Example: Testing for Primality

This section describes two methods for checking the primality of an integer n, one with order of growth , and a ``probabilistic'' algorithm with order of growth . The exercises at the end of this section suggest programming projects based on these algorithms.

Searching for divisors

Since ancient times, mathematicians have been fascinated by problems concerning prime numbers, and many people have worked on the problem of determining ways to test if numbers are prime. One way to test if a number is prime is to find the number's divisors. The following program finds the smallest integral divisor (greater than 1) of a given number n. It does this in a straightforward way, by testing n for divisibility by successive integers starting with 2.

(define (smallest-divisor n)
  (find-divisor n 2))

(define (find-divisor n test-divisor)
  (cond ((> (square test-divisor) n) n)
        ((divides? test-divisor n) test-divisor)
        (else (find-divisor n (+ test-divisor 1)))))

(define (divides? a b)
  (= (remainder b a) 0))

We can test whether a number is prime as follows: n is prime if and only if n is its own smallest divisor.

(define (prime? n)
  (= n (smallest-divisor n)))

The end test for find-divisor is based on the fact that if n is not prime it must have a divisor less than or equal to . This means that the algorithm need only test divisors between 1 and . Consequently, the number of steps required to identify n as prime will have order of growth .

The Fermat test The primality test is based on a result from number theory known as Fermat's Little Theorem.

Fermat's Little Theorem: If n is a prime number and a is any positive integer less than n, then a raised to the nth power is congruent to a modulo n.

(Two numbers are said to be congruent modulo n if they both have the same remainder when divided by n. The remainder of a number a when divided by n is also referred to as the remainder of a modulo n, or simply as a modulo n.)

If n is not prime, then, in general, most of the numbers a< n will not satisfy the above relation. This leads to the following algorithm for testing primality: Given a number n, pick a random number a < n and compute the remainder of aⁿ modulo n. If the result is not equal to a, then n is certainly not prime. If it is a, then chances are good that n is prime. Now pick another random number a and test it with the same method. If it also satisfies the equation, then we can be even more confident that n is prime. By trying more and more values of a, we can increase our confidence in the result. This algorithm is known as the Fermat test.

To implement the Fermat test, we need a procedure that computes the exponential of a number modulo another number:

(define (expmod base exp m)
  (cond ((= exp 0) 1)
        ((even? exp)
         (remainder (square (expmod base (/ exp 2) m))
                    m))
        (else
         (remainder (* base (expmod base (- exp 1) m))
                    m))))        

This is very similar to the fast-expt procedure of section . It uses successive squaring, so that the number of steps grows logarithmically with the exponent.

The Fermat test is performed by choosing at random a number a between 1 and n-1 inclusive and checking whether the remainder modulo n of the nth power of a is equal to a. The random number a is chosen using the procedure random, which we assume is included as a primitive in Scheme. Random returns a nonnegative integer less than its integer input. Hence, to obtain a random number between 1 and n-1, we call random with an input of n-1 and add 1 to the result:

(define (fermat-test n)
  (define (try-it a)
    (= (expmod a n n) a))
  (try-it (+ 1 (random (- n 1)))))

The following procedure runs the test a given number of times, as specified by a parameter. Its value is true if the test succeeds every time, and false otherwise.

(define (fast-prime? n times)
  (cond ((= times 0) true)
        ((fermat-test n) (fast-prime? n (- times 1)))
        (else false)))

Probabilistic methods

The Fermat test differs in character from most familiar algorithms, in which one computes an answer that is guaranteed to be correct. Here, the answer obtained is only probably correct. More precisely, if n ever fails the Fermat test, we can be certain that n is not prime. But the fact that n passes the test, while an extremely strong indication, is still not a guarantee that n is prime. What we would like to say is that for any number n, if we perform the test enough times and find that n always passes the test, then the probability of error in our primality test can be made as small as we like.

Unfortunately, this assertion is not quite correct. There do exist numbers that fool the Fermat test: numbers n that are not prime and yet have the property that aⁿ is congruent to a modulo n for all integers a < n. Such numbers are extremely rare, so the Fermat test is quite reliable in practice. There are variations of the Fermat test that cannot be fooled. In these tests, as with the Fermat method, one tests the primality of an integer n by choosing a random integer a<n and checking some condition that depends upon n and a. (See exercise  for an example of such a test.) On the other hand, in contrast to the Fermat test, one can prove that, for any n, the condition does not hold for most of the integers a<n unless n is prime. Thus, if n passes the test for some random choice of a, the chances are better than even that n is prime. If n passes the test for two random choices of a, the chances are better than 3 out of 4 that n is prime. By running the test with more and more randomly chosen values of a we can make the probability of error as small as we like.

The existence of tests for which one can prove that the chance of error becomes arbitrarily small has sparked interest in algorithms of this type, which have come to be known as probabilistic algorithms. There is a great deal of research activity in this area, and probabilistic algorithms have been fruitfully applied to many fields.

