d4-conf-ecai-2024-0151

· ? (d4-conf-ecai-2024-0151)

mirror foundnew result: NP-hard → NP-hard — proved & adversarially reviewed →
paperd4-conf-ecai-2024-0151
authors—
venue?
filed underunclassified
judged bygpt-6-luna / xhigh (triple__luna__xhigh__c6r1)
judge confidencehigh
authors would recognise ityes

The new result

NP-hard→NP-hard discrete result → its continuous mirror

Answer: \(\mathrm{NP}\text{-hard}\) · discrete side: hardness · adjudication: proved

full proof & adversarial review →

The anchor — hardness

Theorem 1

PDF text-layer excerpt — equations and symbol order may be approximateFor each judgment aggregation rule R ∈ {KEMENY, SLATER, MAXH, YOUNG, DODGSON}, the problems of manipulation and bribery are Σp 2-complete. Hardness holds even when outcome L is restricted to |L| = 1.

statement extracted from the paper’s text layer

Every anchor argued

The continuous mirror question

Given an agenda \(\Phi\), integrity constraint \(\Gamma\), target literal \(\ell\), budget \(\beta\), and rational fixed and eligible masses over complete \(\Gamma\)-consistent judgment sets, decide whether eligible mass can be transferred between judgment types so that at most \(\beta\) mass changes and every continuous Young outcome contains \(\ell\). The resulting mass at type \(K\) is \(\mu'_K=\mu^{\mathrm{fixed}}_K+\sum_J x_{J,K}\), where transfers satisfy \(\sum_{J'}x_{J,J'}=\mu^{\mathrm{eligible}}_J\) and \(\sum_{J\ne J'}x_{J,J'}\le\beta\). Continuous Young outcomes are all complete \(\Gamma\)-consistent judgment sets extending the strict-majority set of any maximum-mass retained subdistribution \(\rho\le\mu'\) whose majority set is \(\Gamma\)-consistent.

The model it lives in

A society over types \((J,\text{eligibility})\), with rational mass \(\mu\); the decision variables transfer eligible mass from reported judgment type \(J\) to \(J'\), the objective is changed mass, and the constraint is that every maximum-retention Young outcome contains the target literal.

What the mirror covers

The mirror covers only the Young bribery case of Theorem 1; it leaves manipulation, the other aggregation rules, and the CEGAR algorithms and experiments aside.

Open questions for a prover

The case FOR (proponent)

The strongest mirror I see is the Young bribery result, because Young’s rule already asks how much of a profile must be removed to restore a consistent majority. In a population model, “how many members?” has a natural counterpart: “what fraction of the population?”

My anchor is Theorem 1, specifically its Young-bribery case. The theorem states that for each of Kemeny, Slater, MaxHamming, Young, and Dodgson, both manipulation and bribery are \(\Sigma_2^p\)-complete, even when the desired outcome is a single literal. For Young, the paper proves manipulation hardness by reducing from \(\exists\forall\)-QBF-SAT, then obtains bribery hardness by reducing manipulation to bribery with one corruptible agent and budget \(k=1\). The theorem’s Kemeny result has prior hardness work behind it; the Young result is part of this paper’s extension.

Here is the continuous problem I would hand to a prover. Call it Young Mass Bribery. Fix an agenda \(\Phi\) and integrity constraint \(\Gamma\), and let \(T\) be the complete \(\Gamma\)-consistent judgment sets. The input gives two distributions over \(T\): \(\mu^{\mathrm{fixed}}\) for members who cannot be bribed, and \(\mu^{\mathrm{eligible}}\) for those who can. Their total mass is \(1\); the distributions have rational masses and combined support size \(\tau\). Eligibility is part of a member’s type, since it affects what intervention is possible. The input also gives a target literal \(\ell\in\Phi\) and a rational budget \(\beta\).

Choose transfers \(x_{J,J'}\geq 0\) from eligible mass of judgment type \(J\) to reported type \(J'\in T\), with \(\sum_{J'}x_{J,J'}=\mu^{\mathrm{eligible}}_J\). The resulting distribution is
\[ \mu'_K=\mu^{\mathrm{fixed}}_K+\sum_{J\in T}x_{J,K}. \]
The changed mass is \(\sum_{J\ne J'}x_{J,J'}\). The question is whether there is a transfer plan changing at most \(\beta\) mass such that every outcome selected by continuous Young aggregation on \(\mu'\) contains \(\ell\). Equivalently, the optimization version minimizes changed mass subject to that guarantee.

