Generator

What the regularisation costs, and what 6 steps of refinement take back

One function in the quasidef library, called 2 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 15 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 the regularisation costs, and what 6 steps of refinement take back. Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σₘᵢₙ(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σₘᵢₙ and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

regularisation-tradeoff is one function in lib/figures/quasidef.js — quasi-definite — every ordering legal, and the fill a symbolic phase can promise exactly. 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 the regularisation costs, and what 6 steps of refinement take backSolving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σₘᵢₙ(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σₘᵢₙ and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.-14-12-10-8-6-4-210⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹log₁₀ δrelative error against the exact answerδ = σₘᵢₙ(K)no refinement6 stepsa left branch that can be removederror ÷ δ, unrefined1489σₘᵢₙ(K)6.8·10⁻⁴refined at δ = 10⁻⁶3.6·10⁻¹⁶refined at δ = 10⁻²0.62the perturbation is known exactlybecause the code chose it

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σₘᵢₙ(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σₘᵢₙ and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

steps: 0

The arguments are the ones The regularisation that legalises every order passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

What the regularisation costs, and what 0 steps of refinement take backSolving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σₘᵢₙ(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σₘᵢₙ and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.-14-12-10-8-6-4-210⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹log₁₀ δrelative error against the exact answerδ = σₘᵢₙ(K)no refinement0 stepsa left branch that can be removederror ÷ δ, unrefined1489σₘᵢₙ(K)6.8·10⁻⁴refined at δ = 10⁻⁶0.0014refined at δ = 10⁻²0.93the perturbation is known exactlybecause the code chose it

Solving [[H + δI, Aᵀ], [A, −δI]] instead of K gives the exact answer to a different problem, so its error is proportional to δ: measured at 1489·δ across six decades, which is a slope of one and not a trend. Refining against the unregularised matrix — the residual formed with K and the correction solved with the regularised factorisation — removes that term entirely, because the perturbation was never in the residual. It works while δ is below σₘᵢₙ(K) = 6.797·10⁻⁴, marked on the axis, and stops working above it: the iteration's contraction factor is δ/σₘᵢₙ and a fixed point needs that under one. So the trade-off curve every regularisation essay on this site has drawn — a term falling in δ against a term rising in it — has, here, a left branch that can simply be removed.

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.

15 distinct claims across 2 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 exact solve can afford

a constraint no larger than the problem

a finite double, since an infinity is not a rational

a pivot rule this routine implements

a refinement count the figure runs

fewer constraints than unknowns, so something is left to minimise

matmul shapes agree

the unrefined error is proportional to δ

δ = 10⁻¹⁰: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻¹²: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻¹⁴: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻⁴: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻⁵: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻⁶: 6 steps contract by δ/σₘᵢₙ each

δ = 10⁻⁸: 6 steps contract by δ/σₘᵢₙ each

Against the rule

It draws a decomposition and prints its residual. It calls svd, regularisationSweep, 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.

The whole library · All essays · What must fail