Answer: Polynomial-time solvable: a certified additive-\(\eta\) approximation is computable, and \(s_\epsilon\) is always finite under the stated strict median hypothesis.
Corollary 1 — class: algorithmic
Given an ordered finite support of \(\tau\) single-peaked types with rational masses \(\mu_i\) and a unique median type \(r\), determine the least interaction time \(s\) at which the triadic mass process places at least \(1-\epsilon\) of its mass on \(r\), and provide a certified bound on that time. The process starts at \(\nu(0)=\mu\); at each infinitesimal interaction, three types are drawn independently from \(\nu\), and their mass is relabeled to the median type. Establish how this accuracy guarantee relates to the finite-population process as population size grows.
A society is given by its support: an ordered list of \(\tau\geq1\) types with positive binary-rational masses \(\mu_i\), with \(\sum_i\mu_i=1\). The support size \(\tau\) counts present types, not possible types. Types and candidates coincide, so \(m=\tau\); type \(i\) has strict single-peaked preferences with peak \(i\).
Assume a unique median type \(r\), meaning \(\sum_{i<r}\mu_i<1/2<\sum_{i\leq r}\mu_i\). Every interaction relabels all three sampled types to their median, including repeated types.
Define the deterministic mass process by \(\nu(0)=\mu\) and \(\dot\nu_i(s)=\mathbb E[3\mathbf1\{\operatorname{med}(X_1,X_2,X_3)=i\}-\sum_{j=1}^3\mathbf1\{X_j=i\}]\), where the samples are independent with distribution \(\nu(s)\).
Given binary rationals \(\epsilon,\eta,\delta\in(0,1)\), put \(s_\epsilon=\inf\{s\geq0:\nu_r(s)\geq1-\epsilon\}\), with \(\inf\varnothing=\infty\). Let \(L\) be the total input bit length.
For comparison, initialize an urn with \(N\mu_i\) balls of type \(i\), taking \(N\) through common denominator multiples. Each step samples three balls uniformly with replacement and relabels each distinct sampled ball to the median sampled type. Let \(\nu^{(N)}(s)\) be its empirical distribution after \(\lfloor Ns\rfloor\) steps.
Can a certified additive-\(\eta\) approximation to \(s_\epsilon\), or detection that it is infinite, be computed in polynomial time in \(m,\tau,L,\log(1/\eta)\), or is this approximation problem NP-hard?
What quantitative guarantees with confidence \(1-\delta\) relate this mass accuracy to \(\nu^{(N)}\) and the eventual winning type as \(N\to\infty\)?
Discrete: with \(n\) ordered participants and \(k=O(a^{-2}\log(1/\delta))\) copies per participant, the winner has rank error \(O(a\sqrt n)\) from the median with probability at least \(1-\delta\); expected total comparisons are \(O(kn\log^2(kn))\). [Goel–Lee](https://arxiv.org/pdf/1210.0664)
Continuous: for a uniform distribution on \([0,1]\), the median of one sampled triad has density \(6x(1-x)\); this concerns one interaction, not the evolving mass process. No result settling the stated computational task was located. [Goel–Lee](https://arxiv.org/pdf/1210.0664)
The stated approximation problem is polynomial-time solvable. Under the strict median hypothesis, \(s_\epsilon\) is always finite. A deterministic algorithm can output rational endpoints \(s_-\) and \(s_+\) satisfying
\[ 0\leq s_-\leq s_\epsilon\leq s_+, \qquad s_+-s_-\leq\eta, \]
in time polynomial in \(\tau,L,\log(1/\eta)\). Since \(m=\tau\), this gives the requested complexity bound.
The finite urn admits both an explicit finite-time concentration bound and an exact formula for its eventual winning type. The population size needed for these guarantees depends on the initial median gap and need not be polynomial in the input bit length.
Define the cumulative masses
\[ F_i(s)=\sum_{j\leq i}\nu_j(s), \qquad F_0(s)=0,\qquad F_\tau(s)=1. \]
A sampled median is at most \(i\) exactly when at least two sampled types are at most \(i\). Independence therefore gives
\[ \Pr(\operatorname{med}(X_1,X_2,X_3)\leq i) =3F_i^2(1-F_i)+F_i^3. \]
Summing the prescribed mass derivatives over types at most \(i\) yields
\[ \dot F_i =3(3F_i^2-2F_i^3-F_i) =f(F_i), \qquad f(x)=3x(1-x)(2x-1). \]
Thus the cumulative masses satisfy independent scalar differential equations.
