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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.

A graph state with edges $13,24,34,45$ simulates $J_2(\gamma )E_{12}(J_1(\alpha )\otimes J_2(\beta ))$ on $|++⟩$. Measure vertices 1 and 2 at angles $\alpha ,\beta$, with results $s_1,s_2$, and vertex 4 at $(-1{)}^{s_2}\gamma$, with result $s_4$. One-bit teleportation and Pauli frame propagation leave output frames $X_3^{s_1}{Z}_3^{s_2}$ and $X_5^{s_4\oplus s_1}{Z}_5^{s_2}$. Computational outputs are therefore corrected by $s_1$ and $s_4\oplus s_1$ respectively.

Label a four-cycle $0-1-2-3-0$. Measure vertex $0$ in the computational basis with result $r$, then vertex $1$ in the equatorial qubit measurement basis at angle $\alpha$ with result $s$. Graph-state vertex deletion and one-bit teleportation leaveFinal computational basis measurements with raw results $d_2,d_3$ therefore simulate the ideal circuit $E_{23}J(\alpha {)}_2$ on $|++⟩$ after the classical correction $k=d_2\oplus r\oplus s$, $l=d_3$. The output probability distribution is correct in every prior branch; no physical Pauli frame correction is needed.

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.

Measurement-based quantum computation prepares an entangled resource quantum state, then applies single-qubit quantum measurements, potentially adapting later bases to earlier outcomes. Pauli frame tracking and classical outcome processing convert random physical branches into a specified logical computation. Graph states and one-bit teleportation provide a concrete construction.

Compile each CNOT gate into a Hadamard gate, a Controlled-Z gate, and another Hadamard gate on its target. Every one-bit teleportation implementing a Hadamard gate uses a fixed angle-zero basis. A Pauli frame propagates by classical binary updates without changing a measurement basis. All internal measurements, and any fixed computational output measurements, therefore share one simultaneous layer. Resource preparation and classical processing are not included in this logical measurement depth.

Write the normalized input quantum state as $|\psi ⟩=a|0⟩+b|1⟩$. The fresh quantum ancilla is $|+⟩$, so the Controlled-Z gate createsApplying the measurement bra $⟨v_k(\theta )|$ to the first qubit leaves the unnormalized second-qubit vectorSince $|+⟩,|-⟩$ form an orthonormal basis,Also $U(\theta )|\psi ⟩=a|+⟩+e^{-i\theta }b|-⟩$, while the Pauli X gate fixes $|+⟩$ and changes the sign of $|-⟩$. Dividing by the branch norm therefore givesThis is one-bit teleportation with the present positive-exponent measurement vectors; the transferred logical gate is $U(\theta )=J(-\theta )$.

Label the chain cluster state from top to bottom by $1,2,3,4$, and let $k,l,m$ be the phase-measurement outcomes on the first three vertices, with raw computational basis outcome $s$ on vertex four. Repeated one-bit teleportation leaves the last qubit, up to global phase, inThis sequential description applies even though the whole graph state is prepared first: a later Controlled-Z gate acts only on unmeasured qubits and commutes with earlier single-qubit measurement operators. It can therefore be moved past those measurements for the purpose of calculating each conditional branch.

Take $\alpha =\theta$ and $\beta =0$, so $U(0)=H$. After the second measurement the logical state isFor binary $p,q$, the identities proved above imply the Pauli-frame propagation along a measurement wire rulewhere $\simeq$ means equality up to global phase. Choose the third angle adaptively as $\gamma =(-1{)}^{l\oplus 1}\varphi$. Then $(-1{)}^l\gamma =-\varphi$, andReading the target circuit in temporal order gives its premeasurement stateConsequentlyThe Pauli Z gate changes only phases in the computational basis, while the Pauli X gate exchanges its two measurement outcomes. Thus the explicit choices areIndeed, for every earlier outcome branch,Hence averaging over the earlier random outcomes leaves exactly the target probability distribution. This implements the desired measurement-based quantum computation without physically removing the final Pauli frame.

Prepare the three-vertex graph-state wire: start all three qubits in $|+⟩$ and apply Controlled-Z gates on edges $1-2$ and $2-3$. In this paper's equatorial qubit measurement convention the basis vectors can be chosen asThe given identity means that measuring an input qubit at angle $\theta$, with outcome $r$, teleports its state to the next qubit as $X^{r}U(\theta )|\psi ⟩$. Each outcome has probability $1/2$, since $X^{r}U(\theta )$ is unitary. This is one-bit teleportation with a known Pauli frame.

