Lazy doesn't just mean deferred; an unnecessary computation might never be performed at all. Take for example a function where the recursive argument is never called:

sample = (\this->(\x->x))

Then, (Y sample) can reduce as follows:

1: (Y sample)

2: (sample (Y sample))

3: ((\this->(\x->x)) (Y sample))

4: (\x->x)

Now, if your evaluation strategy required evaluating arguments to final form before applying an operator to them, then line 2 would expand forever:

2: (sample (Y sample))

3: (sample (sample (Y sample)))

4: (sample (sample (sample (Y sample))))

... And so on.

However, neither call-by-need (lazy) nor call-by-name evaluation strategies require evaluating an argument before applying some operator to it. Thus, in languages that support either of those evaluation strategies, the operator can be evaluated to the point where it throws away the argument, and the argument need never be evaluated.

This is a contrived example but real world recursion works for the same reason; at some point in the evaluation of the function, it no longer needs to be applied recursively, and once it reaches that base case, the recursive function can be discarded.

Thank you!

Ok, I think I see a glimmer of understanding here, it doesn't quite 'click' yet but it is a lot clearer now.

I get the 'never needed' bit, that's like in many other languages where you say for instance

   if x and y then 

And x = false, then y is never evaluated. Not quite the same, but it's a case where you can optimize out the evaluation of something because it's result will be discarded anyway.

The last part of your explanation still has me going around about what makes the decision to discard, it can't be the function itself, so it has to be some criterion outside the function causing that. For instance, maybe if work is no longer done, or if some arbitrary precision level has been reached.

Interesting stuff this!

In 'pure' functional programming you'd not have an 'if' statement to make the decision with, or is that allowed to be used to end recursion ? I just can't see that working using functions alone.

There is very much an "if-statement" in lambda calculus.

The basic idea is, in the Church Booleans, we think of true as a function which takes a then clause and an else clause, and returns the then clause (T=\lambda xy.x), and false as a function which takes a then clause and an else clause and returns the else clause (F=\lambda x y.y). Then we get our if statement just by applying a predicate (which returns F or T) to a then clause and an else clause. Indeed, that is very much a way to get recursion to step -- the typical Church-numeral definition of the factorial function works that way.

By the way, one sure-fire way to get evaluation of a lambda expression to terminate if possible is leftmost-reduction -- that is, reducing the leftmost redex first. Intuitively, the leftmost redex is the "outermost," and any other reduction preserves it, so it must be reduced at some point.

Why wouldn't a pure functional language allow you to have an if statement? Haskell certainly has one!

The key is that given the same input you need to produce the same output every time. Which is great as long as you're not doing I/O. (Haskell addresses that with monads - even though it looks like you're giving the same input, you're really giving a different input each time!)

Ok, so that dilutes the 'pure' of the functional part. I know that in practice functional programs have an if to force termination, but the lambda calculus deals with abstracts. Anything evaluating lambda expressions then has to deal with that or you'd have to expand the syntax of the calculus to contain a termination function which 'magically' decides that enough is enough.

According to the definition all functions are unary functions, an if would have to be a binary function wouldn't it ?

With currying, you can easily make functions of n variables out of functions of one variable. To avoid syntactic confusion, here is a comparison in JavaScript

  // Call this with foo(x, y)
  function foo (x, y) {
      return x*y;
  }

  // Call this with foo(x)(y)
  function foo (x) { return function (y) {
      return x*y;
  }}

The key thing to notice is that you can get from one form to the other with a straight text substitution. So restricting yourself to functions of one variable involves no loss of generality.

Going back to the lambda Calculus, you can do Boolean logic there as well. See http://safalra.com/science/lambda-calculus/boolean-logic/ for details. In fact I believe that the if construct in Haskell is just syntactic sugar around that. So it isn't that there is a decision made within the function, but rather that you call a function which turns out to be one of two different functions.

Ah, right, yes, that's true, I forgot about the Currying bit (I read up on it earlier, promptly forgot it).

I'm coming at this from a 'coders' perspective, not the mathematical side of it, so the code example is much appreciated, that makes it much easier to understand.

The function 'foo' returns a function that has already 'bound' 'x' to whatever value you put in, the 'scope' then allows you to fish out the right 'x' when you're in the inner function doing the multiplication.

So, does that mean that as far as the inner function is concerned 'x' is 'free' ?

So far when reading articles about the lambda calculus it seemed like the 'free' variables where somehow magically present in the scope of the function without a hint about where they came from.

I could easily be wrong, but here is my understanding. Consider this snippet of JavaScript:

  function foo (x) { return function (y) {
      return x*y;
  }}

  bar = foo(5);

Within bar, x is bound and y is free.

Does that make sense to you?

Ok, so now 'bar' contains a function where the 'x' is already bound to the '5', but the 'y' is still free because it has not yet been 'bound' to any parameter, is that what you mean ?

bar contains "function(y) { return 5*y; }"

The 'x' has been replaced by the '5' because of the function call. So 'binding' effectively pins down a function parameter and allows you to replace its instances in 'named' form with its value to reduce the expression ?

That is the way that I understand it. However you should be warned that I have never made a detailed study of the lambda calculus, and so might have basic misconceptions about it.

> Anything evaluating lambda expressions then has to deal with that or you'd have to expand the syntax of the calculus to contain a termination function which 'magically' decides that enough is enough.

This isn't so magic. I don't know all the details, but it has been proved that:

* If some expression can be reduced to normal form, this normal form is unique (modulo variable naming —alpha conversion).

* There is a systematic (mechanic) strategy which guarantee to reach normal form, if it exist at all : normal order reduction.

Normal order reduction reduces the outer-most redex in an expression (instead of the inner-most one when you do eager evaluation —call by name). This is why you don't need `if`.

A small example:

    true  = \t f. t
    false = \t f. f

    if = \b t f. b t f

    nt = something_that_doesn't_reduce_to_normal_form
    id = \x. x

Now, trying to reduce this expression (in normal order):

    if false nt id -- outer redex = if X X X
    false nt id    -- outer redex = false X X
    id             -- normal form = \x. x

Now, without using definitions (except for `nt`):

    (\b t f. b t f) (\t f. f) nt (\x. x)
    (\t f. f) nt (\x. x)
    (\x. x)

Finally, the outer-most thing may not be a redex. (but you can still reduce inner-redexes). This is the weak head normal form. For example:

    \x. ((\x. x) x) -- weak head normal form. Outer redex = (\x. x) X
    \x. x           -- plain normal form

You can decide whether weak head normal form is enough, or go to normal form.