Generator

null-basis-bars

One function in the nullbasis library, called 15 times across 5 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 12 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 three bases for the same null space, at 10 unknowns and 4 constraints. The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10⁸, and its reduced Hessian comes out at 3.801·10¹⁶, which is κ(Z)² to within a factor of 0.956 — the square attained rather than bounded. Its answer is wrong by 0.05179, against 1.07·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 6.71·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.

null-basis-bars is one function in lib/figures/nullbasis.js — the basis of a constraint — three spans of one null space, and the square 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10⁸, and its reduced Hessian comes out at 3.801·10¹⁶, which is κ(Z)² to within a factor of 0.956 — the square attained rather than bounded. Its answer is wrong by 0.05179, against 1.07·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 6.71·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error1.07·10⁻¹⁵first m basic — κ(Z)1.99·10⁸κ(ZᵀHZ)3.8·10¹⁶relative error0.0518pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.9relative error6.71·10⁻¹⁶what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive0.96error, pivoted choice6.7·10⁻¹⁶every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10⁸, and its reduced Hessian comes out at 3.801·10¹⁶, which is κ(Z)² to within a factor of 0.956 — the square attained rather than bounded. Its answer is wrong by 0.05179, against 1.07·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 6.71·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.

gamma: 1

The arguments are the ones A preconditioner that need not know the constraint 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 4.436, and its reduced Hessian comes out at 83.77, which is κ(Z)² to within a factor of 4.26 — the square attained rather than bounded. Its answer is wrong by 1.26·10⁻¹⁵, against 1.29·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.258, an error of 1.41·10⁻¹⁵, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)32relative error1.29·10⁻¹⁵first m basic — κ(Z)4.44κ(ZᵀHZ)83.8relative error1.26·10⁻¹⁵pivoted basic — κ(Z)2.26κ(ZᵀHZ)43.1relative error1.41·10⁻¹⁵what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive4.3error, pivoted choice1.4·10⁻¹⁵every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 4.436, and its reduced Hessian comes out at 83.77, which is κ(Z)² to within a factor of 4.26 — the square attained rather than bounded. Its answer is wrong by 1.26·10⁻¹⁵, against 1.29·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.258, an error of 1.41·10⁻¹⁵, and the same 50 per cent density: all of the sparsity and none of the loss.

gamma: 100000000

The arguments are the ones A preconditioner that need not know the constraint 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10⁸, and its reduced Hessian comes out at 3.801·10¹⁶, which is κ(Z)² to within a factor of 0.956 — the square attained rather than bounded. Its answer is wrong by 0.05179, against 1.07·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 6.71·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error1.07·10⁻¹⁵first m basic — κ(Z)1.99·10⁸κ(ZᵀHZ)3.8·10¹⁶relative error0.0518pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.9relative error6.71·10⁻¹⁶what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive0.96error, pivoted choice6.7·10⁻¹⁶every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10⁸, and its reduced Hessian comes out at 3.801·10¹⁶, which is κ(Z)² to within a factor of 0.956 — the square attained rather than bounded. Its answer is wrong by 0.05179, against 1.07·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 6.71·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.

gamma: 10000

The arguments are the ones The basis nobody chose on purpose 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.997·10⁴, and its reduced Hessian comes out at 5.845·10⁸, which is κ(Z)² to within a factor of 1.47 — the square attained rather than bounded. Its answer is wrong by 8.504·10⁻¹⁰, against 2.25·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 1.33·10⁻¹⁵, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error2.25·10⁻¹⁵first m basic — κ(Z)2·10⁴κ(ZᵀHZ)5.85·10⁸relative error8.5·10⁻¹⁰pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.8relative error1.33·10⁻¹⁵what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive1.5error, pivoted choice1.3·10⁻¹⁵every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.997·10⁴, and its reduced Hessian comes out at 5.845·10⁸, which is κ(Z)² to within a factor of 1.47 — the square attained rather than bounded. Its answer is wrong by 8.504·10⁻¹⁰, against 2.25·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 1.33·10⁻¹⁵, and the same 50 per cent density: all of the sparsity and none of the loss.

