Correcting one corrupted table entry

There are eight cube points, with precisely the zero point corrupted, so $\delta =1\mathrm{/}8<1\mathrm{/}4$. The self-correction theorem gives a unique nearby linear function and success at least $1-2\delta =3\mathrm{/}4$ at each point.

For fixed $x\mathrm{\ne }0$, the query $y$ is corrupted exactly when $y=0$, and the query $x+y$ is corrupted exactly when $y=x$. These are distinct seeds. At either seed exactly one summand is flipped, so the output is wrong; at the other six seeds neither is flipped and linearity gives the right answer. At $x=0$, the sum $f(y)+f(y)$ is zero for every seed, equal to $\mathrm{\ell }(0)$. Thus the lower bound is attained for nonzero requests and exceeded for the zero request.