4 ms·
I don't think that's what they meant. It's the case you can use literal strings of bits to encode a (2^n)-tree node, so you use actual bitstring comparisons an
by mkehrt 10mo ago
I don't think that's what they meant. It's the case you can use literal strings of bits to encode a (2^n)-tree node, so you use actual bitstring comparisons and operations to manipulate them. Rightshift gives you the parent and things like that.
I don't think this is something the article cares about, though.
- senderista 10mo agoThank you, that is exactly what I meant.