Compute the XOR of all the elements in the upper left triangular of a square matrix
You will need to look at \$O(n^2)\$ elements, therefore there is no algorithm that runs faster than \$O(n^2)\$.

That is, unless you have an explicit formula for the solution which abuses the fact that the elements are all consecutive.

Notice what these XOR to:

      cumulative XOR
0 000 000
1 001 001 |
2 010 011 |
3 011 000 |
4 100 100 |
5 101 001
6 110 111
7 111 000


We'll notice a few patterns. The cumulative XOR of all elements \$\lt 2^k\$ for any k is 0.

Here's another, more useful pattern: for any run of \$2^k\$ consecutive elements (e.g. see markers above), all binary digits except the most significant bit will repeat. Therefore we can explicitly compute the XOR of an arbitrarily large sequence \$a \ldots b\$ of length \$2^k\$ in one step (it's all 0s except the most significant digit is b - max(a, largestPowerOf2LessThanB) % 2, I think, where \$\mathrm{largestPowerOf2LessThanB} = 2^{\lfloor\log_2(b)\rfloor}\$).

Therefore we can calculate the sum of any arbitrarily large sequence in roughly \$O(\log(k))\$ steps.

Thus you can make an algorithm with running time \$O(n\log(n))\$ using this explicit formula.

You can maybe make it even faster if you can calculate contiguous blocks of the matrix?