Lecture #11, Tuesday, February 22nd
===================================
 Lexical Scope using Racket Closures
 More Closures (on both levels)
 Types of Evaluators
 Feature Embedding
 Recursion, Recursion, Recursion
 Recursion without the Magic
 The Core of `makerecursive`
 Denotational Explanation of Recursion
 The Y Combinator
 The main property of Y
 Yet another explanation for Y
 Typing the Y Combinator

# Lexical Scope using Racket Closures
> [PLAI §11] (without the last part about recursion)
An alternative representation for an environment.
We've already seen how firstclass functions can be used to implement
"objects" that contain some information. We can use the same idea to
represent an environment. The basic intuition is  an environment is
a *mapping* (a function) between an identifier and some value. For
example, we can represent the environment that maps `'a` to `1` and `'b`
to `2` (using just numbers for simplicity) using this function:
(: mymap : Symbol > Number)
(define (mymap id)
(cond [(eq? 'a id) 1]
[(eq? 'b id) 2]
[else (error ...)]))
An empty mapping that is implemented in this way has the same type:
(: emptymapping : Symbol > Number)
(define (emptymapping id)
(error ...))
We can use this idea to implement our environments: we only need to
define three things  `EmptyEnv`, `Extend`, and `lookup`. If we
manage to keep the contract to these functions intact, we will be able
to simply plug it into the same evaluator code with no other changes.
It will also be more convenient to define `ENV` as the appropriate
function type for use in the `VAL` type definition instead of using the
actual type:
;; Define a type for functional environments
(definetype ENV = Symbol > VAL)
Now we get to `EmptyEnv`  this is expected to be a function that
expects no arguments and creates an empty environment, one that behaves
like the `emptymapping` function defined above. We could define it
like this (changing the `emptymapping` type to return a `VAL`):
(define (EmptyEnv) emptymapping)
but we can skip the need for an extra definition and simply return an
empty mapping function:
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error ...)))
(The unRackety name is to avoid replacing previous code that used the
`EmptyEnv` name for the constructor that was created by the type
definition.)
The next thing we tackle is `lookup`. The previous definition that was
used is:
(: lookup : Symbol ENV > VAL)
(define (lookup name env)
(cases env
[(EmptyEnv) (error 'lookup "no binding for ~s" name)]
[(Extend id val restenv)
(if (eq? id name) val (lookup name restenv))]))
How should it be modified now? Easy  an environment is a mapping: a
Racket function that will do the searching job itself. We don't need to
modify the contract since we're still using `ENV`, except a different
implementation for it. The new definition is:
(: lookup : Symbol ENV > VAL)
(define (lookup name env)
(env name))
Note that `lookup` does almost nothing  it simply delegates the real
work to the `env` argument. This is a good hint for the error message
that empty mappings should throw 
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
Finally, `Extend`  this was previously created by the variant case of
the ENV type definition:
[Extend Symbol VAL ENV]
keeping the same type that is implied by this variant means that the new
`Extend` should look like this:
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
...)
The question is  how do we extend a given environment? Well, first,
we know that the result should be mapping  a `symbol > VAL` function
that expects an identifier to look for:
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
(lambda (name)
...))
Next, we know that in the generated mapping, if we look for `id` then
the result should be `val`:
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
...)))
If the `name` that we're looking for is not the same as `id`, then we
need to search through the previous environment:
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(lookup name restenv))))
But we know what `lookup` does  it simply delegates back to the
mapping function (which is our `rest` argument), so we can take a direct
route instead:
(: Extend : Symbol VAL ENV > ENV)
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name)))) ; same as (lookup name restenv)
To see how all this works, try out extending an empty environment a few
times and examine the result. For example, the environment that we
began with:
(define (mymap id)
(cond [(eq? 'a id) 1]
[(eq? 'b id) 2]
[else (error ...)]))
behaves in the same way (if the type of values is numbers) as
(Extend 'a 1 (Extend 'b 2 (EmptyEnv)))
The new code is now the same, except for the environment code:
#lang pl
#
The grammar:
::=
 { + }
 {  }
 { * }
 { / }
 { with { } }

 { fun { } }
 { call }
Evaluation rules:
eval(N,env) = N
eval({+ E1 E2},env) = eval(E1,env) + eval(E2,env)
eval({ E1 E2},env) = eval(E1,env)  eval(E2,env)
eval({* E1 E2},env) = eval(E1,env) * eval(E2,env)
eval({/ E1 E2},env) = eval(E1,env) / eval(E2,env)
eval(x,env) = lookup(x,env)
eval({with {x E1} E2},env) = eval(E2,extend(x,eval(E1,env),env))
eval({fun {x} E},env) = <{fun {x} E}, env>
eval({call E1 E2},env1)
= eval(Ef,extend(x,eval(E2,env1),env2))
if eval(E1,env1) = <{fun {x} Ef}, env2>
= error! otherwise
#
(definetype FLANG
[Num Number]
[Add FLANG FLANG]
[Sub FLANG FLANG]
[Mul FLANG FLANG]
[Div FLANG FLANG]
[Id Symbol]
[With Symbol FLANG FLANG]
[Fun Symbol FLANG]
[Call FLANG FLANG])
(: parsesexpr : Sexpr > FLANG)
;; parses sexpressions into FLANGs
(define (parsesexpr sexpr)
(match sexpr
[(number: n) (Num n)]
[(symbol: name) (Id name)]
[(cons 'with more)
(match sexpr
[(list 'with (list (symbol: name) named) body)
(With name (parsesexpr named) (parsesexpr body))]
[else (error 'parsesexpr "bad `with' syntax in ~s" sexpr)])]
[(cons 'fun more)
(match sexpr
[(list 'fun (list (symbol: name)) body)
(Fun name (parsesexpr body))]
[else (error 'parsesexpr "bad `fun' syntax in ~s" sexpr)])]
[(list '+ lhs rhs) (Add (parsesexpr lhs) (parsesexpr rhs))]
[(list ' lhs rhs) (Sub (parsesexpr lhs) (parsesexpr rhs))]
[(list '* lhs rhs) (Mul (parsesexpr lhs) (parsesexpr rhs))]
[(list '/ lhs rhs) (Div (parsesexpr lhs) (parsesexpr rhs))]
[(list 'call fun arg)
(Call (parsesexpr fun) (parsesexpr arg))]
[else (error 'parsesexpr "bad syntax in ~s" sexpr)]))
(: parse : String > FLANG)
;; parses a string containing a FLANG expression to a FLANG AST
(define (parse str)
(parsesexpr (string>sexpr str)))
;; Types for environments, values, and a lookup function
(definetype VAL
[NumV Number]
[FunV Symbol FLANG ENV])
;; Define a type for functional environments
(definetype ENV = Symbol > VAL)
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
(: Extend : Symbol VAL ENV > ENV)
;; extend a given environment cache with a new binding
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name))))
(: lookup : Symbol ENV > VAL)
;; lookup a symbol in an environment, return its value or throw an
;; error if it isn't bound
(define (lookup name env)
(env name))
(: NumV>number : VAL > Number)
;; convert a FLANG runtime numeric value to a Racket one
(define (NumV>number val)
(cases val
[(NumV n) n]
[else (error 'arithop "expected a number, got: ~s" val)]))
(: arithop : (Number Number > Number) VAL VAL > VAL)
;; gets a Racket numeric binary operator, and uses it within a NumV
;; wrapper
(define (arithop op val1 val2)
(NumV (op (NumV>number val1) (NumV>number val2))))
(: eval : FLANG ENV > VAL)
;; evaluates FLANG expressions by reducing them to values
(define (eval expr env)
(cases expr
[(Num n) (NumV n)]
[(Add l r) (arithop + (eval l env) (eval r env))]
[(Sub l r) (arithop  (eval l env) (eval r env))]
[(Mul l r) (arithop * (eval l env) (eval r env))]
[(Div l r) (arithop / (eval l env) (eval r env))]
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
[(Id name) (lookup name env)]
[(Fun boundid boundbody)
(FunV boundid boundbody env)]
[(Call funexpr argexpr)
(let ([fval (eval funexpr env)])
(cases fval
[(FunV boundid boundbody fenv)
(eval boundbody
(Extend boundid (eval argexpr env) fenv))]
[else (error 'eval "`call' expects a function, got: ~s"
fval)]))]))
(: run : String > Number)
;; evaluate a FLANG program contained in a string
(define (run str)
(let ([result (eval (parse str) (EmptyEnv))])
(cases result
[(NumV n) n]
[else (error 'run "evaluation returned a nonnumber: ~s"
result)])))
;; tests
(test (run "{call {fun {x} {+ x 1}} 4}")
=> 5)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{call add3 1}}")
=> 4)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{with {add1 {fun {x} {+ x 1}}}
{with {x 3}
{call add1 {call add3 x}}}}}")
=> 7)
(test (run "{with {identity {fun {x} x}}
{with {foo {fun {x} {+ x 1}}}
{call {call identity foo} 123}}}")
=> 124)
(test (run "{with {x 3}
{with {f {fun {y} {+ x y}}}
{with {x 5}
{call f 4}}}}")
=> 7)
(test (run "{call {with {x 3}
{fun {y} {+ x y}}}
4}")
=> 7)
(test (run "{with {f {with {x 3} {fun {y} {+ x y}}}}
{with {x 100}
{call f 4}}}")
=> 7)
(test (run "{call {call {fun {x} {call x 1}}
{fun {x} {fun {y} {+ x y}}}}
123}")
=> 124)

