d4-conf-aaai-2014-0518

· ? (d4-conf-aaai-2014-0518)

mirror foundnew result: other → NP-hard — proved & adversarially reviewed →
paperd4-conf-aaai-2014-0518
authors—
venue?
filed underunclassified
judged bygpt-6-luna / xhigh (triple__luna__xhigh__c6r1)
judge confidencehigh
authors would recognise ityes

The new result

other→NP-hard discrete result → its continuous mirror

Answer: \(\mathrm{coNP}\)-complete under polynomial-time many-one reductions · discrete side: other · adjudication: proved

full proof & adversarial review →

The anchor — other

Proposition 1

PDF text-layer excerpt — equations and symbol order may be approximateCandidate x 2 Y is a robust winner w.r.t. hX, Y, v, ri if there is no destructive control against x by adding candidates, where the spoiler set is U = X \ Y . The proof is immediate: the chair has destructive con- trol against x via adding candidates iff there is a spoiler set S ✓U such that r(v(Y [ S)) 6= x (i.e., x is not a robust winner).

statement extracted from the paper’s text layer

Every anchor argued

The continuous mirror question

Given candidates \(X\), a known-available set \(Y\subseteq X\), a target \(x^\star\in Y\), and a distribution \(\mu\) over a support of complete rankings, does \(x^\star\) win under the paper’s plurality tie-breaking rule for every available set \(A\) with \(Y\subseteq A\subseteq X\), when each candidate’s score is the total mass of rankings that place it first in \(A\)?

The model it lives in

Represent the society by complete ranking types \(\rho\) with rational masses \(\mu_\rho\ge 0\) summing to \(1\). Keep candidate availability as a subset \(A\subseteq X\); for each \(A\), the plurality score of \(c\) is the mass of types whose top-ranked candidate in \(A\) is \(c\). Ask whether the target wins for every \(A\supseteq Y\).

What the mirror covers

The mirrors cover Proposition 1’s plurality robustness result and Corollary 1’s Borda robustness algorithm. They leave the query-policy dynamic program, availability probabilities and costs, and approximate policies outside the population-continuous question.

Open questions for a prover

The case FOR (proponent)

The lead mirror is Continuous Plurality Robust-Winner. Its anchor is Proposition 1, proved in the paper: a candidate \(x\in Y\) is a robust winner exactly when there is no destructive control by adding candidates, using the spoiler set \(U=X\setminus Y\). The paper then states that robust-winner checking is coNP-complete for plurality, deriving this from earlier NP-completeness results for destructive control by adding candidates. The numbering is slightly awkward: Proposition 1 proves the equivalence; the plurality complexity classification appears in the prose immediately after it, and its hardness input is cited from earlier work.

Here is the continuous problem. An instance gives candidates \(X\), a support \(\mathcal T\) of \(\tau\) complete rankings, rational masses \(\mu_\rho>0\) summing to \(1\), a set \(Y\subseteq X\) of known-available candidates, and a target \(x^\star\in Y\). Use the paper’s tie-breaking rule. For any \(A\) with \(Y\subseteq A\subseteq X\), candidate \(c\)’s plurality score is
\[ \operatorname{pl}_\mu(c,A) =\sum_{\rho\in\mathcal T:\,c=\operatorname{top}_{\rho}(A)}\mu_\rho . \]
The question is whether \(x^\star\) wins for every such \(A\). A NO answer has a compact witness: one available set \(A\) where \(x^\star\) loses. This is a decision problem about robust winner determination; it does not introduce a mass-transfer or querying action.

A credible regime is a large electorate ranking public projects or potential sites before costly feasibility studies. Many residents share a small number of complete ranking archetypes, so the population size \(N\) can greatly exceed \(\tau\); pending feasibility results determine which candidates survive. Each type is complete for this problem because its members have the same ranking, while availability and any inspection costs are candidate-level. The model leaves candidate availability discrete and continuizes the society, as required.