For clarity, continuous Young aggregation here retains a subdistribution \(\rho\leq\mu'\) of maximum total mass whose strict-majority judgment set is \(\Gamma\)-consistent. A literal \(l\) is in that majority set when the retained mass supporting \(l\) exceeds half of \(\sum_J\rho_J\). The rule returns every complete \(\Gamma\)-consistent judgment set extending the majority set of some maximum-mass retained subdistribution. Thus the guarantee is robust to ties among optimal retentions, just as the paper requires the target to occur in every selected outcome.

A plausible regime is a large public consultation or membership vote on a fixed set of linked policy propositions. Many participants may share the same complete, coherent position and the same eligibility for contact by an outside campaign. The type is that full position plus eligibility; mass is the fraction of participants with that type. In the intended regime, the number of participants is very large while the number \(\tau\) of represented types is moderate. A campaign’s budget is naturally a share of the eligible population. This preserves the paper’s central question—whether an outside party can make a desired judgment unavoidable under Young aggregation—while expressing both the electorate and the intervention in population shares.

I would expect hardness to survive, most plausibly as Class B, because the paper’s Young reduction puts substantial combinatorics in the agenda and integrity constraint, not just in the number of voters. But Theorem 1 does not establish that transfer: fractional mass can split a type, and the discrete proof may rely on indivisible votes. An answer-preserving reduction, or a rounding result showing when fractional plans can be converted to discrete ones, is a key next question. Others are whether the problem has an algorithm polynomial in \(m,\tau\), and the input bit length; whether fixing \(m\) or \(\tau\) changes its complexity; and whether the nested optimization over minimum-mass deletions can be handled by LP duality or separation.

The weakest point is exactly that fractional split. A finite profile maps naturally to rational proportions, but continuous bribery permits changes involving less than one voter’s share. That can alter the Young outcome and break the paper’s hardness gadget. I would still expect the authors to recognize Young Mass Bribery as their problem’s continuous-population analogue: the types, target, rule, and guarantee are preserved, while the count of altered members becomes altered mass. This case covers the Young bribery result only; it does not claim to mirror the paper’s manipulation results, its other aggregation rules, or its CEGAR experiments.

The case AGAINST (opponent, writing after the proponent)

The strongest case against this anchor is that the theorem’s Young-bribery hardness is less directly tied to a large population than the proponent suggests. The paper gets bribery hardness by reducing from manipulation with a single corruptible agent and budget \(k=1\). That agent can be pivotal. In an atomless population, one person has zero mass, so this mechanism does not carry over. Giving the continuous problem a positive budget \(\beta\) changes the regime: it models bribing a share of the population, roughly \(k=\beta n\), rather than the theorem’s single-agent instance. The theorem therefore supplies no direct reason to expect its complexity result to survive this mirror.

That weakens the claimed bridge, but it does not defeat the mirror. The proposed model is a sensible high-multiplicity version of bribery: eligibility is part of a complete type, judgment sets are the opinions that Young aggregation uses, and changed mass replaces the number of changed agents. A large consultation over a fixed set of linked propositions, with recurring opinion types and a campaign able to reach a fraction of participants, is plausible. Young’s rule is especially well matched to a mass formulation because its defining optimization is already over how much of the profile to retain.

Nor does fractional bribery make the question disappear. It changes the optimization problem, and the authors could reasonably ask whether that change enables continuous-optimization methods or leaves the problem hard. The nested choice of bribed mass and maximum-retained subdistributions makes that a concrete computational question.

So the negative case has one real qualification: this is a new proportional-mass bribery problem, not a limit argument that preserves the paper’s single-agent hardness reduction. But that is not a fundamental modeling objection. I cannot honestly make a strong case that this mirror is not worth asking. The Young-bribery anchor survives.

The adversarial triple: the proponent anchors on up to three named results; the opponent sees that case and must defeat every anchor; the judge decides which case convinced it. These are the pipeline’s own outputs, generated by tools/triple_run.py — no human edited them. The paper’s own text is not reproduced here beyond the quoted statement above.