# More Closures (on both levels)
Racket closures (= functions) can be used in other places too, and as we
have seen, they can do more than encapsulate various values  they can
also hold the behavior that is expected of these values.
To demonstrate this we will deal with closures in our language. We
currently use a variant that holds the three pieces of relevant
information:
[FunV Symbol FLANG ENV]
We can replace this by a functional object, which will hold the three
values. First, change the `VAL` type to hold functions for `FunV`
values:
(definetype VAL
[NumV Number]
[FunV (? > ?)])
And note that the function should somehow encapsulate the same
information that was there previously, the question is *how* this
information is going to be done, and this will determine the actual
type. This information plays a role in two places in our evaluator 
generating a closure in the `Fun` case, and using it in the `Call` case:
[(Fun boundid boundbody)
(FunV boundid boundbody env)]
[(Call funexpr argexpr)
(let ([fval (eval funexpr env)])
(cases fval
[(FunV boundid boundbody fenv)
(eval boundbody ;***
(Extend boundid ;***
(eval argexpr env) ;***
fenv))] ;***
[else (error 'eval "`call' expects a function, got: ~s"
fval)]))]
we can simply fold the marked functionality bit of `Call` into a Racket
function that will be stored in a `FunV` object  this piece of
functionality takes an argument value, extends the closure's environment
with its value and the function's name, and continues to evaluate the
function body. Folding all of this into a function gives us:
(lambda (argval)
(eval boundbody (Extend boundid argval env)))
where the values of `boundbody`, `boundid`, and `val` are known at the
time that the `FunV` is *constructed*. Doing this gives us the
following code for the two cases:
[(Fun boundid boundbody)
(FunV (lambda (argval)
(eval boundbody (Extend boundid argval env))))]
[(Call funexpr argexpr)
(let ([fval (eval funexpr env)])
(cases fval
[(FunV proc) (proc (eval argexpr env))]
[else (error 'eval "`call' expects a function, got: ~s"
fval)]))]
And now the type of the function is clear:
(definetype VAL
[NumV Number]
[FunV (VAL > VAL)])
And again, the rest of the code is unmodified:
#lang pl
(definetype FLANG
[Num Number]
[Add FLANG FLANG]
[Sub FLANG FLANG]
[Mul FLANG FLANG]
[Div FLANG FLANG]
[Id Symbol]
[With Symbol FLANG FLANG]
[Fun Symbol FLANG]
[Call FLANG FLANG])
(: parsesexpr : Sexpr > FLANG)
;; parses sexpressions into FLANGs
(define (parsesexpr sexpr)
(match sexpr
[(number: n) (Num n)]
[(symbol: name) (Id name)]
[(cons 'with more)
(match sexpr
[(list 'with (list (symbol: name) named) body)
(With name (parsesexpr named) (parsesexpr body))]
[else (error 'parsesexpr "bad `with' syntax in ~s" sexpr)])]
[(cons 'fun more)
(match sexpr
[(list 'fun (list (symbol: name)) body)
(Fun name (parsesexpr body))]
[else (error 'parsesexpr "bad `fun' syntax in ~s" sexpr)])]
[(list '+ lhs rhs) (Add (parsesexpr lhs) (parsesexpr rhs))]
[(list ' lhs rhs) (Sub (parsesexpr lhs) (parsesexpr rhs))]
[(list '* lhs rhs) (Mul (parsesexpr lhs) (parsesexpr rhs))]
[(list '/ lhs rhs) (Div (parsesexpr lhs) (parsesexpr rhs))]
[(list 'call fun arg)
(Call (parsesexpr fun) (parsesexpr arg))]
[else (error 'parsesexpr "bad syntax in ~s" sexpr)]))
(: parse : String > FLANG)
;; parses a string containing a FLANG expression to a FLANG AST
(define (parse str)
(parsesexpr (string>sexpr str)))
;; Types for environments, values, and a lookup function
(definetype VAL
[NumV Number]
[FunV (VAL > VAL)])
;; Define a type for functional environments
(definetype ENV = Symbol > VAL)
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
(: Extend : Symbol VAL ENV > ENV)
;; extend a given environment cache with a new binding
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name))))
(: lookup : Symbol ENV > VAL)
;; lookup a symbol in an environment, return its value or throw an
;; error if it isn't bound
(define (lookup name env)
(env name))
(: NumV>number : VAL > Number)
;; convert a FLANG runtime numeric value to a Racket one
(define (NumV>number val)
(cases val
[(NumV n) n]
[else (error 'arithop "expected a number, got: ~s" val)]))
(: arithop : (Number Number > Number) VAL VAL > VAL)
;; gets a Racket numeric binary operator, and uses it within a NumV
;; wrapper
(define (arithop op val1 val2)
(NumV (op (NumV>number val1) (NumV>number val2))))
(: eval : FLANG ENV > VAL)
;; evaluates FLANG expressions by reducing them to values
(define (eval expr env)
(cases expr
[(Num n) (NumV n)]
[(Add l r) (arithop + (eval l env) (eval r env))]
[(Sub l r) (arithop  (eval l env) (eval r env))]
[(Mul l r) (arithop * (eval l env) (eval r env))]
[(Div l r) (arithop / (eval l env) (eval r env))]
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
[(Id name) (lookup name env)]
[(Fun boundid boundbody)
(FunV (lambda (argval)
(eval boundbody (Extend boundid argval env))))]
[(Call funexpr argexpr)
(let ([fval (eval funexpr env)])
(cases fval
[(FunV proc) (proc (eval argexpr env))]
[else (error 'eval "`call' expects a function, got: ~s"
fval)]))]))
(: run : String > Number)
;; evaluate a FLANG program contained in a string
(define (run str)
(let ([result (eval (parse str) (EmptyEnv))])
(cases result
[(NumV n) n]
[else (error 'run "evaluation returned a nonnumber: ~s"
result)])))
;; tests
(test (run "{call {fun {x} {+ x 1}} 4}")
=> 5)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{call add3 1}}")
=> 4)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{with {add1 {fun {x} {+ x 1}}}
{with {x 3}
{call add1 {call add3 x}}}}}")
=> 7)
(test (run "{with {identity {fun {x} x}}
{with {foo {fun {x} {+ x 1}}}
{call {call identity foo} 123}}}")
=> 124)
(test (run "{with {x 3}
{with {f {fun {y} {+ x y}}}
{with {x 5}
{call f 4}}}}")
=> 7)
(test (run "{call {with {x 3}
{fun {y} {+ x y}}}
4}")
=> 7)
(test (run "{with {f {with {x 3} {fun {y} {+ x y}}}}
{with {x 100}
{call f 4}}}")
=> 7)
(test (run "{call {call {fun {x} {call x 1}}
{fun {x} {fun {y} {+ x y}}}}
123}")
=> 124)

