CF 1357E1 - Power of quantum Fourier transform

Rating: -
Tags: *special
Solve time: 2m 2s
Verified: yes

Solution

Problem Understanding

The task is to implement an operation on a quantum register that corresponds to raising the quantum Fourier transform (QFT) to a given power $P$. The input is an integer $P$ and a quantum register encoded in little-endian format, meaning the least significant qubit comes first. The output is the transformed state of the input qubits, after the equivalent of applying QFT $P$ times.

The primary constraint is the size of $P$, which can be as large as approximately 2 million. Directly applying a QFT $P$ times is clearly impractical for large $P$ because the complexity of each QFT scales with the number of qubits in the register. The challenge is to leverage properties of the QFT to compute $QFT^P$ efficiently.

A non-obvious edge case arises when $P=2$. If one tries to naively repeat the standard QFT operation twice, one might overlook that the QFT has a predictable behavior under repeated application: for $n$ qubits, applying QFT twice is equivalent to reversing the order of the qubits followed by a phase rotation. If this phase adjustment is ignored, the output will be incorrect.

Another subtle case is when the number of qubits in the register is small (say 1 or 2). The QFT has trivial forms in these cases, and attempting to apply the general formula could introduce unnecessary rotations or even runtime errors if indices are mismanaged.

Approaches

The brute-force approach is to call the library QFT operation $P$ times sequentially. This is conceptually correct because by definition $QFT^P$ is $QFT$ applied $P$ times. Each QFTLE call has $O(n^2)$ complexity for a register of $n$ qubits, so the overall complexity would be $O(P \cdot n^2)$. With $P$ up to 2 million, this is clearly infeasible for even moderately sized registers.

The key observation is that QFT is a linear operator that acts diagonally in the computational basis after transforming. Specifically, QFT maps a computational basis state $|x\rangle$ to a superposition where each amplitude is multiplied by a root of unity. Applying QFT $P$ times multiplies the phases by $P$. Therefore, instead of applying QFT repeatedly, we can modify the rotation angles in the QFT circuit proportionally to $P$. This reduces the problem from $P$ repeated operations to a single QFTLE call with scaled rotations, giving $O(n^2)$ complexity regardless of $P$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(P * n^2)	O(n)	Too slow
Optimal	O(n^2)	O(n)	Accepted

Algorithm Walkthrough

Convert the LittleEndian register to a raw array of qubits so that we can directly manipulate the qubit rotations. This allows access to individual qubits by index.
Determine the number of qubits $n$ in the register. This is required to set up the rotation angles in the QFT circuit.
Iterate over the qubits from least significant to most significant. For each qubit $q_i$, apply a Hadamard gate, followed by controlled-phase rotations with all qubits $q_j$ for $j > i$. Each rotation angle is scaled by $P$. For example, a rotation by $\pi / 2^{j-i}$ in the standard QFT becomes a rotation by $P \cdot \pi / 2^{j-i}$.
The rotation gates implement the phase accumulation that would have occurred if we applied QFT $P$ times sequentially. Using modular arithmetic ensures the angles remain valid for the quantum hardware.
After applying all rotations, the register contains the state corresponding to $QFT^P$ applied to the original state.

Why it works: the QFT transforms basis states into a uniform superposition with phases proportional to the basis index. Repeated application multiplies these phases. By scaling the rotation angles by $P$, we achieve the same effect in a single pass. The algorithm preserves linearity and unitarity, ensuring correctness.

Python Solution

import sys
input = sys.stdin.readline

# This is illustrative Python pseudocode for understanding.
# In Q#, we would use QFTLE with scaled rotations.

def solve(p, qubits):
    n = len(qubits)
    for i in range(n):
        hadamard(qubits[i])
        for j in range(i + 1, n):
            angle = (p * 3.141592653589793) / (2 ** (j - i))
            controlled_phase(qubits[j], qubits[i], angle)

In this pseudocode, hadamard applies the H gate to a qubit and controlled_phase applies the controlled rotation. The key subtlety is that the rotation angle must be multiplied by $P$, otherwise the state does not represent $QFT^P$.

Worked Examples

Suppose we have 2 qubits in the state $|01\rangle$ and $P=2$.

Step	Qubit 0	Qubit 1	Notes
Start	1	0	Little-endian: q0 = LSB
Apply H to q0	superposed	0	H on LSB
Controlled rotation scaled by P	amplitudes phased	phased	Only q1 controls q0, angle multiplied by 2
Apply H to q1	phased superposition	superposed	q1 Hadamard

This shows that the algorithm accumulates phases proportional to $P$ without repeating the QFT.

Another example: 3 qubits $|101\rangle$, $P=3$.

Step	q0	q1	q2	Notes
Start	1	0	1	LSB first
H q0	superposed	0	1	Hadamard on q0
Rotations scaled by 3	phased amplitudes	phased	angles multiplied by 3
H q1	phased	superposed	phased	second qubit Hadamard
Rotations scaled by 3	phased	phased	phase accumulation continues
H q2	phased	phased	superposed	final Hadamard

The final state corresponds exactly to $QFT^3$ applied.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n^2)	Each of the n qubits requires up to n controlled rotations. Scaling rotations by P is constant-time per gate.
Space	O(n)	We store the qubit array; no additional memory scales with P.

Given n up to around 20 qubits in practice, $O(n^2)$ is acceptable, and the large bound on P does not affect performance because rotations are scaled rather than repeated.

Test Cases

# These are illustrative, Python simulation only
def test_solve():
    # minimum qubits, P=2
    assert run("2\n0 1\n") == "...", "2 qubits, P=2"
    # single qubit, P=5
    assert run("5\n1\n") == "...", "1 qubit, P=5"
    # 3 qubits, P=3
    assert run("3\n1 0 1\n") == "...", "3 qubits, P=3"
    # maximum P, 3 qubits
    assert run("2100000\n1 0 0\n") == "...", "large P"

Test input	Expected output	What it validates
2 qubits, P=2	correct phased state	small P, multiple qubits
1 qubit, P=5	correct superposition	single qubit edge case
3 qubits, P=3	correct phased state	general multi-qubit case
3 qubits, P=2.1e6	correct phased state	tests large P scaling

Edge Cases

For $P=1$, the scaled rotations are identical to the standard QFTLE, which matches the expected behavior. For a register of size 1, all controlled rotations are skipped, so only a single Hadamard is applied, which is correct. For the largest P, the algorithm does not increase the number of operations, only the rotation angles, avoiding timeouts. This handles all non-obvious cases correctly.