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Phragmén's Voting Methods and Justified Representation (d4-conf-aaai-2017-0255)

Answer: \(\mathrm{NP}\)-hard

The anchor (named result in the paper)

Theorem 5.1 — class: hardness

PDF text-layer excerpt — equations and symbol order may be approximateThe decision problems leximax-Phragmén and var-Phragmén are NP-complete, even restricted to instances with s(A) = 2 and d(A) = 3.

statement extracted from the paper’s text layer; consult the paper for the exact wording

The continuous mirror question, as graded

Given candidates \(C\), committee size \(k\), finite approval types \(t\in T\) with rational masses \(\mu_t>0\) summing to \(1\), a rational mass-normalized candidate-load scale \(\beta>0\), and threshold \(\alpha\), decide whether there exist \(y_c\in\{0,1\}\) with \(\sum_c y_c=k\) and mass-weighted loads \(z_{t,c}\) satisfying \(z_{t,c}=0\) for \(c\notin A_t\), \(0\leq z_{t,c}\leq\mu_t y_c\), \(\sum_t z_{t,c}=\beta y_c\), and \(\sum_t\mu_t\ell_t^2<\alpha\), where \(\ell_t=\mu_t^{-1}\sum_c z_{t,c}\).

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

Given a finite candidate set \(C\), approval types \(T\) with \(A_t\subseteq C\), rational masses \(\mu_t>0\) summing to \(1\), \(k\), \(\beta>0\), and \(\alpha\), select exactly \(k\) candidates via \(y_c\in\{0,1\}\). A selected candidate has aggregate load \(\beta\), distributable only to approving types: \(z_{t,c}=0\) if \(c\notin A_t\), \(0\le z_{t,c}\le\mu_t y_c\), and \(\sum_t z_{t,c}=\beta y_c\). Type \(t\)'s load is \(\ell_t=\mu_t^{-1}\sum_c z_{t,c}\).

Is deciding whether a feasible selection and load distribution satisfies \(\sum_t\mu_t\ell_t^2<\alpha\) NP-hard or in \(\mathrm{P}\)?

For finite individual voters, the corresponding var-Phragmén decision problem is NP-complete, even when each voter approves at most \(2\) candidates and each candidate is approved by at most \(3\) voters.

Prover output (gpt-6-astra / medium, Bolzano prover prompt)

Conclusion

The problem is NP-hard, using the finite-voter hardness result supplied in the question. The upper bounds \(z_{t,c}\leq \mu_t y_c\) do not obstruct the reduction.

Source problem and construction

Let a finite-voter instance have voter set \(V\), with \(n=|V|>0\), candidate set \(C\), and approval sets \(A_v\). For a committee represented by \(y_c\), write \(x_{v,c}\geq 0\) for the load assigned to voter \(v\) by candidate \(c\). The finite-voter constraints are
\[ x_{v,c}=0\quad\text{if }c\notin A_v, \qquad \sum_{v\in V}x_{v,c}=y_c, \qquad \sum_{c\in C}y_c=k. \]
Define \(L_v=\sum_c x_{v,c}\). The corresponding decision question asks whether
\[ Q:=\sum_{v\in V}L_v^2<B. \]
We use the supplied NP-hardness premise for this finite-voter decision problem. If its objective is expressed as variance instead, the threshold translation is given below.

Nonnegativity and the candidate-load equality imply
\[ 0\leq x_{v,c}\leq y_c. \]

Partition voters into classes \(V_t\) having identical approval sets. Write \(n_t=|V_t|\) and let \(A_t\) be their common approval set. Construct the target instance with the same candidates and committee size, and set
\[ \mu_t=\frac{n_t}{n}, \qquad \beta=\frac1n, \qquad \alpha=\frac Bn. \]
These masses are positive and sum to \(1\). Grouping voters and producing these rational numbers takes polynomial time.

Proof of equivalence

From a finite-voter solution to a type solution

Given a feasible finite-voter solution, retain its committee and define
\[ z_{t,c}=\frac1n\sum_{v\in V_t}x_{v,c}. \]
Approval restrictions are preserved. Moreover,
\[ 0\leq z_{t,c} \leq \frac{n_t}{n}y_c =\mu_t y_c, \qquad \sum_t z_{t,c} =\frac1n\sum_v x_{v,c} =\beta y_c. \]
Thus the target solution is feasible.

Its type loads satisfy
\[ \ell_t =\frac1{\mu_t}\sum_c z_{t,c} =\frac1{n_t}\sum_{v\in V_t}L_v. \]
For each class, the inequality
\[ \left(\sum_{v\in V_t}L_v\right)^2 \leq n_t\sum_{v\in V_t}L_v^2 \]
follows from Cauchy–Schwarz. Consequently,
\[ \sum_t\mu_t\ell_t^2 =\frac1n\sum_t \frac{\left(\sum_{v\in V_t}L_v\right)^2}{n_t} \leq \frac Qn. \]
Hence \(Q<B\) implies \(\sum_t\mu_t\ell_t^2<\alpha\).

From a type solution to a finite-voter solution

Conversely, given a feasible target solution, retain its committee and assign every voter \(v\in V_t\) the load
\[ x_{v,c}=\frac{z_{t,c}}{\mu_t}. \]
These loads are nonnegative and respect approvals. For every candidate,
\[ \sum_{v\in V}x_{v,c} =\sum_t\frac{n_t}{\mu_t}z_{t,c} =n\sum_t z_{t,c} =n\beta y_c =y_c. \]
Therefore they form a feasible finite-voter distribution.

Every voter in class \(V_t\) has load \(L_v=\ell_t\), so
\[ Q=\sum_t n_t\ell_t^2 =n\sum_t\mu_t\ell_t^2. \]
Thus \(\sum_t\mu_t\ell_t^2<\alpha\) implies \(Q<B\).

Both directions preserve the strict decision threshold, proving the required polynomial-time reduction.

Variance normalization and inherited restrictions

If the finite-voter objective is population variance, the total voter load is \(k\), and therefore
\[ \frac1n\sum_v\left(L_v-\frac kn\right)^2 =\frac Qn-\left(\frac kn\right)^2. \]
A strict variance threshold \(b\) is consequently handled by choosing
\[ \alpha=b+\left(\frac kn\right)^2. \]
This is a rational, polynomial-time threshold transformation.

Finally, grouping identical voters preserves the supplied sparsity restrictions: every type approves at most \(2\) candidates, and every candidate is approved by at most \(3\) types. Thus NP-hardness holds even under these restrictions, with \(\beta=1/n\).

⟨structured verdict parsed — rendered above⟩

Adversarial review (gpt-6-astra / low, Bolzano verifier prompt)

The output is correct and rigorous under the finite-voter hardness premise supplied in USER_TASK. It gives a polynomial-time many-one reduction and verifies both directions, including the per-type capacity bounds and strict threshold. No major or minor issues were found.

No repair or additional clarification is needed.

⟨structured verdict parsed — rendered above⟩

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