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==== The margin is ==== γ=mini∈I⋆(λi−∣∇f(x⋆)i∣).\gamma = \min_{i\in I^\star}(\lambda_i - |\nabla f(x^\star)_i|).γ=i∈I⋆min(λi−∣∇f(x⋆)i∣). Here the tightest inactive coordinate is node 2 (the only one with nonzero gradient), so γ=λ2−∣∇f(x⋆)2∣=ρα2−α(1−α)(1−ρ)(1+α)2.\gamma = \lambda_2 - |\nabla f(x^\star)_2| = \rho\alpha\sqrt{2} - \frac{\alpha(1-\alpha)(1-\rho)}{(1+\alpha)\sqrt{2}}.γ=λ2−∣∇f(x⋆)2∣=ρα2−(1+α)2α(1−α)(1−ρ). This simplifies neatly to a linear function of ρ−ρ0\rho-\rho_0ρ−ρ0: γ=α(3+α)(1+α)2 (ρ−ρ0). \boxed{\; \gamma = \frac{\alpha(3+\alpha)}{(1+\alpha)\sqrt{2}}\;(\rho-\rho_0). \;}γ=(1+α)2α(3+α)(ρ−ρ0). So: * At ρ=ρ0\rho=\rho_0ρ=ρ0, γ=0\gamma=0γ=0. * For ρ=ρ0+δ\rho=\rho_0+\deltaρ=ρ0+δ with tiny δ>0\delta>0δ>0, you get γ=Cδ\gamma = C\deltaγ=Cδ with C=α(3+α)(1+α)2C=\frac{\alpha(3+\alpha)}{(1+\alpha)\sqrt{2}}C=(1+α)2α(3+α). Since ρ\sqrt{\rho}ρ is Θ(1)\Theta(1)Θ(1) here (because ρ0\rho_0ρ0 is a constant depending on α\alphaα), taking δ\deltaδ extremely small makes γ\gammaγ much smaller than ρ\sqrt{\rho}ρ. ===== Take α=0.15\alpha=0.15α=0.15. Then ===== ρ0=0.853.15≈0.26984,ρ0≈0.519.\rho_0 = \frac{0.85}{3.15}\approx 0.26984, \qquad \sqrt{\rho_0}\approx 0.519.ρ0=3.150.85≈0.26984,ρ0≈0.519. Now choose ρ=ρ0+10−6\rho=\rho_0+10^{-6}ρ=ρ0+10−6. Then γ≈0.15⋅3.151.15⋅1.414⋅10−6≈2.9×10−7≪0.519≈ρ.\gamma \approx \frac{0.15\cdot 3.15}{1.15\cdot 1.414}\cdot 10^{-6} \approx 2.9\times 10^{-7} \ll 0.519 \approx \sqrt{\rho}.γ≈1.15⋅1.4140.15⋅3.15⋅10−6≈2.9×10−7≪0.519≈ρ. So this path graph gives you a very explicit, closed-form “bad” case where γ<ρ\gamma < \sqrt{\rho}γ<ρ by an arbitrarily large margin, while the true support is tiny (∣S⋆∣=1≤1/ρ|S^\star|=1\le 1/\rho∣S⋆∣=1≤1/ρ). ===== Algorithm parameters (t,βt,\betat,β, restart, etc.) don’t change γ\gammaγ: γ\gammaγ is a property of the instance (graph + α,ρ,s\alpha,\rho,sα,ρ,s) at the true optimizer. What algorithm parameters affect is how much transient “support thrashing” you might see before you enter the γ\gammaγ-neighborhood. ===== If you want, I can also give a path-prefix support construction (where S⋆={1,2,…,k}S^\star=\{1,2,\dots,k\}S⋆={1,2,…,k}) and show the breakpoint for node k+1k+1k+1, so you can make γ\gammaγ tiny while ∣S⋆∣≈k≈1/ρ|S^\star|\approx k\approx 1/\rho∣S⋆∣≈k≈1/ρ.
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