d4-conf-wine-2015-0011

· ? (d4-conf-wine-2015-0011)

mirror foundnew result: NP-hard → NP-hard — proved & adversarially reviewed →
paperd4-conf-wine-2015-0011
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: \(\mathrm{NP}\text{-hard}\), even with no parents per issue; polynomial-time winner determination would imply \(\mathrm{P}=\mathrm{NP}\). · discrete side: hardness · adjudication: proved_with_minor_gaps

full proof & adversarial review →

The anchor — hardness

Theorem 1

PDF text-layer excerpt — equations and symbol order may be approximateIt is \NP-hard to determine the winner of a given profile when using the voting rule One-Step-$k$-Approval with arbitrary $k$ of polynomial size in $n$ and $m$ (i.e., exponential value).

statement extracted from the paper’s text layer

Every anchor argued

The continuous mirror question

Given \(m\) binary issues, a binary-encoded approval size \(k\), and a support of \(\tau\) compact acyclic CP-net types with issue-order information and rational masses \(\mu_t\) summing to \(1\), find a candidate maximizing the total mass of types that rank it among their top \(k\) candidates.

The model it lives in

A society is a distribution \(\mu\) over complete voter types, each consisting of a compact acyclic CP-net and the issue-order information that induces its ranking. Each unit of mass casts its type's full top-\(k\) approval set; candidates remain binary assignments to the issues, and the winner maximizes total approval mass.

The objection that survived

The reduction relies on exponentially large \(k\) and clause-specific issue orders, which may limit how representative the resulting instances are of product-configuration populations.

fatal: False

What the mirror covers

The mirror covers Theorem 1's winner-determination hardness result; it leaves the paper's destructive-bribery classifications and other tractable bribery cases alone.

Open questions for a prover

The case FOR (proponent)

My lead is a narrow, exact mirror of Theorem 1, proved in this paper. It states that determining a winner under One-Step-\(k\)-Approval is NP-hard when \(k\) is part of the input and may have an exponentially large value. The proof reduces 3SAT with \(M\) clauses and \(m\) variables to an election with \(M\) voters and \(m\) binary issues, using \(k=7\cdot 2^{m-3}\). Each clause voter’s top-\(k\) set consists exactly of the assignments satisfying that clause, with the other issues free. A candidate receives all \(M\) votes exactly when its assignment satisfies every clause.

Call the mirror Continuous One-Step-\(k\)-Approval Winner. An instance has \(m\) binary issues, a binary-encoded integer \(1\le k\le 2^m\), and a support \(T\) of \(\tau\) types. A type is a compact acyclic CP-net together with the issue-order information used to derive its preference order. Each type has rational mass \(\mu_t\ge 0\), with \(\sum_{t\in T}\mu_t=1\). For a candidate \(c\in\{0,1\}^m\), define
\[ S_\mu(c)=\sum_{t\in T}\mu_t\,\mathbf{1}[c\text{ is among type }t\text{’s top }k\text{ candidates}]. \]
The task is to return a candidate maximizing \(S_\mu(c)\), with ties treated as co-winners. The decision is the winning configuration; the objective is its total approval mass. Each unit of mass still casts the type’s full top-\(k\) approval set—the model does not fractionalize anyone’s ballot or the outcome space.

A plausible regime is a large product-configurator or e-commerce population: customers have conditional preferences over configurations, and many share the same complete preference type. The mass records each type’s population share. \(k\) may be large when customers accept many configurations because some product features matter little to them. The paper itself studies this rule and proves Theorem 1 for precisely the large-\(k\) case, so the authors should recognize the mirror. It continuizes the population only; the candidates remain discrete configurations.

I expect Class B: hardness transfers. The reduction carries over by giving each clause type mass \(1/M\): a candidate’s score becomes its fraction of satisfied clauses, and a maximizer scores \(1\) exactly when the formula is satisfiable. Cloning each clause type any number of times makes the electorate arbitrarily large while leaving the support at most \(M\). More generally, clearing the denominators of rational masses recovers a finite high-multiplicity profile with exactly the same winner comparisons and ties. The hard combinatorics live in the exponentially large assignment agenda, not in keeping voters indivisible.

This covers Theorem 1’s winner-determination result, not the paper’s later destructive-bribery classifications. Further questions include whether the hard case becomes tractable under a common issue order, or under additional restrictions on the CP-nets or support. The main weakness is that the anchor is a preliminary result, and its reduction needs exponentially large \(k\); broad approval of that size may be less natural in many applications. Still, the mirror keeps the paper’s exact voting rule and candidate space, and its hardness gives a clear boundary: making the society continuous does not remove computational difficulty that comes from the agenda.

The case AGAINST (opponent, writing after the proponent)

The best case against is one of priority: Theorem 1 is a preliminary winner-determination result, while the paper’s main contribution is its classification of destructive bribery. Its hardness construction also needs exponentially large \(k\) and clause-specific issue orders, which may be an awkward fit for many product-configuration populations. A mirror limited to more natural approval sizes or common orders would not inherit this reduction.

That case does not defeat the mirror the proponent actually proposed, or a stronger version. Clause-specific CP-nets are valid types, and assigning each mass \(1/M\) preserves the reduction: a candidate gets approval mass \(1\) exactly when the formula is satisfiable. Cloning voters leaves the support at \(M\), so this is a genuine continuous-population instance. The objection that the hardness comes from the candidate agenda also fails under the programme’s stated criteria: locating where continuization stops helping is an intended result.

So the honest negative case is weak. I can argue that this mirror deserves lower priority than the paper’s bribery results, but I cannot make a sound case that no worthwhile continuous mirror exists. Theorem 1 survives the challenge.

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.