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diff --git a/frsb_kac-rice.tex b/frsb_kac-rice.tex index dda10f2..51c72bf 100644 --- a/frsb_kac-rice.tex +++ b/frsb_kac-rice.tex @@ -655,6 +655,18 @@ complexity in the ground state are d_d=\hat\beta r_d \end{align} +The supersymmetric solution produces the correct complexity for the ground +state and for a class of minima. Moreover, it produces the correct parameters +for the fields $C$, $R$, and $D$ at those points. This is an important foothold +in the problem of computing the general complexity. The full saddle point +equations at $k$-RSB are not very numerically stable, and a `good' saddle point +has a typically small radius of convergence under methods like Newton's +algorithm. With the supersymmetric solution in hand, it is possible to take +small steps in the parameter space to find non-supersymmetric numeric +solutions, each time ensuring the initial conditions for the solver are +sufficiently close to the correct answer. This is the strategy we use in +\S\ref{sec:examples}. + \section{Full replica symmetry breaking} This reasoning applies equally well to FRSB systems. @@ -810,6 +822,7 @@ that this is the line of stability for the replica symmetric saddle. \section{General solution: examples} +\label{sec:examples} \subsection{1RSB complexity} @@ -819,8 +832,27 @@ to 1RSB in Kac--Rice. For this example, we take \begin{equation} f(q)=\frac12\left(q^3+\frac1{16}q^{16}\right) \end{equation} -With this covariance, the model sees a RS to 1RSB transition at -$\beta_1=1.70615\ldots$ and a 1RSB to 2RSB transition at $\beta_2=6.02198\ldots$. At these points, the average energies are $\langle E\rangle_1=-0.906391\ldots$ and $\langle E\rangle_2=-1.19553\ldots$, and the ground state energy is $E_0=-1.2876055305\ldots$. +established to have a 2RSB ground state \cite{Crisanti_2011_Statistical}. +With this covariance, the model sees a replica symmetric to 1RSB transition at +$\beta_1=1.70615\ldots$ and a 1RSB to 2RSB transition at +$\beta_2=6.02198\ldots$. At these transitions, the average energies are +$\langle E\rangle_1=-0.906391\ldots$ and $\langle E\rangle_2=-1.19553\ldots$, +respectively, and the ground state energy is $E_0=-1.2876055305\ldots$. +Besides these typical equilibrium energies, an energy of special interest for +looking at the landscape topology is the \emph{algorithmic threshold} +$E_\mathrm{alg}$, defined by the lowest energy reached by local algorithms like +approximate message passing \cite{ElAlaoui_2020_Algorithmic, +ElAlaoui_2021_Optimization}. In the spherical models, this has been proven to +be +\begin{equation} + E_{\mathrm{alg}}=-\int_0^1dq\,\sqrt{f''(q)} +\end{equation} +For full RSB systems, $E_\mathrm{alg}=E_0$ and the algorithm can reach the +ground state energy. For the pure $p$-spin models, +$E_\mathrm{alg}=E_\mathrm{th}$, where $E_\mathrm{th}$ is the energy at which +marginal minima are the most common stationary points. Something about the +topology of the energy function is relevant to where this algorithmic threshold +lies. For the $3+16$ model at hand, $E_\mathrm{alg}=1.275140128\ldots$. In this model, the RS complexity gives an inconsistent answer for the complexity of the ground state, predicting that the complexity of minima |