Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

pivot-units

One function in the staticpivot library, called 6 times across 3 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 30 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 the backward error of a solve along a sequence, with the pivot order chosen fresh, kept, and kept after equilibration. A 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 9 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 4.82·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 6.55·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.

pivot-units is one function in lib/figures/staticpivot.js — static pivoting — a pivot order kept, and the units that decide whether it may be. 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.

The backward error of a solve along a sequence, with the pivot order chosen fresh, kept, and kept after equilibrationA 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 9 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 4.82·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 6.55·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.02468101210⁻¹⁸10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸member of the sequencebackward error of the solvefresh order · equilibrated kept orderringed: another pivot replaced hereone order, two sets of unitsfresh order, last member9.5·10⁻¹⁷kept order, last member4.8·10⁻⁹equilibrated, last member6.5·10⁻¹⁷pivots replaced, kept2pivots replaced, equilibrated0the perturbation floor √u1.1·10⁻⁸a reused pivot order is safe or notdepending on what the rows are measured in

A 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 9 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 4.82·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 6.55·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.

decades: 9

The arguments are the ones A factorisation kept past its date 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.

The backward error of a solve along a sequence, with the pivot order chosen fresh, kept, and kept after equilibrationA 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 9 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 4.82·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 6.55·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.02468101210⁻¹⁸10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸member of the sequencebackward error of the solvefresh order · equilibrated kept orderringed: another pivot replaced hereone order, two sets of unitsfresh order, last member9.5·10⁻¹⁷kept order, last member4.8·10⁻⁹equilibrated, last member6.5·10⁻¹⁷pivots replaced, kept2pivots replaced, equilibrated0the perturbation floor √u1.1·10⁻⁸a reused pivot order is safe or notdepending on what the rows are measured in

A 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 9 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 4.82·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 6.55·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.

decades: 4

The arguments are the ones The order that was right last time 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.

The backward error of a solve along a sequence, with the pivot order chosen fresh, kept, and kept after equilibrationA 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 4 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 0 replaced pivots and a backward error of 9.55·10⁻¹⁷. Keeping the same order after dividing each row by its largest entry costs nothing at all: 7.21·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.02468101210⁻¹⁸10⁻¹⁶member of the sequencebackward error of the solvefresh order · equilibrated kept orderringed: another pivot replaced hereone order, two sets of unitsfresh order, last member1.1·10⁻¹⁶kept order, last member9.5·10⁻¹⁷equilibrated, last member7.2·10⁻¹⁷pivots replaced, kept0pivots replaced, equilibrated0the perturbation floor √u1.1·10⁻⁸a reused pivot order is safe or notdepending on what the rows are measured in

A 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 4 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 0 replaced pivots and a backward error of 9.55·10⁻¹⁷. Keeping the same order after dividing each row by its largest entry costs nothing at all: 7.21·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.

decades: 12

The arguments are the ones The order that was right last time 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.

The backward error of a solve along a sequence, with the pivot order chosen fresh, kept, and kept after equilibrationA 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 12 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 5.27·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 5.33·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.02468101210⁻¹⁸10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸member of the sequencebackward error of the solvefresh order · equilibrated kept orderringed: another pivot replaced hereone order, two sets of unitsfresh order, last member7.9·10⁻¹⁷kept order, last member5.3·10⁻⁹equilibrated, last member5.3·10⁻¹⁷pivots replaced, kept2pivots replaced, equilibrated0the perturbation floor √u1.1·10⁻⁸a reused pivot order is safe or notdepending on what the rows are measured in

A 10×10 conflict grid, 100 unknowns, with three of its rows scaled down through 12 decades as the sequence runs — a row that was ordinary becoming a row that is small in the matrix's own units. The sparsity pattern is identical at every member, so a symbolic phase computed once stays valid throughout. Choosing a fresh order each time holds the backward error at the working precision. Keeping the first member's order costs 2 replaced pivots and a backward error of 5.27·10⁻⁹. Keeping the same order after dividing each row by its largest entry costs nothing at all: 5.33·10⁻¹⁷, with no pivot replaced anywhere in the run. The ringed points on the upper curve are the members at which another pivot fell below the floor and was replaced.

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.

30 distinct claims across 4 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 member 0 has exactly the first member's sparsity pattern — asserted 12 times

the equilibrated factorisation at member 0 replaces no pivot — asserted 12 times

a change of units inside the range the family is defined over

a grid the dense elimination is affordable on

a member inside the sequence

a sequence long enough to show where the units bite

matmul shapes agree

while the kept order meets pivots it has to replace once the rows have shrunk far enough

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 113 of 238 generators — 98 print a residual and 15 are exempt with a published reason; 125 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.

When the problem arrives again

A factorisation kept past its date

One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.

 When the problem arrives again

The accuracy that is thrown away

A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.

 Sparsity, and what elimination costs

The order that was right last time

A pivot order computed once and reused across a sequence saves the symbolic phase, and the price is that a pivot which was large may now be small. Replacing it with √u·‖A‖ costs eight orders of backward error and iterative refinement recovers a factor of 8.8 of them. Divide each row by its largest entry first and the same reuse costs nothing at all.

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