1. Pairs and Lists. Data Abstraction. SICP: Sections 2.1.1 – 2.2.1
Lecture notes: Chapter 3

2. Box and Pointer Diagram2 1 a
(define a (cons 1 2))
A pair can be implemented directly using two “pointers”.
Originally on IBM 704:
(car a) Contents of Address part of Register
(cdr a) Contents of Decrement part of Register

3. Pair: A primitive data type.
Constructor: (cons a b)
Selectors: (car p)
(cdr p)
Guarantee: (car (cons a b)) = a
(cdr (cons a b)) = b
Abstraction barrier: We say nothing about the
representation or implementation of pairs.

4. Pairs (define x (cons 1 2)) (define y (cons 3 4))
(define z (cons x y))
(car (car z))  1 ;(caar z)
(car (cdr z))  3 ;(cadr z)

5. Box and pointer diagrams
(cons (cons 1 (cons 2 3)) 4) 4 1 3 2

6. Compound Data
A closure property: The result obtained by creating a compound data structure can itself be treated as a primitive object and thus be input to the creation of another compound object.
Pairs have the closure property:
We can pair pairs, pairs of pairs etc.
(cons (cons 1 2) 3) 3 2 1

7. The empty list (a.k.a. null or nill)
Lists
The empty list (a.k.a. null or nill)
(cons 1 (cons 3 (cons 2 ’() ))) 1 3 2
List are the glue for putting together many values just as pairs are the glue for pairs of values.
Syntactic sugar: (list 1 3 2)

8. Formal Definition of a List
A list is either
’() -- The empty list
A pair whose cdr is a list.
Lists are closed under the operations cons and cdr:
If lst is a non-empty list, then (cdr lst) is a list.
If lst is a list and x is arbitrary, then (cons x lst) is a list.
מבוא מורחב שיעור 7

9. (cons <x1> (cons <x2> ( … (cons <xn> ’() ))))
Lists
(list <x1> <x2> ... <xn>)
is syntactic sugar for
(cons <x1> (cons <x2> ( … (cons <xn> ’() ))))
<x1>
<x2>
<xn> …

10. Lists (examples)
The following expressions all result in the same structure:
(cons 3 (list 1 2))
(cons 3 (cons 1 (cons 2 ’() )))
(list 3 1 2) 3 1 2
and similarly the following
(cdr (list 1 2 3))
(cdr (cons 1 (cons 2 (cons 3 ’() ))))
(cons 2 (cons 3 ’() ))
(list 2 3) 2 3

11. More Elaborate Lists (list 1 2 3 4) Value:(1 2 3 4)
(cons (list 1 2) (list 3 4))
(list (list 1 2) (list 3 4))
Value:( )
Value:((1 2) 3 4)
Value:((1 2) (3 4)) 1 2 3 4 1 3 4 2 1 3 4 2

12. List of Symbols (shorthand)
‘(a b c) translates into (‘a ‘b ‘c)

13. Yet More Examples (define p (cons 1 2)) p (1 . 2)3 p1 p1
( ) 1 2 p p2
(define p2 (list p p)) p2
( (1 . 2) (1 . 2) )

14. The Predicate Null? (null? <z>) null? : anytype -> boolean
#t if <z> evaluates to empty list
#f otherwise
(null? 2)  #f
(null? (list 1))  #f
(null? (cdr (list 1)))  #t
(null? ’())  #t
(null? null)  #t
The beauty of list is that cdr allows us to easily move down a list, and we have a simple test to find out when we have reached the end, we get to the empty list, that is, to a list whose value is null. How do we check that we have reached the end, scheme supplies us with the predicate null?

