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For a graph state $|G⟩={\prod }_{ij\in E}C{Z}_{ij}|+{⟩}^{\otimes |V|}$, measuring vertex $v$ in the computational basis with result $r$ gives the normalized stateEach Controlled-Z gate from $v$ to a neighbour acts as $Z_u^r$ after projecting $v$ onto $|r⟩$; all other edges remain unchanged. The projection contributes $1/\sqrt{2}$, so either result has probability $1/2$. For a four-cycle, deleting one vertex leaves a path and adds byproducts on its two endpoints.

Conjugation by a Controlled-Z gate leaves the $a$ bits unchanged and updates the displayed $b$ bits and phase, using the old bits. The phase occurs when the extra $Z$ from one line is moved past $X$ on the other line. These constant-size updates support Heisenberg propagation of a Pauli observable through a Clifford circuit.

Let $1$ be a leaf with neighbour $2$ and let the graph on the other vertices be generated by the Controlled-Z gate product $E_{\mathrm{rest}}$. If the leaf has Pauli Z gate frame $Z_1^r$, measuring it at angle $\alpha$ with result $s$ applies one-bit teleportation and leavesAny known Pauli Z gate frames on other vertices remain as additional factors before those vertices' input states. The identity $J(\alpha )Z^r=X^{r}J(\alpha )$ combines the leaf byproduct with the measurement result.

For $n\ge 2$, apply a Hadamard gate to one qubit, then controlled-NOTs from it to all others. Replacing each controlled-NOT gate by a target Hadamard, a Controlled-Z gate, and another target Hadamard uses $1+3(n-1)$ allowed gates. Each one-qubit reduced state is $I/2$, proving entanglement across each qubit-versus-rest bipartition.

A two-vertex graph state starts with two $|+⟩$ states joined by a Controlled-Z gate. Measuring the first vertex in the $X$ basis leaves the second in $X^{s}H|+⟩=X^s|0⟩$, where $s$ is the measurement outcome. Thus a known zero input can be supplied to a measurement wire using only a graph-state resource and a known Pauli frame. This preparation link precedes the links implementing the desired logical J gate; later measurement angles and output-bit corrections absorb its byproduct.

Two angle-zero one-bit teleportations with outcomes $p,q$ implement $X^{q}H{X}^{p}H=X^q{Z}^p$ on an arbitrary input, up to global phase. A Controlled-Z gate with a fixed $|1⟩$ quantum ancilla implements $Z$, so $Z^p$ removes the known $Z$ factor. The result is $X^q$ up to phase. With probability one half the desired Pauli X gate is applied; otherwise the input is unchanged. Repeating until $q=1$ gives an exact heralded correction with two attempts on average and almost-sure termination. It has no finite worst-case measurement count. A Pauli frame avoids this repeat-until-success procedure when only logical action or classical output statistics are required.

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

Prepare the two data qubits in $|s⟩=\frac{1}{2}{\sum }_{x\in B_2}|x⟩$ and a phase ancilla qubit in $|-⟩$. A single Boolean quantum oracle query flips only the marked amplitude. Its success fraction is $p=1/4$, so $\theta =\pi /6$ in amplitude amplification. The fixed diffusion reflection $D=2|s⟩⟨s|-I$ gives $\sin (3\theta )=1$ after one iteration.

Directly, after the query the marked amplitude is $-1/2$ and the other three are $1/2$. Their mean is $1/4$. The diffusion reflection replaces each amplitude $c_x$ by $2(1/4)-c_x$, yielding one at the marked input and zero elsewhere. ThusHere $U_f$ is understood with its target ancilla qubit, and $D$ acts only on the data. It is independent of $f$: $D=H^{\otimes 2}(2|00⟩⟨00|-I)H^{\otimes 2}$, with the central diagonal gate implementable using two Pauli Z gates and one Controlled-Z gate. Measuring the data in the computational basis therefore finds the unique marked string with certainty after one oracle query.

An explicit measurement-based quantum computation pattern uses six vertices $u_0,u_1,u_2,v_0,v_1,v_2$. Prepare a graph state with every vertex in $|+⟩$ and apply a Controlled-Z gate for each edgeThe first two links on each wire permit graph-state preparation of a computational-basis input followed by the logical J gate. Use the following single-qubit measurements:

All entangling edges can be made at preparation time because Controlled-Z gates commute. A future edge that does not touch a currently measured vertex can equivalently be deferred, which allows the one-bit teleportation identities to be applied in their logical order.

The two initial $X$ measurements implement $H|+⟩=|0⟩$ with Pauli frames $X^{s_0},X^{t_0}$ on the logical inputs. The first adaptive J gate then has output frame $X^{s_1}{Z}^{s_0}$ on $u_2$. Propagating through $E_{u_2{v}_1}$ gives framesup to branchwise global phase. The second adaptive J gate converts the latter intoA $Z$ correction does not alter a computational-basis measurement, while an $X$ correction flips its bit. Consequently the deterministic classical postprocessing isThis reproduces the output-bit distribution of the original quantum circuit, including its known byproduct corrections.

