Parameterized Analysis of Bribery in Challenge the Champ Tournaments

Juhi Chaudhary, Hendrik Molter, Meirav Zehavi · IJCAI 2024 (ijcai24-00299)

mirror found
paperParameterized Analysis of Bribery in Challenge the Champ Tournaments
authorsJuhi Chaudhary, Hendrik Molter, Meirav Zehavi
venueIJCAI 2024
filed undervoting · bribery-control
judged bygpt-5.6-luna / xhigh (triple__luna__xhigh__c2r1)
judge confidencemedium
authors would recognise ityes

The anchor — hardness

Theorem 2

CBCCT is weakly NP-hard and W[1]-hard when parameterized by the number of players.

statement extracted from the paper’s text layer

Every anchor argued

The continuous mirror question

Given finite types \(T\), rational masses \(\mu_t\) summing to \(1\), and for each type \(t\) a finite menu of per-unit bribes \((b_{t,j},p_{t,j})\) with \(p_{t,j}\in(0,1]\), choose masses \(x_{t,j}\ge0\) satisfying \(\sum_j x_{t,j}=\mu_t\) and \(\sum_{t,j}b_{t,j}x_{t,j}\le\beta\). Decide whether \(G(x)=\exp\!\left(\sum_{t,j}x_{t,j}\ln p_{t,j}\right)\ge\theta\).

The model it lives in

A finite distribution of interchangeable campaign or approval-gatekeeper types \(\mu_t\); splittable mass variables \(x_{t,j}\); linear per-unit budget; and maximization of the normalized log winning probability \(\sum_{t,j}x_{t,j}\ln p_{t,j}\), equivalently \(G(x)\).

The objection that survived

The normalized objective \(G\) is not the paper's literal fixed-threshold probability that \(e^\star\) wins every match, but a per-capita log-survival rate with an \(N\)-dependent underlying threshold.

fatal: False

What the mirror covers

The mirror directly covers CBCCT and the hardness asserted in Theorem 2; it does not claim to cover the cup-tournament result, Product Knapsack, or the paper's other algorithmic and MILP results.

Open questions for a prover

The case FOR (proponent)

My strongest honest case is anchored on Theorem 2, proved in this paper: CBCCT is weakly NP-hard and W[1]-hard when parameterized by the number \(n\) of non-initial players. I would use this as the lead anchor because the paper itself supplies the right bridge to a population model: its campaign-management interpretation treats \(e^\star\) as a candidate, product, or proposal, and each \(e_i\) as a voter or group whose support is required.

The scenario is a large approval campaign. There are \(N\) voters or local gatekeepers, with \(N\gg\tau\) distinct response types. A type records the complete information relevant to bribery: its baseline probability of rejecting \(e^\star\), its finite menu of campaign interventions, the cost of each intervention, and the resulting probability that the type accepts \(e^\star\). The \(\tau\) types might be recurring combinations of sector, jurisdiction, professional role, and baseline support. The original sports story is not the right regime—thirty boxers are not a continuum—but the paper’s own candidate/product interpretation scales naturally to millions of repeated voter or approval-gate types.

I would call the continuous problem Mass-CBCCT\(_\infty\). An instance consists of a finite type set \(T\), rational masses \(\mu_t\ge 0\) with \(\sum_{t\in T}\mu_t=1\), and, for each type \(t\), a bribe vector \(C_t=\{(b_{t,j},p_{t,j}) : j\in J_t\}\), where \(b_{t,j}\ge0\) is cost per unit mass and \(p_{t,j}\in(0,1]\) is the probability that \(e^\star\) defeats an agent receiving option \(j\). There is also a per-capita budget \(\beta\) and a target \(\theta\in(0,1]\).

A solution is a mass allocation \(x_{t,j}\ge0\). The quantity \(x_{t,j}\) is the fraction of type \(t\) receiving bribe option \(j\), so it must satisfy \(\sum_{j\in J_t}x_{t,j}=\mu_t\) for every \(t\). It is feasible when \(\sum_{t,j}b_{t,j}x_{t,j}\le\beta\). Its normalized winning probability is the continuous product \(G(x)=\exp\!\left(\sum_{t,j}x_{t,j}\ln p_{t,j}\right)\). The decision question is whether some feasible \(x\) has \(G(x)\ge\theta\). Zero-probability options can be handled as having log-loss \(+\infty\), or excluded when \(\theta>0\).

This is not an arbitrary replacement of a product by an average. If \(\mu_t=a_t/N\) and \(x_{t,j}=a_{t,j}/N\), then cloning the population gives \(a_t\) players of type \(t\), and the original winning probability is exactly \(G(x)^N=\prod_{t,j}p_{t,j}^{a_{t,j}}\). Likewise, the original budget is recovered by taking \(B=N\beta\). Thus clearing denominators recovers the finite CBCCT instance, with the same feasible bribe allocations and the same ordering of solutions after taking the \(N\)-th root of the probability threshold.

