Transcript Lecture 6: Cons car cdr sdr wdr CS200: Computer Science University of Virginia Computer Science David Evans http://www.cs.virginia.edu/~evans Menu • Recursion Practice: fibo • History of Scheme: LISP • Introducing Lists 28 January 2004 CS.

Lecture 6:
Cons
car
cdr
sdr
wdr
CS200: Computer Science
University of Virginia
Computer Science
David Evans
http://www.cs.virginia.edu/~evans
Menu
• Recursion Practice: fibo
• History of Scheme: LISP
• Introducing Lists
28 January 2004
CS 200 Spring 2004
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Defining Recursive Procedures
1. Be optimistic.
– Assume you can solve it.
– If you could, how would you solve a bigger
problem.
2. Think of the simplest version of the
problem, something you can already
solve. (This is the base case.)
3. Combine them to solve the problem.
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Defining fibo
;;; (fibo n) evaluates to the nth Fibonacci
;;; number
(define (fibo n)
FIBO (1) = FIBO (2) = 1
(if (or (= n 1) (= n 2))
1 ;;; base case
FIBO (n) =
FIBO (n – 1)
(+ (fibo (- n 1))
+ FIBO (n – 2)
(fibo (- n 2)))))
for n > 2
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Fibo Results
> (fibo 2)
1
Why
can’t
our
> (fibo 3)
100,000x Apollo
2
Guidance
> (fibo 4)
3
Computer calculate
> (fibo 10)
(fibo 100)?
55
> (fibo 100)
Still working after 4 hours…
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Tracing Fibo
> (require-library "trace.ss")
> (trace fibo)
(fibo)
> (fibo 3)
|(fibo 3)
|
(fibo 2)
|
1
|
(fibo 1)
|
1
|2
2
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CS 200 Spring 2004
This turns tracing on
6
> (fibo 5)
|(fibo 5)
| (fibo 4)
| |(fibo 3)
| | (fibo 2)
||1
| | (fibo 1)
||1
| |2
| |(fibo 2)
| |1
|3
| (fibo 3)
| |(fibo 2)
| |1
| |(fibo 1)
| |1
|2
|5
5
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(fibo 5) = (fibo 4)
(fibo 3) +
(fibo 2)
(fibo 2) + (fibo 1) +
1
1
+ 1
2
+
1
3
= 5
+ (fibo 3)
+ (fibo 2) + (fibo 1)
+ 1
+ 1
+ 2
+ 2
To calculate (fibo 5) we calculated:
(fibo 4)
1 time
(fibo 3)
2 times
(fibo 2)
3 times
(fibo 1)
2 times
= 8 calls to fibo = (fibo 6)
How many calls to calculate (fibo 100)?
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fast-fibo
(define (fast-fibo n)
(define (fibo-worker a b count)
(if (= count 1)
b
(fibo-worker (+ a b) a (- count 1))))
(fibo-worker 1 1 n))
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Fast-Fibo Results
> (fast-fibo 1)
1
> (fast-fibo 10)
55
> (time (fast-fibo 100))
cpu time: 0 real time: 0 gc time: 0
354224848179261915075
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;;; The Earth's mass is 6.0 x 10^24 kg
> (define mass-of-earth (* 6 (expt 10 24)))
;;; A typical rabbit's mass is 2.5 kilograms
> (define mass-of-rabbit 2.5)
> (/ (* mass-of-rabbit (fast-fibo 100)) mass-of-earth)
0.00014759368674135913
> (/ (* mass-of-rabbit (fast-fibo 110)) mass-of-earth)
0.018152823441189517
> (/ (* mass-of-rabbit (fast-fibo 119)) mass-of-earth)
1.379853393132076
> (exact->inexact (/ 119 12))
9.916666666666666
According to Fibonacci’s model, after less than 10
years, rabbits would out-weigh the Earth!
Beware the Bunnies!!
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History of Scheme
• Scheme [1975]
– Guy Steele and Gerry Sussman
– Originally “Schemer”
– “Conniver” [1973] and “Planner” [1967]
• Based on LISP
– John McCarthy (late 1950s)
• Based on Lambda Calculus
– Alonzo Church (1930s)
– Last few lectures in course
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LISP
“Lots of Insipid Silly Parentheses”
“LISt Processing language”
Lists are pretty important – hard to
write a useful Scheme program
without them.