15. The Predicate Pair? pair? : anytype -> boolean
(pair? <z>) #t if <z> evaluates to a pair
#f otherwise.
(pair? (cons 1 2))  #t
(pair? (cons 1 (cons 1 2)))  #t
(pair? (list 1))  #t
(pair? ’())  #f
(pair? 3)  #f
(pair? pair?)  #f

16. The Predicate Atom? atom? : anytype -> boolean (define (atom? z)
(and (not (pair? z))
(not (null? z))))
(define (square x) (* x x))
(atom? square)  #t
(atom? 3)  #t
(atom? (cons 1 2))  #f
Not a primitive procedure

17. More examples ? ? (define digits (list 1 2 3 4 5 6 7 8 9))
(define digits1 (cons 0 digits))
digits1
(define l (list 0 digits)) l ? ?
( )
(0 ( ))

18. The procedure length (define digits (list 1 2 3 4 5 6 7 8 9))
(length digits) 9
(define l null)
(length l)
(define l (cons 1 l))
(length l) 1
(define (length l)
(if (null? l) 0
(+ 1 (length (cdr l)))))

19. The procedure append (define (append list1 list2)
(cond ((null? list1) list2) ; base
(else
(cons (car list1) ; recursion
(append (cdr list1) list2)))))

20. Constructor Type: Number * T -> LIST(T) > (make-list 7 ’foo)
(foo foo foo foo foo foo foo)
> (make-list 5 1)
( )

21. List type
Pairs:
For every type assignment TA and type expressions S,S1,S2:
TA |- cons:[S1*S2 -> PAIR(S1,S2)]
TA |- car:[PAIR(S1,S2) -> S1]
TA |- cdr:[PAIR(S1,S2) -> S2]
TA |- pair?:[S -> Boolean]
TA |- equal?:[PAIR(S1,S2)*PAIR(S1,S2) -> Boolean]

22. For every type environment TEnv and type expression S: TEnv |- list:[Unit -> LIST(S)] TEnv |- cons:[T*LIST(S) -> LIST(S)] TEnv |- car:[LIST(S) -> S] TEnv |- cdr:[LIST(S) -> LIST(S)] TEnv |- null?:[LIST(S) -> Boolean] TEnv |- list?:[S -> Boolean] TEnv |- equal?:[LIST(S)*LIST(S) -> Boolean]

23. Cont. For every type environment TEnv and type expression S:
TEnv |- list:[Unit -> LIST]
TEnv |- cons:[S*LIST -> LIST]
TEnv |- car:[LIST -> S]
TEnv |- cdr:[LIST -> LIST]
TEnv |- null?:[LIST -> Boolean]
TEnv |- list?:[S -> Boolean]
TEnv |- equal?:[LIST*LIST -> Boolean]

24. Procedural abstraction
Publish: name, number and type of arguments
(and conditions they must satisfy)
type of procedure’s return value
Guarantee: the behavior of the procedure
Hide: local variables and procedures,
way of implementation,
internal details, etc.
Interface
Implementation
Export only what is needed.

25. Data-object abstraction
Publish: constructors, selectors
Guarantee: the behavior
Hide: local variables and procedures,
way of implementation,
internal details, etc.
Interface
Implementation
Export only what is needed.

26. An example: Rational numbers
We would like to represent rational numbers.
A rational number is a quotient a/b of two integers.
Constructor: (make-rat a b)
Selectors: (numer r)
(denom r)
Guarantee: (numer (make-rat a b)) = a
(denom (make-rat a b)) = b

27. An example: Rational numbers
We would like to represent rational numbers.
A rational number is a quotient a/b of two integers.
Constructor: (make-rat a b)
Selectors: (numer r)
(denom r)
A better Guarantee:
(numer (make-rat a b)) a =
(denom (make-rat a b)) b
A weaker condition, but still sufficient!

28. We can now use the constructors and selectors to implement operations on rational numbers:
(add-rat x y)
(sub-rat x y)
(mul-rat x y)
(div-rat x y)
(equal-rat? x y)
(print-rat x)
A form of wishful thinking: we don’t know how make-rat numer and denom are implemented, but we use them.