# Types of Evaluators
What we did just now is implement lexical environments and closures in
the language we implement using lexical environments and closures in our
own language (Racket)!
This is another example of embedding a feature of the host language in
the implemented language, an issue that we have already discussed.
There are many examples of this, even when the two languages involved
are different. For example, if we have this bit in the C implementation
of Racket:
// Disclaimer: not real Racket code
Racket_Object *eval_and(int argc, Racket_Object *argv[]) {
Racket_Object *tmp;
if ( argc != 2 )
signal_racket_error("bad number of arguments");
else if ( racket_eval(argv[0]) != racket_false &&
(tmp = racket_eval(argv[1])) != racket_false )
return tmp;
else
return racket_false;
}
then the special semantics of evaluating a Racket `and` form is being
inherited from C's special treatment of `&&`. You can see this by the
fact that if there is a bug in the C compiler, then it will propagate to
the resulting Racket implementation too. A different solution is to not
use `&&` at all:
// Disclaimer: not real Racket code
Racket_Object *eval_and(int argc, Racket_Object *argv[]) {
Racket_Object *tmp;
if ( argc != 2 )
signal_racket_error("bad number of arguments");
else if ( racket_eval(argv[0]) != racket_false )
return racket_eval(argv[1]);
else
return racket_false;
}
and we can say that this is even better since it evaluates the second
expression in tail position. But in this case we don't really get that
benefit, since C itself is not doing tailcall optimization as a
standard feature (though some compilers do so under some circumstances).
We have seen a few different implementations of evaluators that are
quite different in flavor. They suggest the following taxonomy.
* A ___syntactic evaluator___ is one that uses its own language to
represent expressions and semantic runtime values of the evaluated
language, implementing all the corresponding behavior explicitly.
* A ___meta evaluator___ is an evaluator that uses language features of
its own language to directly implement behavior of the evaluated
language.
While our substitutionbased FLANG evaluator was close to being a
syntactic evaluator, we haven't written any purely syntactic evaluators
so far: we still relied on things like Racket arithmetics etc. The most
recent evaluator that we have studied, is even more of a *meta*
evaluator than the preceding ones: it doesn't even implement closures
and lexical scope, and instead, it uses the fact that Racket itself has
them.
With a good match between the evaluated language and the implementation
language, writing a meta evaluator can be very easy. With a bad match,
though, it can be very hard. With a syntactic evaluator, implementing
each semantic feature will be somewhat hard, but in return you don't
have to worry as much about how well the implementation and the
evaluated languages match up. In particular, if there is a particularly
strong mismatch between the implementation and the evaluated language,
it may take less effort to write a syntactic evaluator than a meta
evaluator. As an exercise, we can build upon our latest evaluator to
remove the encapsulation of the evaluator's response in the VAL type.
The resulting evaluator is shown below. This is a true meta evaluator:
it uses Racket closures to implement FLANG closures, Racket function
application for FLANG function application, Racket numbers for FLANG
numbers, and Racket arithmetic for FLANG arithmetic. In fact, ignoring
some small syntactic differences between Racket and FLANG, this latest
evaluator can be classified as something more specific than a meta
evaluator:
* A ___metacircular evaluator___ is a meta evaluator in which the
implementation and the evaluated languages are the same.
(Put differently, the trivial nature of the evaluator clues us in to the
deep connection between the two languages, whatever their syntactic
differences may be.)

# Feature Embedding
We saw that the difference between lazy evaluation and eager evaluation
is in the evaluation rules for `with` forms, function applications, etc:
eval({with {x E1} E2}) = eval(E2[eval(E1)/x])
is eager, and
eval({with {x E1} E2}) = eval(E2[E1/x])
is lazy. But is the first rule *really* eager? The fact is that the
only thing that makes it eager is the fact that our understanding of the
mathematical notation is eager  if we were to take math as lazy, then
the description of the rule becomes a description of lazy evaluation.
Another way to look at this is  take the piece of code that
implements this evaluation:
(: eval : FLANG > Number)
;; evaluates FLANG expressions by reducing them to numbers
(define (eval expr)
(cases expr
...
[(With boundid namedexpr boundbody)
(eval (subst boundbody
boundid
(Num (eval namedexpr))))]
...))
and the same question applies: is this really implementing eager
evaluation? We know that this is indeed eager  we can simply try it
and check that it is, but it is only eager because we are using an eager
language for the implementation! If our own language was lazy, then the
evaluator's implementation would run lazily, which means that the above
applications of the `eval` and the `subst` functions would also be lazy,
making our evaluator lazy as well.
This is a general phenomena where some of the semantic features of the
language we use (math in the formal description, Racket in our code)
gets *embedded* into the language we implement.
Here's another example  consider the code that implements
arithmetics:
(: eval : FLANG > Number)
;; evaluates FLANG expressions by reducing them to numbers
(define (eval expr)
(cases expr
[(Num n) n]
[(Add l r) (+ (eval l) (eval r))]
...))
what if it was written like this:
FLANG eval(FLANG expr) {
if (is_Num(expr))
return num_of_Num(expr);
else if (is_Add(expr))
return eval(lhs_of_Add(expr)) + eval(rhs_of_Add(expr));
else if ...
...
}
Would it still implement unlimited integers and exact fractions? That
depends on the language that was used to implement it: the above syntax
suggests C, C++, Java, or some other relative, which usually come with
limited integers and no exact fractions. But this really depends on the
language  even our own code has unlimited integers and exact
rationals only because Racket has them. If we were using a language
that didn't have such features (there are such Scheme implementations),
then our implemented language would absorb these (lack of) features too,
and its own numbers would be limited in just the same way. (And this
includes the syntax for numbers, which we embedded intentionally, like
the syntax for identifiers).
The bottom line is that we should be aware of such issues, and be very
careful when we talk about semantics. Even the language that we use to
communicate (semiformal logic) can mean different things.

Aside: read "Reflections on Trusting Trust" by Ken Thompson
(You can skip to the "Stage II" part to get to the interesting stuff.)
(And when you're done, look for "XcodeGhost" to see a relevant example,
and don't miss the leaked document on the wikipedia page...)

