## Euclid, Primes numbers, and a Bit of Algorithm Analysis

Last time we had a look at using the greatest common divisor and Euclid’s algorithm to devise a Las Vegas algorithm for primality testing. We also had a look at how the inclusion exclusion principle helps us determine the proportion of the numbers correctly tested.

However, we finished by asking ourselves if the method is actually efficient compared to, say, simply testing small divisors, one by one. Let us now find out.

A simple way of testing primality against a list of prime numbers is
implemented as follows:

bool alternate_test(unsigned a,const unsigned test_numbers[])
{
for (int i=0;test_numbers[i];i++)
if ((a % test_numbers[i])==0)
return false; // indubitable
return true; // probable.
}


where test_numbers is a null-terminated list of numbers. The procedures tests for the first, and exits if it is divisible by that number. If not, it tests the next one; and so on until 0 is reached, in which case we answer “probably, I don’t know for sure.” (Digression. In the past, I would have insisted for the function to have a single exit point, and would have formulated the loop as dual conditional, with a boolean found or something, and return found; at the end. However, the above violates the single exit point rule, but it is much more readable and its intend is made clearer: you return true only by default, and return false as soon as a match is found.) It is readily verified that the above function returns the same answers as the GCD-based test.

How do they compare speed-wise? Let us repeat last time’s test, where we use the list of factors up to 23 (otherwise it will overflow on 32 bits), and test the complete range from 0 to $2^{32}-1$. The processor is first set at “performance” policy where Speed-Stepping is disabled and the processor runs full blast at about 4.3GHz (with turbo boost). So when we used GCD with 2, we use alternate_test (“loop” in the plot below) with {2,0}, and when we use GCD with $2\times{3}\times{5}\times{7}$ we use the other with {2,3,5,7,0}. The timings are as follows:

Wow.

OK, what’s going on?

Well, let’s analyze. The GCD test uses progressively larger numbers to test against: 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870. If we hypothesize that the CPU does short-cut math depending on the size of the operands, we may suppose that computing % 210 is much faster than computing % 510510. That could account partly for the increase of complexity; while, in contrast, the alternate method uses 23 as the largest test number. But that’s not the major effect, I surmise. The alternate test exits after one test 50% of the times (the even numbers). It exits at the second test one third of one half of the times (multiples of 3 but not 2). The expected complexity is

$C=\displaystyle 1+ (1-\frac{1}{2}) \left(1+(1-\frac{1}{3}) \left(1+(1-\frac{1}{5}) \left(1+(1-\frac{1}{7}) \left(1+ \cdots\right)\right)\right)\right)$,

and this series yields rather small numbers for factors up to 23. (How did I get this? By unrolling how the alternate algorithm proceeds. The algorithm does one iteration in all cases, with one factor. With two factors, it checks the first, and there’s a chance that it doesn’t stop there because the test failed; it continues another iteration. The probability that it continues after testing 2 is $(1-\frac{1}{2})$, because half of the numbers caused the algorithm to stop at the first iteration. Of the $(1-\frac{1}{2})$ remaining, an iteration cost of at least 1 is incurred (testing the next factor) and there’s a probability (compound, that’s why it’s a product form) of $(1-\frac{1}{3})$ that 3 did not divide it; yielding yet another iteration with probability $(1-\frac{1}{2})(1-\frac{1}{3})$ incurring cost of at least 1 to test for five, with the probability of $(1-\frac{1}{5})$ that it continues. You are encouraged to verify it does yield the above expression.) For the GCD method, I expect something like $1+\frac{1}{2}\log_2 d$, where $d$ is the divisor (the complexity of the GCD of two numbers $a$ and $b$ is $O(\log(\min(a,b)))$.

Instrumenting the code allows us to verify this hypothesis:

unsigned gcd(unsigned a, unsigned b, uint64_t & count)
{
count=0;
while (b)
{
count++;
unsigned t=b;
b=a % b;
a=t;
}

return a;
}

////////////////////////////////////////
bool alternative_test(unsigned a,
const unsigned test_numbers[],
uint64_t & count)
{
count=0;
for (int i=0;test_numbers[i];i++)
{
count++;
if ((a % test_numbers[i])==0)
return false;
}
return true;
}


We run the code and gather information. The average number of iterations for each method is as follows:

 Nbfactors GCD Alt 1 1.5 1 2 2.16667 1.5 3 3.36667 1.83333 4 4.81429 2.1 5 6.71385 2.32857 6 8.6944 2.53636 7 10.9449 2.72817 8 13.2974 2.9087 9 15.8115 3.07972

If we plot the above, we get

with the predicted number of iterations of $1+\frac{1}{2}\log_2 d$ for CGD shown in dotted line.

*
* *

At first, testing many divisors at the same time using Euclid’s algorithm for a Las Vegas type algorithm for primality testing seemed like a good idea. It’s elegant. It’s very number-theoretic. However, a bit of algorithmic analysis shows that the number of divisions made by the second algorithm is much smaller; the series $1+(1-\frac{1}{2})(1+(1-\frac{1}{3})(1+\cdots$ diverges, but it grows much slower than $1+\frac{1}{2}\log_2 d$ (as the graph above shows).

### 2 Responses to Euclid, Primes numbers, and a Bit of Algorithm Analysis

1. […] in previous installments, we looked at how to use the euclidean algorithm to devise a Las Vegas-type algorithm for primality […]

2. […] at using the GCD for devising a Las Vegas-type algorithm to test for the primality of a number in a previous entry, we saw that actually testing the number against a list of primes (therefore assuming we have an […]