There is an additional simplification for these particular zero inputs. Since $J(\alpha )|0⟩=|+⟩$, $E(|+⟩|0⟩)=|+⟩|0⟩$, and $J(\beta )|0⟩=|+⟩$, the exact final state is $|+⟩|+⟩$, independently of the angles. The requested bit is therefore fair. A single isolated graph-state vertex measured in $Z$ already simulates that bit distribution; the six-vertex pattern also explicitly realizes the circuit and its corrections.

The J gate is $J(\alpha )=HP(\alpha )$, with $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$. Prepare a fresh qubit in $|+⟩$ and apply the Controlled-Z gate $E$ between it and the input $|\psi ⟩=u|0⟩+v|1⟩$. The resulting state isMeasure the input in the equatorial qubit measurement basis $|{\alpha }_s⟩=(|0⟩+(-1{)}^s{e}^{-i\alpha }|1⟩)/\sqrt{2}$. The unnormalized output isEach outcome has probability $1/2$. Thus one-bit teleportation realizeson the new qubit. Apply the known Pauli X gate correction $X^s$ for the literal $J(\alpha )$ output, or keep the correction in a Pauli frame and adapt later measurements. The old qubit is measured, so this is not cloning the input.

Direct multiplication of the given matrices gives $J(\alpha )X^s=e^{+is\alpha }{Z}^{s}J((-1{)}^s\alpha )$. The displayed negative exponent in the supplied relation has the wrong sign for exact matrix equality. The discrepancy is only a global phase in a fixed measurement branch, so it does not change this measurement implementation or its outcome probabilities. The positive-sign identity is used when tracking exact matrices.

Use the operator norm induced by the usual vector norm, and assume the input quantum state is normalized. Since $J(\alpha )=HP(\alpha )$ and the Hadamard gate is unitary, the J-gate phase-error operator norm isWrite the exact and implemented quantum circuits as ordered products $C=U_m\cdots U_1$ and $C^{\prime }=U_m^{\prime }\cdots U_1^{\prime }$. The quantum circuit gate-error telescoping bound follows fromEvery surrounding factor is unitary, including gates tensored with identities on other qubits, so the triangle inequality and the submultiplicativity of the operator norm giveThe exact Controlled-Z gates contribute zero to that sum. ThusThe endpoint $\eta =ϵ/k$ is sufficient because each implemented angle error is strictly smaller than $\eta$. If $k=0$, the circuits are identical and any positive $\eta$ works. The bound controls the stated vector distance with actual gate phases retained, so no adjustment of the global phase of one output is needed.

Store a tensor-product element of the Pauli group asThe phases of the original single-qubit factors can all be accumulated into $i^s$. This binary phase representation of a Pauli string uses $2n+2$ bits. It retains phases, which must not be dropped when computing probabilities later.

For the conjugation convention $U^{†}PU$ in the question, direct multiplication of the two-by-two matrices givesandThese backward Pauli updates for Hadamard and phase gates use $S^{†}PS$, not $SP{S}^{†}$. On the affected line $j$, the updates areAll phases are reduced modulo four and the updates use the old bits.

For a Controlled-Z gate on lines $j,k$, its diagonal action givesFor the controlled-Z update in the binary phase representation, conjugation respects products, so these determine the rule for every Pauli operator on those lines. In the chosen ordered $XZ$ convention it isThe phase arises when a newly introduced $Z_k$ is moved past $X_k$. For instance $X_j{X}_k$ becomes $-X_j{Z}_j{X}_k{Z}_k$, so this sign matters.

The untouched factors are unchanged. Each Clifford operation therefore needs only a fixed number of local bit updates; reconstructing the requested full list takes $O(n)$ time. Put the final phase into the first factor, which is permitted because the single-qubit Pauli group includes all multiples by $\pm 1,\pm i$. The classical cost is polynomial, despite the exponentially large matrix of the operation on the full state space.

The path graph state is $|{\psi }_3⟩=E_{23}{E}_{12}|+{⟩}_1|+{⟩}_2|+{⟩}_3$. The factor $Z_3^r$ commutes with both the Controlled-Z gates and the quantum measurement of vertex $1$. Treat $Z_1^r|+{⟩}_1$ as the input of one-bit teleportation. The equatorial qubit measurement result $s$ therefore gives the normalized stateHere the subscript on $Z_2^r$ expresses the teleported input factor; it is not an extra operation on the already measured vertex. Use $J(\alpha )Z^r=X^{r}J(\alpha )$ to obtainThe exclusive or exponent is equivalent to the printed sum $r+s$ because $X^2=I$. The conditional probability of $s$ is one half for either $r$, by the one-bit teleportation branch norm. This calculation concerns an equatorial measurement of a graph-state leaf; no third remaining qubit or extra $|-⟩$ factor is present.

Label the measured vertex of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite vertex by $2$. The edges are $01,12,23,30$. By the definition of a graph state,All Controlled-Z gates commute. Conditioning on the computational basis value $r$ of vertex $0$, the factors on edges $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a⟩=(-1{)}^{ra}|r,a⟩$. The remaining edges form the three-vertex path $1-2-3$:The Born rule gives probability one half for either result. Normalizing provesThis is computational-basis measurement of a graph-state vertex: remove the measured vertex and apply a known Pauli Z gate byproduct to each neighbour. The chosen labels specify precisely which factors occur; any vertex of the square can serve as vertex $0$ by relabelling.

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.