Download lecture5-2001

Lecture 5
Material in the textbook on pages
50-53 (1.2.6)
56-66 (1.3.1, 1.3.2)
Of second edition
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Primality Testing - II
n is a prime iff its only divisors are 1 and n
Iff it has no divisors between 2 and (sqrt n)
(define (divides? a b)
(= (remainder b a) 0))
(define (find-smallest-divisor n i)
(cond ((> i (sqrt n)) n)
((divides? i n) i)
(else (find-smallest-divisor n (+ i 1)))))
(define (prime? n)
(= n (find-smallest-divisor n 2)))
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(Prime? 47)
(= 47 (find-smallest-divisor 47 2))
(= 47 (cond (divides? 2 47) 2)
(else (find-smallest-divisor 47 3))))
(= 47 (find-smallest-divisor 47 3))
(=
(=
(=
(=
47
47
47
47
(find-smallest-divisor
(find-smallest-divisor
(find-smallest-divisor
(find-smallest-divisor
47
47
47
47
4))
5))
6))
7))
(= 47 47)
#t
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Analysis
• Correctness: If n is not a prime, then n=a * b for a,b>1.
Then at least one of them is n.
So n must have a divisor smaller then n.
• Time complexity:
first test - (n)
second test  (n) . For a number n, we test at most n
numbers to see if they divide n.
If n is a 800 digit number, that’s very bad.
Absolutely infeasible.
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The Fermat Primality Test
Fermat’s little theorem:
If n is a prime number then:
an = a (mod n) for every 0 < a < n, integer
The Fermat Test:
Do 400 times:
Pick a random a < n and compute an (mod n)
If  a then for sure n is not a prime.
If all 400 tests passed, declare that n is a prime.
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Computing ab (mod m) fast.
(define (expmod a b m) ; computes ab (mod m)
(cond
((= b 0) 1)
((even? b)
(remainder (expmod
(remainder (* a a) m)
(/ b 2)
m) m))
(else
(remainder (* a (expmod a (- b 1) m))
m))))
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Implementing Fermat test
(define (test a n)(= (expmod a n n) a))
(define (one-test n)
(test (+ 1 (random (- n 1))) n))
(define (many-tests n t); calls one-test t times
(cond ((= t 0) true)
((one-test n) (many-test n (- t 1)))
(else false)))
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Time complexity
To test if n is a prime. We run 400 tests.
Each takes about log(n) multiplcations.
T(n) = O(log n)
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Correctness – I (prime numbers)
Fermat’s theorem: Every prime will always pass
the test.
It therefore follows that if n is a prime then for every
a, test(a n) is true, hence we always pass the test,
And we declare n to be a prime.
For a prime n: We are always right.
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Correctness II – Carmichael numbers.
Definition: A Carmichael number, is a number
such that
•n is Composite, and
•n always passes the test.
For every a, an = a (mod n)
If n is a Carmichael number we always pass
the test, hence we always declare that n is
prime.
For a Carmichael number n: We are always wrong.
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Correctness III – any other number
A fact: If n is not prime and not a Carmichael
number then:
for at least half of the choices of a,
an <> a (mod n).
Hence, if we chose a at random, then with probability half
the test fails and we declare that n is composite.
The probability all 100 tests fail is at most 2-400
For such n: We are wrong with probability at most 2-400
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Correctness
Suppose we do the test t=400 times.
• If n is a prime we are never wrong.
• If n is a Carmichael number, we are always wrong
• If n is a composite number and not a Carmichael number
we are wrong with probability at most 2-400 .
Error probability smaller than the chance the hardware
is faulty.
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A probabilistic algorithm
An algorithm that uses random coins,
and for every input gives the right answer
with a good probability.
Even though Carmichael numbers are very rare
Fermat test is not good enough.
There are inputs on which it is wrong.
There are modifications of Fermat’s test,
that for every input give the right answer,
with a high probability.
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Types so far
• Numbers: 1, 7, 1.2
• Boolean: #t , #f
• Strings: “this is a string”
• Procedures: (< 2 3), (even? 7), (+ 6 3),
(define (f x) (if (< x 0)
“x is negative”
“x is not negative”))
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Procedures have types
A procedure
• may have requirements regarding the
number of its arguments,
• may expect each argument to be of a certain type.
The procedure + expects numbers as its arguments.
Can not be applied on strings.
The procedure < expects at least one argument.
Will not accept strings as arguments.
(< “abc” “xyz”)
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Procedures have types
The type of a procedure is a contract:
• If the operands have the specified types,
the procedure will result in a value of the specified type
• otherwise, its behavior is undefined
– maybe an error, maybe random behavior
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Example
The type of the integer-add procedure is
number, number 
two arguments,
both numbers
(+ 7 “xx”)
number
result value of integer-add
is a number
- causes an error.
