Education / Quantum Cryptography

Quantum Cryptography · Chapter 06

Beyond Key Distribution

Which cryptographic tasks can quantum theory enable, and which remain constrained?

In this chapter · 3 sections
  1. 6.1 Encrypting a quantum message
  2. 6.2 Commitment and its limitations
  3. 6.3 Protocol audit and final project
Reading guide · preparation and approach

Starting point: Chapters 1–5 of this course.

Anchor example: Pauli encryption and a delayed-measurement commitment attack.

Method: Trace the protocol, calculate an example, and identify the security assumptions.

Watch for: A cryptographic claim is meaningful only within its stated model.

This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.

The first five chapters asked one question: how do two parties share a secret key? Quantum mechanics supports several other cryptographic tasks, and it also rules some of them out. This chapter treats both directions. Section 6.1 shows that one qubit can be hidden perfectly with two bits of classical key—and that hiding is not the same as protecting against tampering. Section 6.2 shows a task that quantum information cannot deliver unconditionally, and identifies the mechanism behind the obstruction. Section 6.3 turns the whole course into an audit: a specification, a set of reproducible calculations, and an explicit list of what remains unproved.

6.1 Encrypting a quantum message

A one-time pad hides a value by randomizing it. Quantum states carry two kinds of information—bit and phase—so a quantum pad must randomize both. This section proves that the resulting average erases everything, including correlations with a system the sender does not control.

6.1.1 Average over Pauli keys

Randomizing a value is not enough for a qubit, which has two degrees of freedom. Choose independent uniform bits \(a,b\) and apply \(X^aZ^b\). Averaging the resulting density matrices over the four equally likely choices gives \(I/2\) for every input state, so the ciphertext tells an eavesdropper nothing about the plaintext. A receiver who knows \((a,b)\) applies the inverse Pauli and recovers the state exactly, since any overall sign is unobservable.

The cost is two classical key bits per qubit in this construction. That is also known to be optimal for exactly encrypting an unknown qubit, which is why quantum encryption is expensive compared with the classical pad it generalizes.

Two limits belong with the claim. First, the channel hides the input but does not detect tampering: an attacker can still alter the encrypted state, and Section 6.1.3 shows exactly what happens when she does. Second, the strong statement is not about the ciphertext's marginal state but about its correlations. A message entangled with a reference must be hidden together with its correlations, and the averaged channel delivers that stronger property too.

Worked example

Encrypt \(|0\rangle\) with all four equally likely Pauli choices and average, then say what the receiver must be told.

Solution and reasoning

The four ciphertexts are \(|0\rangle\), \(|1\rangle\), \(i|0\rangle\) and \(-|1\rangle\) up to phase, so two outcomes give \(|0\rangle\) and two give \(|1\rangle\) as density matrices. Their average is \(I/2\): no measurement distinguishes the plaintext. The receiver needs both key bits \((a,b)\), because \(a\) alone identifies the bit flip and \(b\) alone the phase flip; knowing only one leaves a state that is still random in the other degree of freedom.

6.1.2 Prove the Pauli average for an arbitrary qubit

Any qubit density operator can be written

\[\rho=\frac{I+r_xX+r_yY+r_zZ}{2},\]

where the real Bloch vector \((r_x,r_y,r_z)\) has length at most one. Conjugation by a Pauli either preserves or flips each component, and the four choices of \(X^aZ^b\) flip them in a pattern that averages to zero:

How each key conjugation acts on the three Bloch components. Every column contains two \(+1\) and two \(-1\) entries, which is why the average vanishes.
Key \((a,b)\)Pauli\(X\) component\(Y\) component\(Z\) component
\((0,0)\)\(I\)\(+\)\(+\)\(+\)
\((1,0)\)\(X\)\(+\)\(-\)\(-\)
\((0,1)\)\(Z\)\(-\)\(-\)\(+\)
\((1,1)\)\(XZ\)\(-\)\(+\)\(-\)

Summing the four rows cancels every component, leaving

\[\mathcal E(\rho)=\frac14\sum_{a,b=0}^{1}X^aZ^b\rho\,Z^bX^a=\frac{I}{2}.\]

The proof works for every Bloch vector, including mixed states, because nothing in it used \(|\mathbf r|=1\). More generally the channel acts on any operator \(A\) as

\[\mathcal E(A)=\operatorname{tr}(A)\,\frac{I}{2},\]

which is a complete depolarization: the output carries no information about the input operator at all. That is stronger than saying the output state is fixed, and it is what makes the next step work.