29. Implementing the operations
(define (add-rat x y) ;n1/d1 + n2/d2 = (n1.d2 + n2.d1) / (d1.d2)
(make-rat (+ (* (numer x) (denom y))
(* (numer y) (denom x)))
(* (denom x) (denom y))))
(define (sub-rat x y) …
(define (mul-rat x y)
(make-rat (* (numer x) (numer y))
(* (denom x) (denom y))))
(define (div-rat x y)
(make-rat (* (numer x) (denom y))
(* (denom x) (numer y))))
(define (equal-rat? x y)
(= (* (numer x) (denom y)) (* (numer y) (denom x))))

30. Using the rational package
(define (print-rat x)
(newline)
(display (numer x))
(display ”/”)
(display (denom x)))
(define one-half (make-rat 1 2))
(print-rat one-half)  1/2
(define one-third (make-rat 1 3))
(print-rat (add-rat one-half one-third))  5/6
(print-rat (add-rat one-third one-third))  6/9

31. Abstraction barriers Programs that use rational numbers
rational numbers in problem domain
add-rat sub-rat mul-rat…
rational numbers as numerators and denumerators
make-rat numer denom

32. Gluing things together
We still have to implement numer, denom, and make-rat
We need a way to glue things together…
A pair:
(define x (cons 1 2))
(car x)  1
(cdr x)  2

33. Implementing make-rat, numer, denom
(define (make-rat n d) (cons n d))
(define (numer x) (car x))
(define (denom x) (cdr x))

34. Abstraction barriers Programs that use rational numbers
rational numbers in problem domain
add-rat sub-rat mul-rat...
rational numbers as numerators and denumerators
make-rat numer denom
rational numbers as pairs
cons car cdr

35. Alternative implementation for add-rat
Abstraction
Violation
(define (add-rat x y)
(cons (+ (* (car x) (cdr y))
(* (car y) (cdr x)))
(* (cdr x) (cdr y))))
If we bypass an abstraction barrier,
changes to one level may affect many levels above it.
Maintenance becomes more difficult.

36. Rationals - Alternative Implementation
In our current implementation we keep 10000/20000
as such and not as 1/2.
This:
Makes the computation more expensive.
Prints out clumsy results.
A solution: change the constructor
(define (make-rat a b)
(let ((g (gcd a b)))
(cons (/ a g) (/ b g))))
No other changes are required!

37. Reducing to lowest terms, another way
(define (make-rat n d)
(cons n d))
(define (numer x)
(let ((g (gcd (car x) (cdr x))))
(/ (car x) g)))
(define (denom x)
(/ (cdr x) g)))

38. How can we implement pairs? (first solution – “lazy” implementation)
(define (cons x y)
(lambda (f) (f x y)))
(define (car z)
(z (lambda (x y) x)))
(define (cdr z)
(z (lambda (x y) y)))

39. How can we implement pairs? (first solution, cont’)
Name Value
(lambda(f) (f 1 2)) p
> (define p (cons 1 2))
> (car p)
( (lambda(f) (f 1 2)) (lambda (x y) x))
( (lambda(x y) x) 1 2 )
(define (cons x y)
(lambda (f) (f x y)))
> 1
(define (car z)
(z (lambda (x y) x)))
(define (cdr z)
(z (lambda (x y) y)))

40. How can we implement pairs? (Second solution: “eager” implementation)
(define (cons x y)
(lambda (m)
(cond ((= m 0) x)
((= m 1) y)
(else (error "Argument not 0 or CONS" m))))))
(define (car z) (z 0))
(define (cdr z) (z 1))

41. Implementing pairs (second solution, cont’)
Name Value
> (define p (cons 3 4)) p
(lambda(m)
(cond ((= m 0) 3)
((= m 1) 4)
(else ..)))
> (car p)
((lambda(m) (cond ..)) 0)
(cond ((= 0 0) 3) ((= 0 1) 4) (else ...)))
(define (cons x y)
(lambda (m)
(cond ((= m 0) x)
((= m 1) y)
(else ...)))
> 3
(define (car z)
(z 0))
(define (cdr z)
(z 1))