= Solution
A binary <Hopfield network> has $N$ recurrently connected units $s_i\in\{-1,+1\}$, symmetric weights $w_{ij}=w_{ji}$, zero self-weights, and thresholds $\theta_i$. A distributed memory is a desired population pattern $\xi^\mu$. One <Hebbian learning> prescription is
$$
\boxed{w_{ij}=\frac1N\sum_{\mu=1}^P\xi_i^\mu\xi_j^\mu\quad(i\ne j),\qquad w_{ii}=0}.
$$
Coactive signs strengthen a connection, and opposite signs weaken it. This is <Hebbian memory weights for a Hopfield network>, rather than a separate address for each memory.
Initialize the units with the input cue, which can be a corrupted or partially specified pattern. Update one unit at a time by $s_i\leftarrow\operatorname{sign}(\sum_jw_{ij}s_j-\theta_i)$, retaining its old sign when the field is exactly zero. A fair asynchronous schedule revisits every unit. The energy
$$
E=-\tfrac12\sum_{i,j}w_{ij}s_is_j+\sum_i\theta_i s_i
$$
changes on a single update by $\Delta E=-(s_i'-s_i)(\sum_jw_{ij}s_j-\theta_i)\leq0$. An actual flip with the tie convention strictly reduces energy; finitely many states imply eventual arrival at a fixed point. This proves <asynchronous Hopfield energy descent>. Synchronous updating lacks this single-unit argument and can cycle.
For one stored pattern and zero thresholds, the field at $s=\xi$ is $(N-1)\xi_i/N$, so the pattern is a fixed point. With multiple patterns, the desired signal is accompanied by cross-talk from other memories; not every arbitrary set is guaranteed to be stable. \b[Retrieval is attraction from a cue to a stored-pattern fixed point], provided the cue lies in its <basin of attraction>. Spurious minima and limited basins are part of the model, not errors in its update algorithm.
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