Here is yet another variation of our evaluator that is even closer to a
metacircular evaluator. It uses Racket values directly to implement
values, so arithmetic operations become straightforward. Note
especially how the case for function application is similar to
arithmetics: a FLANG function application translates to a Racket
function application. In both cases (applications and arithmetics) we
don't even check the objects since they are simple Racket objects  if
our language happens to have some meaning for arithmetics with
functions, or for applying numbers, then we will inherit the same
semantics in our language. This means that we now specify less behavior
and fall back more often on what Racket does.
We use Racket values with this type definition:
(definetype VAL = (U Number (VAL > VAL)))
And the evaluation function can now be:
(: eval : FLANG ENV > VAL)
;; evaluates FLANG expressions by reducing them to values
(define (eval expr env)
(cases expr
[(Num n) n] ;*** return the actual number
[(Add l r) (+ (eval l env) (eval r env))]
[(Sub l r) ( (eval l env) (eval r env))]
[(Mul l r) (* (eval l env) (eval r env))]
[(Div l r) (/ (eval l env) (eval r env))]
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
[(Id name) (lookup name env)]
[(Fun boundid boundbody)
(lambda ([argval : VAL]) ;*** return the racket function
;; note that this requires input type specifications since
;; typed racket can't guess the right one
(eval boundbody (Extend boundid argval env)))]
[(Call funexpr argexpr)
((eval funexpr env) ;*** trivial like the arithmetics!
(eval argexpr env))]))
Note how the arithmetics implementation is simple  it's a direct
translation of the FLANG syntax to Racket operations, and since we don't
check the inputs to the Racket operations, we let Racket throw type
errors for us. Note also how function application is just like the
arithmetic operations: a FLANG application is directly translated to a
Racket application.
However, this does not work quite as simply in Typed Racket. The whole
point of typechecking is that we never run into type errors  so we
cannot throw back on Racket errors since code that might produce them is
forbidden! A way around this is to perform explicit checks that
guarantee that Racket cannot run into type errors. We do this with the
following two helpers that are defined inside `eval`:
(: evalN : FLANG > Number)
(define (evalN e)
(let ([n (eval e env)])
(if (number? n)
n
(error 'eval "got a nonnumber: ~s" n))))
(: evalF : FLANG > (VAL > VAL))
(define (evalF e)
(let ([f (eval e env)])
(if (function? f)
f
(error 'eval "got a nonfunction: ~s" f))))
Note that Typed Racket is "smart enough" to figure out that in `evalF`
the result of the recursive evaluation has to be either `Number` or
`(VAL > VAL)`; and since the `if` throws out on numbers, we're left
with `(VAL > VAL)` functions, not just any function.
#lang pl
(definetype FLANG
[Num Number]
[Add FLANG FLANG]
[Sub FLANG FLANG]
[Mul FLANG FLANG]
[Div FLANG FLANG]
[Id Symbol]
[With Symbol FLANG FLANG]
[Fun Symbol FLANG]
[Call FLANG FLANG])
(: parsesexpr : Sexpr > FLANG)
;; parses sexpressions into FLANGs
(define (parsesexpr sexpr)
(match sexpr
[(number: n) (Num n)]
[(symbol: name) (Id name)]
[(cons 'with more)
(match sexpr
[(list 'with (list (symbol: name) named) body)
(With name (parsesexpr named) (parsesexpr body))]
[else (error 'parsesexpr "bad `with' syntax in ~s" sexpr)])]
[(cons 'fun more)
(match sexpr
[(list 'fun (list (symbol: name)) body)
(Fun name (parsesexpr body))]
[else (error 'parsesexpr "bad `fun' syntax in ~s" sexpr)])]
[(list '+ lhs rhs) (Add (parsesexpr lhs) (parsesexpr rhs))]
[(list ' lhs rhs) (Sub (parsesexpr lhs) (parsesexpr rhs))]
[(list '* lhs rhs) (Mul (parsesexpr lhs) (parsesexpr rhs))]
[(list '/ lhs rhs) (Div (parsesexpr lhs) (parsesexpr rhs))]
[(list 'call fun arg)
(Call (parsesexpr fun) (parsesexpr arg))]
[else (error 'parsesexpr "bad syntax in ~s" sexpr)]))
(: parse : String > FLANG)
;; parses a string containing a FLANG expression to a FLANG AST
(define (parse str)
(parsesexpr (string>sexpr str)))
;; Types for environments, values, and a lookup function
;; Values are plain Racket values, no new VAL wrapper;
;; (but note that this is a recursive definition)
(definetype VAL = (U Number (VAL > VAL)))
;; Define a type for functional environments
(definetype ENV = (Symbol > VAL))
(: EmptyEnv : > ENV)
(define (EmptyEnv)
(lambda (id) (error 'lookup "no binding for ~s" id)))
(: Extend : Symbol VAL ENV > ENV)
;; extend a given environment cache with a new binding
(define (Extend id val restenv)
(lambda (name)
(if (eq? name id)
val
(restenv name))))
(: lookup : Symbol ENV > VAL)
;; lookup a symbol in an environment, return its value or throw an
;; error if it isn't bound
(define (lookup name env)
(env name))
(: eval : FLANG ENV > VAL)
;; evaluates FLANG expressions by reducing them to values
(define (eval expr env)
(: evalN : FLANG > Number)
(define (evalN e)
(let ([n (eval e env)])
(if (number? n)
n
(error 'eval "got a nonnumber: ~s" n))))
(: evalF : FLANG > (VAL > VAL))
(define (evalF e)
(let ([f (eval e env)])
(if (function? f)
f
(error 'eval "got a nonfunction: ~s" f))))
(cases expr
[(Num n) n]
[(Add l r) (+ (evalN l) (evalN r))]
[(Sub l r) ( (evalN l) (evalN r))]
[(Mul l r) (* (evalN l) (evalN r))]
[(Div l r) (/ (evalN l) (evalN r))]
[(With boundid namedexpr boundbody)
(eval boundbody
(Extend boundid (eval namedexpr env) env))]
[(Id name) (lookup name env)]
[(Fun boundid boundbody)
(lambda ([argval : VAL])
(eval boundbody (Extend boundid argval env)))]
[(Call funexpr argexpr)
((evalF funexpr)
(eval argexpr env))]))
(: run : String > VAL) ; no need to convert VALs to numbers
;; evaluate a FLANG program contained in a string
(define (run str)
(eval (parse str) (EmptyEnv)))
;; tests
(test (run "{call {fun {x} {+ x 1}} 4}")
=> 5)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{call add3 1}}")
=> 4)
(test (run "{with {add3 {fun {x} {+ x 3}}}
{with {add1 {fun {x} {+ x 1}}}
{with {x 3}
{call add1 {call add3 x}}}}}")
=> 7)
(test (run "{with {identity {fun {x} x}}
{with {foo {fun {x} {+ x 1}}}
{call {call identity foo} 123}}}")
=> 124)
(test (run "{with {x 3}
{with {f {fun {y} {+ x y}}}
{with {x 5}
{call f 4}}}}")
=> 7)
(test (run "{call {with {x 3}
{fun {y} {+ x y}}}
4}")
=> 7)
(test (run "{with {f {with {x 3} {fun {y} {+ x y}}}}
{with {x 100}
{call f 4}}}")
=> 7)
(test (run "{call {call {fun {x} {call x 1}}
{fun {x} {fun {y} {+ x y}}}}
123}")
=> 124)

