‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performed
At its defaults it draws ‖a − llᵀ‖ ⁄ ‖a‖ for a cholesky computed in the format, against how many truncations it performed. The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 2.119·10⁻¹⁰ to 1.138·10⁻⁹, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.93, 0.92, 0.91, 0.81, never above one. A hundred approximate operations contributed nothing measurable to the answer.
depth-residual is one function in lib/figures/recompress.js —
arithmetic in the format — the one operation it is not closed under, and what a hundred roundings cost. 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.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 2.119·10⁻¹⁰ to 1.138·10⁻⁹, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.93, 0.92, 0.91, 0.81, never above one. A hundred approximate operations contributed nothing measurable to the answer.
logEps: -8
The arguments are the ones The count that is not the budget passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 2.119·10⁻¹⁰ to 1.138·10⁻⁹, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.93, 0.92, 0.91, 0.81, never above one. A hundred approximate operations contributed nothing measurable to the answer.
logEps: -10
The arguments are the ones The count that is not the budget passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 3.856·10⁻¹² to 1.035·10⁻¹¹, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.93, 0.93, 0.91, 0.86, never above one. A hundred approximate operations contributed nothing measurable to the answer.
logEps: -12
The arguments are the ones The count that is not the budget passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 6.134·10⁻¹⁴ to 1.282·10⁻¹³, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.95, 0.90, 0.85, 0.85, never above one. A hundred approximate operations contributed nothing measurable to the answer.
logEps: -2
The arguments are the ones The count that is not the budget passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 3.97·10⁻⁴ to 0.0023, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.95, 0.94, 0.93, 0.83, never above one. A hundred approximate operations contributed nothing measurable to the answer.
logEps: -6
The arguments are the ones The count that is not the budget passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The recursion is the textbook one and every step stays in the format: the off-diagonal factor is exact, because a triangular solve against one factor of a rank-k block leaves a rank-k block, and the Schur complement is a rank-k addition to a hierarchical matrix, which is the only approximate step there is. So the number of truncations is a count — one per admissible block in the subtree being updated, at every level — and it runs from 0 at a leaf of 128 to 98 at a leaf of 8. The residual goes from 6.566·10⁻⁸ to 1.462·10⁻⁷, and all of that movement is the representation, whose own error is the upper curve and moves by the same factor: the ratio between them is 1.00, 0.96, 0.92, 0.88, 0.85, never above one. A hundred approximate operations contributed nothing measurable to the answer.
What it checked while drawing
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.
12 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 factorisation's residual is at the level of the representation's error at leaf 128 — checked 5 times
a positive pivot, which is what positive definite means here
an accuracy the blocks have singular values across
an accuracy, not a rank
and the deepest performs many
LU is for square matrices
matmul shapes agree
the shallowest tree performs no truncation at all
Against the rule
It draws a decomposition and prints its residual. It calls
depthSweep,
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 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.
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.