Finally, note that this is just one way to encode things — other
encodings are possible. One alternative encoding is in the following
code — it uses a list of `N`

falses as the encoding for `N`

. This
encoding makes it easier to `add1`

(just `cons`

another `#f`

), and to
`sub1`

(simply `cdr`

). The tradeoff is that some arithmetics operations
becomes more complicated, for example, the definition of `+`

requires
the fixpoint combinator. (As expected, some people want to see what can
we do with a language without recursion, so they don’t like jumping to Y
too fast.)

▶;; An alternative "Church" encoding: use lists to encode numbers

#lang pl schlac

(define identity (lambda (x) x))

;; Booleans (same as before)

(define #t (lambda (x y) x))

(define #f (lambda (x y) y))

(define if (lambda (c t e) (c t e))) ; not really needed

(test (->bool (if #t #f #t))

=> '#f)

(test (->bool (if #f ((lambda (x) (x x)) (lambda (x) (x x))) #t))

=> '#t)

(define and (lambda (a b) (a b a)))

(define or (lambda (a b) (a a b)))

(define not (lambda (a x y) (a y x)))

(test (->bool (and #f #f)) => '#f)

(test (->bool (and #t #f)) => '#f)

(test (->bool (and #f #t)) => '#f)

(test (->bool (and #t #t)) => '#t)

(test (->bool (or #f #f)) => '#f)

(test (->bool (or #t #f)) => '#t)

(test (->bool (or #f #t)) => '#t)

(test (->bool (or #t #t)) => '#t)

(test (->bool (not #f)) => '#t)

(test (->bool (not #t)) => '#f)

;; Lists (same as before)

(define cons (lambda (x y s) (s x y)))

(define car (lambda (x) (x #t)))

(define cdr (lambda (x) (x #f)))

(define 1st car)

(define 2nd (lambda (l) (car (cdr l))))

(define 3rd (lambda (l) (car (cdr (cdr l)))))

(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))

(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))

(define null (lambda (s) #t))

(define null? (lambda (x) (x (lambda (x y) #f))))

;; Natural numbers (alternate encoding)

(define 0 identity)

(define add1 (lambda (n) (cons #f n)))

(define zero? car) ; tricky

(define sub1 cdr) ; this becomes very simple

;; Note that we could have used something more straightforward:

;; (define 0 null)

;; (define add1 (lambda (n) (cons #t n))) ; cons anything

;; (define zero? null?)

;; (define sub1 (lambda (l) (if (zero? l) l (cdr l))))

(define 1 (add1 0))

(define 2 (add1 1))

(define 3 (add1 2))

(define 4 (add1 3))

(define 5 (add1 4))

(test (->nat* (add1 (add1 5))) => '7)

(test (->nat* (sub1 (sub1 (add1 (add1 5))))) => '5)

(test (->bool (and (zero? 0) (not (zero? 3)))) => '#t)

(test (->bool (zero? (sub1 (sub1 (sub1 3))))) => '#t)

;; list-of-numbers tests

(define l123 (cons 1 (cons 2 (cons 3 null))))

(test (->listof ->nat* l123) => '(1 2 3))

(test (->listof (->listof ->nat*) (cons l123 (cons l123 null)))

=> '((1 2 3) (1 2 3)))

;; Recursive functions

(define Y

(lambda (f)

((lambda (x) (x x)) (lambda (x) (f (x x))))))

(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))

;; note that this example is doing something silly now

(define/rec length

(lambda (l)

(if (null? l)

0

(add1 (length (cdr l))))))

(test (->nat* (length l123)) => '3)

;; addition becomes hard since it requires a recursive definition

;; (define/rec +

;; (lambda (m n) (if (zero? n) m (+ (add1 m) (sub1 n)))))

;; (test (->nat* (+ 4 5)) => '9)

;; faster alternative:

(define/rec +

(lambda (m n)

(if (zero? m) n

(if (zero? n) m

(add1 (add1 (+ (sub1 m) (sub1 n))))))))

(test (->nat* (+ 4 5)) => '9)

;; subtraction is similar to addition

;; (define/rec -

;; (lambda (m n) (if (zero? n) m (- (sub1 m) (sub1 n)))))

;; (test (->nat* (- (+ 4 5) 4)) => '5)

;; but this is not "natural subtraction": doesn't work when n>m,

;; because (sub1 0) does not return 0.

