= Solution
For the symmetric zero-diagonal weights and asynchronous rule just specified, update only unit $k$. In the energy difference, terms involving neither $k$ nor its symmetric partner cancel. Thus <asynchronous Hopfield energy descent> gives
$$
\Delta E=-\frac12\sum_{j\ne k}w_{kj}(x_k'-x_k)x_j
-\frac12\sum_{i\ne k}w_{ik}x_i(x_k'-x_k)
=-(x_k'-x_k)h_k.
$$
If the unit was already aligned, it does not change and $\Delta E=0$. If it flips into alignment, $x_k'-x_k=2\operatorname{sgn}(h_k)$, so
$$
\boxed{\Delta E=-2|h_k|<0\quad\text{on every actual non-tie flip}.}
$$
At $h_k=0$ our tie rule retains the old activity. The network has only $2^M$ configurations. Strict descent on every actual change prevents revisiting a configuration and permits only finitely many changes. With a fair schedule every misaligned unit is eventually updated, so the terminal configuration is a fixed point. At that point each possible single-bit flip has energy cost $2x_kh_k\geq0$. \b[The network reaches a minimum against single-unit changes, not necessarily the global minimum.] Its energy is a <Lyapunov function> for these asynchronous dynamics.
The qualifications are essential to the printed claim. Synchronous updates can cycle: for two units with $w_{12}=w_{21}=1$, the state $(1,-1)$ alternates with $(-1,1)$, keeping energy $+1$. Symmetry alone therefore does not prove convergence under simultaneous updates.
Nor does the valid descent proof guarantee a global minimum. Consider four units with $w_{12}=w_{34}=3$ and all four cross-group weights equal to $1$, symmetrically and with zero diagonal. The state $(1,1,-1,-1)$ has aligned fields $(1,1,-1,-1)$ and energy $-2$, so it is a strict single-flip local minimum. The all-positive state has energy $-10$. This explicit counterexample shows why local minima may trap retrieval even though the energy always descends under the stipulated rule.
Back to article page