# Recursion, Recursion, Recursion
> [PLAI §9]
There is one major feature that is still missing from our language: we
have no way to perform recursion (therefore no kind of loops). So far,
we could only use recursion when we had *names*. In FLANG, the only way
we can have names is through `with` which not good enough for recursion.
To discuss the issue of recursion, we switch to a "broken" version of
(untyped) Racket  one where a `define` has a different scoping rules:
the scope of the defined name does *not* cover the defined expression.
Specifically, in this language, this doesn't work:
#lang pl broken
(define (fact n)
(if (zero? n) 1 (* n (fact ( n 1)))))
(fact 5)
In our language, this translation would also not work (assuming we have
`if` etc):
{with {fact {fun {n}
{if {= n 0} 1 {* n {call fact { n 1}}}}}}
{call fact 5}}
And similarly, in plain Racket this won't work if `let` is the only tool
you use to create bindings:
(let ([fact (lambda (n)
(if (zero? n) 1 (* n (fact ( n 1)))))])
(fact 5))
In the brokenscope language, the `define` form is more similar to a
mathematical definition. For example, when we write:
(define (F x) x)
(define (G y) (F y))
(G F)
it is actually shorthand for
(define F (lambda (x) x))
(define G (lambda (y) (F y)))
(G F)
we can then replace defined names with their definitions:
(define F (lambda (x) x))
(define G (lambda (y) (F y)))
((lambda (y) (F y)) (lambda (x) x))
and this can go on, until we get to the actual code that we wrote:
((lambda (y) ((lambda (x) x) y)) (lambda (x) x))
This means that the above `fact` definition is similar to writing:
fact := (lambda (n)
(if (zero? n) 1 (* n (fact ( n 1)))))
(fact 5)
which is not a wellformed definition  it is *meaningless* (this is a
formal use of the word "meaningless"). What we'd really want, is to
take the *equation* (using `=` instead of `:=`)
fact = (lambda (n)
(if (zero? n) 1 (* n (fact ( n 1)))))
and find a solution which will be a value for `fact` that makes this
true.
If you look at the Racket evaluation rules handout on the web page, you
will see that this problem is related to the way that we introduced the
Racket `define`: there is a handwavy explanation that talks about
*knowing* things.
The big question is: can we define recursive functions without Racket's
magical `define` form?
> Note: This question is a little different than the question of
> implementing recursion in our language  in the Racket case we have
> no control over the implementation of the language. As it will
> eventually turn out, implementing recursion in our own language will
> be quite easy when we use mutation in a specific way. So the question
> that we're now facing can be phrased as either "can we get recursion
> in Racket without Racket's magical definition forms?" or "can we get
> recursion in our interpreter without mutation?".

# Recursion without the Magic
> [PLAI §22.4] (we go much deeper)
> Note: This explanation is similar to the one you can find in "The Why
> of Y", by Richard Gabriel.
To implement recursion without the `define` magic, we first make an
observation: this problem does *not* come up in a dynamicallyscoped
language. Consider the `let`version of the problem:
#lang pl dynamic
(let ([fact (lambda (n)
(if (zero? n) 1 (* n (fact ( n 1)))))])
(fact 5))
This works fine  because by the time we get to evaluate the body of
the function, `fact` is already bound to itself in the current dynamic
scope. (This is another reason why dynamic scope is perceived as a
convenient approach in new languages.)
Regardless, the problem that we have with lexical scope is still there,
but the way things work in a dynamic scope suggest a solution that we
can use now. Just like in the dynamic scope case, when `fact` is
called, it does have a value  the only problem is that this value is
inaccessible in the lexical scope of its body.
Instead of trying to get the value in via lexical scope, we can imitate
what happens in the dynamically scoped language by passing the `fact`
value to itself so it can call itself (going back to the original code
in the brokenscope language):
(define (fact self n) ;***
(if (zero? n) 1 (* n (self ( n 1)))))
(fact fact 5) ;***
except that now the recursive call should still send itself along:
(define (fact self n)
(if (zero? n) 1 (* n (self self ( n 1))))) ;***
(fact fact 5)
The problem is that this required rewriting calls to `fact`  both
outside and recursive calls inside. To make this an acceptable
solution, calls from both places should not change. Eventually, we
should be able to get a working `fact` definition that uses just
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))
The first step in resolving this problem is to curry the `fact`
definition.
(define (fact self) ;***
(lambda (n) ;***
(if (zero? n)
1
(* n ((self self) ( n 1)))))) ;***
((fact fact) 5) ;***
Now `fact` is no longer our factorial function  it's a function that
constructs it. So call it `makefact`, and bind `fact` to the actual
factorial function.
(define (makefact self) ;***
(lambda (n)
(if (zero? n) 1 (* n ((self self) ( n 1))))))
(define fact (makefact makefact)) ;***
(fact 5) ;***
We can try to do the same thing in the body of the factorial function:
instead of calling `(self self)`, just bind `fact` to it:
(define (makefact self)
(lambda (n)
(let ([fact (self self)]) ;***
(if (zero? n)
1
(* n (fact ( n 1))))))) ;***
(define fact (makefact makefact))
(fact 5)
This works fine, but if we consider our original goal, we need to get
that local `fact` binding outside of the `(lambda (n) ...)`  so we're
left with a definition that uses the factorial expression as is. So,
swap the two lines:
(define (makefact self)
(let ([fact (self self)]) ;***
(lambda (n) ;***
(if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact (makefact makefact))
(fact 5)
But the problem is that this gets us into an infinite loop because we're
trying to evaluate `(self self)` too ea(ge)rly. In fact, if we ignore
the body of the `let` and other details, we basically do this:
(define (makefact self) (self self)) (makefact makefact)
reducesugar>
(define makefact (lambda (self) (self self))) (makefact makefact)
replacedefinition>
((lambda (self) (self self)) (lambda (self) (self self)))
renameidentifiers>
((lambda (x) (x x)) (lambda (x) (x x)))
And this expression has an interesting property: it reduces to itself,
so evaluating it gets stuck in an infinite loop.
So how do we solve this? Well, we know that `(self self)` *should* be
the same value that is the factorial function itself  so it must be a
oneargument function. If it's such a function, we can use a value that
is equivalent, except that it will not get evaluated until it is needed,
when the function is called. The trick here is the observation that
`(lambda (n) (add1 n))` is really the same as `add1` (provided that
`add1` is a oneargument function), except that the `add1` part doesn't
get evaluated until the function is called. Applying this trick to our
code produces a version that does not get stuck in the same infinite
loop:
(define (makefact self)
(let ([fact (lambda (n) ((self self) n))]) ;***
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact (makefact makefact))
(fact 5)
Continuing from here  we know that
(let ([x v]) e) is the same as ((lambda (x) e) v)
(remember how we derived `fun` from a `with`), so we can turn that `let`
into the equivalent function application form:
(define (makefact self)
((lambda (fact) ;***
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))
(lambda (n) ((self self) n)))) ;***
(define fact (makefact makefact))
(fact 5)
And note now that the (lambda (fact) ...) expression is everything that
we need for a recursive definition of `fact`  it has the proper
factorial body with a plain recursive call. It's almost like the usual
value that we'd want to define `fact` as, except that we still have to
abstract on the recursive value itself. So lets move this code into a
separate definition for `factstep`:
(define factstep ;***
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makefact self)
(factstep ;***
(lambda (n) ((self self) n))))
(define fact (makefact makefact))
(fact 5)
We can now proceed by moving the `(makefact makefact)` self
application into its own function which is what creates the real
factorial:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makefact self)
(factstep
(lambda (n) ((self self) n))))
(define (makerealfact) (makefact makefact)) ;***
(define fact (makerealfact)) ;***
(fact 5)
Rewrite the `makefact` definition using an explicit `lambda`:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define makefact ;***
(lambda (self) ;***
(factstep
(lambda (n) ((self self) n)))))
(define (makerealfact) (makefact makefact))
(define fact (makerealfact))
(fact 5)
and fold the functionality of `makefact` and `makerealfact` into a
single `makefact` function by just using the value of `makefact`
explicitly instead of through a definition:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makerealfact)
(let ([make (lambda (self) ;***
(factstep ;***
(lambda (n) ((self self) n))))]) ;***
(make make)))
(define fact (makerealfact))
(fact 5)
We can now observe that `makerealfact` has nothing that is specific to
factorial  we can make it take a "core function" as an argument:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makerealfact core) ;***
(let ([make (lambda (self)
(core ;***
(lambda (n) ((self self) n))))])
(make make)))
(define fact (makerealfact factstep)) ;***
(fact 5)
and call it `makerecursive`:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makerecursive core) ;***
(let ([make (lambda (self)
(core
(lambda (n) ((self self) n))))])
(make make)))
(define fact (makerecursive factstep)) ;***
(fact 5)
We're almost done now  there's no real need for a separate
`factstep` definition, just use the value for the definition of `fact`:
(define (makerecursive core)
(let ([make (lambda (self)
(core
(lambda (n) ((self self) n))))])
(make make)))
(define fact
(makerecursive
(lambda (fact) ;***
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))) ;***
(fact 5)
turn the `let` into a function form:
(define (makerecursive core)
((lambda (make) (make make)) ;***
(lambda (self) ;***
(core (lambda (n) ((self self) n)))))) ;***
(define fact
(makerecursive
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))))
(fact 5)
do some renamings to make things simpler  `make` and `self` turn to
`x`, and `core` to `f`:
(define (makerecursive f) ;***
((lambda (x) (x x)) ;***
(lambda (x) (f (lambda (n) ((x x) n)))))) ;***
(define fact
(makerecursive
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))))
(fact 5)
or we can manually expand that first (lambda (x) (x x)) application to
make the symmetry more obvious (not really surprising because it started
with a `let` whose purpose was to do a selfapplication):
(define (makerecursive f)
((lambda (x) (f (lambda (n) ((x x) n)))) ;***
(lambda (x) (f (lambda (n) ((x x) n)))))) ;***
(define fact
(makerecursive
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))))
(fact 5)
And we finally got what we were looking for: a general way to define
*any* recursive function without any magical `define` tricks. This also
work for other recursive functions:
#lang pl broken
(define (makerecursive f)
((lambda (x) (f (lambda (n) ((x x) n))))
(lambda (x) (f (lambda (n) ((x x) n))))))
(define fact
(makerecursive
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))))
(fact 5)
(define fib
(makerecursive
(lambda (fib)
(lambda (n) (if (<= n 1) n (+ (fib ( n 1)) (fib ( n 2))))))))
(fib 8)
(define length
(makerecursive
(lambda (length)
(lambda (l) (if (null? l) 0 (+ (length (rest l)) 1))))))
(length '(x y z))
A convenient tool that people often use on paper is to perform a kind of
a syntactic abstraction: "assume that whenever I write (twice foo) I
really meant to write (foo foo)". This can often be done as plain
abstractions (that is, using functions), but in some cases  for
example, if we want to abstract over definitions  we just want such a
rewrite rule. (More on this towards the end of the course.) The
brokenscope language does provide such a tool  `rewrite` extends the
language with a rewrite rule. Using this, and our `makerecursive`, we
can make up a recursive definition form:
(rewrite (define/rec (f x) E)
=> (define f (makerecursive (lambda (f) (lambda (x) E)))))
In other words, we've created our own "magical definition" form. The
above code can now be written in almost the same way it is written in
plain Racket:
#lang pl broken
(define (makerecursive f)
((lambda (x) (f (lambda (n) ((x x) n))))
(lambda (x) (f (lambda (n) ((x x) n))))))
(rewrite (define/rec (f x) E)
=> (define f (makerecursive (lambda (f) (lambda (x) E)))))
;; examples
(define/rec (fact n) (if (zero? n) 1 (* n (fact ( n 1)))))
(fact 5)
(define/rec (fib n) (if (<= n 1) n (+ (fib ( n 1)) (fib ( n 2)))))
(fib 8)
(define/rec (length l) (if (null? l) 0 (+ (length (rest l)) 1)))
(length '(x y z))
Finally, note that makerecursive is limited to 1argument functions
only because of the protection from eager evaluation. In any case, it
can be used in any way you want, for example,
(makerecursive (lambda (f) (lambda (x) f)))
is a function that *returns itself* rather than calling itself. Using
the rewrite rule, this would be:
(define/rec (f x) f)
which is the same as:
(define (f x) f)
in plain Racket.

