CF 1115U1 - Anti-diagonal unitary

CF 1115U1 - Anti-diagonal unitary

Rating: 1500
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Solve time: 1m 6s
Verified: yes

Solution

Problem Understanding

We are asked to implement a quantum operation on an array of qubits. The operation must correspond to a unitary matrix whose only non-zero entries lie on the anti-diagonal.

For a system of $N$ qubits, the state space has dimension $2^N$. The matrix describing the operation is therefore a $2^N \times 2^N$ matrix. An anti-diagonal matrix maps every basis state to exactly one other basis state, specifically the one whose index is mirrored across the anti-diagonal.

If the basis states are numbered from $0$ to $2^N-1$, then an anti-diagonal matrix sends basis state $i$ to basis state

$$(2^N-1)-i.$$

Looking at the binary representation, this operation flips every bit. For example, with three qubits:

$$000 \leftrightarrow 111$$

$$001 \leftrightarrow 110$$

$$010 \leftrightarrow 101.$$

The task is not to construct the matrix explicitly. Instead, we must implement a quantum circuit whose unitary has exactly this pattern.

The constraints are extremely small, only between two and five qubits, but that is largely irrelevant. The challenge is recognizing the quantum transformation represented by the anti-diagonal structure.

A common mistake is to focus on matrix construction rather than basis-state behavior. Quantum circuits operate on qubits, not on explicit matrices. The intended solution comes from understanding what permutation of basis states the anti-diagonal performs.

Consider a two-qubit system:

$$|00\rangle \to |11\rangle$$

$$|01\rangle \to |10\rangle$$

$$|10\rangle \to |01\rangle$$

$$|11\rangle \to |00\rangle.$$

Every bit is inverted independently. Any approach that only reverses qubit order would produce

$$|01\rangle \to |10\rangle,$$

which works for one state but fails for others. The required operation is bitwise complement, not register reversal.

Another subtle point is the little-endian indexing convention. It affects how matrix rows and columns are numbered, but complementing every qubit still transforms index $i$ into $2^N-1-i$. The solution remains the same regardless of the ordering convention used internally by the simulator.

Approaches

The brute-force viewpoint starts from the matrix itself. Since the desired matrix is anti-diagonal, one could imagine constructing a full $2^N \times 2^N$ permutation matrix and then decomposing it into elementary gates.

This is correct because the anti-diagonal matrix represents a permutation of computational basis states. For $N$ qubits there are $2^N$ basis states, so explicitly reasoning about the matrix requires $O(4^N)$ entries. Even though $N \le 5$ here, such an approach completely misses the underlying structure and would not scale.

The key observation is that the anti-diagonal permutation has a very simple binary interpretation. A basis state with index

$$i=b_{N-1}b_{N-2}\dots b_0$$

is mapped to

$$(2^N-1)-i.$$

The number $2^N-1$ is represented by $N$ ones in binary. Subtracting $i$ from this value flips every bit:

$$b_k \mapsto 1-b_k.$$

Thus the operation is exactly a bitwise NOT on every qubit.

In quantum computing, a bitwise NOT is implemented by applying an X gate to each qubit. Each X gate flips one computational-basis bit, and applying it to all qubits complements the entire basis-state index.

The anti-diagonal matrix is therefore simply

$$X^{\otimes N}.$$

Approach	Time Complexity	Space Complexity	Verdict
Brute Force matrix construction	$O(4^N)$	$O(4^N)$	Unnecessary
Apply X to every qubit	$O(N)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Iterate through all qubits in the input array.
2. Apply the quantum X gate to the current qubit.
3. Continue until every qubit has been flipped.
4. Return.

Why does this produce the desired matrix? Each X gate transforms a basis bit $b$ into $1-b$. After applying X to every qubit, a basis state

$$b_{N-1}b_{N-2}\dots b_0$$

becomes

$$(1-b_{N-1})(1-b_{N-2})\dots(1-b_0).$$

The resulting binary number is exactly

$$(2^N-1)-i.$$

Every basis state is mapped to the unique state paired with it across the anti-diagonal. Since X gates are unitary and the mapping is a permutation, the resulting matrix has exactly one non-zero entry in each row and column, located on the anti-diagonal.