;; a solution is like alternative form of +:

(define/rec -

(lambda (m n)

(if (zero? m) 0

(if (zero? n) m

(- (sub1 m) (sub1 n))))))

(test (->nat* (- (+ 4 5) 4)) => '5)

(test (->nat* (- 2 5)) => '0)

;; alternatively, could change sub1 above:

;; (define sub1 (lambda (n) (if (zero? n) n (cdr n))))

;; we can do multiplication in a similar way

(define/rec *

(lambda (m n)

(if (zero? m) 0

(+ n (* (sub1 m) n)))))

(test (->nat* (* 4 5)) => '20)

(test (->nat* (+ 4 (* (+ 2 5) 5))) => '39)

;; and the rest of the examples

(define/rec fact

(lambda (x)

(if (zero? x) 1 (* x (fact (sub1 x))))))

(test (->nat* (fact 5)) => '120)

(define/rec fib

(lambda (x)

(if (or (zero? x) (zero? (sub1 x)))

1

(+ (fib (sub1 x)) (fib (sub1 (sub1 x)))))))

(test (->nat* (fib (* 5 2))) => '89)

#|

;; Fully-expanded Fibonacci (note: much shorter than the previous

;; encoding, but see how Y appears twice -- two "((lambda" pairs)

(define fib((lambda(f)((lambda(x)(x x))(lambda(x)(f(x x)))))(lambda(

f)(lambda(x)(((((x(lambda(x y)x))(x(lambda(x y)x)))((x(lambda(x y)y)

)(lambda(x y)x)))(lambda(s)(s(lambda(x y)y)(lambda(x)x))))((((lambda

(f)((lambda(x)(x x))(lambda(x)(f(x x))))) (lambda(f)(lambda(m n)((m(

lambda(x y)x))n (((n(lambda(x y)x)) m)(lambda(s)((s (lambda(x y)y))(

lambda(s)((s (lambda(x y)y))((f(m(lambda(x y)y)))(n(lambda(x y)y))))

))))))))(f(x(lambda(x y)y))))(f((x(lambda(x y)y))(lambda(x y)y))))))

)))

|#

#lang pl schlac

(define identity (lambda (x) x))

;; Booleans (same as before)

(define #t (lambda (x y) x))

(define #f (lambda (x y) y))

(define if (lambda (c t e) (c t e))) ; not really needed

(test (->bool (if #t #f #t))

=> '#f)

(test (->bool (if #f ((lambda (x) (x x)) (lambda (x) (x x))) #t))

=> '#t)

(define and (lambda (a b) (a b a)))

(define or (lambda (a b) (a a b)))

(define not (lambda (a x y) (a y x)))

(test (->bool (and #f #f)) => '#f)

(test (->bool (and #t #f)) => '#f)

(test (->bool (and #f #t)) => '#f)

(test (->bool (and #t #t)) => '#t)

(test (->bool (or #f #f)) => '#f)

(test (->bool (or #t #f)) => '#t)

(test (->bool (or #f #t)) => '#t)

(test (->bool (or #t #t)) => '#t)

(test (->bool (not #f)) => '#t)

(test (->bool (not #t)) => '#f)

;; Lists (same as before)

(define cons (lambda (x y s) (s x y)))

(define car (lambda (x) (x #t)))

(define cdr (lambda (x) (x #f)))

(define 1st car)

(define 2nd (lambda (l) (car (cdr l))))

(define 3rd (lambda (l) (car (cdr (cdr l)))))

(define 4th (lambda (l) (car (cdr (cdr (cdr l))))))

(define 5th (lambda (l) (car (cdr (cdr (cdr (cdr l)))))))

(define null (lambda (s) #t))

(define null? (lambda (x) (x (lambda (x y) #f))))

;; Natural numbers (alternate encoding)

(define 0 identity)

(define add1 (lambda (n) (cons #f n)))

(define zero? car) ; tricky

(define sub1 cdr) ; this becomes very simple

;; Note that we could have used something more straightforward:

;; (define 0 null)

;; (define add1 (lambda (n) (cons #t n))) ; cons anything

;; (define zero? null?)

