Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Algorithm selection — where it appears

A library's choice of which implementation to run, made from the size of the input or from the processor it finds itself on. Both implementations are correct and they round differently, so a routine's accuracy is a step function of a size, with the steps at values chosen by whoever tuned it.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

5860626466687010⁻¹⁷10⁻¹⁶terms in the dot productmean relative errorthe kernel changes herea constant in a libraryone accumulator3.2·10⁻¹⁷four accumulators2.1·10⁻¹⁷step at the cutoff1.6cutoff64the problem did not changethe loop did

The length that changes the kernel

A dot product's accuracy steps by a factor of 1.57 between 63 and 64 terms, on vectors drawn identically at both lengths. Nothing about the problem changes there. A library switches from one accumulator to four, at a constant in somebody else's source file.

machine · Algorithm selection the problem you posedA = H12b = A·(1, 2, …, 12)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 1.8·10⁻¹⁷the answer you wantedx = (1, 2, …, 12), exactlythe answer you gotx̂, wrong by 0.02 relativebackward error 1.8·10⁻¹⁷forward error 0.02κ = 1.8·10¹⁶κ · η = 0.33, and the measured forward error is 0.02.The algorithm is not at fault. The problem is.H12, LU with partial pivotingresidual and error differ

The fifth author

Four authors of a wrong answer have been named on this site and each is a statement about one computation. The fifth is not: it is what separates two computations that are both correct, it is a backward error of measurable size, and no residual, bound or condition number contains it.

error · Backward error κ of the suma component of b − Ax8.09·10¹⁷ad − bc, near-degenerate3.6·10¹⁶qᵢᵀqⱼ, an orthogonality check3.26·10¹⁶zᵀAz, a trace probe95.7pᵀAp, a curvature73.7rᵀr, a residual norm1measured, not assumedhighest8.1·10¹⁷lowest1above 10¹⁰3terms128sums of squares are safeand nobody decides anything from one

Two machines, one certificate

Nothing a solver returns says which of its answers you got. Four things could be reported instead — the summation condition number, the partition count, an exactly accumulated residual and a directed-rounding interval — and each costs about one pass over data the routine already has in hand.

machine · Regression tolerance ‖QᵀQ − I‖ of the implied Qκ = 10⁸, Cholesky0.37κ = 10⁸, sweep8.5·10⁻⁹κ = 10¹⁰, Cholesky1.3κ = 10¹⁰, sweep3·10⁻⁷κ = 3·10¹⁰, Choleskyrefusedκ = 3·10¹⁰, sweep6.7·10⁻⁶κ = 10¹², Cholesky1.5κ = 10¹², sweep1.4·10⁻⁴the safe run is the one that failsrefusals in the range1wholly non-orthogonal returns2the pivot it refused on-6.5·10⁻¹⁷one of these outcomes is safeand it is the refusal

The licence is not the boundary

Cholesky QR is licensed by κ²u ≪ 1, which reaches equality at κ = 9.5·10⁷ in double precision. At 10⁸ the factor it returns is already 0.37 away from orthogonal, and it goes on returning factors as far as 10¹³ — refusing at scattered condition numbers in between, at different ones for eight columns and for six.

machine · Algorithm selection

Named alongside it

The objects these essays reach for when they reach for this one.

Bitwise reproducibilityRun-to-run variationBackward errorCondition numberReduction orderApproximation before linearisationBlocked algorithmCertificateDirected roundingError accumulationExact accumulationForward error

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