The scalar flow preserves its initial order: two solutions cannot cross, by uniqueness for the polynomial vector field \(f\). Since \(0\) and \(1\) are stationary solutions, the resulting cumulative masses remain ordered in \([0,1]\). Their successive differences consequently define a distribution solving the original system. This also establishes global existence and uniqueness.
For \(0\leq q<1/2\), let \(Q(s;q)\) solve \(\dot Q=f(Q)\) with \(Q(0;q)=q\). Set \(y=1-2Q\). Direct substitution gives
\[ \dot y=\frac32y(1-y^2). \]
Consequently, \(w=y^{-2}-1\) satisfies \(\dot w=-3w\). Therefore
\[ Q(s;q) =\frac12\left(1-\frac1{\sqrt{1+C(q)e^{-3s}}}\right), \qquad C(q)=\frac{4q(1-q)}{(1-2q)^2}. \]
This formula includes \(q=0\), for which \(C(q)=0\) and \(Q(s;0)=0\). Also, \(f(1-x)=-f(x)\), so the complementary mass of a cumulative solution starting above one half follows this same formula.
Write
\[ a=F_{r-1}(0),\qquad b=1-F_r(0), \qquad A=C(a),\qquad B=C(b). \]
The strict median assumption gives \(a,b<1/2\). Hence
\[ \nu_r(s) =1-Q(s;a)-Q(s;b) =\frac12\left( \frac1{\sqrt{1+Ae^{-3s}}} +\frac1{\sqrt{1+Be^{-3s}}} \right). \]
If \(\tau=1\), then \(\nu_r(s)=1\) and \(s_\epsilon=0\). Otherwise, \(A+B>0\), so \(\nu_r(s)\) is strictly increasing and tends to \(1\). Since \(\epsilon>0\), the hitting time is finite on every promised input.
First test the rational inequality \(\mu_r\geq1-\epsilon\). If it holds, output \(s_-=s_+=0\).
Otherwise, put \(z=e^{-3s}\) and define
\[ G(z)=\frac12\left( (1+Az)^{-1/2}+(1+Bz)^{-1/2} \right). \]
There is a unique \(z_*\in(0,1)\) satisfying \(G(z_*)=1-\epsilon\), and
\[ s_\epsilon=-\frac13\log z_*. \]
We can bracket \(z_*\) by exact rational comparisons, without numerically comparing nearly equal transcendental quantities.
For a rational test point \(z>0\), set
\[ u=\frac1{1+Az},\qquad v=\frac1{1+Bz},\qquad c=2(1-\epsilon),\qquad D=c^2-u-v. \]
Comparing \(G(z)\) with \(1-\epsilon\) is equivalent to comparing \(\sqrt u+\sqrt v\) with \(c\).
If \(D<0\), then \((\sqrt u+\sqrt v)^2>c^2\), so \(G(z)>1-\epsilon\). If \(D\geq0\), squaring quantities whose signs are now known gives
\[ \operatorname{sign}(G(z)-(1-\epsilon)) =\operatorname{sign}(4uv-D^2). \]
Thus each comparison, including equality detection, uses exact rational arithmetic.
For \(x\geq0\),
\[ 1-(1+x)^{-1/2}\leq x/2, \]
because the derivative of the left-hand side is at most \(1/2\). It follows that
\[ 1-G(z)\leq\frac{(A+B)z}{4}. \]
Set \(h=2\epsilon/(A+B)\). Our initial test failed, so
\[ \epsilon<1-G(1)\leq(A+B)/4. \]
Therefore \(0<h<1/2\), and \(1-G(h)\leq\epsilon/2\). In particular,
\[ h<z_*<1. \]
Bisect this bracket until its rational endpoints \(z_-\leq z_*\leq z_+\) satisfy \(z_+-z_-\leq h\eta\). The resulting exact time bracket has width
\[ \frac13\log\frac{z_+}{z_-} \leq\frac{z_+-z_-}{3z_-} \leq\frac{\eta}{3}. \]
Compute outward rational enclosures of its two logarithmic endpoints, each of width at most \(\eta/6\). Taking the outer endpoints, and replacing a negative lower endpoint by zero, produces a certified time interval of width at most \(\eta\).