;; (define sub1 (lambda (l) (if (zero? l) l (cdr l))))

(define 1 (add1 0))

(define 2 (add1 1))

(define 3 (add1 2))

(define 4 (add1 3))

(define 5 (add1 4))

(test (->nat* (add1 (add1 5))) => '7)

(test (->nat* (sub1 (sub1 (add1 (add1 5))))) => '5)

(test (->bool (and (zero? 0) (not (zero? 3)))) => '#t)

(test (->bool (zero? (sub1 (sub1 (sub1 3))))) => '#t)

;; list-of-numbers tests

(define l123 (cons 1 (cons 2 (cons 3 null))))

(test (->listof ->nat* l123) => '(1 2 3))

(test (->listof (->listof ->nat*) (cons l123 (cons l123 null)))

=> '((1 2 3) (1 2 3)))

;; Recursive functions

(define Y

(lambda (f)

((lambda (x) (x x)) (lambda (x) (f (x x))))))

(rewrite (define/rec f E) => (define f (Y (lambda (f) E))))

;; note that this example is doing something silly now

(define/rec length

(lambda (l)

(if (null? l)

0

(add1 (length (cdr l))))))

(test (->nat* (length l123)) => '3)

;; addition becomes hard since it requires a recursive definition

;; (define/rec +

;; (lambda (m n) (if (zero? n) m (+ (add1 m) (sub1 n)))))

;; (test (->nat* (+ 4 5)) => '9)

;; faster alternative:

(define/rec +

(lambda (m n)

(if (zero? m) n

(if (zero? n) m

(add1 (add1 (+ (sub1 m) (sub1 n))))))))

(test (->nat* (+ 4 5)) => '9)

;; subtraction is similar to addition

;; (define/rec -

;; (lambda (m n) (if (zero? n) m (- (sub1 m) (sub1 n)))))

;; (test (->nat* (- (+ 4 5) 4)) => '5)

;; but this is not "natural subtraction": doesn't work when n>m,

;; because (sub1 0) does not return 0.

;; a solution is like alternative form of +:

(define/rec -

(lambda (m n)

(if (zero? m) 0

(if (zero? n) m

(- (sub1 m) (sub1 n))))))

(test (->nat* (- (+ 4 5) 4)) => '5)

(test (->nat* (- 2 5)) => '0)

;; alternatively, could change sub1 above:

;; (define sub1 (lambda (n) (if (zero? n) n (cdr n))))

;; we can do multiplication in a similar way

(define/rec *

(lambda (m n)

(if (zero? m) 0

(+ n (* (sub1 m) n)))))

(test (->nat* (* 4 5)) => '20)

(test (->nat* (+ 4 (* (+ 2 5) 5))) => '39)

;; and the rest of the examples

(define/rec fact

(lambda (x)

(if (zero? x) 1 (* x (fact (sub1 x))))))

(test (->nat* (fact 5)) => '120)

(define/rec fib

(lambda (x)

(if (or (zero? x) (zero? (sub1 x)))

1

(+ (fib (sub1 x)) (fib (sub1 (sub1 x)))))))

(test (->nat* (fib (* 5 2))) => '89)

#|

;; Fully-expanded Fibonacci (note: much shorter than the previous

;; encoding, but see how Y appears twice -- two "((lambda" pairs)

(define fib((lambda(f)((lambda(x)(x x))(lambda(x)(f(x x)))))(lambda(

f)(lambda(x)(((((x(lambda(x y)x))(x(lambda(x y)x)))((x(lambda(x y)y)

)(lambda(x y)x)))(lambda(s)(s(lambda(x y)y)(lambda(x)x))))((((lambda

(f)((lambda(x)(x x))(lambda(x)(f(x x))))) (lambda(f)(lambda(m n)((m(

lambda(x y)x))n (((n(lambda(x y)x)) m)(lambda(s)((s (lambda(x y)y))(

lambda(s)((s (lambda(x y)y))((f(m(lambda(x y)y)))(n(lambda(x y)y))))

))))))))(f(x(lambda(x y)y))))(f((x(lambda(x y)y))(lambda(x y)y))))))

)))

|#

Another interesting way to implement lists follows the pattern matching
approach, where both pairs and the null value are represented by a
function that serves as a kind of a `match`

dispatcher. This function
takes in two inputs — if it is the representation of null then it will
return the first input, and if it is a pair, then it will apply the
second input on the two parts of the pair. This is implemented as
follows:

(define null

(lambda (n p)

n))

(define cons

(lambda (x y)

(lambda (n p)

(p x y))))

(lambda (n p)

n))

(define cons

(lambda (x y)

(lambda (n p)

(p x y))))

This might seem awkward, but it follows the intended use of pairs and null as a match-like construct. Here is an example, with the equivalent Racket code on the side:

;; Sums up a list of numbers

(define (sum l)

(l ; (match l

0 ; ['() 0]

(lambda (x xs) ; [(cons x xs)

(+ x (sum xs))))) ; (+ x (sum xs))])

(define (sum l)

(l ; (match l

0 ; ['() 0]

(lambda (x xs) ; [(cons x xs)

(+ x (sum xs))))) ; (+ x (sum xs))])

In fact, it’s easy to implement our selectors and predicate using this:

