# SICP exercise 1.28 - miller-rabin primality test part II

This is a follow-up to SICP exercise 1.28 - miller-rabin primality test.

Exercise 1.28:

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 $1 \mod n$.

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 $\mod n$ 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 $1 \mod n$,” that is, a number not equal to $1$ or $n-1$ whose square is equal to $1 \mod n$.

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 $1 \mod n$. (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 $0$.

Changes:

• Avoid redundant modulo computations
• Corrected use of base modulo n in expmod to avoid computing extremely large numbers
• for any number greater than 2, prevent use of 1 as random base in the expmod because I think it will always return true when the random chosen number is 1. Is this needed?

While fixing my previous code, I already saw a procedure with recursive process that fixed all my problems above. I can't really call it my own, so to keep the spirit of the exercise, I rewrote my horrible recursive process code and turned it into a procedure that generates an iterative process.

(define (square x) (* x x))

(define (expmod base exp m)
(define (expmod-iter a base exp)
(define squareMod (remainder (square base) m))
(cond ((= exp 0) a)
((and (not (= base (- m 1)))
(not (= base 1))
(= squareMod 1))
0)
((even? exp)
(expmod-iter
a
squareMod
(/ exp 2)))
(else
(expmod-iter
(remainder (* a base) m)
base
(- exp 1)))))
(expmod-iter
1
base
exp))

(define (miller-rabin-test n)
(define (try-it a)
(if (and (= a 1)
(> n 2))
(miller-rabin-test n)
(= (expmod a (- n 1) n) 1)))
(try-it (+ 1 (random (- n 1)))))

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

Did I improve on my previous code? Did I follow all the requirements in the exercise this time? How can I improve this code and make it faster?

Stylistically, I'd write it this way:

;; expmode b e m = b^e modulo m
(define (expmod b e m)
(define (expmod-iter acc b e)      ; iterative accumulation
(define bsq (remainder (* b b) m))
(cond ((= e 0) acc)
((and (not (= b (- m 1)))
(not (= b 1))
(= bsq 1))
0)
((even? exp)
(expmod-iter acc bsq (/ e 2)))
(else
(expmod-iter (remainder (* acc b) m)
b
(- e 1)))))
(expmod-iter 1 b e))               ; initial accumulator = 1

(define (miller-rabin-test n)
(define (try-it b)
(if (and (= b 1)
(> n 2))
(miller-rabin-test n)
(= (expmod b (- n 1) n)
1)))
(try-it (random 1 n)))

(define (fast-prime? n times)
(or (= times 0)
(and (> times 0)
(miller-rabin-test n)
(fast-prime? n (- times 1)))))

You've corrected the error of not taking the remainder of the square.

There's no need to define a function square just to save one call to *.

You've converted your recursive expmod into iterative code; although I wouldn't call the former "horrible" since its complexity is only logarithmic, so recursion depth is kept in check.

Assuming you're using Racket, (+ 1 (random (- n 1))) can be written simply (random 1 n).

You've kept the awkward encoding for fast-prime? which is more naturally coded directly with the logical connectives.

Will think over some more with regards to its correctness, that you ask about.