The cumulative sums, \(A,B,h\), and all initial rational quantities have bit length \(O(L)\). In particular, \(\log(1/h)=O(L)\). The number of bisections is therefore \(O(L+\log(1/\eta))\), and the rational operands retain polynomial bit length.
For completeness, certified logarithms can be computed by binary range reduction and the series
\[ \log u =2\sum_{j=0}^{\infty}\frac{w^{2j+1}}{2j+1}, \qquad w=\frac{u-1}{u+1}, \qquad 1\leq u\leq2. \]
Here \(0\leq w\leq1/3\). After retaining terms through \(j=K-1\), the remaining positive tail is at most
\[ \frac{2w^{2K+1}}{(2K+1)(1-w^2)}. \]
The same series computes \(\log2\). Range reduction introduces only \(O(L)\) multiples of \(\log2\), so polynomially many terms suffice for the required certified precision.
This proves the claimed polynomial-time algorithm. Detection of infinity is unnecessary on valid inputs, because Section 2 excludes it.
Let
\[ F_i^{(N)}(s)=\sum_{j\leq i}\nu_j^{(N)}(s). \]
Assume \(\tau\geq2\), \(N\) is an allowed denominator multiple, \(T>0\), and \(0<\delta_0<1\). Define
\[ E_N(T,\delta_0) =e^{3T}\left( \sqrt{\frac{8T}{N}\log\frac{2(\tau-1)}{\delta_0}} +\frac{9T}{8N} \right) +\frac{3}{4N}. \]
Then, with probability at least \(1-\delta_0\),
\[ \sup_{0\leq s\leq T} \max_{1\leq i<\tau} |F_i^{(N)}(s)-F_i(s)| \leq E_N(T,\delta_0). \]
Consequently, on the same event,
\[ \sup_{0\leq s\leq T} \max_i|\nu_i^{(N)}(s)-\nu_i(s)| \leq2E_N(T,\delta_0), \]
and
\[ \sup_{0\leq s\leq T} \|\nu^{(N)}(s)-\nu(s)\|_1 \leq2\tau E_N(T,\delta_0). \]
For \(\tau=1\), both processes are identical.
Fix a cutoff \(i\), and let \(K_j\) be the number of balls with type at most \(i\) after \(j\) steps. Conditional on \(K_j=k\), put \(p=k/N\).
The median lies on the majority side of the cutoff. Thus
\[ \Pr(K_{j+1}=k+1\mid K_j=k)=3p^2(1-p), \]
\[ \Pr(K_{j+1}=k-1\mid K_j=k)=3p(1-p)^2. \]
These formulas remain exact when a ball is sampled repeatedly. In a sample split between the two sides, the minority side occupies exactly one sampled position, so exactly one ball changes sides. Repeating a majority-side ball does not change that conclusion.
Writing \(Z_j=K_j/N\), we obtain
\[ \mathbb E[Z_{j+1}-Z_j\mid\mathcal F_j]=f(Z_j)/N. \]
The centered increments have absolute value at most \(2/N\). Maximal Azuma–Hoeffding therefore gives, for \(J=\lfloor NT\rfloor\),
\[ \Pr\left(\max_{j\leq J}|M_j|>u\right) \leq2\exp\left(-\frac{N^2u^2}{8J}\right), \]
where \(M_j\) is the cumulative centered increment. For \(J=0\), this martingale is identically zero. A union bound over the \(\tau-1\) cutoffs gives the square-root term in \(E_N\).
On \([0,1]\),
\[ |f(x)|\leq3/4,\qquad |f'(x)|=|-18x^2+18x-3|\leq3. \]
Consequently, each exact ODE step has Euler remainder at most
\[ \left| F_i((j+1)/N)-F_i(j/N)-f(F_i(j/N))/N \right| \leq\frac{9}{8N^2}. \]
Indeed, the integrand changes by at most \(3(3/4)t\) over an interval of length \(1/N\). Summing these remainders and applying discrete Grönwall gives the factor \(e^{3T}\) and the term \(9T/(8N)\). Between grid points, the deterministic cumulative mass changes by at most \(3/(4N)\), while the urn process is constant. This proves the stated bound.
Use the certified upper endpoint \(s_+\) from Section 3. Taking \(T\geq s_+\), the preceding event implies
\[ \nu_r^{(N)}(s_+) \geq1-\epsilon-2E_N(T,\delta_0). \]
Thus the computed time comes with an explicit finite-population mass guarantee.
