Checkpoint: Measure to Compute

This checkpoint pulls the module together. You will build the smallest genuinely two-dimensional cluster state — the graph state on a four-cycle — the resource whose cross-links make measurement-based computation universal. A one-dimensional chain can only teleport single-qubit gates; closing the chain into a loop introduces the connectivity that, with single-qubit measurements, yields entangling gates and hence universal logic.

The resource we are building

The four-qubit ring cluster state $|C_\square\rangle$ is the graph state on the cycle with vertices $\{0,1,2,3\}$ and edges

E = \{(0,1),\,(1,2),\,(2,3),\,(3,0)\}.

By the graph-state recipe, prepare every qubit in $|+\rangle$ and apply a controlled- $Z$ on each edge:

|C_\square\rangle = CZ_{30}\,CZ_{23}\,CZ_{12}\,CZ_{01}\;|+\rangle^{\otimes 4}.

All four $CZ$ gates commute (each is diagonal), so the order in which you lay the edges does not matter.

What the amplitudes look like

Start from the uniform superposition,

|+\rangle^{\otimes 4} = \frac{1}{4}\sum_{b_0 b_1 b_2 b_3 \in \{0,1\}^4} |b_0 b_1 b_2 b_3\rangle,

with every amplitude $+\tfrac14$ . Each $CZ_{ab}$ multiplies a basis string by $(-1)^{b_a b_b}$ — a minus sign exactly when both endpoints of that edge are $1$ . Over the four edges the total sign on the string $b$ is

(-1)^{\,b_0 b_1 + b_1 b_2 + b_2 b_3 + b_3 b_0}.

So $|C_\square\rangle$ is real, every amplitude has magnitude $\tfrac14$ , and the sign of each is set by the parity of the cycle's edges that the string "lights up." For instance $|0011\rangle$ ( $b_2=b_3=1$ ) acquires one minus sign from edge $(2,3)$ ; $|1111\rangle$ has $b_a b_b = 1$ on all four edges, an even number, so it stays positive. The grader checks all sixteen amplitudes against the simulator's derived values — there is nothing to memorise, only the construction to get right.

Why this is the right capstone

Everything in the module converges here. The state is assembled by the cluster-state recipe (Lesson 2), described by the graph-state stabilizers $K_a = X_a\prod_{b\in N(a)} Z_b$ (Lesson 8) — for the ring each vertex has exactly two neighbours — and it is the resource on which measurement patterns (Lesson 4) run gates by measurement and feed-forward (Lessons 3, 5, 6). The square is the unit cell from which large 2D clusters, and thus universal MBQC and its fault-tolerant 3D extension (Lessons 7, 9), are tiled. Build the ring correctly and you have built, in miniature, the substrate of the entire one-way model.

Try it

Construct the four-qubit ring cluster state: a Hadamard on each qubit, then a controlled- $Z$ on each of the four cycle edges $(0,1), (1,2), (2,3), (3,0)$ . The grader verifies the full 16-amplitude state vector against the simulator's own derivation.

Run your code to see the quantum state.