Answer: \(\mathsf{NP}\text{-complete}\), even with \(\tau=1\), \(\mu_t=w_{tt}=1\), and symmetric costs in \(\{0,1\}\).
Theorem 4.4 — class: other
Given a finite type set \(T\), rational masses \(\mu_t\) summing to \(1\), finite action domains \(D_t\), nonnegative rational interaction intensities \(w_{ts}=w_{st}\), asymmetric pairwise cost tables \(f_{ts}(a,b)\), and a rational threshold \(B\), decide whether there are action shares \(\pi_{t,a}\geq 0\), with \(\sum_{a\in D_t}\pi_{t,a}=1\), such that \(J(\pi)=\sum_{t,s\in T}\mu_t\mu_s w_{ts}\sum_{a\in D_t,\,b\in D_s}\pi_{t,a}\pi_{s,b}f_{ts}(a,b)\leq B\).
A society is given by its support \(T\): the list of its \(\tau\geq 1\) types with positive binary-rational masses \(\mu_t\), summing to \(1\). Here \(\tau\) is the support size, not the number of possible types.
Each type has an explicitly listed nonempty finite action domain \(D_t\). Put \(m=\max_t|D_t|\), and let \(L\) be the total binary input length.
Input also comprises symmetric nonnegative rational intensities \(w_{ts}=w_{st}\), explicitly tabulated costs \(f_{ts}:D_t\times D_s\to\mathbb{Q}_{\geq 0}\), and a threshold \(B\in\mathbb{Q}\). The cost \(f_{ts}(a,b)\) is borne by the participant of type \(t\); no equality with \(f_{st}(b,a)\) is required.
A feasible allocation consists of real shares \(\pi_{t,a}\geq 0\) with \(\sum_{a\in D_t}\pi_{t,a}=1\) for every \(t\). Its cost is
\(J(\pi)=\sum_{t,s\in T}\mu_t\mu_s w_{ts}\sum_{a\in D_t,\,b\in D_s}\pi_{t,a}\pi_{s,b}f_{ts}(a,b).\)
The sum is over ordered type pairs, aggregating both participants' costs. Terms with \(t=s\) are allowed and represent interactions between independently sampled members of the same type.
Is deciding whether a feasible allocation satisfies \(J(\pi)\leq B\) solvable in polynomial time in \(m,\tau,L\), or is it NP-hard?
For finite agents with one finite-domain variable each and binary asymmetric nonnegative costs, PT-ISABB terminates and finds a globally optimal assignment; its completeness guarantee gives no polynomial running-time bound. [Source](https://www.ifaamas.org/Proceedings/aamas2019/pdfs/p1506.pdf).
No result settling this exact mass-allocation decision problem was located in the literature search.
The decision problem is \(\mathsf{NP}\)-complete. Hardness holds even with a single supported type, mass \(1\), intensity \(1\), and symmetric costs in \(\{0,1\}\).
Consequently, a polynomial-time algorithm in \(m,\tau,L\) exists if and only if \(\mathsf{P}=\mathsf{NP}\).
We use the standard \(\mathsf{NP}\)-complete problem CLIQUE: given a nonempty finite simple graph \(G=(V,E)\) and a positive integer \(k\), decide whether \(G\) contains a clique of size at least \(k\). This is among the complete problems in [Karp’s original theorem](https://cgi.di.uoa.gr/~sgk/teaching/grad/handouts/karp.pdf).
Construct a society with
\[
\text{}
\]
\(T=\{t\}\), \(\mu_t=1\), \(D_t=V\), and \(w_{tt}=1\).
For actions \(u,v\in V\), define \(f_{tt}(u,v)=0\) when \(\{u,v\}\in E\), and \(f_{tt}(u,v)=1\) otherwise. In particular, every diagonal cost is \(1\). Set \(B=1/k\).
All hypotheses are satisfied: the mass is positive and binary rational, the domain is nonempty, the intensity is symmetric, and all costs are nonnegative rational numbers. Writing \(x_v=\pi_{t,v}\), feasible allocations satisfy \(x_v\geq0\) and \(\sum_{v\in V}x_v=1\). Their cost is
\( \displaystyle q_G(x)=\sum_{v\in V}x_v^2+ 2\sum_{\substack{\{u,v\}\subseteq V\\u\neq v,\ \{u,v\}\notin E}}x_ux_v. \)
Each distinct unordered nonedge occurs with coefficient \(2\), because the original objective sums over ordered action pairs.
Let \(\omega(G)\) denote the maximum clique size. We will prove the exact identity
\( \displaystyle \min_{\substack{x_v\geq0\\\sum_vx_v=1}}q_G(x)=\frac{1}{\omega(G)}. \)
Suppose \(u,v\) are distinct nonadjacent vertices with \(x_u>0\) and \(x_v>0\). Put \(r=x_u+x_v\). Keep all other coordinates fixed and replace these two coordinates by
\( \displaystyle x_u=\alpha,\qquad x_v=r-\alpha,\qquad 0\leq\alpha\leq r. \)
Because \(u,v\) are nonadjacent, their diagonal and mutual contributions together equal
\( \displaystyle \alpha^2+(r-\alpha)^2+2\alpha(r-\alpha)=r^2. \)
Define
\( \displaystyle A=\sum_{z\notin\{u,v\}}f_{tt}(u,z)x_z,\qquad C=\sum_{z\notin\{u,v\}}f_{tt}(v,z)x_z. \)
Symmetry of our cost table implies that the contributions involving exactly one of \(u,v\) equal \(2\alpha A+2(r-\alpha)C\). Thus the whole objective, as a function of \(\alpha\), is affine:
\( \displaystyle q_G(x(\alpha))=K+r^2+2\alpha A+2(r-\alpha)C, \)
where \(K\) is independent of \(\alpha\).
