Where binary32 stops being able to hold the linearisation, and where the scaled problem does not

At its defaults it draws where binary32 stops being able to hold the linearisation, and where the scaled problem does not. Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in binary32, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 1.8447·10¹⁹, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 10¹⁹ and the first that does not is 10²⁰. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

units-overflow is one function in lib/figures/qepscale.js — the units a polynomial is written in — two lines of scaling, a prediction that is half right, and a format's edge. 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.

Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in binary32, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 1.8447·10¹⁹, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 10¹⁹ and the first that does not is 10²⁰. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in binary32, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 1.8447·10¹⁹, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 10¹⁹ and the first that does not is 10²⁰. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in fp16, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 255.94, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 100 and the first that does not is 1000. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in tf32, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 1.8442·10¹⁹, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 10¹⁹ and the first that does not is 10²⁰. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

Forming the first companion linearisation needs γ²M to be representable. The upper row is whether every entry of (γ²M, γC, K) is finite in binary64, at twelve changes of units; the lower row is the same question after Fan–Lin–Van Dooren scaling. The boundary is √(largest finite value ÷ largest entry of M) = 1.3408·10¹⁵⁴, computed from the format's parameters and nothing measured — and the sweep agrees with it at every stop: the last change of units that survives is 10¹⁵⁰ and the first that does not is ∞. The scaled row never fails, because δγ²M has norm about one whatever γ was. In a narrow format the scaling is not an accuracy device: it is what makes the problem exist.

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks 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.

10 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.

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 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.