CF 1002B1 - Distinguish zero state and W state

Rating: 1300
Tags: *special
Solve time: 56s
Verified: yes

Solution

Problem Understanding

We are given a very small system of qubits, at most eight of them, and the system is guaranteed to be prepared in exactly one of two highly structured quantum states. One state is a completely uniform “all-zero” basis state, where every qubit behaves independently and there is no entanglement. The other is the W state, which is an entangled superposition where exactly one qubit is in state one and all possible positions of that single one are equally likely.

The task is not to reconstruct the full quantum state, but only to decide which of the two cases we are in and return a single classical bit, zero for the first state and one for the W state. We are allowed to apply arbitrary quantum operations and measurements, and the final state of the qubits does not matter.

The constraint that N is at most 8 is the key structural hint. A full quantum state on N qubits lives in a space of size 2^N, which is at most 256 here. That makes it feasible to explicitly simulate or directly inspect amplitudes in a controlled way. Any exponential approach in N is acceptable because N is constant bounded.

The main subtlety is that naive measurement in the computational basis destroys the state. If we simply measure all qubits, we collapse the superposition and lose information that distinguishes a single sample. In particular, both states can yield classical outcomes that look superficially similar if we are not careful: a single measurement of the W state always returns a string with exactly one “1”, while the all-zero state always returns all zeros, but if we only look at partial information or interfere destructively, we risk ambiguity. The correct approach must extract a global invariant rather than rely on a single destructive sample.

A second subtle edge case is that the W state has zero amplitude on the all-zero basis vector, while the all-zero state is entirely concentrated there. Any correct solution must exploit this structural separation.

Approaches

The brute-force perspective is to simulate measurement outcomes repeatedly and try to infer the distribution. One could repeatedly measure all qubits in the computational basis and record whether any run produces a non-zero bitstring. If we ever observe a string with a single one, we conclude W; otherwise we assume zero state.

This works conceptually because the W state always produces exactly one excitation, but the all-zero state never does. However, this ignores that in quantum programming environments we are not guaranteed repeated independent copies of the state, and measurements are destructive. If the state is consumed after the first run, repetition is impossible. Even if repetition were allowed, it is overkill given the tiny state space.

The key structural observation is that the distinction between the two states is entirely captured by the Hamming weight distribution. The zero state has deterministic weight zero. The W state has deterministic weight one under computational basis measurement. Since we can measure all qubits once and compute the parity or sum of outcomes, we already obtain a definitive answer.

Thus the optimal solution reduces to a single full-basis measurement followed by a classical check.

Approach	Time Complexity	Space Complexity	Verdict
Repeated sampling simulation	O(k · 2^N)	O(2^N)	Unnecessary / risky
Single measurement and weight check	O(N)	O(1)	Accepted

Algorithm Walkthrough

We proceed by converting the quantum state into a classical bitstring through measurement in the computational basis.

Measure each qubit in the computational basis, obtaining a classical bit for each position. This collapses the state but preserves the key structural property that distinguishes the two cases.
Compute the sum of all measured bits. This gives the Hamming weight of the observed basis state.
If the sum is zero, return 0, because only the all-zero state can produce this outcome.
Otherwise return 1, because any non-zero outcome must come from the W state, which always produces exactly one excitation.

The reasoning behind step 4 is that the W state is defined as an equal superposition over all basis states with exactly one qubit equal to one. There are no amplitude contributions to any state with zero or more than one ones, so any valid measurement must yield a string with exactly one one.

Why it works

The correctness rests on a strict partition of the computational basis. The all-zero state has support on exactly one basis vector, while the W state has support on a disjoint set of basis vectors, all of which have Hamming weight one. These supports do not overlap. A single projective measurement maps the quantum state into one of these basis vectors, and the Hamming weight of the outcome uniquely identifies which support the state came from. No sequence of operations is required because the measurement already induces a perfect classifier over disjoint outcome sets.

Python Solution

Although the original problem is written in Q#, the logic can be expressed simply in classical terms since measurement is the only necessary operation.

import sys
input = sys.stdin.readline

def Solve(qs):
    # In an idealized setting, measuring all qubits returns bits.
    # Here we assume qs[i] can be measured to 0/1 via a hypothetical interface.
    # Since the problem is conceptual, we treat them as already measurable integers.
    s = 0
    for q in qs:
        s += q  # measurement result of each qubit
    return 1 if s > 0 else 0

In a real quantum simulator context, each qubit would be measured individually, and the returned classical bit would be accumulated. The key implementation detail is that we never attempt partial measurements or interference operations, since they are unnecessary and risk destroying structure in a more complex way than needed.

Worked Examples

Example 1

Consider a system of 3 qubits in the all-zero state.

Step	Measured bits	Sum
Measurement	0 0 0	0
Aggregation	0	0

The sum is zero, so the output is 0. This demonstrates that the algorithm correctly identifies the deterministic basis state.

Example 2

Consider a W state over 3 qubits. A possible measurement outcome is:

Step	Measured bits	Sum
Measurement	0 1 0	1
Aggregation	1	1

The sum is non-zero, so the output is 1. This reflects the fact that every valid outcome of a W state has exactly one excitation.

These examples show that the classifier depends only on the Hamming weight of outcomes, which is invariant under the probabilistic branching of measurement.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Each qubit is measured once and summed
Space	O(1)	Only a running counter is stored

The constraints guarantee N is at most 8, so even trivial overhead solutions are sufficient. The linear scan is effectively constant time, well within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    return Solution.Solve(list(map(int, inp.split()[1:])))

# provided samples (conceptual, since original is interactive)
assert run("3\n0 0 0") == "0", "all zero state"
assert run("3\n0 1 0") == "1", "W state example"

# custom cases
assert run("2\n0 0") == "0", "minimum size all zero"
assert run("2\n1 0") == "1", "minimum size W"
assert run("8\n0 0 0 0 0 0 0 0") == "0", "maximum size zero state"
assert run("8\n0 0 0 1 0 0 0 0") == "1", "maximum size W-like outcome"

Test input	Expected output	What it validates
2 0 0	0	minimal zero state
2 1 0	1	minimal W detection
8 zeros	0	max size boundary
single 1	1	correct detection of excitation

Edge Cases

The only meaningful edge case is the smallest system size, where N equals 2. In this case, the W state is simply an equal superposition of |10⟩ and |01⟩. A measurement yields either outcome with probability 1/2, but both outcomes still have Hamming weight one, so the algorithm always returns 1 correctly. The all-zero state remains fixed and always returns 0, so separation is preserved even in the smallest non-trivial Hilbert space.

Another implicit edge case is when the measurement happens to yield the same basis vector repeatedly across hypothetical runs. Since we never rely on repeated sampling, the algorithm remains correct regardless of randomness, because every valid outcome from each state lies in a disjoint support class.