probe-ceiling

At its defaults it draws how many eigenvalues a contour method can return: the probe block is a ceiling and does not say so. Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 4 eigenvalues inside |z| = 4: the rank is 1, 2, 3, 4, 4 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 4. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

probe-ceiling is one function in lib/figures/nlevp.js — infinitely many eigenvalues — a contour that counts, and the probe block that is a ceiling. 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.

Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 4 eigenvalues inside |z| = 4: the rank is 1, 2, 3, 4, 4 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 4. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 4 eigenvalues inside |z| = 4: the rank is 1, 2, 3, 4, 4 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 4. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 2 eigenvalues inside |z| = 2: the rank is 1, 2, 2, 2, 2 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 2. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 12 eigenvalues inside |z| = 5: the rank is 1, 2, 3, 4, 4 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 4, which is the SIZE of the problem rather than the number of eigenvalues — a second ceiling, because an 4 × ℓ block cannot have rank above 4 whatever is inside the contour. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

Beyn's method probes the contour with a random 4 × ℓ block V and reads the eigenvalues out of the rank of ∮T(z)⁻¹V dz. That rank is at most ℓ whatever is inside, so a probe block narrower than the number of eigenvalues present returns ℓ of them — with small residuals, and with nothing in the returned object to say that there were more. Measured with 3 eigenvalues inside |z| = 3: the rank is 1, 2, 3, 3, 3 at ℓ = 1, 2, 3, 4, 5, rising with the probe count and then stopping at 3. A caller who asked for two gets two. The counting integral is the check that says so, and it costs one more pass round the same contour.

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 beynRank, countInside, 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 173 of 325 generators — 158 print a residual and 15 are exempt with a published reason; 152 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.