# C6: asymptotic rate and a finite-time decision differ

Consider the homogeneous six-cycle family P(q)=qI+(1-q)C6, 0<=q<1, with uniform stationary distribution. Let d_q(t) be worst-point-start TV, and tau_q(epsilon) its first threshold time. All point starts are rotations, and TV contracts under the transition matrix. This note compares two specified objectives; it does not replace the standard SLEM objective of fastest-mixing-chain optimization.

## The rate optimum remains correct

Fourier characters on C6 give eigenvalues 1, a,a,b,b,c, where a=(1+q)/2, b=(3q-1)/2 and c=2q-1. Thus beta(q)=max(a,-c), with unique minimum q=1/5 and beta=3/5. This is the optimum of the asymptotic rate objective within this family.

The literature's SLEM optimization criterion is stated explicitly in [Boyd, Diaconis and Xiao, Fastest Mixing Markov Chain on a Graph](https://web.stanford.edu/~boyd/papers/fmmc.html). Strong comparisons at every time under their own hypotheses are also studied in [Fill and Kahn, arXiv:1109.6075v3](https://arxiv.org/abs/1109.6075v3). No novelty claim is made for this example. A path result with particular initial-distribution assumptions is not automatically transferred to this cycle.

## A rational witness that changes the threshold decision

Fix epsilon=19/100. At the rate optimum q=1/5, direct two-step multiplication gives

\[
P^2(0,\cdot)=(9/25,4/25,4/25,0,4/25,4/25),
\qquad d_{1/5}(2)=29/150>19/100.
\]

Three steps give (1/5,6/25,12/125,16/125,12/125,6/25), whose TV is 9/50<19/100. One step has TV 1/2. Therefore tau_(1/5)(19/100)=3.

For the rational alternative q=21/100,

\[
P^2(0,\cdot)=
(7123/20000,1659/10000,6241/40000,0,6241/40000,1659/10000).
\]

Only its initial-state entry exceeds 1/6, so

\[
d_{21/100}(2)=7123/20000-1/6=11369/60000<19/100.
\]

One-step TV is again 1/2, so tau_(21/100)(19/100)=2, even though beta(21/100)=121/200>3/5. The slower asymptotic-rate choice attains this concrete target one step earlier.

For **every** q, the one-step row has support in at most three of the six states. The stationary mass of that support is at most 1/2, so d_q(1)>=1/2. Thus two is the globally smallest possible mixing time at this tolerance within the whole homogeneous-holding family, and q=21/100 attains it. This does not make q=21/100 the unique optimizer, or the minimizer of TV at every time.

## The exact two-step TV minimizer

The generic two-step row is (A,B,D,0,D,B), with

\[
A=q^2+(1-q)^2/2,\qquad B=q(1-q),\qquad D=(1-q)^2/4.
\]

A>=1/3>1/6 for all q. Write alpha=(1-1/sqrt(3))/2 and delta=1-sqrt(2/3), so 0<delta<alpha<1/3. B>=1/6 precisely on [alpha,1-alpha]; D>=1/6 precisely on [0,delta]. Summing positive deviations gives

\[
d_q(2)=
\begin{cases}
2q^2-2q+1/2,&0\le q\le\delta,\\
(3/2)q^2-q+1/3,&\delta\le q\le\alpha,\\
q-q^2/2,&\alpha\le q\le1-\alpha,\\
(3/2)q^2-q+1/3,&1-\alpha\le q\le1.
\end{cases}
\]

The derivatives on the first two intervals are negative; those on the last two are positive. Therefore the unique minimizer of this fixed-time TV objective is

\[
q_*=(1-1/\sqrt3)/2,\qquad d_{q_*}(2)=1/3-1/(4\sqrt3).
\]

It differs from 1/5. Since d_(q*)(2)<19/100<29/150, the rational witness above is consistent with the exact optimum. The first inequality follows from 3<(75/43)^2, with positive quantities. The distinction between optimizing fixed-time distance and minimizing an integer threshold time is retained: many parameters can attain the same optimal integer time.

## Meaning and verification boundary

The research decision now has a concrete target: for this six-cycle family at epsilon=19/100, q=21/100 is a simple rational choice attaining the optimal two-step time, while the unique rate optimum takes three. No source theorem has been refuted. A certificate bound, an asymptotic rate, fixed-time TV and a threshold mixing time answer different questions. In particular the inherited n=4 convenience TV<=beta^t is not silently applied to n=6, whose general point-start spectral coefficient is sqrt(5)/2.

The bundled runner performs exact rational transition iteration. The piecewise algebra and derivative argument above support the parameter-wide optimum; a numerical grid is not a proof. This worked example has no independent scientific acceptance or novelty determination.