| paper | d4-conf-ijcai-2018-0698 |
| authors | — |
| venue | ? |
| filed under | unclassified |
| judged by | gpt-6-luna / xhigh (triple__luna__xhigh__c6r1) |
| judge confidence | medium |
| authors would recognise it | yes |
P→P discrete result → its continuous mirror
Answer: \(\mathrm P\); feasibility is equivalent to \(\sum_{t:b_t=1}\mu_t\ge 2^{-\ell}\), decidable in \(O(\tau L^2)\) bit operations. · discrete side: polynomial · adjudication: proved
full proof & adversarial review →
Theorem 2
statement extracted from the paper’s text layer
Given rational masses \(\mu_t\) over opponent types \(t\), labels \(b_t\) indicating whether the tagged target \(w\) beats type \(t\), and an integer budget \(\ell\), consider sequences of transitive tournaments on \(N\) entrants, where \(N\) tends to infinity through powers of two and the empirical opponent-type masses converge to \(\mu\). Does there exist such a sequence in which at most \(\ell\) match outcomes are reversed and a bracket makes \(w\) win?
A distribution \(\mu\) over opponent types, a tagged target \(w\) outside that mass, and an integer budget \(\ell\) for unit-cost reversals in finite transitive approximants; the decision is whether some bracket and allowed reversals make \(w\) champion along a sequence whose empirical masses converge to \(\mu\).
The target remains a zero-mass tagged entrant, and in a transitive tournament every opponent has a distinct full rank; the proposed \(b_t\) types are therefore a sufficient-statistic quotient rather than identical full tournament types.
fatal: False
This mirrors Theorem 2 for acyclic tournaments with unit reversal costs; it leaves the general-instance algorithms in Theorems 5 and 6 and the paper’s structural results unmirrored.
The strongest positive case is a narrow one: the acyclic-tournament part of bribery, in a large contest with a distinguished target and a high-multiplicity population of opponents.
The anchor is Theorem 2, proved in this paper. For a BTF instance \((D,w,\ell)\) with \(n\) players and acyclic tournament \(D\), it says that \(w\) can be made the winner using at most \(\ell\) reversed match outcomes if and only if
\[
\mathrm{victory}(w)\ge \frac{n}{2^\ell}-1.
\]
When this holds, a solution can be found in polynomial time. The article’s introduction mentions general BTF hardness proved elsewhere, but gives no numbered hardness theorem of its own to anchor here; Theorem 2 is the clean named result for a continuous limit.
Here is the mirror I would put to a prover: Continuous DAG Tournament Bribery (CDTB). Its instance is a finite set \(T\) of opponent archetypes, rational masses \(\mu_t\ge0\) summing to \(1\), a distinguished target player \(w\), a label \(b_t\in\{0,1\}\) for each type indicating whether \(w\) beats that type, and a fixed integer budget \(\ell\). The population mass \(\mu_t\) is the fraction of opponents of type \(t\). Assume finite realizations are acyclic tournaments, with all match reversals costing one. The question is whether there is a sequence of such tournaments, with entrant counts tending to infinity and empirical type masses tending to \(\mu\), in which at most \(\ell\) outcomes are reversed and a bracket makes \(w\) win. A solution is an asymptotic family of those reversal sets and brackets; a no-instance has no such family.
Write \(v=\sum_{t:b_t=1}\mu_t\), the mass of opponents that \(w\) beats. Theorem 2 gives the continuous criterion: CDTB is yes exactly when \(v\ge 2^{-\ell}\). Dividing its finite condition by \(n\) leaves the correction \(1/n\), which vanishes as the population grows. If the task is to minimize the reversal budget, the continuous optimum is \(\lceil\log_2(1/v)\rceil\) for \(v>0\), and no finite budget suffices when \(v=0\). At equality, feasible finite approximations can be chosen from the feasible side. For any chosen finite realization meeting the theorem’s condition, its constructive algorithm supplies the actual bracket and reversals.
A plausible regime is a very large online or annual knockout competition: one designated entrant is the target, while the opponent roster is drawn from a small number of stable skill archetypes. The key assumption is that predicted results are transitive, as in Theorem 2’s DAG restriction. For this decision question, types can be grouped by whether the target beats them: the theorem proves that, in the acyclic case, feasibility depends only on the resulting victory count, not on the opponents’ internal ordering. That makes this grouping sufficient for the continuous yes/no question. The types must also carry any other attributes the problem uses, such as reversal prices; this anchor assumes uniform unit costs.
This is a credible mirror because it keeps the paper’s underlying question—can the target be made to win by choosing a bracket and reversing outcomes?—and takes a high-multiplicity limit of its own acyclic result. The paper’s polynomial characterization becomes an exact mass threshold. It does not continuize the bracket’s outcome space or replace the tournament with a probabilistic model.
The weak point is real: \(w\) remains a single distinguished player, and \(\ell\) remains an integer count of reversals. The continuous object is the opponent population, not every part of the decision. Also, the archetype quotient is sufficient for the decision criterion, while constructing a concrete bracket still uses a finite refinement of the types. A reader who requires a positive-mass target cohort and fractional intervention variables may reject this as too partial a continuization. The case survives only as a focused mirror of Theorem 2, not as a mirror of general BTF.
This covers the acyclic restriction in Theorem 2 only; it does not claim to mirror the paper’s general-instance algorithms in Theorems 5 and 6. Useful next questions are whether the mass threshold has a stable finite-size rounding rule, and whether a similarly meaningful type-mass formulation exists once the tournament has cycles or unequal reversal costs.
The only anchor raised is Theorem 2, and it is a real one: it gives a named computational result, and its condition has a meaningful limit. I would not argue that its answer is dull or already known. The strongest objection is that the proposed limit makes the *opponents* continuous while leaving the paper’s central participant, \(w\), outside the population model.
In an acyclic tournament, the match results form a strict total order. If a type records a player’s full competitive profile, every player has a distinct rank; the instance does not contain repeated types in the high-multiplicity sense. The proposed \(b_t\) grouping—whether \(w\) beats an opponent—does yield the mass \(v\), and Theorem 2 makes that statistic sufficient for the yes/no decision. But this is a compression of the decision problem, not a continuous model of a tournament population: it drops the opponents’ other match relations and retains one named player whose victory is the objective. The limit question is therefore best described as whether a tagged contestant can win against a continuous pool.
The sharper version of that mirror has the same issue. As the entrant count grows, \(w\)’s share tends to zero. Remove the tag and there is no target whose victory can be asked about; keep it and the model is a continuum plus a distinguished individual. The allowed \(\ell\) reversals also concern finitely many individual matches, so their cost per entrant tends to zero. Scaling the budget proportionally with population does not restore a mass-level intervention problem: Theorem 2 notes that a budget of about \(\log_2 n\) already suffices regardless of \(w\)’s strength. The nontrivial threshold \(v\ge 2^{-\ell}\) persists precisely because the tagged individual and finite reversal count remain discrete.
Giving the target a positive-mass type does not repair this while preserving the paper’s question. A cup tournament has one winner, so the winning share of any positive-mass cohort vanishes in the limit. Asking whether some member of that cohort wins still selects an individual and requires enough identity-level match information to distinguish its members. Replacing the objective with a positive mass of winners would be a different problem.
That is a substantial scope objection, but not a conclusive defeat. If ChoCo accepts a tagged-player problem against a continuous opponent pool as a worthwhile population mirror, then the proposed threshold survives: it is well-posed and genuinely depends on opponent mass. My negative case rests on reading the programme’s requirement—make the society itself continuous—as covering all tournament entrants, including the target. Under that reading, Theorem 2 offers a strong singular-limit result, but not a continuous-population tournament problem.
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.