The collection

Every essay — page 13

Essays 289 to 312 of 567, in the same order.

Sparsity, and what elimination costs

Eliminating a variable couples everything it touched to everything else it touched, and every one of those couplings is an entry that was zero in the matrix and is not zero in its factor. On the same matrix one elimination order gives a factor of a thousand entries and another gives ten thousand — the two factorisations equally accurate, one of them fitting in memory. Nothing numerical chooses between them.

8 × 8 gridstatic saving, none small0.31static saving, three in ten0.0300.10.20.30.40.5350400450500550share of diagonal entries made smallentries in the factornatural orderstatic minimum degreesparsest, every stepmedians over four seedsthe order fixed in advance stops fitting

An order fixed before the numbers

On a saddle-point matrix whose constraint rows have no diagonal, taking the sparsest pivot at every step beat the natural order on fill and growth at once, and the reading was deferral: the constraint rows go last and have filled in by then.

5 figures · Sparse pivoting, essay 9
critical pathdepth one, pieces alone5.4·10⁴depth one, separator counted4.1·10⁴minimum degree, whole grid4.1·10⁴01000020000300004000050000dissection depthcritical path, Σ c² along the heaviest chain12348minimum degreepieces ordered aloneseparator countedgrey line: full dissectionthe penalty was the blindness

Pieces ordered blind

One level of nested dissection followed by minimum degree lengthened the factorisation's critical path, and the blame moved from the separator to the halves: minimum degree's path through a 12 × 24 half was longer than through the whole 24 × 24 grid.

5 figures · Ordering, essay 7
share of the saving recovered6 × 6, revised, worst0.978 × 8, revised, worst0.8510 × 10, revised, worst0.6600.250.50.7511.25small diagonal entriesshare of the saving recovered5%10%20%30%6 × 6, static6 × 6, revised8 × 8, static8 × 8, revised10 × 10, static10 × 10, revisedsolid: 8 × 8 · dotted: 6 × 6 · dashed: 10 × 10re-planning at refusals recovers most of it

A plan redrawn where it was refused

A symmetric indefinite factorisation planned once from the pattern loses its fill saving as small pivots are scattered through the diagonal — on an 8 × 8 grid it keeps 74 per cent of what re-reading the degrees at every step saves at one small pivot in twenty, and 20 per cent at three in ten.

5 figures · Sparse pivoting, essay 10
saving with half the diagonal small, %scattered, static, at half-4.3scattered, revised, at half-4.1clustered, static, at half13clustered, revised, at half14-0.100.10.20.3small diagonal entriessaving over the natural order5%10%20%30%50%fixed, scatteredredrawn, scatteredfixed, clusteredredrawn, clusteredsolid: clustered · dashed: scattereda region keeps the plan, not breaks it

A small pivot with a small neighbour

A fill-reducing plan computed from the pattern loses its saving as small pivots are scattered through a symmetric indefinite matrix, and redrawing it where the pivoting test refuses it wins the saving back.

5 figures · Sparse pivoting, essay 11
8 × 8scattered, refused at 30%17along lines, refused at 30%8in a disc, refused at 30%3050100150share of the diagonal that is smallentries added to the fixed plan's factor0%5%10%20%30%50%scatteredalong linesin a discthe badge counts refusals, which the curves do not followthree layouts, one cost

A refusal has no fixed price

A symmetric indefinite factorisation planned once from the pattern is refused by its pivoting test where small diagonal entries sit, and the earlier essays read the refusal count as the plan's cost: scattered small pivots were refused 7 to 19 times and saved little, clustered ones 0 to 6 and saved much more.

5 figures · Sparse pivoting, essay 12
-4004080120160200share of the diagonal that is smallextra entries from interchanges5%10%20%30%50%scatteredalong rowsalong columnsin a discsolid: 10 × 10 · dashed: 8 × 8the row lines pay for the natural order's interchanges

The baseline was the interchanges

Small pivots along grid rows gave a fixed fill-reducing plan twice the saving over the natural order that scattered ones did, for a factor of the same size, and the reading was that the natural order did badly on rows.

5 figures · Sparse pivoting, essay 13
longest ÷ shortesttwo rows, 12²1.5two rows, 24²1.311.11.21.31.41.5longest ÷ shortest critical path12²16²20²24²12²16²20²24²g1g2g312²16²20²24²grid, one rowgrid, level setgeometric graphgrid, two rowsdashed: every shuffle gives the same factorfree unless the separator is thick

A clique has no order

Ordering a dissection's pieces with their separator in view cut the critical path by a quarter, and the next proposal was to order the separator's own vertices by how much fill each carries in.