There is also a quantitative guarantee for the urn hitting time
\[ S_\epsilon^{(N)} =\inf\{s\geq0:\nu_r^{(N)}(s)\geq1-\epsilon\}. \]
Let
\[ \gamma=\min\{1/2-a,\;1/2-b\}>0, \qquad \kappa=2E_N(T,\delta_0). \]
If \(\kappa<\min\{\epsilon,1-\epsilon\}\) and \(T\geq s_{\epsilon-\kappa}\), then, on the same event,
\[ s_{\epsilon+\kappa} \leq S_\epsilon^{(N)} \leq s_{\epsilon-\kappa}. \]
Before the lower endpoint, the deterministic median mass plus its error allowance is below the target. At the upper endpoint, the deterministic mass minus that allowance reaches the target. This argument does not require the urn mass to be monotone.
To convert this bracket into a time error, let \(D(s)=1-\nu_r(s)\). Each outside tail \(q\) is nonincreasing and satisfies
\[ -\dot q=3q(1-q)(1-2q)\geq3\gamma q, \]
since \(1-q\geq1/2\) and \(1-2q\geq2\gamma\). Hence \(-\dot D\geq3\gamma D\). Integrating between successive target levels gives
\[ |S_\epsilon^{(N)}-s_\epsilon| \leq\frac1{3\gamma} \log\frac{\epsilon}{\epsilon-\kappa}. \]
If \(s_\epsilon=0\), the urn also starts above the target and its hitting time is exactly zero.
For fixed input and fixed \(T\), \(E_N(T,\delta_0)\) tends to zero along the allowed denominator multiples.
The cutoff reduction and binomial absorption law appear in [Goel–Lee, Sections 3.1–3.3](https://arxiv.org/pdf/1210.0664). Here they apply to arbitrary initial masses, as the following direct argument verifies.
Let \(W_N\) denote the eventual winning type, and set
\[ k_i=N F_i(0),\qquad k_0=0,\qquad k_\tau=N. \]
Define
\[ H_N(k) =2^{-(N-1)} \sum_{j=0}^{k-1}\binom{N-1}{j}, \qquad H_N(0)=0. \]
Then
\[ \Pr(W_N\leq i)=H_N(k_i), \qquad \Pr(W_N=i)=H_N(k_i)-H_N(k_{i-1}). \]
Equivalently, if \(B_N\) has distribution \(\operatorname{Bin}(N-1,1/2)\), the winner has the type containing initial ball rank \(B_N+1\).
For a cutoff count, both transition probabilities are positive at every interior state. From every such state, consecutive upward moves can reach \(N\). Since there are finitely many states, there is a uniform positive probability of reaching a boundary within \(N\) steps. Iterating this bound proves almost-sure absorption.
Every cutoff therefore absorbs almost surely. There are finitely many cutoffs, and absorbed cutoffs remain absorbed. Once all have absorbed, all balls have one type.
Let \(h(k)\) be the probability that a cutoff absorbs at \(N\). Its harmonic equation gives
\[ p_+(k)(h(k+1)-h(k)) =p_-(k)(h(k)-h(k-1)). \]
Writing \(d_k=h(k)-h(k-1)\), we obtain
\[ d_{k+1}=\frac{N-k}{k}d_k. \]
Thus \(d_k=d_1\binom{N-1}{k-1}\). The boundary conditions \(h(0)=0\), \(h(N)=1\) imply \(d_1=2^{-(N-1)}\), proving \(h=H_N\). Applying this to each cutoff proves the winner law.
The two ways to miss \(r\) are
\[ W_N<r \iff B_N\leq Na-1, \]
\[ W_N>r \iff B_N\geq N(1-b), \]
where these equivalences describe the binomial representation of the winner law.
The binomial mean is \((N-1)/2\). The corresponding distances from that mean are
\[ N(1/2-a)+1/2, \qquad N(1/2-b)+1/2. \]
Both are at least \(N\gamma\). Hoeffding’s inequality for the \(N-1\) independent Bernoulli variables therefore yields
\[ \Pr(W_N\neq r) \leq2\exp\left(-\frac{2N^2\gamma^2}{N-1}\right) \leq2e^{-2N\gamma^2}. \]
In particular,
\[ N\geq\frac{\log(2/\delta)}{2\gamma^2} \]
suffices for \(r\) to win with probability at least \(1-\delta\). Consequently, \(\Pr(W_N=r)\) tends to \(1\) as \(N\) tends to infinity.
