SICP - Solution: Exercise 1.24

# SICP - Solution: Exercise 1.24

## Exercise 1.24 #

Modify the timed-prime-test procedure of Exercise 1.22 to use fast-prime? (the Fermat method), and test each of the 12 primes you found in that exercise. Since the Fermat test has ${\mathrm\Theta(\log n)}$ 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?

## Solution #

(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))))

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

(define (fast-prime? n times)
(cond ((= times 0) #t)
((fermat-test n)
(fast-prime? n (- times 1)))
(else #f)))


I choosed to run each test on 100 random numbers, but this is somewhat a guessed value:

(define (start-prime-test n start-time)
(if (fast-prime? n 100)
(report-prime (- (runtime) start-time))
"nothing"))


Here too, in order to increase precision of the execution time measurement, all computation are run 1000 times on each prime number for each of the algorithm.

I ran in a few issues with the random number generator for the largest prime in my table. I just removed them as it doesn’t change the conclusion.

### DrRacket #

log(prime) prime time prime? (µs) time fast-prime? (µs)
3 1,009 2.8029 175.80
3 1,013 2.3579 160.85
3 1,019 2.2888 179.21
4 10,007 8.1271 218.31
4 10,009 7.7409 193.58
4 10,037 7.6669 241.30
5 100,003 25.369 238.16
5 100,019 24.364 279.69
5 100,043 24.239 240.71
6 1,000,003 76.642 388.47
6 1,000,033 91.104 277.81
6 1,000,037 83.062 304.60
7 10,000,019 245.59 364.91
7 10,000,079 253.24 372.02
7 10,000,103 264.88 358.31
8 100,000,007 855.93 386.96
8 100,000,037 974.68 392.14
8 100,000,039 829.26 401.62
9 1,000,000,007 2588.9 438.92
9 1,000,000,009 2728.9 425.29
9 1,000,000,021 2848.4 455.42

Which can be summarized:

log(prime) average time prime? (µs) delta for 10x (µs) average time fast-prime? (µs) delta for 10x (µs)
3 2,4832 171,95
4 7,8450 5,3618 217,73 45,7
5 24,658 16,813 252,85 35,1
6 83,603 58,945 323,63 70,7
7 254,57 170,97 365,08 41,4
8 886,62 632,04 393,57 28,4
9 2722,1 1835,5 439,88 46,3

### Chicken Scheme (compiled) #

log(prime) prime time prime? (µs) time fast-prime? (µs)
3 1,009 4.0 172.0
3 1,013 3.0 180.0
3 1,019 4.0 193.0
4 10,007 11.0 221.0
4 10,009 13.0 218.0
4 10,037 10.0 224.0
5 100,003 35.0 256.0
5 100,019 35.0 261.0
5 100,043 38.0 276.0
6 1,000,003 113.0 297.0
6 1,000,033 111.0 296.0
6 1,000,037 114.0 311.0
7 10,000,019 353.0 355.0
7 10,000,079 355.0 381.0
7 10,000,103 362.0 384.0
8 100,000,007 1157.0 435.0
8 100,000,037 1126.0 444.0
8 100,000,039 1126.0 450.0
9 1,000,000,007 3516.0 462.0
9 1,000,000,009 3744.0 477.0
9 1,000,000,021 3641.0 499.0

Which can be summarized:

log(prime) average time prime? (µs) time multiplication for 10x average time fast-prime? (µs) time increases for 10x (µs)
3 3,67 181,67
4 11,33 3,09 221 39,3
5 36 3,18 264,33 43,3
6 112,67 3,13 301,33 37,0
7 356,67 3,17 373,33 72,0
8 1136,33 3,19 443 69,7
9 3633,67 3,20 479,33 36,3

### Conclusion #

When the size of prime is 10 times larger, prime? require 3 times more computation.

When the size of prime is 10 times larger, fast-prime? require a constant increase of 50µs. ${\mathrm\Theta(\log n)}$ growth means that when the prime are ten times larger, the time will increase by a constant amount ${\log(100)-\log(10)=1}$. This is quite clear here.

Although that for smaller number prime? is faster, fast-prime? become much faster with larger and larger prime.