Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

forcing-trail

One function in the sequence library, called 8 times across 3 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 7 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 what each newton step asked of its linear solve, and what the solve cost, under the adaptive policy. The lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.9 and costs 1 iteration, and the last asks for 0.0042 and costs 271. The whole solve costs 1009 inner iterations across 10 Newton steps and ends at a relative residual of 3.37·10⁻¹¹.

forcing-trail 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.

What each Newton step asked of its linear solve, and what the solve cost, under the adaptive policyThe lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.9 and costs 1 iteration, and the last asks for 0.0042 and costs 271. The whole solve costs 1009 inner iterations across 10 Newton steps and ends at a relative residual of 3.37·10⁻¹¹.0123456789101110⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110³Newton steptolerance asked for, and iterations paiditerations paidtolerance asked forouter residualthe adaptive policy, step by stepNewton steps10inner iterations, total1009first step's cost1last step's cost271final outer residual3.4·10⁻¹¹the rule reads the last two residualsand asks for nothing it cannot use

The lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.9 and costs 1 iteration, and the last asks for 0.0042 and costs 271. The whole solve costs 1009 inner iterations across 10 Newton steps and ends at a relative residual of 3.37·10⁻¹¹.

which: "adaptive"

The arguments are the ones A factorisation kept past its date 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.

What each Newton step asked of its linear solve, and what the solve cost, under the adaptive policyThe lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.9 and costs 1 iteration, and the last asks for 0.0042 and costs 271. The whole solve costs 1009 inner iterations across 10 Newton steps and ends at a relative residual of 3.37·10⁻¹¹.0123456789101110⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110³Newton steptolerance asked for, and iterations paiditerations paidtolerance asked forouter residualthe adaptive policy, step by stepNewton steps10inner iterations, total1009first step's cost1last step's cost271final outer residual3.4·10⁻¹¹the rule reads the last two residualsand asks for nothing it cannot use

The lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.9 and costs 1 iteration, and the last asks for 0.0042 and costs 271. The whole solve costs 1009 inner iterations across 10 Newton steps and ends at a relative residual of 3.37·10⁻¹¹.

which: "tight"

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.

What each Newton step asked of its linear solve, and what the solve cost, under the tight policyThe lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 10⁻¹⁴ and costs 1775 iterations, and the last asks for 10⁻¹⁴ and costs 1128. The whole solve costs 9358 inner iterations across 9 Newton steps and ends at a relative residual of 1.43·10⁻¹³.01234567891010⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110³Newton steptolerance asked for, and iterations paiditerations paidtolerance asked forouter residualthe tight policy, step by stepNewton steps9inner iterations, total9358first step's cost1775last step's cost1128final outer residual1.4·10⁻¹³the rule reads the last two residualsand asks for nothing it cannot use

The lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 10⁻¹⁴ and costs 1775 iterations, and the last asks for 10⁻¹⁴ and costs 1128. The whole solve costs 9358 inner iterations across 9 Newton steps and ends at a relative residual of 1.43·10⁻¹³.

which: "loose"

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.

What each Newton step asked of its linear solve, and what the solve cost, under the loose policyThe lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.1 and costs 4 iterations, and the last asks for 0.1 and costs 133. The whole solve costs 980 inner iterations across 10 Newton steps and ends at a relative residual of 4.09·10⁻¹¹.0123456789101110⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110³Newton steptolerance asked for, and iterations paiditerations paidtolerance asked forouter residualthe loose policy, step by stepNewton steps10inner iterations, total980first step's cost4last step's cost133final outer residual4.1·10⁻¹¹the rule reads the last two residualsand asks for nothing it cannot use

The lower series is the tolerance handed to the inner solve at each step and the upper one is the number of conjugate gradient iterations it took. Under the adaptive rule the first step asks for 0.1 and costs 4 iterations, and the last asks for 0.1 and costs 133. The whole solve costs 980 inner iterations across 10 Newton steps and ends at a relative residual of 4.09·10⁻¹¹.

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.

7 distinct claims across 4 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 forcing policy the trail is drawn for

a size the dense factorisations are affordable at

an outer tolerance that is a relative residual

matmul shapes agree

one tolerance and one cost for every step

the run reaches the outer tolerance

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 factorisation kept past its date

One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.

 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.

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