Generator

error-triangle

One function in the error library, called 6 times across 5 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 5 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 backward error, forward error and the condition number, with measured values. Two boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.

error-triangle is one function in lib/figures/error.js — error — the backward one, the forward one, and the number between them. 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.

Backward error, forward error and the condition number, with measured valuesTwo boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.the problem you posedA = H10b = A·(1, 2, …, 10)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.3·10⁻¹⁷the answer you wantedx = (1, 2, …, 10), exactlythe answer you gotx̂, wrong by 2.7·10⁻⁴ relativebackward error 2.3·10⁻¹⁷forward error 2.7·10⁻⁴κ = 1.6·10¹³κ · η = 3.6·10⁻⁴, and the measured forward error is 2.7·10⁻⁴.The algorithm is not at fault. The problem is.H10, LU with partial pivotingresidual and error differ

Two boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.

n: 13

The arguments are the ones A small residual is not a small error 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.

Backward error, forward error and the condition number, with measured valuesTwo boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.the problem you posedA = H13b = A·(1, 2, …, 13)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.2·10⁻¹⁷the answer you wantedx = (1, 2, …, 13), exactlythe answer you gotx̂, wrong by 0.75 relativebackward error 2.2·10⁻¹⁷forward error 0.75κ = 1.7·10¹⁸κ · η = 38, and the measured forward error is 0.75.The algorithm is not at fault. The problem is.H13, LU with partial pivotingresidual and error differ

Two boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.

n: 10

The arguments are the ones Buying the accuracy back 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.

Backward error, forward error and the condition number, with measured valuesTwo boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.the problem you posedA = H10b = A·(1, 2, …, 10)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.3·10⁻¹⁷the answer you wantedx = (1, 2, …, 10), exactlythe answer you gotx̂, wrong by 2.7·10⁻⁴ relativebackward error 2.3·10⁻¹⁷forward error 2.7·10⁻⁴κ = 1.6·10¹³κ · η = 3.6·10⁻⁴, and the measured forward error is 2.7·10⁻⁴.The algorithm is not at fault. The problem is.H10, LU with partial pivotingresidual and error differ

Two boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.

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.

5 distinct claims across 3 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.

and the answer is wrong anyway

LU is for square matrices

the algorithm did its job

the forward error is within κ·η

the residual is nowhere near the error

Against the rule

It calls a factoriser without drawing a factorisation (solve), so the rule is written down as not applying, with the reason: the factorisation is a step towards the solution; the figure is the error identity

The exemption list is the interesting half of the rule rather than an escape hatch — it is where a decision about a figure had to be argued in one line. residualcheck refuses an exemption that is not doing work, and rejected ten of the fifteen written for the expansion's figures on exactly that ground: a figure whose vertical axis is a residual satisfies the rule by construction, and touching a factoriser does not by itself require an entry.

Across the library: the rule bites on 52 of 99 generators — 37 print a residual and 15 are exempt with a published reason; 47 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.

Two errors, and whose fault they are

A small residual is not a small error

Substituting the answer back and finding that it fits is the most natural check there is, and it verifies the wrong thing. A residual of 10⁻¹⁷ is entirely compatible with an answer whose second digit is wrong.

Two errors, and whose fault they are

An answer that is known

Almost every demonstration of numerical error estimates the error by computing the same thing more carefully. The Hilbert matrix does not need that: its inverse is a closed form in integers, so the true answer is available exactly and the error is measured rather than approximated.

The arithmetic underneath

Buying the accuracy back

Factorise in single precision, then correct the answer using residuals computed in double, and the result is what a full double-precision solve would have given. Compute those residuals in single instead and the identical algorithm, at identical cost, recovers nothing.

Two errors, and whose fault they are

The exact answer to a nearby problem

A good algorithm does not give an approximate answer to your problem. It gives the exact answer to a problem very close to yours — and once that is the definition, a wrong result has two possible authors and they can be measured apart.

Iterating, instead of factorising

The rate the condition number predicts

Conjugate gradients converge at a rate governed by the square root of the condition number. That is a bound rather than an estimate, it is provable, and it is loose enough that provisioning iterations from it wastes nine out of ten.

The whole library · All essays · What must fail