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Making Lists
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Making a Pair
> (cons 1 2)
(1 . 2)
1 2
cons constructs a pair
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Splitting a Pair
> (car (cons 1 2))
1
> (cdr (cons 1 2))
2
cons
1 2
car
cdr
car extracts first part of a pair
cdr extracts second part of a pair
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Why “car” and “cdr”?
• Original (1950s) LISP on IBM 704
– Stored cons pairs in memory registers
– car = “Contents of the Address part of the
Register”
– cdr = “Contents of the Decrement part of the
Register” (“could-er”)
• Doesn’t matter unless you have an IBM 704
• Think of them as first and rest
(define first car)
(define rest cdr)
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(The DrScheme “Pretty Big” language
already defines these, but they are not part
of standard Scheme)
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Implementing cons, car and cdr
Using PS2:
(define cons make-point)
(define car x-of-point)
(define cdr y-of-point)
As we implemented make-point, etc.:
(define (cons a b) (lambda (w) (if (w) a b)))
(define (car pair) (pair #t)
(define (cdr pair) (pair #f)
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Pairs are fine, but how do
we make threesomes?
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Threesome?
(define (threesome a b c)
(lambda (w)
(if (= w 0) a (if (= w 1) b c))))
(define (first t) (t 0))
(define (second t) (t 1))
(define (third t) (t 2))
Is there a better way of thinking about our triple?
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Triple
A triple is just a pair where one of the
parts is a pair!
(define (triple a b c)
(cons a (cons b c)))
(define (t-first t) (car t))
(define (t-second t) (car (cdr t)))
(define (t-third t) (cdr (cdr t)))
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Quadruple
A quadruple is a pair where the second
part is a triple
(define (quadruple a b c d)
(cons a (triple b c d)))
(define (q-first q) (car q))
(define (q-second q) (t-first (cdr t)))
(define (q-third t) (t-second (cdr t)))
(define (q-fourth t) (t-third (cdr t)))
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Multuples
• A quintuple is a pair where the second part is
a quadruple
• A sextuple is a pair where the second part is a
quintuple
• A septuple is a pair where the second part is a
sextuple
• An octuple is group of octupi
• A list (any length tuple) is a pair where the
second part is a …?
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Lists
List ::= (cons Element List)
A list is a pair where the second part is a list.
One big problem: how do we stop?
This only allows infinitely long lists!
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From Lecture 5
Recursive Transition Networks
ORNATE NOUN
begin
ARTICLE
ADJECTIVE
NOUN
end
ORNATE NOUN ::= ARTICLE ADJECTIVE NOUN
ORNATE NOUN ::= ARTICLE ADJECTIVE ADJECTIVE NOUN
ORNATE NOUN ::= ARTICLE ADJECTIVE ADJECTIVE ADJECTIVE NOUN
ORNATE NOUN ::= ARTICLE ADJECTIVE ADJECTIVE ADJECTIVE ADJECTIVE NOUN
ORNATE NOUN ::= ARTICLE ADJECTIVE ADJECTIVE ADJECTIVE ADJECTIVE ADJECTIVE NOUN
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Recursive Transition Networks
ORNATE NOUN
begin
ARTICLE
ADJECTIVE
NOUN
end
ORNATE NOUN ::= ARTICLE ADJECTIVES NOUN
ADJECTIVES
::= ADJECTIVE ADJECTIVES
ADJECTIVES
::=
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Lists
List ::= (cons Element List)
List ::=
It’s hard to write this!
A list is either:
a pair where the second part is a list
or, empty
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Null
List ::= (cons Element List)
List ::= null
A list is either:
a pair where the second part is a list
or, empty (null)
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List Examples
> null
()
> (cons 1 null)
(1)
> (list? null)
#t
> (list? (cons 1 2))
#f
> (list? (cons 1 null))
#t
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More List Examples
> (list? (cons 1 (cons 2 null)))
#t
> (car (cons 1 (cons 2 null)))
1
> (cdr (cons 1 (cons 2 null)))
(2)
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Charge
• Next Time: List Recursion
• PS2 Due Monday
– Lots of the code we provided uses lists
– But, you are not expected to use lists in
your code (but you can if you want)
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