Exactly the same as for eq, instead the "xor" operator for example is the amalgamation of 4 partial results, x0 to x3.
- x0 = input[0:7].xor()
- x1 = input[8:15].xor()
- x2 = input[16:23].xor()
- x3 = input[24:31].xor()
+ partition: P P P (3 bits)
+ a : .... .... .... .... (32 bits)
+ xor-a : x0 P x1 P x2 P x3 (4+3 bits, P=1 if no partition)
+
+ x0 = input[0:7].xor()
+ x1 = input[8:15].xor()
+ x2 = input[16:23].xor()
+ x3 = input[24:31].xor()
table showing how to combine x0-3 based on partitions p0-2 to produce result o0-3
0 0 1 | x0 | ^(x1-3) | 0 | 0
0 1 0 | ^(x0-1) | 0 | ^(x2-3) | 0
0 1 1 | x0 | x1 | ^(x2-3) | 0
-1 0 0 | ^(x0-2) | 0 | 0 | eq3
-1 0 1 | x0 | ^(x1-2) | 0 | eq3
-1 1 0 | ^(x0-1) | 0 | x2 | eq3
-1 1 1 | x0 | x1 | x2 | eq3
+1 0 0 | ^(x0-2) | 0 | 0 | x3
+1 0 1 | x0 | ^(x1-2) | 0 | x3
+1 1 0 | ^(x0-1) | 0 | x2 | x3
+1 1 1 | x0 | x1 | x2 | x3
"""]]
Example: