| paper | d4-conf-ijcai-2020-0257 |
| authors | — |
| venue | ? |
| filed under | unclassified |
| judged by | gpt-6-luna / xhigh (triple__luna__xhigh__c6r1) |
| judge confidence | high |
| authors would recognise it | yes |
NP-hard→NP-hard discrete result → its continuous mirror
Answer: The decision problem is \(\mathrm{NP}\)-complete, even when all masses are dyadic. · discrete side: hardness · adjudication: proved
full proof & adversarial review →
Theorem 2
statement extracted from the paper’s text layer
Given candidates \(C\), a candidate graph \(H\), committee size \(k\), a common axis \(\pi\), a rational threshold \(q\), and a finite-support distribution \(\mu\) with binary-rational masses over rankings single-peaked on \(\pi\), is there a set \(S\subseteq C\) with \(|S|=k\), \(H[S]\) connected, and \(\sum_t \mu_t \max_{c\in S}(m-\operatorname{pos}_t(c))\ge q\)?
Types are complete rankings \(t\) over \(C\), supported on rankings single-peaked on \(\pi\), with population masses \(\mu_t\). Choose a discrete committee \(S\) of size \(k\) whose induced graph \(H[S]\) is connected, maximizing \(\sum_t \mu_t\max_{c\in S}(m-\operatorname{pos}_t(c))\).
This mirror covers the decision question in Theorem 2 for single-peaked profiles; it does not settle the other connected-committee or conflict-free results, or their parameterized cases.
The strongest honest case is that this paper has a natural high-multiplicity mirror for its connected-committee problems, even though the hardness results suggest that continuity may not make them easier.
Consider a large professional association selecting \(k\) candidates for a committee whose members need communication links between them. Association members rank the candidates; edges in the candidate graph represent those links. With \(100{,}000\) or more members but, say, dozens or hundreds of recurring complete rankings, a voter type is a ranking \(t\), and its mass \(\mu_t\) is the fraction of members holding it. This gives many agents but relatively small support \(\tau\), while preserving the paper’s committee, preference, and candidate-network questions. It is a more convincing population regime than the paper’s example where “voters” are a finite set of required skills.
My lead anchor is Theorem 3. “CONN-k-BORDA is NP-complete.” This is the paper’s own theorem, rather than a result attributed to another source; its proof is not included in the supplied text. It says that deciding whether there is a size-\(k\), connected committee with Borda score at least the given threshold is NP-complete.
The continuous problem I would hand to a prover is \(\mathrm{CONN}\text{-}k\text{-Borda}_{\infty}\). An instance gives candidates \(C\), an undirected candidate graph \(H\), a committee size \(k\), a rational threshold \(q\), and a finite-support distribution \(\mu\) over complete rankings of \(C\), with binary-rational masses summing to \(1\). For ranking \(t\), let \(\operatorname{pos}_t(c)\) be the position of candidate \(c\), and set
\[
\mathrm{sc}_{\mu}(c)=\sum_t \mu_t\bigl(m-\operatorname{pos}_t(c)\bigr).
\]
The question is whether there is a set \(S\subseteq C\), with \(|S|=k\), such that \(H[S]\) is connected and
\[
\sum_{c\in S}\mathrm{sc}_{\mu}(c)\ge q.
\]
A solution is such a committee \(S\); the optimization version maximizes the left-hand side over connected size-\(k\) committees.
Theorem 3’s hardness transfers directly. Given a discrete election with \(n\) voters, set \(\mu_t\) to the fraction of voters with ranking \(t\), and set \(q=s/n\) for the original score threshold \(s\). Every committee has continuous score exactly its discrete score divided by \(n\), and connectedness is unchanged. This is a polynomial-size transformation with \(\tau\le n\). The continuous decision problem is also in NP: a proposed committee can be checked for size, connectivity, and threshold in polynomial time using the rational input. So the expected outcome is Class B: NP-completeness survives through the discrete embedding, not continuum-specific hardness. The paper’s Theorem 4 also gives a natural follow-up: determine which of its parameterized algorithms carry over with dependence on \(k\), \(\tau\), and the mass bit length.
