/// <summary>
/// Creates an instance whose range is the k-element subsets of
/// {0, ..., n - 1} represented as <c>int[]</c> arrays.
/// <para>
/// If the <c>iterationOrder</c> argument is set to
/// <see cref="IterationOrder.LEXICOGRAPHIC"/>, the arrays returned by the
/// <see cref="iterator()"/> are sorted in descending order and
/// they are visited in lexicographic order with significance from
/// right to left.
/// For example, <c>new Combinations(4, 2).iterator()</c> returns
/// an iterator that will generate the following sequence of arrays
/// on successive calls to
/// <c>next()</c>:<para/>
/// <c>[0, 1], [0, 2], [1, 2], [0, 3], [1, 3], [2, 3]</c>
/// </para>
/// If <c>k == 0</c> an iterator containing an empty array is returned;
/// if <c>k == n</c> an iterator containing [0, ..., n - 1] is returned.
///
/// </summary>
/// <param name="n">Size of the set from which subsets are selected.</param>
/// <param name="k">Size of the subsets to be enumerated.></param>
/// <param name="iterationOrder">Specifies the <see cref="iterator()"/>.</param>
/// <exception cref="NotPositiveException"> if <c>n < 0</c>.</exception>
/// <exception cref="NumberIsTooLargeException"> if <c>k > n</c>.</exception>
private Combinations(int n, int k, IterationOrder iterationOrder)
{
CombinatoricsUtils.checkBinomial(n, k);
this.n = n;
this.k = k;
this.iterationOrder = iterationOrder;
}
/// <summary>
/// Returns a <c>double</c> representation of the <a
/// href="http://mathworld.wolfram.com/BinomialCoefficient.html"> Binomial
/// Coefficient</a>, "<c>n choose k</c>", the number of
/// <c>k</c>-element subsets that can be selected from an
/// <c>n</c>-element set.
/// <para>
/// Preconditions:
/// <list type="bullet">
/// <item> <c>0 <= k <= n</c> (otherwise
/// <c>IllegalArgumentException</c> is thrown)</item>
/// <item> The result is small enough to fit into a <c>double</c>. The
/// largest value of <c>n</c> for which all coefficients are <
/// Double.MaxValue is 1029. If the computed value exceeds Double.MaxValue,
/// Double.PositiveInvinifty is returned</item>
/// </list></para>
/// </summary>
/// <param name="n">the size of the set</param>
/// <param name="k">the size of the subsets to be counted</param>
/// <returns><c>n choose k</c></returns>
/// <exception cref="NotPositiveException"> if <c>n < 0</c>.</exception>
/// <exception cref="NumberIsTooLargeException"> if <c>k > n</c>.</exception>
/// <exception cref="MathArithmeticException"> if the result is too large to be
/// represented by a double.</exception>
public static double binomialCoefficientDouble(int n, int k)
{
CombinatoricsUtils.checkBinomial(n, k);
if ((n == k) || (k == 0))
{
return(1d);
}
if ((k == 1) || (k == n - 1))
{
return(n);
}
if (k > n / 2)
{
return(binomialCoefficientDouble(n, n - k));
}
if (n < 67)
{
return(binomialCoefficient(n, k));
}
double result = 1d;
for (int i = 1; i <= k; i++)
{
result *= (double)(n - k + i) / (double)i;
}
return(FastMath.floor(result + 0.5));
}
/// <summary>
/// Returns the natural <c>log</c> of the <a
/// href="http://mathworld.wolfram.com/BinomialCoefficient.html"> Binomial
/// Coefficient</a>, "<c>n choose k</c>", the number of
/// <c>k</c>-element subsets that can be selected from an
/// <c>n</c>-element set.
/// <para>
/// Preconditions:
/// <list type="bullet">
/// <item> <c>0 <= k <= n</c> (otherwise
/// <c>IllegalArgumentException</c> is thrown)</item>
/// </list></para>
/// </summary>
/// <param name="n">the size of the set</param>
/// <param name="k">the size of the subsets to be counted</param>
/// <returns><c>n choose k</c></returns>
/// <exception cref="NotPositiveException"> if <c>n < 0</c>.</exception>
/// <exception cref="NumberIsTooLargeException"> if <c>k > n</c>.</exception>
/// <exception cref="MathArithmeticException"> if the result is too large to be
/// represented by a double.</exception>
public static double binomialCoefficientLog(int n, int k)
{
CombinatoricsUtils.checkBinomial(n, k);
if ((n == k) || (k == 0))
{
return(0);
}
if ((k == 1) || (k == n - 1))
{
return(FastMath.log(n));
}
/*
* For values small enough to do exact integer computation,
* return the log of the exact value
*/
if (n < 67)
{
return(FastMath.log(binomialCoefficient(n, k)));
}
/*
* Return the log of binomialCoefficientDouble for values that will not
* overflow binomialCoefficientDouble
*/
if (n < 1030)
{
return(FastMath.log(binomialCoefficientDouble(n, k)));
}
if (k > n / 2)
{
return(binomialCoefficientLog(n, n - k));
}
/*
* Sum logs for values that could overflow
*/
double logSum = 0;
// n!/(n-k)!
for (int i = n - k + 1; i <= n; i++)
{
logSum += FastMath.log(i);
}
// divide by k!
for (int i = 2; i <= k; i++)
{
logSum -= FastMath.log(i);
}
return(logSum);
}
/// <summary>
/// Compute n!, the <a href="http://mathworld.wolfram.com/Factorial.html">
/// factorial</a> of <c>n</c> (the product of the numbers 1 to n), as a
/// <c>double</c>.
