6.001 SICP 1

6.001 SICP – September 22

6001-Introduction

Trevor Darrell trevor@csail.mit.edu32-D512

6.001 web page: http://sicp.csail.mit.edu/section web page: http://www.csail.mit.edu/trevor/6001/

Office Hours W11, 32-D512

• Orders of Growth– Linear– Exponential– Logarithmic

• Let

6.001 SICP 2

6.001 SICP

• Orders of Growth• Linear• Exponential• Logarithmic

• let

6.001 SICP 3

Orders of growth of processes

• Suppose n is a parameter that measures the size of a problem

• Let R(n) be the amount of resources needed to compute a procedure of size n.

• We say R(n) has order of growth f(n)if there are constants k1 and k2 such that k1f(n)<= R(n)<= k2f(n)

for large n • Two common resources are space, measured by the

number of deferred operations, and time, measured by the number of primitive steps.

6.001 SICP 4

Orders of Growth

N=2 N=10 N=100• Constant (1) 1 1 1• Logarithmic (log N) 1 3.33 6.66• Linear (n) 2 10 100• Quadratic (n2) 4 100 10000• Exponential (2n) 4 1024 1.26x1030

• At 1 billion operations per second (current state of the art), if you were to run an exponential time algorithm in the lab on a data set of size n=100, you would be waiting for approximately 4x1011 centuries for the code to finish running!

6.001 SICP 5

Order of growth examples

For each, find simplest and slowest growing f for which R(n)=(f(n))

• R(n)=6

• R(n)=n2 + 3

• R(n)=6*n3 + 3*n2 + 7n + 100

• R(n)=23n+7

6.001 SICP 6

Order of growth examples

For each, find simplest and slowest growing f for which R(n)=(f(n))

• R(n)=6

(1) 1 * 1 <= 6 <= 6 * 1 for all n• R(n)=n2 + 3

(n2) 1*n2 <= n2 + 3 <= 2*n2 for all n > 2• R(n)=6*n3 + 3*n2 + 7n + 100

(n3) 1*n3 <= 6*n3 + 3*n2 + 7n + 100 <= 7*n3 for all n>100• R(n)=23n+7

(8n) 1*8n <= 23n+7 <= 28*8n for all n>0

6.001 SICP 7

Examples

(define (factorial n)(if (= n 1)

1(* n (factorial (- n 1)))))

(factorial 5)(* 5 (factorial 4))(* 5 (* 4 (factorial 3)))(* 5 (* 4 (* 3 (factorial 2))))(* 5 (* 4 (* 3 (* 2 (factorial 1)))))(* 5 (* 4 (* 3 (* 2 1))))(* 5 (* 4 (* 3 2)))(* 5 (* 4 6))120

Time =(n) Space = (n)

6.001 SICP 8

Examples

(define (fact2 n)(define (helper cur k)

(if (= k 1)cur(helper (* cur k) (- k 1))))

(helper 1 n))

(fact2 5)(helper 1 5)(helper 5 4)(helper 20 3)(helper 60 2)(helper 120 1)

Time =(n) Space =(1)

6.001 SICP 9

Towers of Hanoi

• Three posts, and a set of different size disks• any stack must be sorted in decreasing order from bottom

to top• the goal is to move the disks one at a time, while

preserving these conditions, until the entire stack has moved from one post to another

6.001 SICP 10

Towers of Hanoi

(define move-tower (lambda (size from to extra) (cond ((= size 0) true) (else (move-tower (- size 1) from extra to) (print-move from to) (move-tower (- size 1) extra to from)))))

(define print-move (lambda (from to) (write-line ``Move top disk from ``) (write-line from) (write-line `` to ``) (write-line to)))

6.001 SICP 11

A tree recursion

Move 4

Move 3 Move 3

Move 2Move 2Move 2Move 2

Move 1

Move 1Move 1

Move 1

Move 1

Move 1

Move 1

Move 1

6.001 SICP 12

Orders of growth for towers of Hanoi

• Let tn be the number of steps that we need to take to solve the case for n disks. Then

•  tn = 2tn-1 + 1 = 2(2tn-2 +1) + 1 ….= 2n -1

•  So in time we have (2n) -- exponential• In space, we have one deferred operation for each

increment of the stack of disks -- (n) -- linear

6.001 SICP 13

Quiz

What is simplest expression for the order of growth of running time of procedure mul1 & mul2?

