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Contents
  1. C73-11: Universal convergence of simultaneous fictitious play
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 5) VERIFICATION
  7. 6) FINAL
  8. Reproducibility and scope.
  9. References

Research attempt and adversarial verification

9 October 2026 Original-target status: UNRESOLVED

C73-11: Universal convergence of simultaneous fictitious play

Input: C73-11.tex.  Tools: web browsing [YES]; Python computation [YES].

1) FORMAL RESTATEMENT

Let $N=\{1,\ldots,n\}$, with $n\geq1$, and let every finite action set $S_i$ be nonempty. Let $X=\prod_i\Delta(S_i)$, and extend each payoff array $u_i$ multilinearly to independent mixed profiles. For every $x^1\in X$, and every sequence of legal simultaneous best replies, set \[ a_i^{t+1}\in{\operatorname*{arg\,max}}_{a\in S_i}u_i(a,x_{-i}^t),\qquad x_i^{t+1}=x_i^t+\frac{e_{a_i^{t+1}}-x_i^t}{t+1},\quad t\geq1. \] The class to characterize is \[ \mathcal C_{\rm FP}=\{G:\ \forall x^1\in X\ \forall\text{ legal reply sequences}, \ \operatorname{dist}(x^t,{\operatorname{NE}}(G))\longrightarrow0\}. \] Euclidean distance is used. Boundary initializations and arbitrary choices at ties are included. The output sought is a noncircular condition on payoff arrays, not another formula quantifying over the same trajectories. This is a classification agenda; a nonconvergent game excludes that game, not the classification question itself.

2) QUICK LITERATURE/CONTEXT CHECK

The input survey [li-survey] supplies the learning-classification context. Monderer and Shapley [monderer-fp] prove a fictitious-play property for identical-interest games. The proof below independently treats positive weighted potential games under the precise simultaneous update and tie convention in the input. A specific nonconvergent $3\times3$ array is analyzed directly, rather than relying on an unspecified version of a cyclic-play example.

3) ATTACK PLAN

Proof track. Use a continuous Nash residual; seek a payoff-derived Lyapunov function with a summable discretization error; turn weighted summability of residuals into pointwise residual convergence.

Construction track. Test a cyclic $3\times3$ interaction, first with synchronized beliefs and then with distinct pure initializations. Track constant best-reply blocks exactly to distinguish an actual counterexample from a finite simulation that only appears to cycle.

Tools and purpose. Multilinear extension identifies directional gains; compactness equates residual and Nash-set convergence; Taylor's theorem controls simultaneous-update errors; harmonic step sums prevent isolated residual spikes; strict payoff comparisons make block lengths exact; permutation support arguments identify the equilibrium set; integer count arithmetic eliminates floating-point tie errors in the search.

4) WORK

For later use, define \[ r_i(x)=\max_{a_i}u_i(a_i,x_{-i})-u_i(x),\qquad R(x)=\sum_i r_i(x). \] Every $r_i$ is nonnegative and Lipschitz on $X$, because it is a finite maximum of polynomials minus a polynomial. Also $R(x)=0$ exactly on ${\operatorname{NE}}(G)$: no mixed deviation can outperform the best pure deviation. Since $X$ is compact and ${\operatorname{NE}}(G)$ is nonempty by finite-game Nash existence [nash], any sequence has $R(x^t)\to0$ if and only if its distance to ${\operatorname{NE}}(G)$ tends to zero. To verify the nontrivial direction, a sequence with distance bounded below has a convergent subsequence whose limit is outside ${\operatorname{NE}}(G)$ but has zero residual, a contradiction.
Theorem (Positive weighted potential games belong to $\mathcal C_{\rm FP}$): Suppose there are $w_i>0$ and an array $\Phi$ such that, for every pure opponents' profile and every $a_i,b_i$, \[ u_i(a_i,s_{-i})-u_i(b_i,s_{-i}) =w_i\left(\Phi(a_i,s_{-i})-\Phi(b_i,s_{-i})\right). \] Then every process in the formal restatement converges in distance to ${\operatorname{NE}}(G)$.
Proof: Extend $\Phi$ multilinearly and put $\alpha_t=1/(t+1)$ and $d^t=(e_{a_i^{t+1}}-x_i^t)_i$. The potential identity extends by multilinearity, so \[ \langle\nabla\Phi(x^t),d^t\rangle =\sum_i\frac{r_i(x^t)}{w_i}=:Q(x^t)\geq0. \] The norm of $d^t$ is bounded by a constant $D$, and the Hessian of $\Phi$ is bounded on the compact convex set $X$. Taylor's theorem on the line segment of the update therefore gives a constant $C$ with \[ \Phi(x^{t+1})-\Phi(x^t)\geq\alpha_tQ(x^t)-C\alpha_t^2. \] Writing $e_t=C\alpha_t^2$, the bounded sequence $\Phi(x^t)-\sum_{s=t}^{\infty}e_s$ is nondecreasing. It converges, and summing the displayed inequality gives $\sum_t\alpha_tQ(x^t)<\infty$.