The expected result is coNP-completeness: hardness transfers. Map a discrete profile to \(\mu_\rho=n_\rho/n\), where \(n_\rho\) is the number of voters with ranking \(\rho\). For every available set, weighted plurality scores are the original scores divided by \(n\), so the winners are identical. This is a polynomial-size mapping from the paper’s discrete instances. Moreover, cloning every voter the same number of times leaves \(\mu\) unchanged: the underlying electorate can be made arbitrarily large relative to its number of distinct ranking types. The hard choices still concern subsets of candidates, so population continuization does not remove them. This is a useful Class B boundary, not a claim that the continuous representation makes the problem tractable.

A second, tractable mirror is Continuous Borda Robust-Winner, anchored by Corollary 1, proved in the paper from its characterizations. The corollary says that checking whether \(x\) is a robust winner can be done in polynomial time for top cycle, uncovered set, and Borda; I rely on its Borda entry. It uses the same instance schema, asking whether \(x^\star\) wins under Borda for every \(A\) containing \(Y\), with the paper’s unfavorable tie-breaking convention. Define
\[ B_\mu(c,A)=\sum_{\rho\in\mathcal T}\mu_\rho \bigl(|A|-\operatorname{pos}_{\rho|_A}(c)\bigr), \qquad N_\mu(a,b)=\sum_{\rho\in\mathcal T}\mu_\rho\,\mathbf 1[a\succ_\rho b]. \]
For each rival \(z\ne x^\star\), the paper’s Borda characterization reduces the check to one set:
\[ A_\mu(x^\star,z)=Y\cup\{z\}\cup \{u\in X\setminus(Y\cup\{z\}): N_\mu(z,u)>N_\mu(x^\star,u)\}. \]
The target must strictly outscore \(z\) at that set. Computing these sets and scores takes polynomial time in \(m,\tau\), and the input bit length \(L\); rational masses preserve the score comparisons. The expected direction is therefore tractable, Class A. This is an exact weighted-population extension of the paper’s result, not a claim that Borda robustness itself is newly tractable.

These mirrors cover robust-winner verification only. They do not continuize the paper’s decision-tree policies or availability probabilities. Those generate a further question: given \(\mu\), candidate availability probabilities, and query costs, can an optimal policy be computed substantially faster than the paper’s \(O(3^{|U|})\) dynamic program, perhaps using structure in the ranking support?

The weak point is that both mirrors retain the paper’s candidate-subset problem, and both can be expressed through rational type frequencies. The plurality mirror transfers hardness rather than dissolving it; the Borda mirror carries over an existing polynomial characterization. The case is strongest as a faithful high-multiplicity classification of where robustness is hard and where it is tractable, not as evidence that this paper already contains a new continuous complexity result.

The case AGAINST (opponent, writing after the proponent)

The strongest case against these mirrors is a priority argument: both are unusually direct weighted-profile versions of results the paper already establishes. For any fixed available set \(A\), plurality and Borda scores are finite sums over ranking types. The hard branching in robust-winner checking is over candidate subsets; changing voter counts to rational masses does not bring in the continuous-optimization machinery that motivates ChoCo.

That argument does not defeat the plurality mirror. A large population with a small set of shared ranking archetypes is a sensible regime for the paper’s public-project or feasibility-study examples. A discrete profile maps to masses \(\mu_\rho=n_\rho/n\) with polynomial bit length, and every candidate subset has the same winners. Thus the coNP-hardness carries over. Its source being candidate choices makes this a Class B result, which the programme explicitly treats as valuable. Calling it “only a weighted-profile reformulation” is a concern about novelty, not a reason the continuous question is ill-posed.

Borda is harder to dismiss. The paper’s characterization extends directly: replace voter counts in its pairwise comparisons with rational masses, then check the resulting candidate sets and scores. That gives a polynomial-time algorithm in \(m,\tau,L\). It may require little new technique, but it is a valid high-multiplicity classification, and the paper itself does not provide a continuous computational treatment. The continuous optimization toolbox need not be used for the question to count.

The proposed query-policy direction should carry little weight in this case. Its probabilities and tests concern candidate availability, not population mass. It is not evidence for a further population-continuous result unless one separately shows that the weighted society representation changes the complexity of finding policies.

So the best negative case is that these mirrors may be low in novelty and methodologically modest. I cannot make that into the requested universal claim: the type space is natural, population multiplicity is plausible, and neither anchor has a fundamental modeling obstruction. Under the stated rules, the plurality hardness transfer and the Borda algorithm both survive.

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.