2 ms·
No, QFT requires only O(log(n)^2) gates. It can be described by using a recursion. That is, QFT on 2^n values (n qubits) can be computed using QFT on 2^(n-1) va
by dandanua 4y ago
No, QFT requires only O(log(n)^2) gates. It can be described by using a recursion. That is, QFT on 2^n values (n qubits) can be computed using QFT on 2^(n-1) values (n-1 qubits).