Expand a joint state on message \(Q\) and reference \(R\) in matrix units on \(Q\) and apply the identity term by term. Every term with a non-identity matrix unit is annihilated, and only the identity survives:

\[(\mathcal E_Q\otimes\mathrm{id}_R)(\rho_{QR})=\frac{I_Q}{2}\otimes\rho_R.\]

This is the statement a quantum encryption proof actually needs. The ciphertext is not merely uniform; it is uncorrelated with everything the sender kept. An adversary holding the reference learns nothing about the plaintext, and the receiver's inverse Pauli restores the original joint state exactly.

Pauli encryption decorrelates a message from its reference Before encryption the message system Q is entangled with a reference R. The message passes through a Pauli pad chosen by two secret bits, while the reference is untouched. Afterwards the two systems share no information. Decryption with the key restores the entanglement. Q R entangled XᵃZᵇ a, b secret Q′ : I/2 R : unchanged after encryption the joint state is a product, so Q′ and R share no information; the key restores the original state.
Pauli encryption as a decoupling operation. The message passes through the pad while the reference does not, and the resulting joint state factorises. The useful claim is about the pair, not just about the ciphertext: an adversary who holds \(R\) learns nothing about \(Q\) until the key is used.

Worked example

Verify the average for the Y-basis state \(|{+}i\rangle\), whose Bloch vector is \((0,1,0)\).

Solution and reasoning

Conjugating \(Y\) gives \(XYX=-Y\), \(ZYZ=-Y\), and \((XZ)Y(XZ)^{\dagger}=+Y\), because \(XZ\) anticommutes with \(Y\) twice. The four contributions are therefore \(+Y,-Y,-Y,+Y\), which sum to zero, leaving \(I/2\). The cancellations are independent of the particular vector: no matter which component is nonzero, two keys flip it and two preserve it.

6.1.3 Show why encryption does not authenticate a quantum state

Suppose an attacker applies \(X\) to the encrypted qubit. After the receiver decrypts with the secret Pauli \(P\), the effective operation on the plaintext is

\[P^{\dagger}XP.\]

Paulis either commute or anticommute, so \(P^{\dagger}XP=\pm X\) for every possible key: the sign is a global phase and vanishes from the density matrix. The attacker has therefore enacted a logical \(X\) on the plaintext without knowing the key, and the receiver cannot tell that anything happened.

What each tampering Pauli becomes after decryption. Confidentiality survives every row; integrity survives none of them.
Attacker appliesAfter decryptionEffect on the plaintextDetectable by the pad?
\(I\)\(I\)none—
\(X\)\(\pm X\)bit flipno
\(Z\)\(\pm Z\)phase flipno
\(Y\)\(\pm Y\)bothno

So the output remains hidden but may be wrong. Hiding and integrity are different requirements, and a pad that satisfies the first supplies none of the second.

A quantum authentication scheme adds a test whose acceptance and rejection behaviour is covered by a stronger security definition. It must protect a state the sender may not even know, including its entanglement with a reference; that rules out simply measuring the message to see whether it "looks right", since a measurement would destroy exactly the correlations the definition is about. Encryption and authentication use the same kinds of key material but answer different questions.

The distinction reappears in quantum key distribution. Eve can damage signals, and the protocol's response must be either to reject or to produce a key meeting its definition. No-cloning is genuinely useful in quantum cryptography, but it does not create a universal authentication mechanism: the object that needs an argument is the full encode, transmit, decode, and verify process.

6.2 Commitment and its limitations

Commitment is the cryptographic task of choosing now and revealing later. It is the natural test case for what quantum information can and cannot do, because it asks for two properties that pull in opposite directions.

6.2.1 Specify the cheating parties

Bit commitment asks Alice to commit to a bit now and open it later. Hiding means Bob cannot learn the bit before the opening; binding means Alice cannot change it afterwards. Each property protects a different honest party, which is why an argument that establishes one says nothing about the other.

