two-sided-census

At its defaults it draws what happens to the two-sided recurrence on 4000 random 5×5 integer matrices. Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 4. 295 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 532 end in a serious one, where the divisor vanishes and neither vector does: 13.3 per cent. Of those, 495 are cleared by a look-ahead block of length two, 29 need a longer one, and 8 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

two-sided-census is one function in lib/figures/breakdown.js — breakdown — the zero that is an answer, and the zero that is nothing. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 4. 295 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 532 end in a serious one, where the divisor vanishes and neither vector does: 13.3 per cent. Of those, 495 are cleared by a look-ahead block of length two, 29 need a longer one, and 8 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 4. 295 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 532 end in a serious one, where the divisor vanishes and neither vector does: 13.3 per cent. Of those, 495 are cleared by a look-ahead block of length two, 29 need a longer one, and 8 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 6. 33 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 337 end in a serious one, where the divisor vanishes and neither vector does: 8.4 per cent. Of those, 322 are cleared by a look-ahead block of length two, 15 need a longer one, and 0 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 3. 625 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 663 end in a serious one, where the divisor vanishes and neither vector does: 16.6 per cent. Of those, 579 are cleared by a look-ahead block of length two, 40 need a longer one, and 44 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

Entries drawn uniformly from −2 to 2, starting vectors both e₁, and the recurrence run to step 5. 112 of the 4000 end in a lucky breakdown — a Krylov space closed, which is the event that is an answer. 414 end in a serious one, where the divisor vanishes and neither vector does: 10.3 per cent. Of those, 398 are cleared by a look-ahead block of length two, 15 need a longer one, and 1 have no j at all with w̃ᵀAʲṽ ≠ 0 and cannot be cleared at any block length. On Gaussian matrices every one of the last three counts is zero, which is the usual reason given for not worrying about this.

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

5 distinct claims across 5 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

It draws a decomposition and prints its residual. It calls census, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 113 of 219 generators — 98 print a residual and 15 are exempt with a published reason; 106 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.