I expect Mass-CBCCT\(_\infty\) to be Class A. Taking logarithms turns the objective into the linear expression \(\sum_{t,j}x_{t,j}\ln p_{t,j}\), while the mass-conservation constraints and budget constraint are linear. The explicit formulation has \(\sum_t|J_t|\) variables and is solvable by linear programming, with running time polynomial in the number of types, menu entries, and encoding length under the usual computable-real or log-loss representation. The paper’s own proof of Theorem 5 already works with logarithms of probabilities and groups players by probability profiles and bribe-value sets; the continuous formulation makes those grouped quantities genuine masses and removes the need to recover an integral optimum.

The resulting interpretation of Theorem 2 is informative rather than disappointing. Its weak NP-hardness and W[1]-hardness arise from choosing one discrete menu entry for every individual player, via product-knapsack-style combinatorics. In the high-multiplicity regime, fractions of a type can receive different options, so that combinatorics becomes a transportation/knapsack LP. The hardness is therefore population-multiplicity hardness that continuization can dissolve. This directly illustrates the programme’s Class A mechanism.

The mirror covers only CBCCT and Theorem 2. I would not claim it covers Theorem 3 on cup tournaments: cross-player matches and seeding introduce identity and bracket structure that are not determined by type masses. Natural follow-up questions are whether the LP admits an exact bit-model algorithm when probabilities are rational but \(\ln p\) is not explicitly encoded; how to round an optimal mass solution to a finite \(N\)-agent campaign with an additive loss bound; and what happens if every type must receive one uniform campaign package rather than being splittable. That last variant would preserve more of the product-knapsack choice structure and might be Class B or W[1]-hard.

The weakest point is the normalization. The raw probability that \(e^\star\) defeats every member of an atomless population is either \(0\) or degenerate, since \(\prod_i p_i\) collapses as \(N\) grows. The geometric-mean, or per-capita log-probability, objective is therefore an extension of the paper’s literal threshold question. I think it is defensible because it exactly preserves every finite rational-clone instance after taking an \(N\)-th root, and because the paper already treats the product through its logarithm. But I would present Mass-CBCCT\(_\infty\) as an author-recognizable high-multiplicity continuation of CBCCT, not pretend that the unnormalized probability has a nontrivial continuum limit.

The case AGAINST (opponent, writing after the proponent)

Theorem 2 is a genuine computational anchor, so the negative case cannot rely on the sports interpretation alone. The difficulty is that CBCCT’s objective is not population-extensive. The initial champ wins exactly when every challenger loses, with probability

\[ P_N=\prod_{i=1}^{N}p_i. \]

A distribution of types does not determine this quantity: the same \(\mu\) with \(N=100\) and \(N=10^9\) has radically different winning probabilities. In an atomless population, independent all-win probability is \(0\) whenever \(p<1\) on a positive-mass set, and is \(1\) only in the degenerate case \(p=1\) almost everywhere. This is not merely a technical failure of the literal formalism; the cardinality of the population is part of the event being measured.

The proposed repair,

\[ G(x)=\exp\!\left(\sum_{t,j}x_{t,j}\ln p_{t,j}\right), \]

is mathematically clean, and \(P_N=G(x)^N\) for a finite cloned population. But \(G\) is not the probability that the champ wins. It is a per-capita log-survival rate. It corresponds to a family of changing thresholds \(t_N=\theta^N\), not to the paper’s fixed-threshold decision problem. Thus the LP is a valid fractional allocation problem, but it is a new reliability-rate or campaign-intensity problem rather than a continuous version of CBCCT’s winner-probability question.

The obvious alternatives do not repair this. Retaining the literal probability gives a degenerate limit. Asking for the expected fraction of challengers defeated, or for a positive mass of supporters, replaces unanimous challenge-the-champ victory by an aggregate voting rule. Keeping a finite \(N\) while grouping identical players gives a useful high-multiplicity encoding, but not a scale-free continuous society; the population size remains indispensable. Treating each type as one group-level challenger makes mass irrelevant to the probability and therefore gives continuity no substantive role.

The campaign interpretation faces the same dichotomy. If repeated voters are interchangeable and only their aggregate support matters, the natural mirror is an approval or scoring problem, not CBCCT. If every voter or gatekeeper is individually essential, then the all-win event retains finite-population sensitivity: one failed individual matters, while every individual has measure zero in the continuum. Splitting a type across bribe options is defensible only after replacing that event by the geometric-mean objective.

This is the strongest case against the anchor, but it is not airtight. If ChoCo accepts normalized log-probability as a natural extensive observable—especially given the paper’s own use of logarithms—then Mass-CBCCT is a legitimate Class A mirror, and the proponent’s case survives. I therefore cannot honestly support the universal claim that no worthwhile mirror exists; I can defend only the narrower claim that no nondegenerate continuous mirror preserves CBCCT’s literal winner-probability semantics.

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.