Use \(\delta_0=\delta/2\) in the finite-time mass bound, and require
\[ N\geq\frac{\log(4/\delta)}{2\gamma^2}. \]
A union bound then shows that, with probability at least \(1-\delta\), both of the following hold:
There is also a guarantee obtained directly from the mass at \(s_+\). Put \(c=\epsilon+2E_N(T,\delta_0)\). If \(c<1/2\), then on the mass-accuracy event each outside tail at \(s_+\) is at most \(c\). Applying the absorption bound conditionally on that urn state gives
\[ \Pr(W_N\neq r) \leq\delta_0+ 2\exp\left(-2N(1/2-c)^2\right). \]
This explicitly connects certified near-consensus mass to the eventual winner.
The deterministic hitting-time computation is polynomial in the encoded input and requested precision. The finite-population requirements have different scaling: the displayed trajectory estimate contains \(e^{3T}\), and winner concentration requires a population proportional to \(\gamma^{-2}\log(1/\delta)\). Since the median gap can be exponentially small in \(L\), these population requirements can be exponential in the input length.
All guarantees above concern the specified strict-median, finite-support model, with replacement sampling and relabeling of each distinct sampled ball. No interchange of the infinite-time and large-population limits is needed: the trajectory estimate handles fixed time horizons, and the exact absorption law handles the eventual winner separately.
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The output resolves the computational task and establishes the stated finite-time and eventual-winner guarantees, with one minor edge-case omission. I found no major gap or false central step. The appropriate status is proved_with_minor_gaps.
The probability that a sampled median lies at or below cutoff \(i\) is correctly computed as \(3F_i^2(1-F_i)+F_i^3\). Summing the prescribed derivatives therefore gives
\( \dot F_i=3F_i(1-F_i)(2F_i-1) \).
The factor of three agrees with the interaction rule in USER_TASK.
The existence and uniqueness argument is valid. The polynomial vector field is locally Lipschitz; its stationary endpoints and uniqueness prevent solutions from leaving \([0,1]\) or crossing one another. Boundedness then permits global continuation. Successive differences of the ordered cumulative solutions define nonnegative masses summing to one and recover the original system.
The substitutions \(y=1-2Q\) and \(w=y^{-2}-1\) give the stated equations and solution. The restriction \(q<1/2\) ensures that \(y>0\), so the substitution is legitimate. The formula also correctly covers \(q=0\).
The complementary-flow argument is valid because \(f(1-x)=-f(x)\). Consequently, the displayed formula for \(\nu_r(s)\) follows from the two outside tails.
The strict median assumption supplies both required inequalities \(a<1/2\) and \(b<1/2\). For \(\tau>1\), positivity of all present masses ensures \(A+B>0\), even when the median is an endpoint type. Thus \(\nu_r(s)\) is strictly increasing toward one. For \(\tau=1\), it is identically one. The conclusion that every promised hitting time is finite is justified.
The initial rational test correctly handles every input whose target is already met, including equality.
Otherwise, the root is unique in \((0,1)\). In the comparison oracle, \(u,v,c\) are positive. When \(D<0\), the claimed strict inequality follows. When \(D\geq0\), both quantities being squared have known nonnegative signs, and comparison with \(4uv-D^2\) is valid. Equality detection is therefore exact; there is no extraneous-root problem.
The bound \(1-G(z)\leq(A+B)z/4\) and the failed initial test imply \(0<h<1/2\). At \(h\), the deficit is at most \(\epsilon/2\), so the asserted bracket \(h<z_*<1\) is correct.
The logarithmic interval-width estimate uses \(z_-\geq h>0\) correctly. Outward endpoint enclosures preserve certification, and their stated error budgets fit within \(\eta\). Clamping a negative lower time endpoint to zero is safe.
The cumulative sums and the rational quantities \(A,B,h\) have bit length \(O(L)\). The strict rational median gaps may be very small, but their reciprocals still have polynomial bit length. Hence \(\log(1/h)=O(L)\), as claimed.
The bisection count is \(O(L+\log(1/\eta))\), and reduced rational arithmetic keeps operand lengths polynomial.
