After all lazy is just another way of saying 'deferred', and call-by-name is a way to identify what it is that you are calling, in either case you'd sooner or later have to do the evaluation and then you're back to that loop again.
So, what am I missing ?
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.
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
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.
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.
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!)
According to the definition all functions are unary functions, an if would have to be a binary function wouldn't it ?
// 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;
}}
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.
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.
function foo (x) { return function (y) {
return x*y;
}}
bar = foo(5);
Does that make sense to you?
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 ?
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
if false nt id -- outer redex = if X X X
false nt id -- outer redex = false X X
id -- normal form = \x. x
(\b t f. b t f) (\t f. f) nt (\x. x)
(\t f. f) nt (\x. x)
(\x. x)
\x. ((\x. x) x) -- weak head normal form. Outer redex = (\x. x) X
\x. x -- plain normal form