Example #1
0
  /**
   * Returns {@code n!}, that is, the product of the first {@code n} positive integers, or {@code 1}
   * if {@code n == 0}.
   *
   * <p><b>Warning:</b> the result takes <i>O(n log n)</i> space, so use cautiously.
   *
   * <p>This uses an efficient binary recursive algorithm to compute the factorial with balanced
   * multiplies. It also removes all the 2s from the intermediate products (shifting them back in at
   * the end).
   *
   * @throws IllegalArgumentException if {@code n < 0}
   */
  public static BigInteger factorial(int n) {
    checkNonNegative("n", n);

    // If the factorial is small enough, just use LongMath to do it.
    if (n < LongMath.factorials.length) {
      return BigInteger.valueOf(LongMath.factorials[n]);
    }

    // Pre-allocate space for our list of intermediate BigIntegers.
    int approxSize = IntMath.divide(n * IntMath.log2(n, CEILING), Long.SIZE, CEILING);
    ArrayList<BigInteger> bignums = new ArrayList<BigInteger>(approxSize);

    // Start from the pre-computed maximum long factorial.
    int startingNumber = LongMath.factorials.length;
    long product = LongMath.factorials[startingNumber - 1];
    // Strip off 2s from this value.
    int shift = Long.numberOfTrailingZeros(product);
    product >>= shift;

    // Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;
    int bits = LongMath.log2(startingNumber, FLOOR) + 1;
    // Check for the next power of two boundary, to save us a CLZ operation.
    int nextPowerOfTwo = 1 << (bits - 1);

    // Iteratively multiply the longs as big as they can go.
    for (long num = startingNumber; num <= n; num++) {
      // Check to see if the floor(log2(num)) + 1 has changed.
      if ((num & nextPowerOfTwo) != 0) {
        nextPowerOfTwo <<= 1;
        bits++;
      }
      // Get rid of the 2s in num.
      int tz = Long.numberOfTrailingZeros(num);
      long normalizedNum = num >> tz;
      shift += tz;
      // Adjust floor(log2(num)) + 1.
      int normalizedBits = bits - tz;
      // If it won't fit in a long, then we store off the intermediate product.
      if (normalizedBits + productBits >= Long.SIZE) {
        bignums.add(BigInteger.valueOf(product));
        product = 1;
        productBits = 0;
      }
      product *= normalizedNum;
      productBits = LongMath.log2(product, FLOOR) + 1;
    }
    // Check for leftovers.
    if (product > 1) {
      bignums.add(BigInteger.valueOf(product));
    }
    // Efficiently multiply all the intermediate products together.
    return listProduct(bignums).shiftLeft(shift);
  }
Example #2
0
  /**
   * Returns {@code n} choose {@code k}, also known as the binomial coefficient of {@code n} and
   * {@code k}, that is, {@code n! / (k! (n - k)!)}.
   *
   * <p><b>Warning:</b> the result can take as much as <i>O(k log n)</i> space.
   *
   * @throws IllegalArgumentException if {@code n < 0}, {@code k < 0}, or {@code k > n}
   */
  public static BigInteger binomial(int n, int k) {
    checkNonNegative("n", n);
    checkNonNegative("k", k);
    checkArgument(k <= n, "k (%s) > n (%s)", k, n);
    if (k > (n >> 1)) {
      k = n - k;
    }
    if (k < LongMath.biggestBinomials.length && n <= LongMath.biggestBinomials[k]) {
      return BigInteger.valueOf(LongMath.binomial(n, k));
    }
    BigInteger accum = BigInteger.ONE;
    long numeratorAccum = n;
    long denominatorAccum = 1;
    int bits = LongMath.log2(n, RoundingMode.CEILING);
    int numeratorBits = bits;
    for (int i = 1; i < k; i++) {
      int p = n - i;
      int q = i + 1;

      // log2(p) >= bits - 1, because p >= n/2
      if (numeratorBits + bits >= Long.SIZE - 1) {
        // The numerator is as big as it can get without risking overflow.
        // Multiply numeratorAccum / denominatorAccum into accum.
        accum =
            accum
                .multiply(BigInteger.valueOf(numeratorAccum))
                .divide(BigInteger.valueOf(denominatorAccum));
        numeratorAccum = p;
        denominatorAccum = q;
        numeratorBits = bits;
      } else {
        // We can definitely multiply into the long accumulators without overflowing them.
        numeratorAccum *= p;
        denominatorAccum *= q;
        numeratorBits += bits;
      }
    }
    return accum
        .multiply(BigInteger.valueOf(numeratorAccum))
        .divide(BigInteger.valueOf(denominatorAccum));
  }