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Your turn
• The following expressions evaluate to values of what type?
(lambda (a b c) (if (> a 0) (+ b c) (- b c)))
number, number, number
number
(lambda (p) (if p "hi" "bye"))
Boolean
string
(* 3.14 (* 2 5))
number
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Types (summary)
• type: a set of values
• every value has a type
• procedure types (types which include ) indicate
• number of arguments required
• type of each argument
• type of result of the procedure
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Can procedures get and return procedures?
• Can a procedure return a procedure as its value?
• Can a procedure get a procedure as an argument?
• Can this be useful?
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Consider the following three sums
•1 + 2 + … + 100 = (100 * 101)/2
•1 + 4 + 9 + … + 1002 = (100 * 101 * 201)/6
•1 + 1/32 + 1/52 + … + 1/1012 = p2/8
100
k
k 1
100
k
2
k 1
In mathematics they are all captured
by the notion of a sum:
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k
2
k 1,odd
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Let’s have a look at the three programs
(define (sum-integers a b)
(if (> a b)
k

0
k 1
(+ a (sum-integers (+ 1 a) b))))
100
100
k
2
k 1
(define (sum term a next b)
(define (sum-squares a b)
(if (> a b)
(if (> a b)
0
0
(+ (square a)
(+ (term
(sum-squares
(+ 1 a)a)b))))
(sum term (next a) next b))))
101
k
k 1,odd
2
(define (pi-sum a b)
(if (> a b)
0
(+ (/ 1 (square a))
(pi-sum (+ a 2) b))))
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Let’s check this new procedure out!
(define (sum term a next b)
(if (> a b)
0
(+ (term a)
(sum term (next a) next b))))
What is the type of this procedure?
(number  number, number, number number, number)  number
procedure
procedure
procedure
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Higher order procedures
A higher order procedure:
takes a procedure as an argument or
returns one as a value
Examples:
1. (define (sum-integers1 a b)
(sum (lambda (x) x) a (lambda (x) (+ x 1)) b))
2. (define (sum-squares1 a b)
(sum square a (lambda (x) (+ x 1)) b))
3. (define (pi-sum1 a b)
(sum (lambda (x) (/ 1 (square x))) a (lambda (x) (+ x 2)) b))
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Does it work?
(define (sum term a next b)
(if (> a b) 0
(+ (term a) (sum term (next a) next b))))
100
(sum square 1 (lambda (x) (+ x 1)) 100)
k
2
k 1
(+ (square 1)
(sum square ((lambda (x) (+ x 1)) 1) (lambda (x) (+ x 1)) 100))
(+ 1 (sum square 2 (lambda (x) (+ x 1)) 100))
(+ 1 (+ (square 2) (sum square 3 (lambda (x) (+ x 1)) 100)))
(+ 1 (+ 4 (sum square 3 (lambda (x) (+ x 1)) 100)))
(+ 1 (+ 4 (+ 9 (sum square 4 (lambda (x) (+ x 1)) 100)))
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Integration as a procedure
Integration under a curve f is given roughly by
dx (f(a) + f(a + dx) + f(a + 2dx) + … + f(b))
f
a
dx
b
(define (integral f a b)
(* (sum f a (lambda (x) (+ x dx)) b) dx))
(define dx 1.0e-3)
(define atan (lambda (a)
(integral (lambda (x) (/ 1 (+ 1 (square x)))) 0 a)))
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A moment of reflection
It is nice that procedures can be treated as any other value.
It can help abstract our thinking as with the sum example.
Sometime we would actually send a program rather than execute
E.g., if the data is not under our control.
In fact, it happens quite a lot with web (and other highly
distributed) settings. It is nice we can send a mobile agent free to
Wonder around and execute somewhere else.
What does it cost us?
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The syntactic sugar “Let”
Suppose we wish to implement the function
f(x,y) = x(1+x*y)2 + y(1-y) + (1+x*y)(1-y)
We can also express this as
a = 1+x*y
b = 1-y
f(x,y) = xa2 + yb + ab
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The syntactic sugar “Let”
(define (f x y)
(define (f-helper a b)
(+ (* x (square a))
(* y b)
(* a b)))
(f-helper (+ 1 (* x y))
(- 1 y)))
(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))))
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The syntactic sugar “Let”
(Let ((<var1> <exp1>)
(<var2> <exp2>)
..
(<varn> <expn>))
<body>)
((lambda (<var1> ….. <varn>)
<body>)
<exp1>
<exp2>
…
<expn>)
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