The two commitment properties, and the failure each one forbids.
PropertyProtectsAgainstBroken when
HidingAn honest AliceA cheating Bob, who wants the bit earlyBob's two possible states are distinguishable
BindingAn honest BobA cheating Alice, who wants to change the bitAlice holds a state that opens either way

In the standard non-relativistic two-party quantum model without extra restrictions, unconditional security for both properties is impossible. A simple illustration uses two ensembles: the equally weighted Z states and the equally weighted X states. Both give Bob \(I/2\), so the scheme is hiding as an average preparation. But Alice can keep half of \(|\Phi^+\rangle\) and postpone her basis measurement, obtaining a compatible opening in either basis.

That example exhibits a cheating strategy; it is not the general impossibility proof. And the no-go statement must not be exported to every model. Extra resources or restrictions—bounded quantum storage, computational assumptions, trusted hardware, relativistic signalling constraints—change the analysis, and each requires its own argument.

Worked example

Compare Bob's density matrices for equal mixtures of Z eigenstates and of X eigenstates.

Solution and reasoning

Both mixtures give \(I/2\), since each is an average over an orthonormal basis. No measurement Bob performs on his half can distinguish them, so hiding holds for this scheme. Note carefully what was shown: the ensembles are indistinguishable, which is a statement about Bob's marginal state. It is not a statement about Alice's ability to keep multiple options open, and it is that second ability which fails in Section 6.2.2.

6.2.2 Construct a coherent cheating strategy

Consider a concrete proposal. To commit to 0, Alice sends a uniformly random Z eigenstate; to commit to 1, she sends a uniformly random X eigenstate. At opening she announces the basis and the bit, and Bob measures in that basis and checks agreement. Honest openings always pass, and both commitment ensembles average to \(I/2\), so the scheme appears perfectly hiding.

A cheating Alice does not send a random state at all. She prepares \(|\Phi^+\rangle\), sends Bob one half, and keeps the other without measuring it. At the opening she chooses which commitment to reveal:

In both cases Bob's conditional state matches the label she announces, and the check passes. Because she chooses between the two openings only after the commitment stage, the scheme is not binding.

A coherent attack on a commitment scheme At commitment, Alice sends Bob one half of a Bell pair and keeps the other; Bob's half alone is maximally mixed. At opening, Alice chooses to measure her half in the Z basis and claim zero, or in the X basis and claim one. Both branches leave Bob with a state consistent with the announced label. commit Alice holds half A of Φ⁺, unmeasured Bob holds half B state I/2 open measure A in Z, claim 0 measure A in X, claim 1 Bob checks the named basis Z branch: agrees X branch: agrees either opening passes
The cheating strategy. Alice's half stays unmeasured through the commitment stage, so Bob's marginal state is \(I/2\) and hiding is preserved. Only at opening does she choose a basis, and each choice produces a state consistent with the label she announces. Binding fails even though hiding holds perfectly.

No superluminal signal is involved. Bob's unconditioned state is \(I/2\) throughout, and he learns something only when Alice speaks—at which point the announcement itself determines what he can check. The attack uses coherent postponement, not any physical change in Bob's marginal state, which is why it survives every test Bob can perform on his own half.

6.2.3 Explain the mechanism behind the general obstruction

In the perfectly hiding case, Bob's reduced density operators for the two commitments are equal. Write two purifications of that shared state in a Schmidt basis aligned with Bob's system:

\[|\Psi_b\rangle=\sum_j\sqrt{\lambda_j}\,|a_j^{(b)}\rangle_A|j\rangle_B,\qquad b\in\{0,1\}.\]

The coefficients \(\lambda_j\) are the same in both rows because the reduced state on Bob is the same. Only the vectors \(|a_j^{(b)}\rangle\) in Alice's private space differ, and any two orthonormal families of the same size are related by a unitary. So a unitary \(U\) on Alice's space satisfies

\[U|a_j^{(0)}\rangle=|a_j^{(1)}\rangle\quad\text{for all }j,\]

converting one purification into the other while leaving Bob's side untouched.

Two purifications of one hiding state Two rows, one for each commitment. Bob's factors are identical in both rows, with the same Schmidt coefficients. Alice's factors differ, and a single unitary on her private space maps the top row to the bottom row without touching Bob's system. Alice: |a⁽⁰⁾ⱼ⟩ same λⱼ Bob: |j⟩ Alice: |a⁽¹⁾ⱼ⟩ same λⱼ Bob: |j⟩ one unitary U on A Bob cannot tell which row he is in; Alice can choose.
Why hiding leaves room to cheat. Because Bob's reduced states coincide, the two purifications share the same Schmidt coefficients and live in the same Bob basis. The difference sits entirely in Alice's private space, where a unitary can convert one into the other. Alice's ability to hold an unmeasured purification is exactly the freedom that breaks binding.