The logarithm series is valid on the stated range. Its positive-tail estimate is correct, and binary range reduction requires only \(O(L)\) multiples of \(\log 2\), because the relevant endpoints remain between \(h\) and one. The needed precision and rational-series arithmetic therefore have polynomial cost.
This establishes certified additive-\(\eta\) approximation in polynomial time under the promised input conditions. Infinity detection is correctly resolved by proving that infinity cannot occur on those inputs.
The cutoff transition probabilities remain exact with replacement sampling and relabeling of distinct sampled balls. In a mixed sample, the minority side occupies one sampled position and therefore contributes exactly one ball that changes sides. Repetition on the majority side does not alter that count.
The conditional drift is consequently \(f(Z_j)/N\). The centered increments are bounded by \(2/N\), which supports the stated maximal Azuma–Hoeffding bound. The union bound over \(\tau-1\) cutoffs gives the displayed square-root term. The separate treatment of \(J=0\) is appropriate.
Both deterministic bounds, \(|f|\leq3/4\) and \(|f'|\leq3\), hold on \([0,1]\). They give the Euler remainder \(9/(8N^2)\). Summing remainders and applying discrete Grönwall produces the stated \(e^{3T}\) factor. The additional \(3/(4N)\) correctly handles times between urn updates.
Taking successive cumulative differences gives the type-mass bound \(2E_N\), and summing those bounds gives the valid, though conservative, bound \(2\tau E_N\).
The guarantee at \(s_+\) follows from deterministic certification and the uniform type-mass error.
The stochastic hitting-time bracket uses uniform approximation in both directions and does not assume monotonicity of the urn mass. The conditions on \(\kappa\) keep both perturbed targets in the valid range, and the horizon condition ensures that the upper comparison time lies within the controlled interval.
The tail differential inequality is valid: each outside tail remains below its initial value, so the factors used in the lower bound remain valid throughout the evolution. The resulting logarithmic time-error bound has the correct direction and constant. A comparison hitting time truncated at zero does not invalidate the bound. The separate assertion that both hitting times equal zero when the initial target is met is also correct.
These guarantees are conditional on the displayed horizon and population conditions; the output does not assert an unconditional finite-\(N\) hitting-time estimate.
Each cutoff has positive upward and downward transition probabilities at every interior state. The finite-state argument establishes almost-sure absorption. Since there are finitely many cutoffs and absorbed cutoffs remain absorbed, eventual consensus follows.
The harmonic equation, increment recurrence, binomial coefficients, and normalization are all correct. The formula also handles cutoffs initially at either boundary.
The binomial rank representation gives exactly the two failure thresholds stated in the output. Both distances from the binomial mean are correctly computed, including the additional \(1/2\).
The cited cutoff reduction and winner law are consistent with [Goel–Lee, Sections 3.1–3.3](https://arxiv.org/pdf/1210.0664). The output supplies its own derivation for arbitrary initial masses, so its extension does not depend on an unstated hypothesis from that reference.
Where: Section 5, “Exponential concentration on the median type,” in the expression containing \(N-1\) in the denominator.
Issue: The promised inputs allow \(\tau=1\) and \(N=1\). For that case, the displayed quantity \(2N^2\gamma^2/(N-1)\) is undefined, and the cited binomial concentration step must be restricted to \(N\geq2\).
Severity: Minor. Positive initial counts imply \(N\geq\tau\), so every nontrivial support already has \(N\geq2\). The output separately recognizes that the single-type process is constant. Its exact winner formula remains valid at \(N=1\), and the final confidence conclusions survive.
Minimum repair: State that the Hoeffding calculation applies for \(N\geq2\), and explicitly exempt the already-treated single-ball case.
Splitting the failure budget into two parts and taking a union bound requires no independence. The stated population threshold gives the desired eventual-winner confidence.
The final conditional absorption estimate also follows from the mass event: when \(c<1/2\), both outside tails are at most \(c\), placing the current median gaps at least \(1/2-c\). Applying the winner bound conditional on that urn state and allowing for failure of the mass event gives the displayed unconditional estimate.
For fixed input and fixed finite horizon, the trajectory error tends to zero along allowed denominator multiples. Separately, the winner probability tends to one. The output correctly avoids an interchange of these limits. The displayed population conditions are sufficient guarantees; no optimal population-size lower bound is established.
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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.