Let's say i have two array of integers a and b with n integers each. I want to know the sum of the xor on all combinations of two integers in the two different subsets.
for example,
if n == 3:
i want to know the value of:
a1^b1 + a1^b2 + a1^b3 + a2^b1 + a2^b2 + a2^b3 + a3^b1 + a3^b2 + a3^b3
is there a way to this efficiently do this in a similar way as with + and x
i.e. 1*2 + 1*3 + 2*2 + 2*3 = (1+2)*(2+3)
There is a formula that works if there is only one non-zero value in the arrays. Because of this, you can do this one bit-value at a time, and then add up the results for each bit-value.
If you know that a contains x ones and n-x zeros, and b contains y ones and n-y zeros, then every a^b is either 1 or 0, and the number of 1s is exactly x * (n-y) + y * (n-x).
If you isolate the 1 bits, in the subsets, then you can calculate how many 1 bits are set in the XOR pairs. Similarly if you isolate the 2 bits, you can calculate how many 2 bits are set in the XOR pairs. Adding the results for each bit value give the final answer:
int total = 0;
for (int bit=1; bit>0 && (bit < a.length || bit < b.length); bit<<=1) {
int acount = 0;
for (int val : a) {
acount += val & bit;
}
acount /= bit;
int bcount = 0;
for (int val: b) {
bcount += val & bit;
}
bcount /= bit;
total += bit * ( acount * (b.length-bcount) + bcount * (a.length-acount) );
}