Yes, but more importantly is that different nodes will compute the same superhub, given the same i and the same input set of nodes. If for the current bit only 1 or 2 remain in the set, there is a very high probability that matching the previous bit only will be a set of 3->5 nodes, which is reasonable. This is because each bit decimates the set of nodes by about half. If 1 or 2 remain in the set, then it was decimated (with high probability) from 3 to 5 nodes.

• Returning only 1 node to the set may return a different node depending on the order in which nodes were procured from the working set/
• if you have 2 or fewer nodes left in the working set, then you have 1 or 2 nodes only (you cannot have 0 nodes, as your own node, which should be in the original working set, will match your own hash in all bits). If you put back only one node, you get 2->3 nodes only, and 2 nodes are not enough for a cyclic superhub (notwithstanding the previous bullet point, that the order in which you remove items from the set becomes significant and may diverge depending on implementation details on different nodes).