# The Core of `makerecursive`
As in Racket, being able to express recursive functions is a fundamental
property of the language. It means that we can have loops in our
language, and that's the essence of making a language powerful enough to
be TMequivalent  able to express undecidable problems, where we
don't know whether there is an answer or not.
The core of what makes this possible is the expression that we have seen
in our derivation:
((lambda (x) (x x)) (lambda (x) (x x)))
which reduces to itself, and therefore has no value: trying to evaluate
it gets stuck in an infinite loop. (This expression is often called
"Omega".)
This is the key for creating a loop  we use it to make recursion
possible. Looking at our final `makerecursive` definition and ignoring
for a moment the "protection" that we need against being stuck
prematurely in an infinite loop:
(define (makerecursive f)
((lambda (x) (x x)) (lambda (x) (f (x x)))))
we can see that this is almost the same as the Omega expression  the
only difference is that application of `f`. Indeed, this expression
(the result of (makerecursive F) for some `F`) reduces in a similar way
to Omega:
((lambda (x) (x x)) (lambda (x) (F (x x))))
((lambda (x) (F (x x))) (lambda (x) (F (x x))))
(F ((lambda (x) (F (x x))) (lambda (x) (F (x x)))))
(F (F ((lambda (x) (F (x x))) (lambda (x) (F (x x))))))
(F (F (F ((lambda (x) (F (x x))) (lambda (x) (F (x x)))))))
...
which means that the actual value of this expression is:
(F (F (F ...forever...)))
This definition would be sufficient if we had a lazy language, but to
get things working in a strict one we need to bring back the protection.
This makes things a little different  if we use `(protect f)` to be a
shorthand for the protection trick,
(rewrite (protect f) => (lambda (x) (f x)))
then we have:
(define (makerecursive f)
((lambda (x) (x x)) (lambda (x) (f (protect (x x))))))
which makes the (makerecursive F) evaluation reduce to
(F (protect (F (protect (F (protect (...forever...)))))))
and this is still the same result (as long as `F` is a singleargument
function).
(Note that `protect` cannot be implemented as a plain function!)

