Material in the textbook on pages Lecture 5 Material in the textbook on pages 50-53 (1.2.6) 56-66 (1.3.1, 1.3.2) Of second edition מבוא מורחב

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))) מבוא מורחב

(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 (find-smallest-divisor 47 4)) (= 47 (find-smallest-divisor 47 5)) (= 47 (find-smallest-divisor 47 6)) (= 47 (find-smallest-divisor 47 7)) (= 47 47) #t מבוא מורחב

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. מבוא מורחב

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. מבוא מורחב

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))))

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))) מבוא מורחב

Time complexity To test if n is a prime. We run 400 tests. Each takes about log(n) multiplcations. T(n) = O(log n) מבוא מורחב

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. מבוא מורחב

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. מבוא מורחב

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 מבוא מורחב

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. מבוא מורחב

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. We want an algorithm that is good for every input, because the input can be chosen by an adversary, and he will chose the wrost input for us. We therefore can not assume any distribution over the inputs, certainly not that they are chosen at random. Given the input, however, we chose random coins, and we control the probability distribution we use. מבוא מורחב

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”)) מבוא מורחב

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”) מבוא מורחב

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 מבוא מורחב

Example The type of the integer-add procedure is two arguments, both numbers result value of integer-add is a number number, number  number (+ 7 “xx”) - causes an error. מבוא מורחב

number, number, number number Your turn The following expressions evaluate to values of what type? (lambda (a b c) (if (> a 0) (+ b c) (- b c))) (lambda (p) (if p "hi" "bye")) (* 3.14 (* 2 5)) number, number, number number Boolean string number

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 מבוא מורחב

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? מבוא מורחב

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 In mathematics they are all captured by the notion of a sum: מבוא מורחב

Let’s have a look at the three programs (define (pi-sum a b) (if (> a b) 0 (+ (/ 1 (square a)) (pi-sum (+ a 2) b)))) (define (sum-integers a b) (if (> a b) 0 (+ a (sum-integers (+ 1 a) b)))) (define (sum-squares a b) (if (> a b) 0 (+ (square a) (sum-squares (+ 1 a) b)))) (define (sum term a next b) (if (> a b) (+ (term a) (sum term (next a) next b)))) מבוא מורחב

Let’s check this new procedure out! (define (sum term a next b) (if (> a b) (+ (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 מבוא מורחב

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)) מבוא מורחב

Does it work? (define (sum term a next b) (if (> a b) 0 (+ (term a) (sum term (next a) next b)))) (sum square 1 (lambda (x) (+ x 1)) 100) (+ (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))) מבוא מורחב

Integration as a procedure Integration under a curve f is given roughly by dx (f(a) + f(a + dx) + f(a + 2dx) + … + f(b)) a b dx f (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)))

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 it. 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? מבוא מורחב

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

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))))

The syntactic sugar “Let” (Let ((<var1> <exp1>) (<var2> <exp2>) .. (<varn> <expn>)) <body>) ((lambda (<var1> ….. <varn>) <body>) <exp1> <exp2> … <expn>)