Python Solution

This problem is actually a Q# implementation task rather than a traditional competitive-programming problem. There is no stdin/stdout interface.

The required operation is:

namespace Solution {
    open Microsoft.Quantum.Intrinsic;

    operation Solve (qs : Qubit[]) : Unit {
        for (q in qs) {
            X(q);
        }
    }
}

For completeness, if we model the same transformation in Python as a basis-state permutation, it would look like:

import sys
input = sys.stdin.readline

def anti_diagonal_index(i, n):
    return ((1 << n) - 1) ^ i

The implementation is minimal because the mathematical structure is minimal. Each qubit is flipped independently. There are no interactions between qubits, no controlled operations, and no dependence on the value of any other qubit.

The only subtle point is recognizing that complementing all bits corresponds exactly to reflection across the anti-diagonal of the unitary matrix.

Worked Examples

Example 1

Let $N=2$.

Original state	Binary index	After X on both qubits	New index
(	00\rangle)	0	(
(	01\rangle)	1	(
(	10\rangle)	2	(
(	11\rangle)	3	(

The mapping is

$$0\leftrightarrow3,\quad1\leftrightarrow2,$$

which is exactly the anti-diagonal permutation of a $4\times4$ matrix.

Example 2

Let $N=3$.

Original state	Index	After flipping all bits	New index
000	0	111	7
001	1	110	6
010	2	101	5
011	3	100	4
100	4	011	3
101	5	010	2
110	6	001	1
111	7	000	0

Every index $i$ is transformed into $7-i$, which is the required anti-diagonal correspondence.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(N)$	One X gate is applied to each qubit
Space	$O(1)$	No auxiliary storage is required

The number of qubits is at most five, so the circuit is trivially within all limits. Even for much larger registers, the solution remains linear in the number of qubits.

Test Cases

Since this is a Q# kata task rather than a standard input/output problem, conventional assert-based tests are not applicable. The unitary pattern can instead be verified conceptually.

def anti_diagonal_index(i, n):
    return ((1 << n) - 1) ^ i

# N = 2
assert anti_diagonal_index(0, 2) == 3
assert anti_diagonal_index(1, 2) == 2
assert anti_diagonal_index(2, 2) == 1
assert anti_diagonal_index(3, 2) == 0

# N = 3
assert anti_diagonal_index(0, 3) == 7
assert anti_diagonal_index(5, 3) == 2

# N = 5, first and last basis states swap
assert anti_diagonal_index(0, 5) == 31
assert anti_diagonal_index(31, 5) == 0

# applying twice returns original state
for n in range(2, 6):
    for i in range(1 << n):
        j = anti_diagonal_index(i, n)
        assert anti_diagonal_index(j, n) == i

Test input	Expected output	What it validates
$N=2, i=0$	3	First basis state maps to last
$N=2, i=3$	0	Last basis state maps to first
$N=3, i=5$	2	General bitwise complement behavior
Apply operation twice	Original index	Unitary is self-inverse

Edge Cases

Consider the smallest valid register size, $N=2$. The state $|00\rangle$ must become $|11\rangle$. Applying X to each qubit produces exactly that transformation. Any solution that only swaps qubit positions would leave $|00\rangle$ unchanged, which is incorrect.

Consider the largest valid register size, $N=5$. The index range is $0$ through $31$. Flipping all five bits maps $0$ to $31$, $1$ to $30$, and so on. The anti-diagonal structure remains identical because the transformation is defined independently for each bit.

Consider a state such as $|10101\rangle$. Applying X to every qubit yields $|01010\rangle$. The corresponding indices satisfy

$$21 \mapsto 10 = 31-21.$$

This confirms that the operation works for arbitrary basis states, not only for symmetric or extreme cases.

The little-endian indexing convention does not create a special case. Complementing every qubit flips every bit of the binary index regardless of which qubit is considered least significant. The resulting matrix is still anti-diagonal.