6 figures · Ordering, essay 8

Structure, and the solver that cannot see it

The sparsity field is about a matrix most of whose entries are zero. This is the other way a matrix can be small: a circulant is n numbers and a Toeplitz matrix is 2n − 1, and neither of them has a single zero entry, so nothing in the sparsity field applies. What the structure buys is an exact spectrum, a condition number in closed form and a solve in n log n — and what it does not buy is a well-conditioned problem, which is the half that gets summarised away.

024681012141610⁻¹1index kλthe transformthe eigensolverC = F* Λ F is a factorisationworst relative disagreement2.6·10⁻¹⁵‖Cx − b‖/‖b‖ from the transform solve4.7·10⁻¹⁶imaginary part of a real spectrum1.2·10⁻¹⁶n = 16, and the whole matrix is 16 numberseigenvectors known in advance

The matrix that is one row

A circulant of size 16 is sixteen numbers, has no zero entry anywhere, and hands over its entire spectrum in closed form — the discrete Fourier transform of its first column, exactly.

5 figures · Circulant, essay 1
10¹10²10²size ncondition numberlimit 81measureda limit, as a fraction of itselfreached at n = 1280.99still to go0.011κ at n = 8, as a fraction0.52every point is below the line and none of them is on itthe limit is not a value

A limit the matrix never reaches

Szegő's theorem gives a Toeplitz family's condition number in closed form — ((1+ρ)/(1−ρ))², which is 81 at ρ = 0.8.

9 figures · Toeplitz, essay 1
10²110¹10²size niterations to 10⁻¹⁰λₘᵢₙ(C) changes signno preconditionerStrang's circulantthe preconditioner's own spectrumλₘᵢₙ(C) at n = 16-0.4λₘᵢₙ(C) at n = 32-0.14λₘᵢₙ(C) at n = 640.016λₘᵢₙ(C) at n = 1280.051λₘᵢₙ(C) at n = 2560.053left of the line the repair costs stepsright of it, the count stops counting n

A preconditioner that changes sign

Strang's circulant preconditioner takes Toeplitz conjugate gradients from 179 steps to 10 at n = 256.

8 figures · Preconditioning, essay 2
10²110¹10²size niterations to 10⁻¹⁰λₘᵢₙ(C) changes signno preconditionerStrang's circulantthe preconditioner's own spectrumλₘᵢₙ(C) at n = 16-0.4λₘᵢₙ(C) at n = 32-0.14λₘᵢₙ(C) at n = 640.016λₘᵢₙ(C) at n = 1280.051λₘᵢₙ(C) at n = 2560.053left of the line the repair costs stepsright of it, the count stops counting n

The circulant that cannot be indefinite

The previous essay found a preconditioner taking 117 steps against an unpreconditioned 59, because its smallest eigenvalue was −0.173.

7 figures · Toeplitz, essay 2
10²020406080100120unknownsiterations2D, no preconditioner2D, preconditioned1D, preconditionedone construction, two dimensions2D steps at 16 unknowns102D steps at 100 unknowns211D steps at 100 unknowns10the same averaging, the same transformand a count that no longer stops growing

Two dimensions, and the cluster that thins

The same kernel, the same averaging, the same transform — applied along two axes instead of one.

7 figures · Toeplitz, essay 3
01234567810⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹μ, the entry above the diagonalsep, and the eigenvalue gapmin |λᵢ + μⱼ|sep(A, B)the spectra never moveeigenvalue gap, throughout2sep at μ = 02sep at μ = 89.5·10⁻⁴amplification there1056solvability is the eigenvaluesand conditioning is not

An equation whose unknown is a matrix

AX + XB = C is linear in X, so it has a coefficient matrix, and writing it down is the obvious thing to do.

5 figures · Matrix equation, essay 1
the smallest perturbation of any kind — 1.89·10⁻¹⁷the smallest Toeplitz one — 2.65·10⁻¹³both exact for the same x̂smallest of any kind1.9·10⁻¹⁷smallest Toeplitz one2.7·10⁻¹³the price of the constraint1.4·10⁴diagonal defect, unconstrained0.97an exact answer to a nearby problemof a kind nobody posed

A nearby problem of the wrong kind

A good algorithm returns the exact answer to a nearby problem.