Here is the required no-spike argument. Let $L$ be a Lipschitz constant for $Q$. If $LD=0$, then $Q(x^t)$ is constant along the process and the divergent sum $\sum_t\alpha_t$ forces that constant to be zero. Otherwise suppose infinitely many $t$ have $Q(x^t)\geq\delta>0$. Starting at any sufficiently large such time, let $k>t$ be the first index such that $\sum_{s=t}^{k-1}\alpha_s\geq\delta/(2LD)$. For $t\leq j<k$, the minimality of $k$ gives $Q(x^j)\geq\delta/2$. Thus this interval contributes at least $\delta^2/(4LD)$ to $\sum_s\alpha_sQ(x^s)$. If there are infinitely many spikes, choose these finite intervals successively with disjoint interiors. Their contributions contradict the finite sum. Hence $Q(x^t)\to0$. Positive weights make this equivalent to $R(x^t)\to0$, and the residual criterion finishes the proof. ∎

This is a sufficient array condition, not a necessary one. For example, strict dominance gives another class: if every player has a strictly dominant pure action, all selected replies are those actions, and $x_i^t=e_{a_i^*}+(x_i^1-e_{a_i^*})/t$. The arrays \[ A=\begin{pmatrix}3&2\\1&0\end{pmatrix},\qquad B=\begin{pmatrix}3&0\\2&0\end{pmatrix} \] have strictly dominant first actions but cannot have a positive weighted potential: the row player's rectangular payoff difference is zero, while the column player's is one; multiplying either by a positive weight cannot reconcile them.

Theorem (An explicit game outside $\mathcal C_{\rm FP}$): Let \[ A=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},\qquad B=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}. \] Start from $x^1=e_1$, $y^1=e_2$ and choose the lowest numbered best action at every tie. This legal process does not approach the Nash set.
Proof: Each player best responds by selecting the cyclic successor of a largest coordinate of the other's empirical counts. First, the only equilibrium is $x=y=(1/3,1/3,1/3)$. Indeed, a singleton support for one player forces a singleton support for the other; their cyclic-successor requirements would give a pure action equal to its successor twice, which is impossible. If a support has size two, the other player's largest-coordinate set must contain at least two actions. The other support cannot have size one; if it had size three, best-response indifference would force the first player to be uniform with full support. Consequently both supports have size two and both distributions put weight $1/2$ on their supports. Each support must be the cyclic successor of the other, so a two-element subset would be invariant under a three-cycle squared. No such subset exists. Full supports force both distributions to be uniform.

Use integer counts $c=tx^t$ and $d=ty^t$. These are integers for the chosen initial condition. Consecutive constant reply blocks have the following forms; the final column is the condition at which the next block begins: \[ \begin{array}{c|c|c} k\bmod6&(\text{row action},\text{column action})&\text{switch condition}\\\hline 0&(3,2)&c_3=c_1\\ 1&(3,1)&d_1=d_2\\ 2&(2,1)&c_2=c_3+1\\ 3&(2,3)&d_3=d_1\\ 4&(1,3)&c_1=c_2\\ 5&(1,2)&d_2=d_3+1 \end{array} \] We verify this table and its repetition, including all tie offsets. At the start of a six-block cycle write $c=(a,b,c_0)$ and $d=(d_0,e,f)$, with $a>b\geq c_0$, $e=f+1$, and $f\geq d_0$. Initially these are $(1,0,0)$ and $(0,1,0)$. The six positive lengths are \begin{align*} A_0&=a-c_0, &B_0&=e+A_0-d_0, &C_0&=a+B_0-b+1,\\ D_0&=C_0+A_0+1, &E_0&=D_0+B_0+1, &F_0&=E_0+C_0+1. \end{align*} In the first block, the third row count catches the first; at equality the column's first action takes precedence over its second. In the second, the first column count catches the second; the row's second action takes precedence over its third. In the third, the second row count must exceed the third, since equality still favors the column's first action. In the fourth and fifth, equality suffices by the same lowest-action rule; in the sixth the second column count must exceed the third. These are exactly the six conditions in the table. The untouched third competitor is never larger than the two being compared, by the displayed initial inequalities and the positive preceding increments.