(define null? (lambda (l) (l #t (lambda (x xs) #f))))

(define car (lambda (l) (l #f (lambda (x y) x))))

(define cdr (lambda (l) (l #f (lambda (x y) y))))

;; in the above `#f' is really any value, since it

;; should be an error alternatively:

(define car (lambda (l)

(l ((lambda (x) (x x)) (lambda (x) (x x))) ; "error"

(lambda (x y) x))))

(define car (lambda (l) (l #f (lambda (x y) x))))

(define cdr (lambda (l) (l #f (lambda (x y) y))))

;; in the above `#f' is really any value, since it

;; should be an error alternatively:

(define car (lambda (l)

(l ((lambda (x) (x x)) (lambda (x) (x x))) ; "error"

(lambda (x y) x))))

The same approach can be used to define any kind of new data type in a
way that looks like our own `define-type`

definitions. For example,
consider a much-simplified definition of the AE type we’ve seen early in
the semester, and a matching `eval`

definition as an example for using
`cases`

:

(define-type AE

[Num Number]

[Add AE AE])

(: eval : AE -> Number)

(define (eval expr)

(cases expr

[(Num n) n]

[(Add l r) (+ (eval l) (eval r))]))

[Num Number]

[Add AE AE])

(: eval : AE -> Number)

(define (eval expr)

(cases expr

[(Num n) n]

[(Add l r) (+ (eval l) (eval r))]))

We can follow the above approach now to write Schlac code that more than being equivalent, it’s also very similar in nature. Note that the type definition is replaced by two definitions for the two constructors:

(define Num (lambda (n) (lambda (num add) (num n ))))

(define Add (lambda (l r) (lambda (num add) (add l r))))

(define eval

(lambda (expr) ; `expr` is always a (lambda (num add) ...), and it

; expects a unary `num` argument and a binary `add`

(expr (lambda (n) n)

(lambda (l r) (+ (eval l) (eval r))))))

(define Add (lambda (l r) (lambda (num add) (add l r))))

(define eval

(lambda (expr) ; `expr` is always a (lambda (num add) ...), and it

; expects a unary `num` argument and a binary `add`

(expr (lambda (n) n)

(lambda (l r) (+ (eval l) (eval r))))))

What we really need for recursion, is a special kind of an environment,
one that can refer to itself. So instead of doing (note: `call`

s
removed for readability):

{with {fact {fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}

{fact 5}}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}

{fact 5}}

which does not work for the usual reasons, we want to use some

{rec {fact {fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}

{fact 5}}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}

{fact 5}}

that will do the necessary magic.

One way to achieve this is using the Y combinator as we have seen — a
kind of a “constructor” for recursive functions. We can do that in a
similar way to the `rewrite`

rule that we have seen in Schlac —
translate the above expression to:

{with {fact {make-rec

{fun {fact}

{fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}}}

{fact 5}}

{fun {fact}

{fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}}}

{fact 5}}

or even:

{with {fact {{fun {f} {{fun {x} {f {x x}}} {fun {x} {f {x x}}}}}

{fun {fact}

{fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}}}

{fact 5}}

{fun {fact}

{fun {n}

{if {zero? n} 1 {* n {fact {- n 1}}}}}}}}

{fact 5}}

Now, we will see how it can be used in *our* code to implement a
recursive environment.

If we look at what `with`

does in

{with {fact {fun {n}

{if {zero? n} 1 {* n {call fact {- n 1}}}}}}

{call fact 5}}

{if {zero? n} 1 {* n {call fact {- n 1}}}}}}

{call fact 5}}

then we can say that to evaluate this expression, we evaluate the body
expression in an extended environment that contains `fact`

, even if a
bogus one that is good for `0`

only — the new environment is created
with something like this:

extend("fact", make-fact-closure(), env)

so we can take this whole thing as an operation over `env`

add-fact(env) := extend("fact", make-fact-closure(), env)

This gives us the first-level fact. But `fact`

itself is still
undefined in `env`

, so it cannot call itself. We can try this:

add-fact(add-fact(env))

but that still doesn’t work, and it will never work no matter how far we go:

add-fact(add-fact(add-fact(add-fact(add-fact(...env...)))))

