (the AllSubstrings function) a one-liner?
  3. Whether or not the Y combinator is not possible in C#, are there any other methods of programming that would allow for one-lining AllSubstrings?
Hypabyssal answered 4/8, 2015 at 21:31 Comment(3)
Y g = g(Y g) is only good with lazy evaluation. With eager evalution, the workaround is to use eta-expansion: Y g = g (\x-> (Y g) x). Or maybe Y g x = g (\x-> (Y g) x) x will be easier.Adapa
@WillNess Thanks! That worked when I wrote it as Lambda y = null; y = g => g(new Lambda(x => (y(g))(x))); Well I guess that answers the first question.Hypabyssal
will it help you if I gave you a Haskell semi-one-liner? it's concatMap permutations . sequences with sequences (x:xs) = [a | b<-sequences xs, a<-[x:b,b] ] ; sequences [] = [[]] and permutations a standard function.Adapa
H
29

Here's the implementation of the Y-combinator that I use in c#:

public delegate T S<T>(S<T> s);

public static T U<T>(S<T> s)
{
    return s(s);
}

public static Func<A, Z> Y<A, Z>(Func<Func<A, Z>, Func<A, Z>> f)
{
    return U<Func<A, Z>>(r => a => f(U(r))(a));
}

I can use it like:

var fact = Y<int, int>(_ => x => x == 0 ? 1 : x * _(x - 1));
var fibo = Y<int, int>(_ => x => x <= 1 ? 1 : _(x - 1) + _(x - 2));

It truly scares me, so I'll leave the next two parts of your question to you, given this as a starting point.


I've had a crack at the function.

Here it is:

var allsubstrings =
    Y<string, IEnumerable<string>>
        (_ => x => x.Length == 1
            ? new [] { x }
            : Enumerable
                .Range(0, x.Length)
                .SelectMany(i =>
                    _(x.Remove(i, 1))
                    .SelectMany(z => new [] { x.Substring(i, 1) + z, z }))
                .Distinct());

Of course, you run it like this:

allsubstrings("abcd");

From which I got this result:

abcd 
bcd 
acd 
cd 
abd 
bd 
ad 
d 
abdc 
bdc 
adc 
dc 
abc 
bc 
ac 
c 
acbd 
cbd 
acdb 
cdb 
adb 
db 
acb 
cb 
ab 
b 
adbc 
dbc 
adcb 
dcb 
bacd 
bad 
badc 
bac 
bcad 
cad 
bcda 
cda 
bda 
da 
bca 
ca 
ba 
a 
bdac 
dac 
bdca 
dca 
cabd 
cadb 
cab 
cbad 
cbda 
cba 
cdab 
dab 
cdba 
dba 
dabc 
dacb 
dbac 
dbca 
dcab 
dcba 

It seems that your non-Y-Combinator code in your question missed a bunch of permutations.

Happ answered 4/8, 2015 at 23:52 Comment(4)
I would be much happier about upvoting/marking as answer/awarding bounty if you answered more than just part 1 of my question. At this point, it seems you are going to receive half the bounty, but I'm sure you can do better. :)Hypabyssal
Note that the question, and this answer, are the subject of a Meta question.Chatav
@Hypabyssal - I've added an implementation.Happ
@Happ Beautiful. As you can tell, I've marked as answer and awarded the bounty! Thanks for answering fully. (Also, yeah, I realized that my permutations code actually didn't quite work, but that doesn't affect this question or your answer.)Hypabyssal

© 2022 - 2024 — McMap. All rights reserved.