Measure vertex 1 at angle $\alpha$ and record $r$. Then measure vertex 2 at angle $\gamma =(-1{)}^r\beta$, taken modulo $2\pi$, and record $s$. The controlled-Z on edge $2-3$ commutes with the first measurement, so the two teleportation steps can be applied successively. The remaining normalized state isFor $r=0$ this is $X^{s}U(\beta )U(\alpha )|+⟩$. For $r=1$, the relation proved above gives $U(-\beta )X=e^{i\beta }ZU(\beta )$. Thus in every branch the remaining state isThe scalar is a global phase and has no effect on probabilities.

Finally measure vertex 3 in the computational basis, obtaining $z$, and reportThe $Z^r$ byproduct changes only computational-basis phases; $X^s$ changes the output label by $s$. Hence the corrected bit has exactly the ideal circuit's probability distribution in every branch, not merely on average. The procedure uses only single-qubit measurements after preparing the graph state, adaptive classical angle selection, and deterministic classical output processing.

Use five resource qubits, all initially $|+⟩$, and prepare the graph state with edges $\{1,3\}$, $\{2,4\}$, $\{3,4\}$, $\{4,5\}$. The labels here identify resource vertices; logical wire 1 travels from vertex 1 to 3, and logical wire 2 travels from 2 to 4 to 5. Each edge is a Controlled-Z gate, and these entangling gates commute, so all can be applied when preparing the resource.

Measure vertex 1 in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s_1$, and vertex 2 at angle $\beta$, with outcome $s_2$. These measurements can be simultaneous. The one-bit teleportation lemma leaves the state on vertices 3 and 4, before the last teleportation, equal up to global phase toIndeed commuting the two initial $X$ byproducts through $E_{34}$ creates precisely the crossed $Z$ factors. The unused edge from 4 to the fresh vertex 5 is the last one-bit teleportation link.

Measure vertex 4 at angle $\theta =(-1{)}^{s_2}\gamma$, with outcome $s_4$. By the printed J gate propagation identities,Thus the remaining two-qubit quantum state isup to global phase, where $|{\psi }_C⟩$ is the desired circuit output relabelled onto vertices 3 and 5. The sign choice cancels the unwanted angle reversal caused by the $X$ frame on vertex 4.

Finally measure vertices 3 and 5 in the computational basis, obtaining raw results $t_3,t_5$. The Pauli Z gate affects only phases, while the Pauli X gate flips a computational result. The five-vertex graph-state circuit simulation therefore outputsThis deterministic classical correction reproduces the ideal joint probability distribution for every preceding measurement branch. The final fixed-basis measurements may share the second layer with the adaptive measurement of vertex 4, because their projectors act on different vertices; measuring them afterwards is also valid.

Replace every CNOT gate by $H_j{E}_{ij}{H}_j$. The resulting quantum circuit contains only Hadamard gates and Controlled-Z gates. Realize each Hadamard gate by a fresh one-bit teleportation link measured at angle zero, so every internal measurement is in the fixed $X$ basis. Insert a graph edge between the current wire vertices for each Controlled-Z gate. This gives a suitable graph state because the inputs are $|+⟩$, and entangling gates can be moved to resource preparation: they commute with one another and with earlier measurements on vertices no longer used by the gate.

Track a Pauli frame ${⨂}_i{X}_i^{p_i}{Z}_i^{q_i}$. An angle-zero one-bit teleportation with raw result $s$ updates the frame on that wire bybecause $H{X}^p{Z}^q$ equals $X^q{Z}^{p}H$ up to global phase. A Controlled-Z gate updates $q_i\leftarrow q_i\oplus p_j$ and $q_j\leftarrow q_j\oplus p_i$, leaving the $p$ bits unchanged. These are classical binary updates; no measurement angle depends on them.

