Source
Czarnecki, Gidel, Tracey, Tuyls, Omidshafiei, Balduzzi & Jaderberg, LIT-1504, read in NOTE-1146 from arXiv:2004.09468v2: Definition 3 and Theorems 2–3 (p. 7, proofs pp. 13–14). RPP is Definition 3 of Balduzzi et al., LIT-1509.
What was actually shown
- Nash clustering. C₁ is the support of a Nash equilibrium of the whole game. C₂ is the support of a Nash of the game restricted to what remains, and so on until nothing remains. With maximum-entropy equilibria at each step the clustering is unique.
- Theorem 2. With X_i the strategies left at step i, and C_j ⊂ X_i for j > i: RPP(C_i, C_j) ≥ RPP(C_i, X_i), since a smaller opposing set cannot lower the value. RPP(C_i, X_i) = 0, because C_i can play the Nash of X_i and so holds X_i to zero, while X_i can mirror it.
- Theorem 3. Let the population contain C_i, and let π beat every member. If π lay in a later cluster C_k, then RPP(C_i, C_k) ≤ max_{π_j ∈ C_i} f(π_j, π) < 0, contrary to Theorem 2. π is not in C_i, because f(π, π) = 0. So π lies in an earlier cluster.
The reader re-derived both. Each is a few lines on the value of a zero-sum game.
What this does not say
- That every member of an earlier cluster beats every member of a later one. The order is between clusters as populations.
- That the order is strict. Clusters can tie at RPP = 0.
- Anything about real games. Neither theorem uses the spinning-top shape. Both hold in rock-paper-scissors, which is one cluster, and in the disc game.
- Anything ordinal. Clusters are Nash supports, which depend on payoff magnitudes. A monotone transform of payoffs can change them.
Connections
A cardinal layering of a cyclic record: the cycling stays inside clusters and the order runs between them. That parallels the ordinal condensation reading in NOTE-1156. Theorem 3 is a sufficient condition for escaping a cycle: beat its whole class, not its latest member. That bears on CLAIM-451’s accommodation (NOTE-1146).
Note of 2026-10-30: the second check
The second check at the status review (ADR-065 §5) followed each step.
In a symmetric zero-sum game with antisymmetric payoff matrix A the value is 0, and a mixed strategy p is an equilibrium strategy for either player exactly when Ap ≤ 0: it then scores at least 0 against every pure strategy.
Theorem 2. Let p_i be the equilibrium of the game restricted to X_i whose support is C_i. It scores at least 0 against every member of X_i, so RPP(C_i, X_i) ≥ 0. The X_i player can play p_i itself, since C_i ⊂ X_i, and p_iᵀAp_i = 0, so RPP(C_i, X_i) ≤ 0. For j > i, C_j ⊂ X_i, and shrinking the minimiser’s set cannot lower the max–min, so RPP(C_i, C_j) ≥ RPP(C_i, X_i) = 0.
Theorem 3. Let the population S contain C_i, and let π beat every member of S, so f(π, s) > 0 for every s in C_i. The clusters partition the strategies, so π lies in some C_k. If k > i, the C_k player can play π, and RPP(C_i, C_k) ≤ max_{s ∈ C_i} f(s, π) = −min_{s ∈ C_i} f(π, s) < 0, against Theorem 2. If k = i, π is itself in S, and f(π, π) = 0 contradicts its beating every member. So k < i.
A script (numpy, with an exact linear-programming solver, in the session’s scratchpad) checked both theorems on 300 random games of 3 to 8 strategies, mixing transitive and cyclic parts, a third of them rounded to signs so as to include ties. RPP(C_i, C_j) ≥ 0 held in all 2349 cluster pairs, and all 884 strategies that beat a population containing a whole cluster lay in an earlier cluster. As a control, populations holding only one member of a cluster of two or more were beaten by 557 strategies in the same or a later cluster, so the whole cluster is what the theorem uses.