Exercise. Use the smallest-divisor procedure to find the smallest divisor of each of the following numbers: 199, 1999, 19999.

Exercise. Most Lisp implementations include a primitive called runtime that returns an integer that specifies the amount of time the system has been running (measured, for example, in microseconds). The following timed-prime-test procedure, when called with an integer n, prints n and checks to see if n is prime. If n is prime, the procedure prints three asterisks followed by the amount of time used in performing the test.

(define (timed-prime-test n)
  (newline)
  (display n)
  (start-prime-test n (runtime)))

(define (start-prime-test n start-time)
  (if (prime? n)
      (report-prime (- (runtime) start-time))))

(define (report-prime elapsed-time)
  (display " *** ")
  (display elapsed-time))

Using this procedure, write a procedure search-for-primes that checks the primality of consecutive odd integers in a specified range. Use your procedure to find the three smallest primes larger than 1000; larger than 10,000; larger than 100,000; larger than 1,000,000. Note the time needed to test each prime. Since the testing algorithm has order of growth of , you should expect that testing for primes around 10,000 should take about times as long as testing for primes around 1000. Do your timing data bear this out? How well do the data for 100,000 and 1,000,000 support the prediction? Is your result compatible with the notion that programs on your machine run in time proportional to the number of steps required for the computation?  

Exercise. The smallest-divisor procedure shown at the start of this section does lots of needless testing: After it checks to see if the number is divisible by 2 there is no point in checking to see if it is divisible by any larger even numbers. This suggests that the values used for test-divisor should not be 2, 3, 4, 5, 6, ..., but rather 2, 3, 5, 7, 9, .... To implement this change, define a procedure next that returns 3 if its input is equal to 2 and otherwise returns its input plus 2. Modify the smallest-divisor procedure to use (next test-divisor) instead of (+ test-divisor 1). With timed-prime-test incorporating this modified version of smallest-divisor, run the test for each of the 12 primes found in exercise . Since this modification halves the number of test steps, you should expect it to run about twice as fast. Is this expectation confirmed? If not, what is the observed ratio of the speeds of the two algorithms, and how do you explain the fact that it is different from 2?

Exercise. Modify the timed-prime-test procedure of exercise  to use fast-prime? (the Fermat method), and test each of the 12 primes you found in that exercise. Since the Fermat test has growth, how would you expect the time to test primes near 1,000,000 to compare with the time needed to test primes near 1000? Do your data bear this out? Can you explain any discrepancy you find?  

Exercise. Alyssa P. Hacker complains that we went to a lot of extra work in writing expmod. After all, she says, since we already know how to compute exponentials, we could have simply written

(define (expmod base exp m)
  (remainder (fast-expt base exp) m))

Is she correct? Would this procedure serve as well for our fast prime tester? Explain.  

Exercise. Louis Reasoner is having great difficulty doing exercise . His fast-prime? test seems to run more slowly than his prime? test. Louis calls his friend Eva Lu Ator over to help. When they examine Louis's code, they find that he has rewritten the expmod procedure to use an explicit multiplication, rather than calling square:

(define (expmod base exp m)
  (cond ((= exp 0) 1)
        ((even? exp)
         (remainder (* (expmod base (/ exp 2) m)
                       (expmod base (/ exp 2) m))
                    m))
        (else
         (remainder (* base (expmod base (- exp 1) m))
                    m))))

``I don't see what difference that could make,'' says Louis. ``I do.'' says Eva. ``By writing the procedure like that, you have transformed the process into a process.'' Explain.

Exercise. Demonstrate that the Carmichael numbers listed in footnote  really do fool the Fermat test. That is, write a procedure that takes an integer n and tests whether aⁿ is congruent to a modulo n for every a<n, and try your procedure on the given Carmichael numbers.

Exercise. One variant of the Fermat test that cannot be fooled is called the Miller-Rabin test (Miller 1976; Rabin 1980). This starts from an alternate form of Fermat's Little Theorem, which states that if n is a prime number and a is any positive integer less than n, then a raised to the (n-1)st power is congruent to 1 modulo n. To test the primality of a number n by the Miller-Rabin test, we pick a random number a<n and raise a to the (n-1)st power modulo n using the expmod procedure. However, whenever we perform the squaring step in expmod, we check to see if we have discovered a ``nontrivial square root of 1 modulo n,'' that is, a number not equal to 1 or n-1 whose square is equal to 1 modulo n. It is possible to prove that if such a nontrivial square root of 1 exists, then n is not prime. It is also possible to prove that if n is an odd number that is not prime, then, for at least half the numbers a<n, computing a^n-1 in this way will reveal a nontrivial square root of 1 modulo n. (This is why the Miller-Rabin test cannot be fooled.) Modify the expmod procedure to signal if it discovers a nontrivial square root of 1, and use this to implement the Miller-Rabin test with a procedure analogous to fermat-test. Check your procedure by testing various known primes and non-primes. Hint: One convenient way to make expmod signal is to have it return 0.