Generator

ordering-legality

One function in the quasidef library, called 17 times across 8 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 10 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 500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproduces. A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 136 of the 500. They span 1.56·10⁻¹⁶ to 1.02·10⁻⁷ — 9 orders — and the growth factor across them runs from 1 to 6.43·10⁵ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

ordering-legality 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.

500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 136 of the 500. They span 1.56·10⁻¹⁶ to 1.02·10⁻⁷ — 9 orders — and the growth factor across them runs from 1 to 6.43·10⁵ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6.4·10⁵growth × δ0.64the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 136 of the 500. They span 1.56·10⁻¹⁶ to 1.02·10⁻⁷ — 9 orders — and the growth factor across them runs from 1 to 6.43·10⁵ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

delta: 1e-8

The arguments are the ones A condition number sent to infinity 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.

500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁸ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 130 of the 500. They span 1.65·10⁻¹⁶ to 6.17·10⁻⁴ — 13 orders — and the growth factor across them runs from 1 to 6.43·10⁷ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6.4·10⁷growth × δ0.64the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁸ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 130 of the 500. They span 1.65·10⁻¹⁶ to 6.17·10⁻⁴ — 13 orders — and the growth factor across them runs from 1 to 6.43·10⁷ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

delta: 0.000001

The arguments are the ones An eigenvalue count that cannot be slightly wrong 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.

500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 136 of the 500. They span 1.56·10⁻¹⁶ to 1.02·10⁻⁷ — 9 orders — and the growth factor across them runs from 1 to 6.43·10⁵ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6.4·10⁵growth × δ0.64the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 136 of the 500. They span 1.56·10⁻¹⁶ to 1.02·10⁻⁷ — 9 orders — and the growth factor across them runs from 1 to 6.43·10⁵ — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

delta: 0.0001

The arguments are the ones An ordering that does not wait for the numbers 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.

500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁴ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 151 of the 500. They span 1.34·10⁻¹⁶ to 1.23·10⁻¹¹ — 5 orders — and the growth factor across them runs from 1 to 6428 — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6428growth × δ0.64the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁴ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 151 of the 500. They span 1.34·10⁻¹⁶ to 1.23·10⁻¹¹ — 5 orders — and the growth factor across them runs from 1 to 6428 — which is 0.643/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

delta: 0.01

The arguments are the ones The regularisation that legalises every order 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.

500 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 0.01 — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 301 of the 500. They span 1.17·10⁻¹⁶ to 6.76·10⁻¹⁵ — 2 orders — and the growth factor across them runs from 1 to 63.29 — which is 0.633/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth63growth × δ0.63the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 0.01 — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 500 of 500 orderings factorise here, against 69.0 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 301 of the 500. They span 1.17·10⁻¹⁶ to 6.76·10⁻¹⁵ — 2 orders — and the growth factor across them runs from 1 to 63.29 — which is 0.633/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

delta: 0.000001, trials: 120

The arguments are the ones The regularisation that legalises every order 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.

120 random symmetric orderings of a regularised saddle-point matrix, and what each factorisation reproducesA quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 120 of 120 orderings factorise here, against 67.5 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 28 of the 120. They span 1.71·10⁻¹⁶ to 2.01·10⁻⁸ — 8 orders — and the growth factor across them runs from 1.03 to 5.112·10⁵ — which is 0.511/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.68worst growth5.1·10⁵growth × δ0.51the ordering is free to chooseand not free of consequence

A quasi-definite matrix — [[H + δI, Aᵀ], [A, −δI]] with δ = 10⁻⁶ — has an LDLᵀ factorisation with diagonal D for every symmetric permutation, so the ordering can be chosen for fill with no numerical veto at all. That is the count: 120 of 120 orderings factorise here, against 67.5 per cent for the same matrix with the zero block left where the problem put it. Each bar counts the orderings whose factorisation left a relative residual ‖PKPᵀ − LDLᵀ‖/‖K‖ in that decade, with the tallest holding 28 of the 120. They span 1.71·10⁻¹⁶ to 2.01·10⁻⁸ — 8 orders — and the growth factor across them runs from 1.03 to 5.112·10⁵ — which is 0.511/δ, an inverse law that holds at every δ this figure is drawn at. Existence is not stability, and the theorem says only the first.

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.

10 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 constraint no larger than the problem

a pivot rule this routine implements

a regularisation small enough to be one

a trial count the figure can afford

and the unregularised matrix breaks down on some of them

and the worst growth factor is a constant over δ

every ordering factorises the quasi-definite matrix

fewer constraints than unknowns, so something is left to minimise

matmul shapes agree

with the best and worst orderings far apart in residual

Against the rule

It draws a decomposition and prints its residual. It calls permutationTrial, 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 165 of 306 generators — 150 print a residual and 15 are exempt with a published reason; 141 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 matrix a constraint makes

A condition number sent to infinity

An interior-point method manufactures an ill-conditioned matrix on every iteration, deliberately, because the separating of a diagonal is how it discovers which constraints are active. Written one way the answer keeps fifteen digits at a condition number of 3·10¹⁵. Written the other way — the way almost every code writes it — it has none left.

Eigenvalues, singular values, rank

An eigenvalue count that cannot be slightly wrong

Every spectral computation on this site returns floats with errors in them. Counting eigenvalues below a shift by the signs of an unpivoted elimination returns an integer, and an integer cannot be 6.9999999997 — so the answer is exactly right, or wrong by a whole eigenvalue, and where the second happens is a band of measurable width.

Sparsity, and what elimination costs

An ordering that does not wait for the numbers

A sparse factorisation's memory is decided by an ordering computed from the graph, and its stability by pivots computed from the values, and the two decisions fight. On one family of matrices they do not — the ordering can be chosen for fill alone, and the fill the symbolic phase predicts is the fill the factorisation produces — exactly, not as a bound.

The matrix a constraint makes

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 makes

Three eigenvalues, and two are the golden ratio

Precondition a saddle-point system by the block diagonal of its own two definite pieces and the preconditioned matrix has exactly three distinct eigenvalues — 1, and the two roots of λ² − λ − 1. A minimal polynomial of degree three means three steps, at every conditioning, and the preconditioner nobody can afford turns out to be the statement the affordable ones are measured against.

Two errors, and whose fault they are

Two condition numbers of one matrix

κ₂ is a worst case over perturbations of a given norm, and a normwise perturbation may put its whole budget on the smallest entry. The componentwise number is a worst case over perturbations proportional to the entries, which is what a backward-stable factorisation actually makes. On one matrix they are 3·10¹³ and 13.3, and the error obeys the second.

The matrix a constraint makes

Two ways to remove a constraint

A constrained system can be reduced by eliminating the multipliers or by eliminating the constrained directions. Both give the same answer in exact arithmetic and inherit different condition numbers — one of them squares the constraint's, and the other does not contain it at all.

When the problem arrives again

What survives one step of the barrier

An interior-point method solves the same system dozens of times with the same pattern and different numbers, and exactly p entries change between one step and the next. The pattern is reusable for ever. The factorisation is reusable for none of them, and the threshold that says so is a reduction factor of about a per cent against schedules that use ten.

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