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What is a Y-combinator? [closed]

A Y-combinator is a computer science concept from the “functional” side of things. Most programmers don't know much at all about combinators, if they've even heard about them.

  • What is a Y-combinator?
  • How do combinators work?
  • What are they good for?
  • Are they useful in procedural languages?
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Chris Ammerman Avatar asked Sep 18 '08 15:09

Chris Ammerman


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2 Answers

A Y-combinator is a "functional" (a function that operates on other functions) that enables recursion, when you can't refer to the function from within itself. In computer-science theory, it generalizes recursion, abstracting its implementation, and thereby separating it from the actual work of the function in question. The benefit of not needing a compile-time name for the recursive function is sort of a bonus. =)

This is applicable in languages that support lambda functions. The expression-based nature of lambdas usually means that they cannot refer to themselves by name. And working around this by way of declaring the variable, refering to it, then assigning the lambda to it, to complete the self-reference loop, is brittle. The lambda variable can be copied, and the original variable re-assigned, which breaks the self-reference.

Y-combinators are cumbersome to implement, and often to use, in static-typed languages (which procedural languages often are), because usually typing restrictions require the number of arguments for the function in question to be known at compile time. This means that a y-combinator must be written for any argument count that one needs to use.

Below is an example of how the usage and working of a Y-Combinator, in C#.

Using a Y-combinator involves an "unusual" way of constructing a recursive function. First you must write your function as a piece of code that calls a pre-existing function, rather than itself:

// Factorial, if func does the same thing as this bit of code... x == 0 ? 1: x * func(x - 1); 

Then you turn that into a function that takes a function to call, and returns a function that does so. This is called a functional, because it takes one function, and performs an operation with it that results in another function.

// A function that creates a factorial, but only if you pass in // a function that does what the inner function is doing. Func<Func<Double, Double>, Func<Double, Double>> fact =   (recurs) =>     (x) =>       x == 0 ? 1 : x * recurs(x - 1); 

Now you have a function that takes a function, and returns another function that sort of looks like a factorial, but instead of calling itself, it calls the argument passed into the outer function. How do you make this the factorial? Pass the inner function to itself. The Y-Combinator does that, by being a function with a permanent name, which can introduce the recursion.

// One-argument Y-Combinator. public static Func<T, TResult> Y<T, TResult>(Func<Func<T, TResult>, Func<T, TResult>> F) {   return     t =>  // A function that...       F(  // Calls the factorial creator, passing in...         Y(F)  // The result of this same Y-combinator function call...               // (Here is where the recursion is introduced.)         )       (t); // And passes the argument into the work function. } 

Rather than the factorial calling itself, what happens is that the factorial calls the factorial generator (returned by the recursive call to Y-Combinator). And depending on the current value of t the function returned from the generator will either call the generator again, with t - 1, or just return 1, terminating the recursion.

It's complicated and cryptic, but it all shakes out at run-time, and the key to its working is "deferred execution", and the breaking up of the recursion to span two functions. The inner F is passed as an argument, to be called in the next iteration, only if necessary.

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Chris Ammerman Avatar answered Oct 16 '22 14:10

Chris Ammerman


If you're ready for a long read, Mike Vanier has a great explanation. Long story short, it allows you to implement recursion in a language that doesn't necessarily support it natively.

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Nicholas Mancuso Avatar answered Oct 16 '22 14:10

Nicholas Mancuso