What we really want is infinity: a place where add-fact works and the result is the same as what we’ve started with — we want to create a “magical” environment that makes this possible:

let magic-env = ???

such that:

add-fact(magic-env) = magic-env

such that:

add-fact(magic-env) = magic-env

which basically gives us the illusion of being at the infinity point.
This magic-env thing is exactly the *fixed-point* of the `add-fact`

operation. We can use:

magic-env = rec(add-fact)

and following the main property of the Y combinator, we know that:

magic-env = rec(add-fact) ; def. of magic-env

= add-fact(rec(add-fact)) ; Y(f) = f(Y(f))

= add-fact(magic-env) ; def. of magic-env

= add-fact(rec(add-fact)) ; Y(f) = f(Y(f))

= add-fact(magic-env) ; def. of magic-env

What does all this mean? It means that if we have a fixed-point operator at the level of the implementation of our environments, then we can use it to implement a recursive binder. In our case, this means that a fixpoint in Racket can be used to implement a recursive language. But we have that — Racket does have recursive functions, so we should be able to use that to implement our recursive binder.

There are two ways that make it possible to write recursive functions in
Racket. One is to define a function, and use its name to do a recursive
call — using the Racket formal rules, we can see that we said that we
mark that we now *know* that a variable is bound to a value. This is
essentially a side-effect — we modify what we know, which corresponds
to modifying the global environment. The second way is a new form:
`letrec`

. This form is similar to `let`

, except that the scope that is
established includes the named expressions — it is exactly what we
want `rec`

to do. A third way is using recursive local definitions, but
that is equivalent to using `letrec`

, more on this soon.

`letrec`

So we want to add recursion to our language, practically. We already
know that Racket makes it possible to write recursive functions, which
is possible because of the way it implements its “global environment”:
our evaluator can only *extend* an environment, while Racket *modifies*
its global environment. This means that whenever a function is defined
in the global environment, the resulting closure will have it as its
environment “pointer”, but the global environment was not extended —
it stays the same, and was just modified with one additional binding.

But Racket has another, a bit more organized way of using recursion:
there is a special local-binding construct that is similar to `let`

, but
allows a function to refer to itself. It is called `letrec`

:

(letrec ([fact (lambda (n)

(if (zero? n)

1

(* n (fact (- n 1)))))])

(fact 5))

(if (zero? n)

1

(* n (fact (- n 1)))))])

(fact 5))

Some people may remember that there was a third way for creating recursive functions: using local definition in function bodies. For example, we have seen things like:

(define (length list)

(define (helper list len)

(if (null? list)

len

(helper (rest list) (+ len 1))))

(helper list 0))

(define (helper list len)

(if (null? list)

len

(helper (rest list) (+ len 1))))

(helper list 0))

This looks like the same kind of environment magic that happens with a
global `define`

— but actually, Racket defines the meaning of internal
definitions using `letrec`

— so the above code is exactly the same as:

(define (length list)

(letrec ([helper (lambda (list len)

(if (null? list)

len

(helper (rest list) (+ len 1))))])

(helper list 0)))

(letrec ([helper (lambda (list len)

(if (null? list)

len

(helper (rest list) (+ len 1))))])

(helper list 0)))

The scoping rules for a `letrec`

is that the scope of the bound name
covers both the body *and* the named expression. Furthermore, multiple
names can be bound to multiple expressions, and the scope of each name
covers all named expression as well as the body. This makes it easy to
define mutually recursive functions, such as:

(letrec ([even? (lambda (n) (if (zero? n) #t (odd? (- n 1))))]

[odd? (lambda (n) (if (zero? n) #f (even? (- n 1))))])

(even? 99))

[odd? (lambda (n) (if (zero? n) #f (even? (- n 1))))])

(even? 99))

But it is not a required functionality — it could be done with a single recursive binding that contains several functions:

(letrec ([even+odd

(list (lambda (n)

(if (zero? n) #t ((second even+odd) (- n 1))))

(lambda (n)

(if (zero? n) #f ((first even+odd) (- n 1)))))])

((first even+odd) 99))

(list (lambda (n)

(if (zero? n) #t ((second even+odd) (- n 1))))

(lambda (n)

(if (zero? n) #f ((first even+odd) (- n 1)))))])

((first even+odd) 99))

This is basically the same problem we face if we want to use the Y
combinator for mutually recursive bindings. The above solution is
inconvenient, but it can be improved using more `let`

s to have easier
name access. For example:

(letrec ([even+odd

(list (lambda (n)

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(if (zero? n) #t (odd? (- n 1)))))

(lambda (n)

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(if (zero? n) #f (even? (- n 1))))))])

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(even? 99)))

(list (lambda (n)

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(if (zero? n) #t (odd? (- n 1)))))

(lambda (n)

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(if (zero? n) #f (even? (- n 1))))))])

(let ([even? (first even+odd)]

[odd? (second even+odd)])

(even? 99)))