/// <summary>
/// The Miller primality test uses a witness sub-routine to determine if a value may be prime
/// </summary>
/// <param name="value">The value to be tested for primality.</param>
/// <param name="powersOfTwo">The powers of 2 from the value - 1.</param>
/// <param name="factor">The factor from the value - 1.</param>
/// <param name="witness">The witness value.</param>
/// <returns>False if value is composite. True if value *could* be prime.</returns>
/// <remarks>Although this is a translation of the pseudo code from Wikipedia, for prime number 4447483681 witness 2 will exhaust the loop and claim it's a composite.</remarks>
private static bool WitnessTest(long value, long powersOfTwo, long factor, long witness)
{
// First iteration 0 starting power modulo is (witness ^ ((2 ^ 0) * factor)) % value
// i.e. just (witness ^ factor) % value
var powerModulo = MathAlgorithms.PowerModulo(_base: witness, exponent: factor, modulus: value);
// First condition for being identified as a composite is (witness ^ factor) % value != 1
// If that condition is not met, stop right away.
if (powerModulo == 1 || powerModulo == value - 1)
{
return(true);
}
// value - 1 was decomposed in (2 ^ powersOfTwo) * factor
// Iterate powersOfTwo times to compute all (witness ^ ((2 ^ iteration) * factor)) % value - let's call then f(iteration)
for (var iteration = 0; iteration < powersOfTwo - 1; iteration++)
{
// f(iteration + 1) == (f(iteration) ^ 2) % value
powerModulo = (powerModulo * powerModulo) % value;
if (powerModulo == 1)
{
return(false);
}
if (powerModulo == value - 1)
{
return(true);
}
}
return(false);
}
/// <summary>
/// Calculates the count of unique ways of making a target sum given set of coins denominations
/// </summary>
/// <param name="targetSum">The target sum.</param>
/// <param name="coins">The coins denominations available.</param>
/// <returns>The count of unique ways of making the target sum.</returns>
public static int OrderedCombinationsOfCoins(int targetSum, Stack <int> coins)
{
// When only a single coin denomination remains, check if we can make the target sum with such coins.
if (coins.Count == 1)
{
return((targetSum % coins.Peek()) == 0 ? 1 : 0);
}
// Compute a hash of the target and coins denominations to check if we've learnt this result yet.
var hash = (targetSum.GetHashCode() * 769) ^ coins.ToArray().Aggregate((partialHash, coinToHash) => (partialHash * 769) ^ coinToHash);
if (learntCoinCombinations.TryGetValue(hash, out int countOfOrderedCombinations))
{
return(countOfOrderedCombinations);
}
var coin = coins.Pop();
var runningSum = 0;
while (runningSum < targetSum)
{
countOfOrderedCombinations += MathAlgorithms.OrderedCombinationsOfCoins(targetSum - runningSum, coins);
runningSum += coin;
}
if (runningSum == targetSum)
{
countOfOrderedCombinations++;
}
coins.Push(coin);
learntCoinCombinations.Add(hash, countOfOrderedCombinations);
return(countOfOrderedCombinations);
}
public static bool TryFindFactor(this long value, int offset, ref long factor)
{
long pseudoRandomSequence = 2; long doubleSpeedPseudoRandomSequence = 2; factor = 1;
do
{
pseudoRandomSequence = pseudoRandomSequence.NextPseudoRandomValue(offset: offset, modulus: value);
doubleSpeedPseudoRandomSequence = doubleSpeedPseudoRandomSequence
.NextPseudoRandomValue(offset: offset, modulus: value)
.NextPseudoRandomValue(offset: offset, modulus: value);
factor = MathAlgorithms.GreatestCommonDivisor(Math.Abs(pseudoRandomSequence - doubleSpeedPseudoRandomSequence), value);
}while (factor == 1);
return(!(factor == value));
}
/// <summary>
/// Gets the greatest common divisor using the binary algorithm.
/// </summary>
public static long GreatestCommonDivisor(long leftValue, long rightValue)
{
if (leftValue < 0)
{
throw new ArgumentOutOfRangeException(nameof(leftValue), leftValue, "Only natural numbers are supported.");
}
if (rightValue < 0)
{
throw new ArgumentOutOfRangeException(nameof(rightValue), rightValue, "Only natural numbers are supported.");
}
if (leftValue == rightValue)
{
return(leftValue);
}
// Special case zeroes to ensure loop termination going forward.
if (leftValue == 0)
{
return(rightValue);
}
if (rightValue == 0)
{
return(leftValue);
}
var shiftCount = 0;
// While both values are even, keep binary shifting right i.e. divide by 2 as it is a common divisor.
// Both values are even if their right most bit is zero.
// We store the common factors of 2 so we can apply it back to the GCD found on the remaining values.
// In lower level languages than C# this can be optimized with built-in trailing zero count function.
for (; ((leftValue | rightValue) & 1) == 0; shiftCount++)
{
leftValue >>= 1;
rightValue >>= 1;
}
// If one value is still even, then the other one must be odd, as per the exit condition of the previous iteration.
// When one is even and the other odd, 2 is not a common divisor, hence the even number can be divided by 2 while the other remains the same.
while ((leftValue & 1) == 0)
{
leftValue >>= 1;
}
// From here on, leftValue is always odd.
// Now that we have an odd value, we are looking for the odd number which is the gcd of the remaining values
do
{
// Remove all factors of 2 in rightValue, as we ensured current leftValue is odd, hence 2 is not a common divisor.
while ((rightValue & 1) == 0)
{
rightValue >>= 1;
}
// Here both values are odd.
// Swap if necessary so leftValue <= rightValue.
MathAlgorithms.SwapIfGreater(ref leftValue, ref rightValue);
// Setting rightValue to the difference guarantees it is even, while leftValue remains odd.
// If the difference is zero, we stop as the remaining greatest common divisor is met (and in left value).
// If the difference is not zero, we need to seek the gcd between it and the smallest value (in left value).
rightValue = rightValue - leftValue;
} while (rightValue != 0);
// Re-apply the common factors of 2.
return(leftValue << shiftCount);
}