xor without xor

Author: ZapDos7

README.md

@@ -58,7 +58,7 @@
 - [x] [Bit Hacks — Part 3 (Playing with rightmost set bit of a number)](http://www.techiedelight.com/bit-hacks-part-3-playing-rightmost-set-bit-number/)
 - [x] [Bit Hacks — Part 4 (Playing with letters of English alphabet)](http://www.techiedelight.com/bit-hacks-part-4-playing-letters-english-alphabet/)
 - [x] [Bit Hacks — Part 5 (Find absolute value of an integer without branching)](http://www.techiedelight.com/bit-hacks-part-5-find-absolute-value-integer-without-branching/)
-- [ ] [Bit Hacks — Part 6 (Random Problems)](http://www.techiedelight.com/bit-hacks-part-6-random-problems/)
+- [x] [Bit Hacks — Part 6 (Random Problems)](http://www.techiedelight.com/bit-hacks-part-6-random-problems/)
 
 ## Algorithms
 

src/binary/xorFinder/FindXORWithoutXOR.java

@@ -0,0 +1,30 @@
+package binary.xorFinder;
+
+// Given two integers, find their XOR without using the XOR operator.
+// (x | y) - (x & y) is equivalent to x ^ y
+// XOR sets the bits in either num
+// then unsets the ones both have
+
+// e.g. 01000001 XOR 01010000 :
+// set all: 01010001
+//           ^ ^   ^
+//    (S1, S2) S1   S2
+// now unset common aka leftmost set: 00010001
+
+import utils.InputReader;
+
+public class FindXORWithoutXOR {
+    public static void main(String[] args) {
+        System.out.println("Number 1: ");
+        int x1 = InputReader.readInt();
+        System.out.println("Number 2: ");
+        int x2 = InputReader.readInt();
+        System.out.println("N1 in binary: " + Integer.toBinaryString(x1));
+        System.out.println("N2 in binary: " + Integer.toBinaryString(x2));
+        System.out.println("XOR: " + Integer.toBinaryString(xor(x1, x2)));
+    }
+
+    private static int xor(int x, int y) {
+        return (x | y) - (x & y);
+    }
+}