All the internal quantum measurements consequently have predetermined bases. Their projectors act on distinct resource vertices and commute, so all internal measurements can be performed simultaneously: the logical measurement depth is at most one. If the output is to be read in the computational basis, those fixed-basis measurements can occur in the same layer; a raw output $t_i$ is relabelled $t_i\oplus p_i$. If quantum outputs are retained, the same calculation gives the desired output in a known Pauli frame. This nonadaptive Hadamard–CNOT measurement pattern concerns measurement depth, not the depth of resource preparation or classical frame computation.

Write the single-qubit gate as $R(\alpha )=HP(\alpha )H$, where $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$. Thus it is, up to global phase, a rotation about the x-axis, and it commutes with $X$. Compile it as a Hadamard gate followed by the J gate $J(\alpha )$. Compile each CNOT gate as before into $H_j{E}_{ij}{H}_j$. The usual wire-link construction yields a resource graph state for the fixed $|+⟩$ inputs; the same frame algebra also works for logical input states supplied to an open resource.

The point is stronger than counting two gates per rotation: a long circuit must not acquire a new adaptive layer for every rotation. We show that every nonzero-angle basis depends only on outcomes of the angle-zero measurements.

For an $R(\alpha )$ gadget, let the incoming Pauli frame be $X^p{Z}^q$. Denote the angle-zero outcome by $s$, and the subsequent angle-$\theta$ outcome by $t$. The two one-bit teleportations givewhere $\simeq$ omits a global phase. Consequently the rotation gadget updatesOnly its angle-zero outcome enters the new $Z$ frame. Its arbitrary-angle outcome enters only the $X$ frame, which commutes through all later $R$ gates.

For a CNOT gate from $i$ to $j$, call the outcomes of its two angle-zero target-wire measurements $r,u$, in that order. The Pauli frame update isThese follow by applying the preceding Hadamard gate and Controlled-Z gate frame updates twice. In particular, the new $Z$ frames depend on old $Z$ frames and angle-zero outcomes only; they never depend on old $X$ frames or arbitrary-angle outcomes. This is the measurement-level version of the given fact that a CNOT gate propagates $X$ operators into products of $X$ operators.

Starting with no frame, induction in the original circuit order now computes every $q$ solely from angle-zero results. Therefore every sign $(-1{)}^{s\oplus q}$ is known after one simultaneous layer containing all angle-zero measurements, including those appearing late in the circuit. Compute those signs classically, and perform all remaining equatorial measurements in a second simultaneous layer. An angle zero that occurs accidentally among these remaining choices may stay in that layer. Reordering is valid because, once a branch's bases have been fixed, the projectors on different vertices commute; no second-layer result is needed to choose another second-layer basis.

The two-layer measurement pattern for CNOT and x-axis rotations thus hasindependently of the number or order of gates. Fixed computational-basis measurements of classical outputs can be included in layer two, followed by relabelling $t_i\mapsto t_i\oplus p_i$. For quantum outputs the final Pauli frame gives the required correction. The depth statement excludes graph preparation and classical processing, as in the definition in the paper.

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.

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.

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.

With equatorial qubit measurement vectors $(|0⟩+(-1{)}^s{e}^{i\eta }|1⟩)/\sqrt{2}$, the gate in one-bit teleportation is $U(\eta )=H\mathrm{diag}(1,e^{-i\eta })=J(-\eta )$. Direct multiplication gives $U(\eta )Z=XU(\eta )$ and $U(\eta )X=e^{-i\eta }ZU(-\eta )$. Moving a Pauli frame through this gate therefore exchanges its $X$ and $Z$ exponents and reverses the angle if the old $X$ exponent is one, as displayed up to global phase. A measurement outcome $s$ adds a further $X^s$ factor. This calculation supplies adaptive angle choices and the final classical output-bit correction of a linear cluster state computation.

A path graph state on three qubits simulates two successive gates $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$. Measure vertex one in the equatorial qubit measurement basis $|v_r(\alpha )⟩=(|0⟩+(-1{)}^r{e}^{i\alpha }|1⟩)/\sqrt{2}$, then vertex two at angle $(-1{)}^r\beta$, obtaining bits $r,s$. Two one-bit teleportations and $U(\theta )X=e^{-i\theta }ZU(-\theta )$ leave vertex three in $X^s{Z}^{r}U(\beta )U(\alpha )|+⟩$, up to global phase. A computational output bit $z$ is corrected to $z\oplus s$; the $Z$ factor affects only phase. Adaptive sign choice is required for general angles, although the final bit correction is purely classical.