gamma: 100

The arguments are the ones The basis nobody chose on purpose 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 205.7, and its reduced Hessian comes out at 8.39·10⁴, which is κ(Z)² to within a factor of 1.98 — the square attained rather than bounded. Its answer is wrong by 3.074·10⁻¹³, against 1.15·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.061, an error of 2.88·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error1.15·10⁻¹⁵first m basic — κ(Z)206κ(ZᵀHZ)8.39·10⁴relative error3.07·10⁻¹³pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.7relative error2.88·10⁻¹⁶what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive2error, pivoted choice2.9·10⁻¹⁶every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 205.7, and its reduced Hessian comes out at 8.39·10⁴, which is κ(Z)² to within a factor of 1.98 — the square attained rather than bounded. Its answer is wrong by 3.074·10⁻¹³, against 1.15·10⁻¹⁵ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.061, an error of 2.88·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.

gamma: 10000000000

The arguments are the ones The basis nobody chose on purpose 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.

Three bases for the same null space, at 10 unknowns and 4 constraintsThe same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10¹⁰, and its reduced Hessian comes out at 4.871·10¹⁸, which is κ(Z)² to within a factor of 0.0123 — the square attained rather than bounded. Its answer is wrong by 0.3173, against 8.67·10⁻¹⁶ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 8.07·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error8.67·10⁻¹⁶first m basic — κ(Z)1.99·10¹⁰κ(ZᵀHZ)4.87·10¹⁸relative error0.317pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.9relative error8.07·10⁻¹⁶what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive0.012error, pivoted choice8.1·10⁻¹⁶every one of them is a basisand one of them loses fourteen digits

The same constrained problem solved three times, differing only in which basis Z is used for the null space of A. The orthonormal basis, from a QR of Aᵀ, has κ(Z) = 1 exactly and is dense — 100 per cent of its entries are nonzero. The fundamental basis built on the first 4 columns has κ(Z) = 1.994·10¹⁰, and its reduced Hessian comes out at 4.871·10¹⁸, which is κ(Z)² to within a factor of 0.0123 — the square attained rather than bounded. Its answer is wrong by 0.3173, against 8.67·10⁻¹⁶ for the orthonormal one. Choosing the same kind of basis by pivoting instead gives κ(Z) = 2.06, an error of 8.07·10⁻¹⁶, and the same 50 per cent density: all of the sparsity and none of the loss.

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.

12 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 badness the double can hold

a basis rule this file implements

a constraint no larger than the problem

a finite double, since an infinity is not a rational

a problem the exact solve can afford

and pivoting improves on the naive choice

fewer constraints than unknowns

fewer constraints than unknowns, so something is left to minimise

LU is for square matrices

matmul shapes agree

the orthonormal basis has condition number one exactly

while the fundamental bases are the sparse ones

Against the rule

It draws a decomposition and prints its residual. It calls basisComparison, 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 preconditioner that need not know the constraint

Keep the constraint block exactly and replace the objective block by anything positive definite on the null space. The preconditioned matrix then has 2m eigenvalues at exactly one, and its remaining n − m are the generalised eigenvalues of a pencil in which the constraint does not appear. Sweep its condition number over six decades and they do not move in six digits.

Orthogonality, measured

The basis nobody chose on purpose

A method that eliminates a constraint has to pick a basis for its null space, and every basis is correct. Their condition numbers are eight orders apart, the reduced problem inherits the square, and the choice is usually made by a one-line rule nobody thought of as a numerical decision.

Randomised, and the guarantee that changes kind

The half of a problem a sketch may touch

A sketch guarantees that a norm is preserved to within a factor. An equality constraint is a statement that a quantity is zero, and no multiplicative guarantee says anything about zero. Sketch a constrained problem written as a weighted one and the constraint is not destroyed — it is demoted, from a violation of 1/τ² to one of ε/τ, exactly half the exponent.

The matrix a constraint makes

The 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.

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.

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