= Solution
Two independent cyclic intervals intersect with <probability> at most $\min\{1,(2k-1)/d\}\leq2k/d$. For $k\leq d/2$, precisely the $2k-1$ starting-point offsets $-(k-1),\ldots,k-1$ can overlap; for larger $k$, the upper bound is automatic. The <cyclic interval overlap bound> and the preceding <chi-squared divergence> estimate therefore give
$$
\chi^2(P_{\theta,n}\Vert P_0^{\otimes n})\leq\frac{2k}{d}(e^{n\theta^2}-1)\leq2d^{-\varepsilon}(\alpha d)^\varepsilon=2\alpha^\varepsilon.
$$
As in 1(e), the square-root hypothesis is meaningful when $\alpha d\geq1$. The <chi-squared testing lower bound> bounds the <total variation distance> by $\sqrt{2\alpha^\varepsilon}/2$. The worst individual <Type II error> dominates the error under the uniform <mixture model>. Thus
$$
\boxed{\inf_\psi\max\{P_0^{\otimes n}(\psi=1),\max_{S\in\mathcal S}P_{\theta,S}^{\otimes n}(\psi=0)\}\geq\tfrac12-\nu_{\varepsilon,\alpha},\quad\nu_{\varepsilon,\alpha}=\frac{\sqrt{2\alpha^\varepsilon}}4.}
$$
Taking $\alpha$ small makes this lower bound informative. This argument uses the actual cyclic-interval alternative, without substituting a different family of subsets.
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