/* Compute the greatest common divisor of positive integers */
int gcd(int x, int y) {
  int g;
  g = y;
  while (x > 0) {
    g = x;
    x = y % x;
    y = g;
  }
  return g;
}


-- main.go --
package main

import (
	"fmt"
)

func main() {
	// Call our gcd() function
	x := 42
	y := 105
	g := Gcd(x, y)
	fmt.Println("The gcd of", x, "and", y, "is", g)

	}
	var Y *Int
	if y != nil {
		Y = new(Int)
	}

	D := new(Int).GCD(X, Y, a, b)
	if D.Cmp(d) != 0 {
		t.Errorf("GCD(%s, %s, %s, %s): got d = %s, want %s", x, y, a, b, D, d)
	}
	if x != nil && X.Cmp(x) != 0 {
		t.Errorf("GCD(%s, %s, %s, %s): got x = %s, want %s", x, y, a, b, X, x)
	}
	if y != nil && Y.Cmp(y) != 0 {
		t.Errorf("GCD(%s, %s, %s, %s): got y = %s, want %s", x, y, a, b, Y, y)
	}

    int bTwos = Integer.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

      for (int b : POSITIVE_INTEGER_CANDIDATES) {
        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b)));
      }
    }
  }

  public void testGCDZero() {
    for (int a : POSITIVE_INTEGER_CANDIDATES) {
      assertEquals(a, IntMath.gcd(a, 0));
      assertEquals(a, IntMath.gcd(0, a));
    }
    assertEquals(0, IntMath.gcd(0, 0));
  }

  public void testGCDNegativePositiveThrows() {

1024 drives. In this scenario 16 becomes the erasure set size. This is decided based on the greatest common divisor (GCD) of acceptable erasure set sizes ranging from *4 to 16*.

- *If total drives has many common divisors the algorithm chooses the minimum amounts of erasure sets possible for a erasure set size of any N*.  In the example with 1024 drives - 4, 8, 16 are GCD factors. With 16 drives we get a total of 64 possible sets, with 8 drives we get a total of 128 possible sets, with 4...

        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b)));
      }
    }
  }

  @GwtIncompatible // TODO
  public void testGCDZero() {
    for (long a : POSITIVE_LONG_CANDIDATES) {
      assertEquals(a, LongMath.gcd(a, 0));
      assertEquals(a, LongMath.gcd(0, a));
    }
    assertEquals(0, LongMath.gcd(0, 0));
  }

  @GwtIncompatible // TODO

    int bTwos = Integer.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b)));
      }
    }
  }

  @GwtIncompatible // TODO
  public void testGCDZero() {
    for (long a : POSITIVE_LONG_CANDIDATES) {
      assertEquals(a, LongMath.gcd(a, 0));
      assertEquals(a, LongMath.gcd(0, a));
    }
    assertEquals(0, LongMath.gcd(0, 0));
  }

  @GwtIncompatible // TODO

    int bTwos = Long.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

    int bTwos = Long.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from