The residual a rank budget leaves, against the budget, on two right-hand sides of the same operator

At its defaults it draws the residual a rank budget leaves, against the budget, on two right-hand sides of the same operator. Both runs use the same method on the same matrix for 70 steps and differ only in b. With a constant right-hand side the floor falls 0.118, 0.0107, 3.61·10⁻⁴, 5.4·10⁻⁶, 1.07·10⁻¹³, 1.07·10⁻¹³, 1.07·10⁻¹³ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.874, 0.801, 0.735, 0.656, 0.47: 8 columns of a 10-column answer buy a factor of 2.06. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

rank-floor is one function in lib/figures/ttrain.js — the train — one rank per cut of the index list, and the iterate that must be cut back. 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.

Both runs use the same method on the same matrix for 70 steps and differ only in b. With a constant right-hand side the floor falls 0.118, 0.0107, 3.61·10⁻⁴, 5.4·10⁻⁶, 1.07·10⁻¹³, 1.07·10⁻¹³, 1.07·10⁻¹³ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.874, 0.801, 0.735, 0.656, 0.47: 8 columns of a 10-column answer buy a factor of 2.06. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

Both runs use the same method on the same matrix for 60 steps and differ only in b. With a constant right-hand side the floor falls 0.118, 0.0108, 3.66·10⁻⁴, 5.55·10⁻⁶, 5.9·10⁻¹⁴, 5.9·10⁻¹⁴, 5.9·10⁻¹⁴ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.874, 0.801, 0.735, 0.655, 0.47: 8 columns of a 10-column answer buy a factor of 2.06. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

Both runs use the same method on the same matrix for 30 steps and differ only in b. With a constant right-hand side the floor falls 0.119, 0.0116, 4.11·10⁻⁴, 9.36·10⁻⁶, 7.17·10⁻¹⁴, 7.17·10⁻¹⁴, 7.17·10⁻¹⁴ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.872, 0.8, 0.732, 0.652, 0.469: 8 columns of a 10-column answer buy a factor of 2.07. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

Both runs use the same method on the same matrix for 40 steps and differ only in b. With a constant right-hand side the floor falls 0.119, 0.0112, 3.85·10⁻⁴, 6.32·10⁻⁶, 1.55·10⁻¹⁴, 1.55·10⁻¹⁴, 1.55·10⁻¹⁴ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.873, 0.8, 0.733, 0.654, 0.469: 8 columns of a 10-column answer buy a factor of 2.07. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

Both runs use the same method on the same matrix for 50 steps and differ only in b. With a constant right-hand side the floor falls 0.118, 0.011, 3.73·10⁻⁴, 5.79·10⁻⁶, 2.28·10⁻¹⁴, 2.28·10⁻¹⁴, 2.28·10⁻¹⁴ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.873, 0.801, 0.734, 0.655, 0.469: 8 columns of a 10-column answer buy a factor of 2.06. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

Both runs use the same method on the same matrix for 90 steps and differ only in b. With a constant right-hand side the floor falls 0.118, 0.0105, 3.54·10⁻⁴, 5.22·10⁻⁶, 1.99·10⁻¹³, 1.99·10⁻¹³, 1.99·10⁻¹³ — about two decades per column — and disappears at rank 5, because the solution *is* a train of that rank and the truncation has stopped removing anything. With independent normal entries the same ladder gives 0.969, 0.925, 0.874, 0.802, 0.736, 0.657, 0.47: 8 columns of a 10-column answer buy a factor of 2.06. So a floor is not a property of the truncation. It is the distance from the answer to the set the truncation projects onto, and the same budget is free on one problem and useless on the next.

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.

4 distinct claims across 6 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.