[dzaima]

A property of the initial XOR trick for 2 different elements is that it guarantees finding a way to partition in one pass (and with very trivial code; no hashing involved!), which is lost by replacing that with hashing. (the original trick does take two passes - finding the bit to partition on, and doing the actual partitioning, whereas hashing is 1+ε passes, but the first pass in the original is just an xor-fold, and the partitioning really only needs to be a "accumulator ^= (current_val & mask) ? current_val : 0" (other partition is just xoring the results of both passes), both of which can be trivially parallelized and SIMD'd with O(1) extra memory usage)

The approach in my comment achieves guaranteeing finding partitions, and still avoids actual hashing or anything strictly-probabilistic, but does still lose the extreme triviality and mechanical sympathy of the original approach.