// We know a sieve is pretty fast, especially only up to a million. And it can be modified easily
// to allow retrieving the prime factors of a given number. So combine that with Euler's product
// formula (see https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler.27s_product_formula).
// That's a lot better than my first idea for using a sieve, which was getting the prime factorization
// of every number below n and comparing it to n's prime factorization.
public static int Solve(int n)
{
int[] distinctPrimeFactors = _factorizer.GetDistinctPrimeFactors(n).ToArray();
double totient = n;
foreach (int primeFactor in distinctPrimeFactors)
{
totient *= (1 - 1 / (double)primeFactor);
}
return((int)Math.Round(totient));
}
public void GetDistinctPrimeFactors_AgreesWithNaiveFactorizer()
{
var sieveFactorizer = new SieveOfEratosthenesFactorizer(3481);
var trialDivisionFactorizer = new TrialDivisionFactorizer(3481);
for (int n = 0; n <= 3481; ++n)
{
CollectionAssert.AreEquivalent(
NaivePrimeDeciderProviderFactorizer.GetDistinctPrimeFactors(n).ToArray(),
sieveFactorizer.GetDistinctPrimeFactors(n).ToArray());
CollectionAssert.AreEquivalent(
NaivePrimeDeciderProviderFactorizer.GetDistinctPrimeFactors(n).ToArray(),
trialDivisionFactorizer.GetDistinctPrimeFactors(n).ToArray());
}
}
public void GetDistinctPrimeFactors_AgreesWithKnownOutput()
{
var sieveFactorizer = new SieveOfEratosthenesFactorizer(1000);
var trialDivisionFactorizer = new TrialDivisionFactorizer(1000);
foreach (var numberPrimeFactorsPair in _numberPrimeFactorsPairs)
{
int number = numberPrimeFactorsPair.Item1;
int[] distinctPrimeFactors = numberPrimeFactorsPair.Item2.Distinct().ToArray();
CollectionAssert.AreEquivalent(distinctPrimeFactors, sieveFactorizer.GetDistinctPrimeFactors(number).ToArray());
CollectionAssert.AreEquivalent(distinctPrimeFactors, trialDivisionFactorizer.GetDistinctPrimeFactors(number).ToArray());
CollectionAssert.AreEquivalent(distinctPrimeFactors, NaivePrimeDeciderProviderFactorizer.GetDistinctPrimeFactors(number).ToArray());
}
}