# Denotational Explanation of Recursion
> Note: This explanation is similar to the one you can find in "The
> Little Schemer" called "(Y Y) Works!", by Dan Friedman and Matthias
> Felleisen.
The explanation that we have now for how to derive the `makerecursive`
definition is fine  after all, we did manage to get it working. But
this explanation was done from a kind of an operational point of view:
we knew a certain trick that can make things work and we pushed things
around until we got it working like we wanted. Instead of doing this,
we can reapproach the problem from a more declarative point of view.
So, start again from the same broken code that we had (using the
brokenscope language):
(define fact
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))
This is as broken as it was when we started: the occurrence of `fact` in
the body of the function is free, which means that this code is
meaningless. To avoid the compilation error that we get when we run
this code, we can substitute *anything* for that `fact`  it's even
better to use a replacement that will lead to a runtime error:
(define fact
(lambda (n) (if (zero? n) 1 (* n (777 ( n 1)))))) ;***
This function will not work in a similar way to the original one  but
there is one case where it *does* work: when the input value is `0`
(since then we do not reach the bogus application). We note this by
calling this function `fact0`:
(define fact0 ;***
(lambda (n) (if (zero? n) 1 (* n (777 ( n 1))))))
Now that we have this function defined, we can use it to write `fact1`
which is the factorial function for arguments of `0` or `1`:
(define fact0
(lambda (n) (if (zero? n) 1 (* n (777 ( n 1))))))
(define fact1
(lambda (n) (if (zero? n) 1 (* n (fact0 ( n 1))))))
And remember that this is actually just shorthand for:
(define fact1
(lambda (n)
(if (zero? n)
1
(* n ((lambda (n)
(if (zero? n)
1
(* n (777 ( n 1)))))
( n 1))))))
We can continue in this way and write `fact2` that will work for n<=2:
(define fact2
(lambda (n) (if (zero? n) 1 (* n (fact1 ( n 1))))))
or, in full form:
(define fact2
(lambda (n)
(if (zero? n)
1
(* n ((lambda (n)
(if (zero? n)
1
(* n ((lambda (n)
(if (zero? n)
1
(* n (777 ( n 1)))))
( n 1)))))
( n 1))))))
If we continue this way, we *will* get the true factorial function, but
the problem is that to handle *any* possible integer argument, it will
have to be an infinite definition! Here is what it is supposed to look
like:
(define fact0 (lambda (n) (if (zero? n) 1 (* n (777 ( n 1))))))
(define fact1 (lambda (n) (if (zero? n) 1 (* n (fact0 ( n 1))))))
(define fact2 (lambda (n) (if (zero? n) 1 (* n (fact1 ( n 1))))))
(define fact3 (lambda (n) (if (zero? n) 1 (* n (fact2 ( n 1))))))
...
The true factorial function is `factinfinity`, with an infinite size.
So, we're back at the original problem...
To help make things more concise, we can observe the repeated pattern in
the above, and extract a function that abstracts this pattern. This
function is the same as the `factstep` that we have seen previously:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact0 (factstep 777))
(define fact1 (factstep fact0))
(define fact2 (factstep fact1))
(define fact3 (factstep fact2))
...
which is actually:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact0 (factstep 777))
(define fact1 (factstep (factstep 777)))
(define fact2 (factstep (factstep (factstep 777))))
...
(define fact
(factstep (factstep (factstep (... (factstep 777) ...)))))
Do this a little differently  rewrite `fact0` as:
(define fact0
((lambda (mk) (mk 777))
factstep))
Similarly, `fact1` is written as:
(define fact1
((lambda (mk) (mk (mk 777)))
factstep))
and so on, until the real factorial, which is still infinite at this
stage:
(define fact
((lambda (mk) (mk (mk (... (mk 777) ...))))
factstep))
Now, look at that `(lambda (mk) ...)`  it is an infinite expression,
but for every actual application of the resulting factorial function we
only need a finite number of `mk` applications. We can guess how many,
and as soon as we hit an application of `777` we know that our guess is
too small. So instead of `777`, we can try to use the maker function to
create and use the next.
To make things more explicit, here is the expression that is our
`fact0`, without the definition form:
((lambda (mk) (mk 777))
factstep)
This function has a very low guess  it works for 0, but with 1 it
will run into the `777` application. At this point, we want to somehow
invoke `mk` again to get the next level  and since `777` *does* get
applied, we can just replace it with `mk`:
((lambda (mk) (mk mk))
factstep)
The resulting function works just the same for an input of `0` because
it does not attempt a recursive call  but if we give it `1`, then
instead of running into the error of applying `777`:
(* n (777 ( n 1)))
we get to apply `factstep` there:
(* n (factstep ( n 1)))
and this is still wrong, because `factstep` expects a function as an
input. To see what happens more clearly, write `factstep` explicitly:
((lambda (mk) (mk mk))
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
The problem is in what we're going to pass into `factstep`  its
`fact` argument will not be the factorial function, but the `mk`
function constructor. Renaming the `fact` argument as `mk` will make
this more obvious (but not change the meaning):
((lambda (mk) (mk mk))
(lambda (mk)
(lambda (n) (if (zero? n) 1 (* n (mk ( n 1)))))))
It should now be obvious that this application of `mk` will not work,
instead, we need to apply it on some function and *then* apply the
result on `( n 1)`. To get what we had before, we can use `777` as a
bogus function:
((lambda (mk) (mk mk))
(lambda (mk)
(lambda (n) (if (zero? n) 1 (* n ((mk 777) ( n 1)))))))
This will allow one recursive call  so the definition works for both
inputs of `0` and `1`  but not more. But that `777` is used as a
maker function now, so instead, we can just use `mk` itself again:
((lambda (mk) (mk mk))
(lambda (mk)
(lambda (n) (if (zero? n) 1 (* n ((mk mk) ( n 1)))))))
And this is a *working* version of the real factorial function, so make
it into a (nonmagical) definition:
(define fact
((lambda (mk) (mk mk))
(lambda (mk)
(lambda (n) (if (zero? n) 1 (* n ((mk mk) ( n 1))))))))
But we're not done  we "broke" into the factorial code to insert that
`(mk mk)` application  that's why we dragged in the actual value of
`factstep`. We now need to fix this. The expression on that last line
(lambda (n) (if (zero? n) 1 (* n ((mk mk) ( n 1)))))
is close enough  it is `(factstep (mk mk))`. So we can now try to
rewrite our `fact` as:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact
((lambda (mk) (mk mk))
(lambda (mk) (factstep (mk mk)))))
... and would fail in a familiar way! If it's not familiar enough, just
rename all those `mk`s as `x`s:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact
((lambda (x) (x x))
(lambda (x) (factstep (x x)))))
We've run into the eagerness of our language again, as we did before.
The solution is the same  the `(x x)` is the factorial function, so
protect it as we did before, and we have a working version:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define fact
((lambda (x) (x x))
(lambda (x) (factstep (lambda (n) ((x x) n))))))
The rest should not be surprising now... Abstract the recursive making
bit in a new `makerecursive` function:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
(define (makerecursive f)
((lambda (x) (x x))
(lambda (x) (f (lambda (n) ((x x) n))))))
(define fact (makerecursive factstep))
and now we can do the first reduction inside `makerecursive` and write
the `factstep` expression explicitly:
#lang pl broken
(define (makerecursive f)
((lambda (x) (f (lambda (n) ((x x) n))))
(lambda (x) (f (lambda (n) ((x x) n))))))
(define fact
(makerecursive
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1))))))))
and this is the same code we had before.

# The Y Combinator
Our `makerecursive` function is usually called the *fixpoint operator*
or the *Y combinator*.
It looks really simple when using the lazy version (remember: our
version is the eager one):
(define Y
(lambda (f)
((lambda (x) (f (x x)))
(lambda (x) (f (x x))))))
> Note that if we *do* allow a recursive definition for Y itself, then
> the definition can follow the definition that we've seen:
>
> (define (Y f) (f (Y f)))
And this all comes from the loop generated by:
((lambda (x) (x x)) (lambda (x) (x x)))
This expression, which is also called *Omega* (the `(lambda (x) (x x))`
part by itself is usually called *omega* and then `(omega omega)` is
*Omega*), is also the idea behind many deep mathematical facts. As an
example for what it does, follow the next rule:
I will say the next sentence twice:
"I will say the next sentence twice".
(Note the usage of colon for the first and quotes for the second 
what is the equivalent of that in the lambda expression?)
By itself, this just gets you stuck in an infinite loop, as Omega does,
and the Y combinator adds `F` to that to get an infinite chain of
applications  which is similar to:
I will say the next sentence twice:
"I will hop on one foot and then say the next sentence twice".
Sidenote:
[see this SO question](https://stackoverflow.com/q/25228394/128595)
and my answer, which came from the PLQ implementation.