At the next cycle the counts are \[ c'=(a+E_0+F_0,\ b+C_0+D_0,\ a+B_0),\quad d'=(e+A_0+C_0,\ e+A_0+F_0,\ e+A_0+C_0+E_0). \] They satisfy the same inequalities, now strictly where needed, and $d'_2=d'_3+1$. This proves repetition of the table. If $\ell_k$ is the length of block $k$, the formulas also give \[ \ell_0=1,\quad\ell_1=2,\quad\ell_2=4,\qquad \ell_{k+3}=\ell_{k+2}+\ell_k+1\quad(k\geq0). \] For clarity, the cross-cycle identities follow by subtraction: $A'_0=D_0+F_0+1$, $B'_0=A'_0+E_0+1$, and $C'_0=B'_0+F_0+1$. Thus the recurrence covers every index, not just the three within-cycle transitions.

The lengths increase and $\ell_{k+3}\geq2\ell_k$. Grouping the earlier lengths in triples consequently gives $\sum_{j=0}^k\ell_j\leq6\ell_k$. If block $k$ begins at time $t_k=1+\sum_{j<k}\ell_j$, its averaging weight is \[ \lambda_k=\frac{\ell_k}{t_k+\ell_k}\geq\frac17. \] During this block the row player repeats one action $a$, so $x^{t_k+\ell_k}=(1-\lambda_k)x^{t_k}+\lambda_ke_a$. If both endpoint profiles were within Euclidean distance $1/30$ of the uniform profile, their $a$ coordinates could differ by at most $1/15$. But the update would increase that coordinate by at least \[ \frac17\left(\frac23-\frac1{30}\right)=\frac{19}{210}>\frac1{15}. \] At least one endpoint is therefore a distance at least $1/30$ from the unique Nash equilibrium. The endpoint times tend to infinity. Set convergence fails. ∎

5) VERIFICATION

For a one-player game, every chosen action maximizes its fixed payoff and initial mass on suboptimal actions is divided by $t$. The same argument handles singleton action sets. In the two-player identical-payoff anti-coordination game with matrix $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$, synchronized pure-start fictitious play has total residual $(t+1)/t^2$ at odd $t$, consistent with the potential theorem. Actual actions need not converge in this example.

The integer implementation of the cyclic counterexample records total residuals $0.3054$, $0.282011$, $0.27850335$, and $0.278920207$ at times $10^2,10^3,10^4,10^5$. The analytical block proof, not those four numbers, certifies nonconvergence. An initial attempted test with both initial beliefs $e_1$ instead produced residuals tending to zero; the universal quantifier requires only one legal bad initialization, not that every initialization fail.

Minimal search pseudocode is 

\begin{lstlisting} c, d = (1,0,0), (0,1,0) for t = 1,2,...:     row = first_argmax(A*d)     col = first_argmax(transpose(B)*c)     record_switches_and_residual_before_update()     c[row] += 1; d[col] += 1 \end{lstlisting}
The potential proof uses $\sum\alpha_t^2<\infty$ and $\sum\alpha_t=\infty$ separately. It neither assumes strict potential improvement at a tie nor interchanges a maximum and an infinite-time limit.

\Needspace{8\baselineskip}

6) FINAL

LABEL: UNRESOLVED.

Strongest checked partial result. A complete convergence theorem for positive weighted potential games, a distinct dominance class, and a fully verified explicit $3\times3$ exclusion with an infinite subsequence bounded away from its unique equilibrium.

Exact first gap. No noncircular necessary-and-sufficient payoff-array criterion for arbitrary finite games is established.

Next three moves.

  1. Characterize which nonpotential perturbations preserve the Lyapunov inequality with an error that remains summable along every legal path.
  2. Search for cyclic-block certificates with rational inequalities and prove completeness or incompleteness of that certificate family in small dimensions.
  3. Test proposed criteria against both the strictly dominant nonpotential array and the synchronized versus asymmetric paths of the cyclic array.
Counterexample perspective. The displayed game falsifies universal convergence across all games. It does not falsify a yet-unspecified characterization, and it does not show that all nonpotential games fail.

Reproducibility and scope.
This is one of eleven companion reports. The bundle includes verify.py and its executed results. Finite experiments supplement the proofs; they are not proofs of asymptotic claims. Literature coverage is targeted, and no novelty or formal proof-assistant certification is claimed.

References