This is the mathematical reason hiding gives Alice room to change her opening. Approximate hiding requires a quantitative closeness argument—if the two states are only nearly equal, the unitary only nearly works—and interactive protocols require careful treatment of transcripts and coherent strategies. Mayers' impossibility result supplies that broader analysis within its model.

The theorem constrains unconditional two-party commitment with unrestricted coherent adversaries in the standard setting. It does not say that QKD is impossible: the honest parties' goals and the adversary's position are different, which is why one no-go result says nothing about the other task. Nor does it settle commitment under storage bounds, computational assumptions, trusted hardware, or relativistic arrangements; each of those models needs its own analysis.

Other quantum primitives—randomness generation, secret sharing, signatures, delegated computation—pose different security questions: who may collude, whether messages are classical or quantum, what must stay private, and whether verification is transferable. Before importing a QKD intuition, write the new ideal functionality and check which properties it demands. The final project asks for exactly that kind of disciplined comparison.

Worked example

Both \(|\Phi^+\rangle\) and \(|\Phi^-\rangle\) are purifications of \(I/2\) on Bob. Exhibit the unitary that maps one to the other and check that Bob's factors are unchanged.

Solution and reasoning

Both states have \(\lambda_0=\lambda_1=1/2\) and the same Bob basis. Alice's two families are the Z basis for \(|\Phi^+\rangle\) and the pair \(\{|0\rangle,-|1\rangle\}\) for \(|\Phi^-\rangle\), and \(U=Z\) maps the first family to the second. Acting on Alice's half only, \((Z\otimes I)|\Phi^+\rangle\) gives \((|00\rangle-|11\rangle)/\sqrt2\), which is \(|\Phi^-\rangle\). Bob's marginal is \(I/2\) before and after, so he has no way to detect which purification Alice holds—the concrete version of the general argument.

6.3 Protocol audit and final project

The last section is an exercise in accounting. Anyone can simulate a protocol; the skill this course aims to teach is knowing what a simulation did and did not establish, and being able to say so precisely.

6.3.1 Make a reviewable security argument

A reviewable claim has three kinds of support, and mixing them is the most common failure in student projects. Every statement in a dossier should be traceable to a calculation you performed, an assumption you stated, or a theorem you cited with its hypotheses.

The audit chain and the step that must be cited Counts and a test sample give a classical error bound. Turning that bound into a min-entropy bound is a separate theorem, drawn as a dashed step. Only then does a key length follow, after leakage and margins are subtracted. counts and test sample classical error bound missing step min-entropy bound key length minus leakage the dashed step is a theorem with attack and device assumptions, not something a simulation can supply.
The audit chain. Counts and a sample give a classical bound, and a separate theorem must convert that bound into a statement about Eve; only then does a key length follow. A dossier that draws the dashed arrow without naming its theorem has not made a security argument.

Concretely: specify an ideal BB84 teaching implementation by defining its sources, bases, bit mapping, random choices, public messages, sample selection, reconciliation, verification, extractor seed, authentication resource, and abort rules. Reproduce the intercept–resend error calculation and the asymptotic entropy curve, labelling both as model calculations. Then write a separate finite-block ledger that starts from an explicitly assumed entropy bound, and identify the theorem you would need to replace that assumption with an experimental claim.

The comparison with post-quantum cryptography belongs in the same ledger, because the two approaches are often ranked by name rather than by resource.

Two different routes to key establishment. Neither is uniformly stronger; they fail in different ways.
AspectQuantum key distributionPost-quantum cryptography
Resource usedQuantum channel plus authenticated classical channelClassical channel only
Security basisPhysical law plus a device and protocol modelComputational hardness believed to resist quantum algorithms
AdversaryUnbounded computation, constrained only by physicsBounded computation, including quantum algorithms
Typical failureImplementation or device flaw, side channel, model mismatchCryptanalytic advance, parameter error, bad randomness
What it does not giveAuthentication without an initial secretEverlasting secrecy against an unbounded adversary

A strong submission makes every claim traceable to one of the three categories and never lets a successful simulation stand in for a theorem.

