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d4-conf-ecai-2020-0253 (d4-conf-ecai-2020-0253)

Answer: Both add mode and delete mode are in \(\mathrm P\), with running time polynomial in \(m,\tau,L\).

The anchor (named result in the paper)

Theorem 4.3 — class: hardness

The continuous mirror question, as graded

Given a candidate set \(C\), a supplied partition of \(C\) into three blocks, supported complete rankings whose restrictions to each block are single-peaked, registered masses \(\mu_t\), a target \(p\), a budget \(b\), and an add or delete mode (with available masses \(\nu_t\) in add mode), does there exist an action vector \(x\) with \(0\le x_t\le\nu_t\) for adding or \(0\le x_t\le\mu_t\) for deleting, and \(\sum_t x_t\le b\), such that the resulting society makes \(p\) beat every \(c\ne p\)? The resulting mass is \(\rho_t=\mu_t+x_t\) in add mode and \(\rho_t=\mu_t-x_t\) in delete mode; require \(\sum_t\rho_t(\mathbf 1[p\succ_t c]-\mathbf 1[c\succ_t p])>0\) for each opponent.

Packaged proof task (what the prover saw — nothing else)

Let \(C\) have \(m\) candidates, with target \(p\in C\) and a supplied partition \((C_1,C_2,C_3)\) into nonempty blocks. Types are strict complete rankings. For each block, there must exist one common axis on which every supported ranking restricted to that block is single-peaked; axes need not be supplied. In add mode this promise covers both registered and available types.
A society is given by its support: the list of types with positive mass, each mass encoded as a binary rational. Let \(\tau\) be the number of distinct types in the union of the supplied supports, not the number of possible rankings, and \(L\) the total input bit length. Registered masses \(\mu_t\) and, in add mode, available masses \(\nu_t\) are nonnegative, with absent types assigned zero. Masses are finite measures and need not sum to one. The budget \(b\) is a positive binary rational.
An action may split type masses: choose \(x_t\in\mathbb R_{\geq0}\) with \(\sum_t x_t\leq b\). In add mode require \(x_t\leq\nu_t\) and set \(\rho_t=\mu_t+x_t\); in delete mode require \(x_t\leq\mu_t\) and set \(\rho_t=\mu_t-x_t\).
The action succeeds precisely when \(p\) is the strict Condorcet winner: for every \(c\in C\setminus\{p\}\), require \(\sum_t\rho_t(\mathbf1[p\succ_t c]-\mathbf1[c\succ_t p])>0\). Pairwise ties do not suffice.

For each of add mode and delete mode, is this decision problem in \(\mathrm P\), with running time polynomial in \(m,\tau,L\), or is it \(\mathrm{NP}\)-hard?

Discrete constructive control by adding votes and by deleting votes to obtain a unique Condorcet winner is \(\mathrm{NP}\)-hard even with a supplied partition into three single-peaked blocks ([Yang, 2020](https://arxiv.org/pdf/2002.03539)). No result settling this exact divisible-mass problem was located in the literature search.

Modeling choices made when packaging the task

Prover output (gpt-6.1-sol / high, Bolzano prover prompt)

Result

Both add mode and delete mode are in \(\mathrm P\), with running time polynomial in \(m,\tau,L\). The proof works for arbitrary strict rankings, so the three-block single-peakedness promise is unnecessary.

1. Pairwise margins and action constraints

Let \(T\) be the union of the supplied supports, with \(|T|=\tau\). Assign mass zero to absent types. For each \(c\in C\setminus\{p\}\) and \(t\in T\), define

\[ a_{ct}=\mathbf 1[p\succ_t c]-\mathbf 1[c\succ_t p]\in\{-1,1\}, \qquad d_c=\sum_{t\in T}\mu_ta_{ct}. \]

Thus \(d_c\) is the registered pairwise margin of \(p\) against \(c\).

