# Deriving the Y Combinator in 7 Easy Steps

The Y Combinator is a method of implementing recursion in a programming language that does not support it natively (actually, it’s used more for exercising programming brains). The requirement, though, is that language to support anonymous functions.

I chose JavaScript for deriving the Y Combinator, starting from the definition of a recursive factorial function, using a step-by-step transformation over the initial function.

## Update

There’s now a Chinese translation of this article as well as a Clojure port.

## Step 1

The initial implementation, using JavaScript’s built-in recursion mechanism.

```
var fact = function (n) {
if (n < 2) return 1;
return n * fact(n - 1);
};
```

## Step 2

What would be the simplest thing to do to obtain basic recursion? We could just define a function which receives itself as an argument and calls that argument with the same argument. That’s an infinite loop of course, and would cause a stack overflow.

```
(function (f) {
f(f);
})(function (f) {
f(f);
});
```

Let’s use the above pattern for our factorial function. There is however a small difference. The factorial function receives an argument which we don’t know yet, so what we want is to return a function which takes that argument. That function can then be used to compute factorial numbers. Also, this is what makes our implementation to not result into an infinite loop.

```
var fact = (function (f) {
return function (n) {
// termination condition
if (n < 2) return 1;
// because f returns a function, we have a double function call.
return n * f(f)(n - 1);
};
})(function (f) {
return function (n) {
// termination condition
if (n < 2) return 1;
// because f returns a function, we have a double function call.
return n * f(f)(n - 1);
};
});
```

## Step 3

At this point we have some ugly duplication in there. Let’s hide it away into a helper function called `recur`

.

```
var recur = function (f) {
return f(f);
};
var fact = recur(function (f) {
return function (n) {
if (n < 2) return 1;
// because f returns a function, we have a double function call.
return n * f(f)(n - 1);
};
});
```

## Step 4

The problem with the above version is that double function call. We want to eliminate it so that the implementation of this factorial is similar with the recursive version. How can we do that?

We can use a helper function that takes a numeric argument and performs the double call. The trick is though to keep this helper function in the same environment where `f`

is visible, so that `g`

can actually call `f`

.

```
var recur = function (f) {
return f(f);
};
var fact = recur(function (f) {
var g = function (n) {
return f(f)(n);
};
return function (n) {
if (n < 2) return 1;
// no more double call, g is a function which takes a numeric arg
return n * g(n - 1);
};
});
```

## Step 5

The above works nice, but the definition contains so much clutter code. We can hide it away inside yet another helper function, keeping almost just the definition of factorial.

```
var recur = function (f) {
return f(f);
};
var wrap = function (h) {
return recur(function (f) {
var g = function (n) {
return f(f)(n);
};
return h(g);
});
};
var fact = wrap(function (g) {
return function (n) {
if (n < 2) return 1;
return n * g(n - 1);
};
});
```

## Step 6

Let’s inline the definition of `g`

inside `wrap`

because we only call it once.

```
var recur = function (f) {
return f(f);
};
var wrap = function (h) {
return recur(function (f) {
return h(function (n) {
return f(f)(n);
});
});
};
var fact = wrap(function (g) {
return function (n) {
if (n < 2) return 1;
return n * g(n - 1);
};
});
```

## Step 7

Now, if we also inline the definition of `recur`

function inside `wrap`

we end up with the famous Y Combinator.

```
var Y = function (h) {
return (function (f) {
return f(f);
})(function (f) {
return h(function (n) {
return f(f)(n);
});
});
};
var fact = Y(function (g) {
return function (n) {
if (n < 2) return 1;
return n * g(n - 1);
};
});
```

## The End

I hope you enjoyed it!