rank-floor
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.
At its defaults
Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.
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.
steps: 70
The arguments are the ones An iterate that must be made smaller passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.
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.
What it checked while drawing
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.
4 distinct claims across 2 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.
a run long enough to have reached its floor
and the unstructured answer is nowhere near reached
matmul shapes agree
one more column never raises the floor
Against the rule
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 146
of 287 generators —
131 print a residual and
15 are exempt with a published reason;
141 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.
Where it is called
Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.
An iterate that must be made smaller
Applying a Kronecker-sum operator to a low-rank iterate multiplies its ranks by d and adding two of them adds their ranks, so a solver in a compressed format cannot keep what it produces. Every step is followed by a truncation — and whether that truncation is a floor on the residual depends on the right-hand side rather than on the truncation.
When the index is a tupleThe format that does not notice the dimension
A Tucker core is r^d numbers, so the format that repaired the definition still cannot go past five indices. Cutting between the indices rather than across them gives d − 1 ranks instead of d, storage linear in the number of indices, and a family whose ranks are two everywhere by an addition formula.
Neither sparse nor denseThe rounding that was not the problem
A rank-k block plus a rank-k block is a rank-2k block, exactly, so every arithmetic in this format truncates after every addition. A Cholesky performed inside it does ninety-eight of those and its residual is 1.14·10⁻⁹ against a representation error of 1.40·10⁻⁹ — the roundings cost nothing measurable.