/// The result should be small enough to fit into a <c>double</c>: The
/// largest <c>n</c> for which <c>n! < Double.MaxValue</c> is 170.
/// If the computed value exceeds <c>Double.MaxValue</c>,
/// <c>Double.PositiveInfinity</c> is returned.
/// </summary>
/// <param name="n">Argument.</param>
/// <returns><c>n!</c></returns>
/// <exception cref="NotPositiveException"> if <c>n < 0</c>.</exception>
public static double factorialDouble(int n)
{
if (n < 0)
{
throw new NotPositiveException <Int32>(new LocalizedFormats("FACTORIAL_NEGATIVE_PARAMETER"), n);
}
if (n < 21)
{
return(FACTORIALS[n]);
}
return(FastMath.floor(FastMath.exp(CombinatoricsUtils.factorialLog(n)) + 0.5));
}
public static long stirlingS2(int n, int k)
{
return(CombinatoricsUtils.stirlingS2(n, k));
}
public static long binomialCoefficient(int n, int k)
{
return(CombinatoricsUtils.binomialCoefficient(n, k));
}
public static double factorialLog(int n)
{
return(CombinatoricsUtils.factorialLog(n));
}
public static long factorial(int n)
{
return(CombinatoricsUtils.factorial(n));
}
public static double binomialCoefficientLog(int n, int k)
{
return(CombinatoricsUtils.binomialCoefficientLog(n, k));
}
/// <summary>
/// Returns an exact representation of the <a
/// href="http://mathworld.wolfram.com/BinomialCoefficient.html"> Binomial
/// Coefficient</a>, "<c>n choose k</c>", the number of
/// <c>k</c>-element subsets that can be selected from an
/// <c>n</c>-element set.
/// <para>
/// Preconditions:
/// <list type="bullet">
/// <item> <c>0 <= k <= n</c> (otherwise
/// <c>MathIllegalArgumentException</c> is thrown)</item>
/// <item> The result is small enough to fit into a <c>long</c>. The
/// largest value of <c>n</c> for which all coefficients are
/// <c> < Int64.MaxValue</c> is 66. If the computed value exceeds
/// <c>Int64.MaxValue</c> an <c>ArithMeticException</c> is
/// thrown.</item>
/// </list></para>
/// </summary>
/// <param name="n">the size of the set</param>
/// <param name="k">the size of the subsets to be counted</param>
/// <returns><c>n choose k</c></returns>
/// <exception cref="NotPositiveException"> if <c>n < 0</c>.</exception>
/// <exception cref="NumberIsTooLargeException"> if <c>k > n</c>.</exception>
/// <exception cref="MathArithmeticException"> if the result is too large to be
/// represented by a long integer.</exception>
public static long binomialCoefficient(int n, int k)
{
CombinatoricsUtils.checkBinomial(n, k);
if ((n == k) || (k == 0))
{
return(1);
}
if ((k == 1) || (k == n - 1))
{
return(n);
}
// Use symmetry for large k
if (k > n / 2)
{
return(binomialCoefficient(n, n - k));
}
// We use the formula
// (n choose k) = n! / (n-k)! / k!
// (n choose k) == ((n-k+1)*...*n) / (1*...*k)
// which could be written
// (n choose k) == (n-1 choose k-1) * n / k
long result = 1;
if (n <= 61)
{
// For n <= 61, the naive implementation cannot overflow.
int i = n - k + 1;
for (int j = 1; j <= k; j++)
{
result = result * i / j;
i++;
}
}
else if (n <= 66)
{
// For n > 61 but n <= 66, the result cannot overflow,
// but we must take care not to overflow intermediate values.
int i = n - k + 1;
for (int j = 1; j <= k; j++)
{
// We know that (result * i) is divisible by j,
// but (result * i) may overflow, so we split j:
// Filter out the gcd, d, so j/d and i/d are integer.
// result is divisible by (j/d) because (j/d)
// is relative prime to (i/d) and is a divisor of
// result * (i/d).
long d = ArithmeticUtils.gcd(i, j);
result = (result / (j / d)) * (i / d);
i++;
}
}
else
{
// For n > 66, a result overflow might occur, so we check
// the multiplication, taking care to not overflow
// unnecessary.
int i = n - k + 1;
for (int j = 1; j <= k; j++)
{
long d = ArithmeticUtils.gcd(i, j);
result = ArithmeticUtils.mulAndCheck(result / (j / d), i / d);
i++;
}
}
return(result);
}