(define (mul1 n m) (if (= n 0) 0 (+ m (mul1 (- n 1) m))))

(define (mul2 n m) (define (help count ans) (if (= count 0) ans (help (- count 1) (+ m ans)))) (help n 0))

1. Theta(1) 2. Theta(2n) 3. Theta(n) 4. Theta(n/2)

6.001 SICP 14

Mul

Now write a procedure mul4 that computes m*n in (log n) time

Hint: Compare with a^b = (a^2)^(b/2) if b is even, and a^b = a*a^(b-1) if b is odd,

n*m = 2 * ( (n / 2) * m) if n even, n*m = m + ((n-1) * m) if n odd.

6.001 SICP 15

Local Variables

• Suppose we want to compute

F(x,y)=x(1+xy)2 + y(1-y) + (1+xy)(1-y)• Useful to name intermediate quantities:

A = 1 + xy

B = 1 - y

F(x,y) = xA2 + yB + AB • How can we do this?

Using a “helper function?”

Using a nested lambda?

Using Let…

6.001 SICP 16

Local variables with helper fcn. / lambda

A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB• Can use an auxiliary procedure to bind the intermediate

variables:

(define (f x y)

(define (f-helper a b) (+ (* x (square a)) (* y b) (* a b)))

(f-helper (+ 1 (* x y)) (- 1 y)))

6.001 SICP 17

Local variables with helper fcn. / lambda

A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB

Or use lambda directly….

(define (f x y) ((lambda (a b) (+ (* x (square a)) (* y b) (* a b))) (+ 1 (* x y)) (- 1 y)))

(define (f x y) (define (f-helper a b) (+ (* x (square a)) (* y b) (* a b))) (f-helper (+ 1 (* x y) ) (- 1 y)) )

6.001 SICP 18

Local variables with let

• Let is a special form which is syntactic sugar for this use of lambda to create local variables

(let (( <var1> <exp1> )

( <var2> <exp2> )

…

( <varn> <expn> ))

<body> )

Which means: let <var1> be <exp1>, <var2> be <exp2>, <var3> be <exp3> in <body> , and return the value of the last expression in <body>

6.001 SICP 19

Local variables with let

A = 1 + xy; B = 1 - y; F(x,y) = xA2 + yB + AB

(define (f x y) ((lambda (a b) (+ (* x (square a)) (* y b) (* a b))) (+ 1 (* x y)) (- 1 y)))

(define (f x y) (let ((a (+ 1 (* x y)) (b (- 1 y)) ) (+ (* x (square a)) (* y b) (* a b))))

6.001 SICP 20

Let vs. nested define

Use of nested defines for local variable definition is considered bad form, but is technically legal:

(define (f x) (define k 2) (+ x k))

Let can be used inside another expression:(define x 5)(+ (let ((x 3)) (+ x (* x 10))) 5)

(define (f x) (let ((k 2)) (+ x k)))

6.001 SICP 21

Let*

•Let first evaluates all of the values, then binds each to the variable names.

(define x 0) (define y 0)(let ((x 1) (y 2) (z (+ x y))) z) === 0!

•Let* evaluates and binds each name/value pair in sequence, so that each definition is available to the next(let* ((x 1) (y 2) (z (+ x y))) z) === 3!

6.001 SICP 22

Let/Let*

Write a single function using let or let* to solve (x- 5)^3 + 2(x-5)^2 where x = y – 1 and y = K*z and K = 39.2, where z is given.

(let* ((k 39.2)

(y (* k z))

(x (- y 1))

(a (- x 5)))

(+ (* a a a) (* 2 a a)))

6.001 SICP 23

Remember…

• calendar…