An affine function on \([0,r]\) attains its minimum at an endpoint. Choosing such an endpoint does not increase the cost and removes one vertex from the positive support.
Repeat this operation whenever the positive support contains a nonadjacent pair. Each operation strictly decreases its size, so the process terminates with a feasible vector \(y\) supported on a clique, and \(q_G(y)\leq q_G(x)\).
If that clique has size \(h\), then
\( \displaystyle q_G(y)=\sum_{v\in\operatorname{supp}(y)}y_v^2 \geq\frac{\bigl(\sum_{v\in\operatorname{supp}(y)}y_v\bigr)^2}{h} =\frac1h\geq\frac1{\omega(G)}. \)
The first inequality is Cauchy–Schwarz. Therefore every feasible \(x\) satisfies \(q_G(x)\geq1/\omega(G)\).
Conversely, assigning share \(1/\omega(G)\) to every vertex of a maximum clique, and zero elsewhere, gives cost \(1/\omega(G)\). This proves the identity.
It follows that
\( \displaystyle \exists\pi:\ J(\pi)\leq B \quad\Longleftrightarrow\quad \frac1{\omega(G)}\leq\frac1k \quad\Longleftrightarrow\quad \omega(G)\geq k. \)
The construction explicitly lists \(|V|^2\) binary costs; its size and construction time are polynomial in the graph input length and the encoding length of \(k\). Here \(\tau=1\) and \(m=|V|\).
This proves \(\mathsf{NP}\)-hardness while fully accounting for arbitrary real shares.
We now prove membership in \(\mathsf{NP}\), including instances with asymmetric costs and same-type interactions.
Index all type-action pairs by a coordinate set \(I\), and put \(N=|I|=\sum_t|D_t|\). Define the rational matrix
\( \displaystyle H_{(t,a),(s,b)}=\mu_t\mu_s w_{ts}f_{ts}(a,b), \qquad Q=\frac{H+H^{\mathsf T}}2. \)
Then \(Q\) is symmetric, its encoding length is polynomial in \(L\), and the objective is exactly \(J(x)=x^{\mathsf T}Qx\).
The feasible set is a nonempty compact product of simplices. Hence the continuous objective attains a global minimum at some \(x^*\).
Let \(S=\{i:x_i^*>0\}\), let \(z=x_S^*\), and let \(R=Q_{S,S}\). Every type has at least one coordinate in \(S\). For each type \(t\), choose one such coordinate \(i_t\).
For any \(i\in S\) belonging to type \(t\), sufficiently small perturbations in both directions along \(e_i-e_{i_t}\) remain feasible. Differentiating at the minimum therefore gives
\( \displaystyle (Rz)_i=(Rz)_{i_t}. \)
Consider the rational polytope
\( \displaystyle P_S=\left\{y\in\mathbb R^S: y\geq0,\quad \sum_{i\in S:\,\operatorname{type}(i)=t}y_i=1\ \text{for every }t,\quad (Ry)_i=(Ry)_{i_t}\ \text{for every }i\in S \right\}. \)
It contains \(z\) and is bounded.
For any \(y\in P_S\), put \(d=y-z\). The coordinates of \(d\) sum to zero within each type. Both \(Rz\) and \(Ry\) are constant within each type, so
\( \displaystyle d^{\mathsf T}Rz=d^{\mathsf T}Ry=0. \)
Using symmetry of \(R\), we obtain
\( \displaystyle y^{\mathsf T}Ry-z^{\mathsf T}Rz =(y-z)^{\mathsf T}R(y+z)=0. \)
Thus every point of \(P_S\), extended by zeros outside \(S\), achieves the global minimum.
A nonempty bounded polytope has a vertex. Moreover, every vertex of \(P_S\) has polynomial-size rational coordinates. Here is the encoding argument. Write \(d_S=|S|\), and let \(h\) bound the numerator and denominator bit lengths of its constraint coefficients. At a vertex, \(d_S\) linearly independent active constraints determine the coordinates. Clearing denominators separately in each row produces integer coefficients with bit length \(O(d_Sh)\). Cramer’s rule expresses each coordinate as a ratio of determinants. The determinant expansion bounds their bit lengths by \(O(d_S^2h+d_S\log d_S)\), which is polynomial in \(L\).
Consequently, every yes-instance has a feasible rational allocation of polynomial encoding length attaining the minimum, and hence satisfying \(J\leq B\). A verifier checks nonnegativity, each type’s normalization, and the objective inequality using exact rational arithmetic in polynomial time. The decision problem belongs to \(\mathsf{NP}\).
The reduction and certificate argument together establish \(\mathsf{NP}\)-completeness. The hardness already arises from independently sampled members of one type: the quadratic terms encode maximum clique size.
The cited finite-agent completeness guarantee is compatible with this conclusion. [PT-ISABB’s Theorem 4.4](https://www.ifaamas.org/Proceedings/aamas2019/pdfs/p1506.pdf) establishes completeness, while Section 4.3 describes exponential growth of search-phase messages with the agent count.
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The output resolves the exact mass-allocation decision problem. The reduction proves \(\mathsf{NP}\)-hardness for the stated single-type restriction, and the rational-certificate argument establishes membership in \(\mathsf{NP}\). I found no mathematical gaps requiring repair.
No additional mathematical clarification is necessary. The empty display near the construction is merely a removable formatting artifact.
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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.