Clear Lowest Bit
by Isai Damier, Android Engineer @ Google

/***************************************************************************
 * Author: Isai Damier
 * Title: Clear Lowest Bit
 * Project: geekviewpoint
 * Package: algorithms
 *
 * Statement:
 *   Clear the least significant (set) bit in the given sequence of bits.
 *
 * Sample Input: integer representation of bit sequence 1111111110
 * Sample Output: 1111111100 after clearing the lowest set bit
 *
 * Technical Details: When counting in binary, the difference
 *    between two consecutive numbers (i.e. n and n-1) is twofold:
 *     1] The least significant set bit in the greater number
 *       is not set in the lesser number.
 *     2] All bits to the left of said least significant set bit
 *        are the same for both numbers.
 *    (7) 0000_0111,  (6) 0000_0110,  (5) 0000_0101,  (4) 0000_0100,
 *    (6) 0000_0110,  (5) 0000_0101,  (4) 0000_0100,  (3) 0000_0011.
 *
 *     Hence, taking the bitwise AND of n and n-1 is sufficient for
 *     clearing the least significant bit.
 *
 **************************************************************************/ 
 public int clearLowestBit(int n) {
  return n & (n - 1);
}

import org.junit.Test;
import static org.junit.Assert.*;

public class BitwiseTest {

  /**
   * Test of clearLowestBit method, of class Bitwise.
   */
  @Test
  public void testClearLowestBit() {
    Bitwise bits = new Bitwise();
    System.out.println(""clearLowestBit"");
    String[] set = {""1111111111"", ""1111111110"", ""1111111100"",
      ""1111111000"", ""1111110000"", ""1111100000"", ""1111000000"",
      ""1110000000"", ""1100000000"", ""1000000000"", ""000000000"",
      ""000000000""};
    int x, y;
    for (int i = 1; i < set.length; i++) {
      x = Integer.parseInt(set[i - 1], 2);
      y = Integer.parseInt(set[i], 2);
      assertEquals(y, bits.clearLowestBit(x));
    }
  }
}