What the regularisation costs, and what 6 steps of refinement take back
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.
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.
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 regularisation that legalises every order
Perturb a saddle-point matrix's two blocks in opposite directions and it acquires a factorisation with a diagonal D under every symmetric permutation — not under a good one, under all of them. Five hundred random orderings, five hundred successes, and a growth factor that spans six orders across them.
The matrix a constraint makesThe zero that is not a missing entry
A constrained minimisation produces a matrix with a zero block, and the zero is a theorem rather than a sparsity pattern. No pivot order makes it positive definite, no precision changes that, and Cholesky does not fail somewhere on it — it fails at the first constraint row, on a number the problem already contained.