Worked example

A project simulates BB84 and observes 25% QBER under intercept–resend. What has it demonstrated?

Solution and reasoning

It has reproduced a prediction of one specified attack model. That is a real but narrow result: it confirms the arithmetic of Section 2.2.1 and the correctness of the simulator. It has not established security against arbitrary attacks, validated physical devices, or derived a finite-key entropy bound—the three things the audit chain requires before "25% QBER" can be turned into a statement about Eve.

6.3.2 Run a reproducible experiment and diagnose its limits

The downloadable BB84 laboratory uses only the Python standard library. It simulates ideal detections, uniform basis choices, random-basis intercept–resend on a chosen fraction of rounds, sifting, and public testing. A fixed seed makes a run reproducible; the pseudorandom generator is a teaching tool, not a source of cryptographic randomness.

python3 education/cryptography/examples/bb84_lab.py --signals 20000 --attack 0 --seed 7
python3 education/cryptography/examples/bb84_lab.py --signals 20000 --attack 0.2 --seed 7
python3 education/cryptography/examples/bb84_lab.py --signals 20000 --attack 1 --seed 7
python3 education/cryptography/tools/check_examples.py

Predict before running: the expected sifted QBERs are zero, 0.05, and 0.25. Then compare the predictions with the reported sample estimates and ask how far each one moved, given the size of the test sample. Three diagnostic points are worth keeping in view.

First, the program also prints the hidden raw-block error rate. That number is available only because this is a simulation; a real Alice and Bob cannot inspect all their disagreements without publishing the data they disagree about. Second, the finite-population calculation bounds a classical error string. It is explicitly not a phase-error certificate, which is the dashed step of the audit chain. Third, the module exposes syndrome decoding, Toeplitz hashing, a decoy-yield bound, and a key-budget function so that the calculations of Chapters 2, 4, and 5 can be reproduced, but a deterministic transcript plus hashing is not a secure end-to-end implementation: there is no quantum entropy bound and no authenticated channel.

Worked example

The lab's budget function is called with an entropy allowance of 7500 bits, leakage 1300, verification 40, a privacy-amplification exponent of 40, and 128 authentication bits. Compute the output length and the net expansion.

Solution and reasoning

The exponent 40 means a privacy-amplification error of \(2^{-40}\), and the standard accounting charges twice the exponent, so the output is \(7500-1300-40-2\cdot40=6080\) bits. Subtracting the 128 authentication bits gives a net expansion of 5952 bits. The arithmetic is easy; the assumption is not. The 7500-bit entropy figure is an input, and nothing in this calculation certifies it—that is the theorem the dossier must name, with its attack model and device assumptions.

6.3.3 Submit a protocol dossier with explicit acceptance criteria

The project is a dossier with five sections and a fixed rubric, so that "it worked" is never the criterion.

  1. Specification. Give a diagram and an ordered transcript. State the basis probabilities, bit mapping, source model, block size, sample rule, acceptance threshold, and initial authentication resource. Identify exactly which systems Eve may control.
  2. Attack study. Reproduce the three attack settings above with at least five seeds each. Report counts and rates separately, and compare the spread with the size of the test sample. Explain why an attack simulation cannot prove security.
  3. Reconciliation and extraction. Trace the seven-bit syndrome example, demonstrate a two-error miscorrection, and verify a Toeplitz collision count by exhaustive enumeration. State which entropy premise would be needed to use the hash as an extractor.
  4. Finite-block analysis. Reproduce the exact \(N=20\) sampling example and the 6080-bit budget. Name the missing protocol-specific theorem connecting your observations to the entropy input, with its attack and device assumptions.
  5. Extension. Choose decoy-state estimation, MDI correlations, a CHSH visibility sweep, or the commitment attack. Derive one result and reproduce it numerically. Explain one assumption whose failure invalidates it.

Assessment rubric. Allocate 25% to a complete and consistent protocol specification, 25% to correct derivations and reproducible results, 25% to entropy, leakage, and error accounting, and 25% to the assumption audit and interpretation. A project that labels simulated agreement as a proof of secrecy has not met the central learning objective, however many trials succeeded.

