 Hopfield recall from a corrupted cue

Store only $\xi =(1,-1,1,-1,1)$ with $M=5$, zero thresholds, and Hebbian memory weights for a Hopfield network. If $d$ bits differ from $\xi$, their Hamming distance is $d$ and ${\sum }_i{\xi }_i{x}_i=M-2d$. Squaring this sum and subtracting the $M$ diagonal terms yields the displayed energy. Thus $d=0,1,2,3,4,5$ gives $E=-2,-2/5,2/5,2/5,-2/5,-2$.

For the cue $x=(-1,1,1,-1,1)$, the first unit has input $h_1=2/5$ and flips to $1$. The second then has $h_2=-4/5$ and flips to $-1$, completing recall. The energy drops $2/5\to -2/5\to -2$. Every cue with at most two wrong bits is corrected under a fair asynchronous schedule: a wrong bit sees the correct majority among the other four, while a correct bit either sees agreement or retains its value at a tie. The opposite pattern $-\xi$ is also stored implicitly, so cues with at least three errors retrieve it. This is a limitation of this symmetric one-pattern construction.