Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

inner-plateau

One function in the sequence library, called 12 times across 6 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 9 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 single newton step solved to eleven inner tolerances, against how far the resulting point is from the root. The iterate is 0.0372 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1126 conjugate gradient iterations, and the resulting point is 0.005319 and 0.002497 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.001383 — however exactly it is computed.

inner-plateau is one function in lib/figures/sequence.js — sequences — a solve inside an outer loop, and the accuracy the loop throws away. 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 single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 0.0372 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1126 conjugate gradient iterations, and the resulting point is 0.005319 and 0.002497 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.001383 — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻⁴10⁻³10⁻²10⁻¹1inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 0.00138distance before the step, 0.03721093547231126the number by each point is the iterations it costeleven decades, one landing placedistance before the step0.037its square0.0014where η = 10⁻³ lands0.0025where η = 10⁻¹⁴ lands0.0025iterations for the first354iterations for the second1126the accuracy that is thrown awaymeasured against a root that is known

The iterate is 0.0372 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1126 conjugate gradient iterations, and the resulting point is 0.005319 and 0.002497 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.001383 — however exactly it is computed.

at: 3

The arguments are the ones A tolerance that reads its own residual 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.

A single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 1.03 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 53 and 849 conjugate gradient iterations, and the resulting point is 0.5725 and 0.5071 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 1.055 — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻¹110¹inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 1.05distance before the step, 1.0353219543849the number by each point is the iterations it costeleven decades, one landing placedistance before the step1its square1.1where η = 10⁻³ lands0.51where η = 10⁻¹⁴ lands0.51iterations for the first219iterations for the second849the accuracy that is thrown awaymeasured against a root that is known

The iterate is 1.03 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 53 and 849 conjugate gradient iterations, and the resulting point is 0.5725 and 0.5071 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 1.055 — however exactly it is computed.

at: 7

The arguments are the ones A tolerance that reads its own residual 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.

A single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 0.0025 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1129 conjugate gradient iterations, and the resulting point is 2.837·10⁻⁴ and 1.2·10⁻⁵ from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 6.234·10⁻⁶ — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 6.23·10⁻⁶distance before the step, 0.00251093397131129the number by each point is the iterations it costeleven decades, one landing placedistance before the step0.0025its square6.2·10⁻⁶where η = 10⁻³ lands1.2·10⁻⁵where η = 10⁻¹⁴ lands1.2·10⁻⁵iterations for the first339iterations for the second1129the accuracy that is thrown awaymeasured against a root that is known

The iterate is 0.0025 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1129 conjugate gradient iterations, and the resulting point is 2.837·10⁻⁴ and 1.2·10⁻⁵ from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 6.234·10⁻⁶ — however exactly it is computed.

at: 6

The arguments are the ones The accuracy that is thrown away 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.

A single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 0.0372 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1126 conjugate gradient iterations, and the resulting point is 0.005319 and 0.002497 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.001383 — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻⁴10⁻³10⁻²10⁻¹1inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 0.00138distance before the step, 0.03721093547231126the number by each point is the iterations it costeleven decades, one landing placedistance before the step0.037its square0.0014where η = 10⁻³ lands0.0025where η = 10⁻¹⁴ lands0.0025iterations for the first354iterations for the second1126the accuracy that is thrown awaymeasured against a root that is known

The iterate is 0.0372 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 109 and 1126 conjugate gradient iterations, and the resulting point is 0.005319 and 0.002497 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.001383 — however exactly it is computed.

at: 4

The arguments are the ones The accuracy that is thrown away 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.

A single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 0.507 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 78 and 1005 conjugate gradient iterations, and the resulting point is 0.2069 and 0.1787 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.2572 — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻²10⁻¹110¹inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 0.257distance before the step, 0.507782576211005the number by each point is the iterations it costeleven decades, one landing placedistance before the step0.51its square0.26where η = 10⁻³ lands0.18where η = 10⁻¹⁴ lands0.18iterations for the first257iterations for the second1005the accuracy that is thrown awaymeasured against a root that is known

The iterate is 0.507 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 78 and 1005 conjugate gradient iterations, and the resulting point is 0.2069 and 0.1787 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 0.2572 — however exactly it is computed.

at: 2

The arguments are the ones The accuracy that is thrown away 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.

A single Newton step solved to eleven inner tolerances, against how far the resulting point is from the rootThe iterate is 1.8 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 33 and 700 conjugate gradient iterations, and the resulting point is 1.138 and 1.027 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 3.238 — however exactly it is computed.10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻¹110¹inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 3.24distance before the step, 1.833167416700the number by each point is the iterations it costeleven decades, one landing placedistance before the step1.8its square3.2where η = 10⁻³ lands1where η = 10⁻¹⁴ lands1iterations for the first167iterations for the second700the accuracy that is thrown awaymeasured against a root that is known

The iterate is 1.8 from a root that is known exactly by construction. The same linear system is solved to relative residuals from 0.3 down to 10⁻¹⁴, at 33 and 700 conjugate gradient iterations, and the resulting point is 1.138 and 1.027 from the root. The curve is flat below about 10⁻³: the linearisation is wrong at second order, so the step cannot land closer than the square of the distance it started at — 3.238 — however exactly it is computed.

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.

9 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.

a conditioning the SPD construction can hold

a size the dense factorisations are affordable at

an iterate the exact Newton sequence reaches

and a tolerance of 10⁻³ reaches the same point as one of 10⁻¹⁴

enough tolerances to see a plateau rather than two points

LU is for square matrices

matmul shapes agree

the exact Newton step exists at every iterate walked to

the tightest tolerance costs the most iterations

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 113 of 238 generators — 98 print a residual and 15 are exempt with a published reason; 125 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.

When the problem arrives again

A tolerance that reads its own residual

The cheapest constant forcing term costs 980 inner iterations and arrives with a hundred times the forward error of the dearest, which costs 9,358. A rule that sets each step's tolerance from the ratio of the last two residuals costs 1,009 and arrives with neither problem — and it is not a constant, so it does not appear on the curve the constants are compared on.

 When the problem arrives again

The accuracy that is thrown away

A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.

 Sparsity, and what elimination costs

The order that was right last time

A pivot order computed once and reused across a sequence saves the symbolic phase, and the price is that a pivot which was large may now be small. Replacing it with √u·‖A‖ costs eight orders of backward error and iterative refinement recovers a factor of 8.8 of them. Divide each row by its largest entry first and the same reuse costs nothing at all.

 When the problem arrives again

The problem that arrives again

A hundred and thirty essays have solved a system once and measured how wrong the answer was. Almost no computation is shaped like that. A solve is one step of an outer loop, its answer is an input rather than a deliverable, and four quantities this collection has treated as accuracy requirements turn out to be assets with a shelf life.

 Iterating, instead of factorising

The residual the method reports

Conjugate gradients prints a relative residual of 6.9·10⁻²¹. The unit roundoff is 1.1·10⁻¹⁶, so that is not a small residual and not a large one — it is not a residual. The vector the method is holding at that step has ‖b − Ax‖/‖b‖ = 5.1·10⁻¹⁰, and nothing in the run says so.

 Randomised, and the guarantee that changes kind

The sketch that is spent

Every other object a sequence carries has a shelf life. A random sketch has one use. Deflate what its first round found and apply it again, and it returns the zero matrix — 9.0·10⁻¹⁵ where the first round saw 2.62 — because the input has been made orthogonal to the very draw the guarantee is over.

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