Use the following mode-dependent constants:

In either mode, the admissible actions are precisely the vectors satisfying

\[ 0\leq x_t\leq u_t\quad(t\in T), \qquad \sum_{t\in T}x_t\leq b. \]

The resulting mass is \(\rho_t=\mu_t+s x_t\). Consequently, the resulting margin against \(c\) is

\[ M_c(x)=\sum_{t\in T}\rho_ta_{ct} =d_c+s\sum_{t\in T}a_{ct}x_t. \]

The required success condition is \(M_c(x)>0\) for every opponent \(c\).

2. A linear program handling strict victory

Introduce a real variable \(\varepsilon\) and solve the following rational linear program:

\[ \text{maximize }\varepsilon \]

subject to

\[ 0\leq x_t\leq u_t\quad(t\in T), \]

\[ \sum_{t\in T}x_t\leq b, \]

\[ 0\leq\varepsilon\leq1, \]

\[ d_c+s\sum_{t\in T}a_{ct}x_t\geq\varepsilon \quad(c\in C\setminus\{p\}). \]

If the program is infeasible, reject. Otherwise, let \(\varepsilon^\ast\) be its optimal value and accept exactly when \(\varepsilon^\ast>0\).

Correctness. Suppose first that a successful action \(x\) exists. The supplied partition has three nonempty blocks, so \(m\geq3\), and the opponent set is finite and nonempty. Every margin is positive; hence

\[ \delta=\min_{c\in C\setminus\{p\}}M_c(x)>0. \]

Taking \(\varepsilon=\min\{1,\delta\}\) gives a feasible point with positive objective value. Therefore \(\varepsilon^\ast>0\).

Conversely, a feasible point with \(\varepsilon>0\) satisfies all action constraints and, for every opponent,

\[ M_c(x)\geq\varepsilon>0. \]

Its action therefore makes \(p\) a strict Condorcet winner.

Finally, whenever the program is feasible, its feasible region is closed and bounded: every \(x_t\) lies in the finite interval \([0,u_t]\), and \(\varepsilon\in[0,1]\). It is therefore compact, so the objective attains its maximum. Thus accepting when \(\varepsilon^\ast>0\) is equivalent to the existence of a successful action, including when actions are allowed to have irrational coordinates.

3. Polynomial bit complexity

The program has \(\tau+1\) variables and \(2\tau+m+2\) scalar inequalities. The coefficients \(a_{ct}\) can be computed by reading the supplied rankings in time polynomial in \(m,\tau,L\).

The only newly computed rational constants are the margins \(d_c\). To bound their encoding lengths, let \(Q\) be the product of the denominators of all registered masses. Because their encodings are part of the input, \(Q\) has \(O(L)\) bits. Writing each registered mass over denominator \(Q\) produces integer numerators of \(O(L)\) bits. Summing at most \(\tau\) signed numerators produces a numerator with \(O(L+\log(\tau+1))\) bits. Thus every \(d_c\), and consequently the entire linear program, has encoding length polynomial in \(m,\tau,L\).

We apply the standard polynomial-time theorem for rational linear programming: given explicitly encoded rational constraints and objective coefficients, one can decide infeasibility and, when a finite optimum exists, compute its exact rational value in time polynomial in the encoding length. See [Khachiyan, *Polynomial algorithms in linear programming*](https://doi.org/10.1016/0041-5553(80)90061-0).

All hypotheses hold here: the program is explicit, its data are rational, and any feasible instance has an attained finite optimum. Comparing the exact rational value \(\varepsilon^\ast\) with zero therefore decides strict victory in polynomial time.

4. Relation to discrete control

The reduction uses divisibility exactly by allowing continuous variables \(x_t\). Discrete vote control imposes integrality constraints, yielding a different feasibility problem. The discrete hardness results in [Yang (2020)](https://arxiv.org/pdf/2002.03539) therefore do not contradict the polynomial-time algorithms proved here.

⟨structured verdict parsed — rendered above⟩

Adversarial review (gpt-6.1-sol / medium, Bolzano verifier prompt)

The output resolves both modes correctly. The linear program exactly captures divisible actions and strict Condorcet victory, and its encoding has polynomial length. No major or minor mathematical issues were found.

No additional clarification or repair is needed.

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Tier: machine-proved, adversarially reviewed by an independent model. No human has checked this proof. Generated from pipeline artifacts by tools/atlas_build.py.