Checkpoint: Distillation Math

This checkpoint pulls together the quantitative core of the module: the $15$-to-$1$ error map, the threshold, and the suppression that makes distillation worthwhile.

The map, once more

A single round of $15$-to-$1$ distillation transforms the input error into

The cubic power comes from the distance-$3$ punctured Reed–Muller code: post-selecting on a trivial syndrome rejects every weight-$1$ and weight-$2$ input error, so the leading surviving failure is a weight-$3$ correlated event, of which there are $35$ relevant patterns.

Working the numbers

Take a factory operating at $\varepsilon_{\text{in}} = 0.01$, comfortably below the threshold $\varepsilon^{*} = 1/\sqrt{35} \approx 0.169$. One round gives

The suppression factor is

so a single round improves the magic-state error by nearly a factor of $300$ at this input quality. Note how the suppression factor grows as $\varepsilon_{\text{in}}$ shrinks — distillation gets relatively more effective the cleaner your raw states already are, which is the quantitative face of "operate well below threshold."

Your task

Compute the single-round output error $\varepsilon_{\text{out}} = 35\,\varepsilon_{\text{in}}^3$ for $\varepsilon_{\text{in}} = 0.01$ and return it. The grader also verifies that your value is an improvement ($\varepsilon_{\text{out}} < \varepsilon_{\text{in}}$) and that the implied suppression factor matches $1/(35\,\varepsilon_{\text{in}}^2)$. Derive everything from the map — nothing is hard-coded.

Sign in on the full site to ask questions and join the discussion.