A second, genuinely useful anchor is Theorem 2. “CONN-CC is NP-complete for SP profiles.” This is also the paper’s own result, not a cited theorem; the supplied text gives the statement but not its proof. It establishes NP-completeness even when the rankings are single-peaked.
Its mirror, \(\mathrm{CONN}\text{-CC}_{\infty}^{\mathrm{SP}}\), has the same \(C,H,k\) setup, plus a witnessed common axis \(\pi\), a distribution \(\mu\) supported only on rankings single-peaked with respect to \(\pi\), and a rational threshold \(q\). Define
\[
\mathrm{CC}_{\mu}(S)
=\sum_t\mu_t\max_{c\in S}\bigl(m-\operatorname{pos}_t(c)\bigr).
\]
The question is whether some connected size-\(k\) set \(S\) has \(\mathrm{CC}_{\mu}(S)\ge q\); the solution is the committee \(S\). For a discrete single-peaked profile, its empirical distribution and the threshold \(q=s/n\) preserve every yes/no answer, so NP-hardness transfers; the committee remains a polynomial-size certificate. The expected classification is again Class B. This anchor makes the case sturdier by showing the population mirror can retain a meaningful preference restriction rather than requiring arbitrary rankings.
The soft spot is that a continuous population is a good description only in the large-electorate regime. The paper also emphasizes settings with few voters, and its skill-based story need not describe a population of people at all. In the association scenario, though, voters really are numerous, and the committee and graph constraints remain exactly the paper’s. These mirrors therefore seem like questions the authors could recognize as their problems. Their main value is a precise high-multiplicity complexity question; the cited hardness results suggest that, for these anchors, continuization preserves the challenge rather than dissolving it.
The strongest objection is that the paper’s computational choices are committees, while voter mass only supplies their scores. For Theorem 3, define
\[
w(c)=\sum_t\mu_t\bigl(m-\operatorname{pos}_t(c)\bigr).
\]
Then \(\mathrm{CONN}\text{-}k\text{-Borda}_{\infty}\) is exactly the problem of finding a connected \(k\)-candidate set of maximum total node weight. The distribution over rankings vanishes after computing \(m\) rational weights. The paper itself reduces this problem to a weighted tree problem and gives an FPT algorithm in \(k\). So the mirror adds little population-side structure: its most natural contribution would be to recognize a familiar graph optimization problem as a high-multiplicity election.
That is a real case against Theorem 3 as a *ChoCo* priority, but it does not defeat the mirror. Rational weights are a sensible representation of a large electorate, and the resulting question has a clean discrete embedding. The programme explicitly treats hardness transfer as a worthwhile result; that the graph constraint may be doing most of the computational work is not enough to rule it out.
Theorem 2 is harder to argue against. Its stronger mirror uses an electorate with a common single-peaked axis and the mass-weighted Chamberlin–Courant score
\[
\mathrm{CC}_{\mu}(S)=\sum_t\mu_t\max_{c\in S}\bigl(m-\operatorname{pos}_t(c)\bigr).
\]
Unlike Borda, this generally cannot be reduced to one weight per candidate: which representative a voter type gets depends on the whole committee. A large association with recurring ideological rankings is a plausible high-multiplicity setting, and the empirical distribution of a discrete single-peaked election preserves its score after normalization. The fact that the theorem’s hardness may come from the graph constraint does not invalidate that question.
The general negative case is therefore weak. The Borda anchor looks like a low-yield mirror because the population compresses to candidate weights. But the single-peaked CC anchor survives the stronger objections: its population interpretation is sensible, its objective retains type-level information, and its computational question is well-posed. I cannot honestly defend the universal claim that neither anchor has a worthwhile continuous mirror.
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.