State preparation does not print all coordinates

For each $x$, the ratios $(b_x{)}_j\mathrm{/}(b_x{)}_1=c_j\mathrm{/}a$ are $1$ or $4$ when $a=1$, and $1\mathrm{/}2$ or $2$ when $a=2$. Since $N\ge 2$, at least one ratio determines $a$; each ratio then determines its corresponding $c_j$. Thus the $2^N$ inputs give $2^N$ distinct solution rays, with no ambiguity from a global phase.

Let $z$ be a classical coordinate output for one of these rays, phase-fixed so that its first coordinate is positive real. If $\mathrm{\mid }{z}_j-(b_x{)}_j\mathrm{\mid }<\delta :=1\mathrm{/}(100\sqrt{N})$ for every $j$, then $(b_x{)}_1\ge 1\mathrm{/}(4\sqrt{N})$ because $a\ge 1$ and $\mathrm{\parallel }a{e}_1+{\sum }_{j\ge 2}{c}_j{e}_j\mathrm{\parallel }\le 4\sqrt{N}$. Hence $\mathrm{\mid }{z}_1\mathrm{\mid }\ge 6\mathrm{/}(25\sqrt{N})$. For each $j\ge 2$, $\mid \frac{z_j}{z_1}-\frac{(b_x{)}_j}{(b_x{)}_1}\mid \le \frac{\delta }{\mathrm{\mid }{z}_1\mathrm{\mid }}(1+\frac{(b_x{)}_j}{(b_x{)}_1})\le \frac{5}{24}<\frac{1}{4}\mathrm{.}$ The four candidate ratios have pairwise separation at least $1\mathrm{/}2$, so nearest-ratio decoding recovers $a$ and every $c_j$, hence all $N$ bits. Any unambiguous classical output code for all these rays therefore needs at least $2^N$ codewords and at least $N$ bits. This output-length requirement is independent of the ability to prepare or measure the corresponding quantum state.