## The main property of Y
`factstep` is a function that given any limited factorial, will
generate a factorial that is good for one more integer input. Start
with `777`, which is a factorial that is good for nothing (because it's
not a function), and you can get `fact0` as
fact0 == (factstep 777)
and that's a good factorial function only for an input of `0`. Use that
with `factstep` again, and you get
fact1 == (factstep fact0) == (factstep (factstep 777))
which is the factorial function when you only look at input values of
`0` or `1`. In a similar way
fact2 == (factstep fact1)
is good for `0`...`2`  and we can continue as much as we want, except
that we need to have an infinite number of applications  in the
general case, we have:
factn == (factstep (factstep (factstep ... 777)))
which is good for `0`...`n`. The *real* factorial would be the result
of running `factstep` on itself infinitely, it *is* `factinfinity`.
In other words (here `fact` is the *real* factorial):
fact = factinfinity == (factstep (factstep ...infinitely...))
but note that since this is really infinity, then
fact = (factstep (factstep ...infinitely...))
= (factstep fact)
so we get an equation:
fact = (factstep fact)
and a solution for this is going to be the real factorial. The solution
is the *fixedpoint* of the `factstep` function, in the same sense that
`0` is the fixed point of the `sin` function because
0 = (sin 0)
And the Y combinator does just that  it has this property:
(makerecursive f) = (f (makerecursive f))
or, using the more common name:
(Y f) = (f (Y f))
This property encapsulates the real magical power of Y. You can see how
it works: since `(Y f) = (f (Y f))`, we can add an `f` application to
both sides, giving us `(f (Y f)) = (f (f (Y f)))`, so we get:
(Y f) = (f (Y f)) = (f (f (Y f))) = (f (f (f (Y f)))) = ...
= (f (f (f ...)))
and we can conclude that
(Y factstep) = (factstep (factstep ...infinitely...))
= fact

# Yet another explanation for Y
Here's another explanation of how the Y combinator works. Remember that
our `factstep` function was actually a function that generates a
factorial function based on some input, which is supposed to be the
factorial function:
(define factstep
(lambda (fact)
(lambda (n) (if (zero? n) 1 (* n (fact ( n 1)))))))
As we've seen, you can apply this function on a version of factorial
that is good for inputs up to some n, and the result will be a factorial
that is good for those values up to n+1. The question is *what is the
fixpoint of `factstep`*? And the answer is that if it maps factₙ
factorial to factₙ₊₁, then the input will be equal to the output on the
*infinitieth* `fact`, which is the *actual* factorial. Since Y is a
fixpoint combinator, it gives us exactly that answer:
(define therealfactorial (Y factstep))

# Typing the Y Combinator
Typing the Y combinator is a tricky issue. For example, in standard ML
you must write a new type definition to do this:
datatype 'a t = T of 'a t > 'a
val y = fn f => (fn (T x) => (f (fn a => x (T x) a)))
(T (fn (T x) => (f (fn a => x (T x) a))))
> Can you find a pattern in the places where `T` is used?
>  Roughly speaking, that type definition is
>
> ;; `t' is the type name, `T' is the constructor (aka the variant)
> (definetype (RecTypeOf t)
> [T ((RecTypeOf t) > t)])
>
> First note that the two `fn a => ...` parts are the same as our
> protection, so ignoring that we get:
>
> val y = fn f => (fn (T x) => (f (x (T x))))
> (T (fn (T x) => (f (x (T x)))))
>
> if you now replace `T` with `Quote`, things make more sense:
>
> val y = fn f => (fn (Quote x) => (f (x (Quote x))))
> (Quote (fn (Quote x) => (f (x (Quote x)))))
>
> and with our syntax, this would be:
>
> (define (Y f)
> ((lambda (qx)
> (cases qx
> [(Quote x) (f (x qx))]))
> (Quote
> (lambda (qx)
> (cases qx
> [(Quote x) (f (x qx))])))))
>
> it's not really quotation  but the analogy should help: it uses
> `Quote` to distinguish functions as values that are applied (the `x`s)
> from functions that are passed as arguments.
In OCaml, this looks a little different:
# type 'a t = T of ('a t > 'a) ;;
type 'a t = T of ('a t > 'a)
# let y f = (fun (T x) > x (T x))
(T (fun (T x) > fun z > f (x (T x)) z)) ;;
val y : (('a > 'b) > 'a > 'b) > 'a > 'b =
# let fact = y (fun fact n > if n < 1 then 1 else n * fact(n1)) ;;
val fact : int > int =
# fact 5 ;;
 : int = 120
but OCaml has also a `rectypes` command line argument, which will make
it infer the type by itself:
# let y f = (fun x > x x) (fun x > fun z > f (x x) z) ;;
val y : (('a > 'b) > 'a > 'b) > 'a > 'b =
# let fact = y (fun fact n > if n < 1 then 1 else n * fact(n1)) ;;
val fact : int > int =
# fact 5 ;;
 : int = 120
The translation of this to `#lang pl` is a little verbose because we
don't have autocurrying, and because we need to declare input types to
functions, but it's essentially a direct translation of the above:
(definetype (RecTypeOf t)
[T ((RecTypeOf t) > t)])
(: Y : (All (A B) ((A > B) > (A > B)) > (A > B)))
(define (Y f)
((lambda ([x : (RecTypeOf (A > B))])
(cases x
[(T x) (x (T x))]))
(T (lambda ([x : (RecTypeOf (A > B))])
(cases x
[(T x) (lambda ([z : A])
((f (x (T x))) z))])))))
(define fact
(Y (lambda ([fact : (Integer > Integer)])
(lambda ([n : Integer])
(if (< n 1) 1 (* n (fact (sub1 n))))))))
(fact 5)
It is also possible to write this expression in "plain" Typed Racket,
without a userdefined type  and we need to start with a proper type
definition. First of all, the type of Y should be straightforward: it
is a fixpoint operation, so it takes a `T > T` function and produces
its fixpoint. The fixpoint itself is some `T` (such that applying the
function on it results in itself). So this gives us:
(: makerecursive : (T > T) > T)
However, in our case `makerecursive` computes a *functional* fixpoint,
for unary `S > T` functions, so we should narrow down the type
(: makerecursive : ((S > T) > (S > T)) > (S > T))
Now, in the body of `makerecursive` we need to add a type for the `x`
argument which is behaving in a weird way: it is used both as a function
and as its own argument. (Remember  I will say the next sentence
twice: "I will say the next sentence twice".) We need a recursive type
definition helper (not a new type) for that:
(definetype (Tau S T) = (Rec this (this > (S > T))))
This type is tailored for our use of `x`: it is a type for a function
that will *consume itself* (hence the `Rec`) and spit out the value that
the `f` argument consumes  an `S > T` function.
The resulting full version of the code:
(: makerecursive : (All (S T) ((S > T) > (S > T)) > (S > T)))
(definetype (Tau S T) = (Rec this (this > (S > T))))
(define (makerecursive f)
((lambda ([x : (Tau S T)]) (f (lambda (z) ((x x) z))))
(lambda ([x : (Tau S T)]) (f (lambda (z) ((x x) z))))))
(: fact : Number > Number)
(define fact (makerecursive
(lambda ([fact : (Number > Number)])
(lambda ([n : Number])
(if (zero? n)
1
(* n (fact ( n 1))))))))
(fact 5)