Generator

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performed

One function in the recompress library, called 7 times across 1 essay. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 12 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations2.1·10⁻¹⁰residual, 98 truncations1.1·10⁻⁹representation, deepest1.4·10⁻⁹residual ⁄ representation0.81levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations2.1·10⁻¹⁰residual, 98 truncations1.1·10⁻⁹representation, deepest1.4·10⁻⁹residual ⁄ representation0.81levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻¹²10⁻¹¹10⁻¹⁰truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations3.9·10⁻¹²residual, 98 truncations10⁻¹¹representation, deepest1.2·10⁻¹¹residual ⁄ representation0.86levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻¹⁴10⁻¹³10⁻¹²truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations6.1·10⁻¹⁴residual, 98 truncations1.3·10⁻¹³representation, deepest1.5·10⁻¹³residual ⁄ representation0.85levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻⁴10⁻³10⁻²10⁻¹truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations4·10⁻⁴residual, 98 truncations0.0023representation, deepest0.0028residual ⁄ representation0.83levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

‖A − LLᵀ‖ ⁄ ‖A‖ for a Cholesky computed in the format, against how many truncations it performedThe 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.02040608010010⁻⁸10⁻⁷10⁻⁶10⁻⁵truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations6.6·10⁻⁸residual, 98 truncations1.5·10⁻⁷representation, deepest1.7·10⁻⁷residual ⁄ representation0.85levels, deepest5a hundred approximate stepsand an exact-looking factorisation

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.

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