Answer: \(\mathrm{NP}\)-hard
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
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 \(\ell_t=\mu_t^{-1}\sum_c z_{t,c}\).
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.
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.
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.
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 \(\sum_t\mu_t\ell_t^2<\alpha\).
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.
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⟩
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.