For a reading extension, choose one linked primary paper and identify a theorem in it, its hypotheses, and the corresponding objects in your dossier. Distinguish a theorem you implemented from one you merely cited. That mapping—from a classroom model to a stated theorem and back—is the bridge to research-level cryptographic analysis.

Further reading

These sources cover the encryption primitive, the commitment no-go results, and the security framing the final project asks you to adopt.

Summary

Questions and worked problems

These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.

1. Worked problem §

Why is random X alone insufficient for arbitrary qubits?

Hint 1

Try input |+⟩.

Worked solution

X leaves |+⟩ unchanged, so that input is not randomized; Z randomization is also needed.

Revisit the relevant explanation

2. Worked problem §

How many key bits does this construction use for n qubits?

Hint 1

Use two independent bits for each qubit.

Worked solution

2n classical bits, with fresh independent choices for the stated one-time construction.

Revisit the relevant explanation

3. Worked problem §

What do hiding and binding protect against?

Hint 1

Identify each possible cheater.

Worked solution

Hiding protects an honest Alice from a cheating Bob; binding protects an honest Bob from a cheating Alice.

Revisit the relevant explanation

4. Worked problem §

Does the toy attack prove every proposed commitment scheme insecure?

Hint 1

Distinguish example from theorem.

Worked solution

No. It defeats this scheme. General impossibility requires a theorem with explicit model assumptions.

Revisit the relevant explanation

5. Worked problem §

List the minimum deliverables for this project.

Hint 1

Separate specification, computation, and argument.

Worked solution

A protocol/transcript specification; reproducible rate and attack calculations; an entropy and failure-budget ledger; and an assumption audit with sources.

Revisit the relevant explanation

6. Worked problem §

What is a useful extension after this course?

Hint 1

Choose one assumption to replace with a theorem.

Worked solution

Implement a finite-key analysis from a chosen primary paper, documenting its model and checking that the estimator, leakage, and failure terms match the protocol.

Revisit the relevant explanation

7. Worked problem §

Prove that averaging over the four Pauli encryptions sends every qubit state to I/2, then extend the result to a qubit entangled with a reference.

Hint 1

Expand first in the Pauli basis, then use linearity on matrix units.

Worked solution

Each nonidentity Pauli component appears with two positive and two negative signs under conjugation and cancels, while I survives. Thus the map on any operator A is Tr(A)I/2. Applying this to the Q matrix units of a joint QR state gives I_Q/2 tensor ρ_R. A fixed marginal output is insufficient by itself; the joint identity proves decoupling from the reference.

Revisit the relevant explanation

8. Worked problem §

Show algebraically that an attacker’s Pauli X on a Pauli-encrypted qubit becomes a logical X after decryption, regardless of the secret key.

Hint 1

Commute X through each possible encryption Pauli.

Worked solution

For each encryption Pauli P, P†XP equals X or −X because Paulis commute or anticommute. The sign is a global phase and disappears from density operators. The receiver therefore obtains XρX instead of ρ. This is a deliberate modification despite perfect confidentiality, so the pad alone is not authentication.

Revisit the relevant explanation

9. Worked problem §

Write the two conditional states that Bob receives when a dishonest committer measures her half of |Φ+⟩ in Z or X at the opening stage. Explain why hiding survives while binding fails.

Hint 1

Project Alice’s half onto a basis vector and normalize Bob’s state.

Worked solution

A Z outcome a occurs with probability 1/2 and leaves Bob in |a⟩; an X outcome ± occurs with probability 1/2 and leaves Bob in |±⟩. Before receiving the opening, Bob has I/2 in both cases, so hiding remains. Alice can select the basis only at opening and pass either commitment test, so the proposal is not binding.

Revisit the relevant explanation

10. Extension §

Audit the claim: “Our simulation produced identical random-looking strings, so our device distributes information-theoretically secure keys.” Give three independent missing arguments and a corrected conclusion.

Hint 1

Consider Eve’s information, the physical model, and post-processing.

Worked solution

Agreement does not establish independence from Eve; simulation of selected attacks does not bound arbitrary attacks; and an ideal signal model does not validate the real source, detector, authentication, and leakage assumptions. A corrected conclusion is that the program reproduces specified correlations under its model. A security claim needs a theorem linking observed data to conditional entropy, validated assumptions, and composable processing/error accounting.

Revisit the relevant explanation

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