8 figures · Structured backward error, essay 1
110¹10²10³10⁻¹⁸10⁻¹⁴10⁻¹⁰10⁻⁶10⁻²10²10⁶numbers that describe the matrixcondition number, and backward errordensetoeplitzsymmetricρ aloneno such problemcondition numberbackward errorone matrix, four descriptionsκ, all n² entries7.9·10⁴as one number, ρ62backward error, unconstrained2·10⁻¹⁷as a symmetric Toeplitz matrix5.6·10⁻¹²fewer numbers, better conditionedand no nearby problem left

The condition number of the model

Describe a 40×40 Toeplitz matrix by its 1,600 entries and its condition number is 78,800.

7 figures · Structured backward error, essay 2
1357910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴log₁₀ b, the couplingrelative errorsmall half, directstructure-preservingsmall half, by 1/λone divisionlarge half, direct5.5·10⁻¹⁵small half, direct10⁻⁷small half, by 1/λ4.9·10⁻¹⁵structure-preserving6.5·10⁻⁶the structure is not decorationit is where half the accuracy is

A perturbation that keeps the symmetry

The smallest perturbation that makes a computed answer exact is the backward error.

5 figures · Structured backward error, essay 3
10²02040size niterationsno preconditionerwrapped (Strang)averaged (T. Chan)both are circulant approximations‖C − T‖/‖T‖, averaged0.22‖C − T‖/‖T‖, wrapped0.23smallest eigenvalue, wrapped, n = 160.23one of them is positive definiteand it is the one that is nearer

A speedup with a ceiling of its own

At ρ = 0.5 the averaged circulant takes 5 conjugate gradient steps at n = 512 against an unpreconditioned 30 — and that 30 is where the unpreconditioned count stops.

7 figures · Preconditioning, essay 3
357911110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017the cluster grows with the sideand the spectrum with the area

Four orders of conditioning, and four steps

On a 10×10 grid the two-dimensional kernel's condition number runs from 62 at ρ = 0.5 to 818,561 at ρ = 0.98.

7 figures · Toeplitz, essay 4
A6×6B6×6I ⊗ A + Bᵀ ⊗ I36×36one equation, two objectsentries in A and B72entries in the coefficient matrix1296two routes, relative gap1.7·10⁻¹⁶‖AX + XB − C‖/‖C‖1.4·10⁻¹⁶the small squares are the problemand the large one is the notation

The elimination the matrix does not need

The Kronecker form of AX + XB = C is dismissed with a hundred million entries and (2/3)n⁶ operations.

7 figures · Matrix equation, essay 2
10⁻³10⁻²10⁻¹110⁻¹⁹10⁻¹⁷10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹1 − ρbackward errorLevinson, Toeplitzelimination, ToeplitzLevinson, any kindelimination, any kindat ρ = 0.999Levinson, any perturbation2.9·10⁻¹⁷Levinson, Toeplitz only6·10⁻¹²the ratio between them2·10⁵diagonal defect of the first0.97the number that is reportedand the number that is asked about

The number that cannot rank them

Levinson and Gaussian elimination are indistinguishable on the backward error a library reports — every one of ninety-six measurements between 1.16·10⁻¹⁷ and 5.73·10⁻¹⁷.

7 figures · Structured backward error, essay 4
10⁻¹⁷10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹10¹10³10⁵σ — how far the periodic wrap is from singularrelative forward error10⁻¹10⁻²10⁻³10⁻⁴10⁻⁵10⁻⁶10⁻⁷10⁻⁸10⁻⁹10⁻¹⁰10⁻¹¹10⁻¹²periodic correctionκ(C)·uelimination on Tn = 64, median of five answersperiodic correction, σ = 10⁻⁸1.2·10⁻⁴elimination on T, σ = 10⁻⁸1.1·10⁻¹⁴κ(C) at σ = 10⁻⁸4·10⁸κ(T) at σ = 10⁻⁸1712the same matrix T in every columnonly the wrap it is solved through changes

The circulant the problem did not contain

A matrix that differs from a circulant in two corner entries can be solved through the circulant, by a transform and a two-by-two correction, and the cost claim is exact.

7 figures · Circulant, essay 2
-7-6-5-4-3-210⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110²10⁴log₁₀ of the twist's distance from landing, in turnsrelative forward errorsimple zero, landed on at 0simple zero, landed on at 0.3double zero, d = 2n = 64, twist 10⁻⁵ of a turn from landingpair: error, κ(A) = 4.1·10⁶5.4·10⁻¹²single: error, κ(A) = 3.8·10⁶3·10⁻⁸double: error, κ(A) = 4.2·10¹²7515the same condition number of the wrapa different number of modes near zero

Two near-zeros cost less than one

Solve a well-conditioned tridiagonal matrix through a nearly singular wrap and the correction's accuracy is not set by how singular the wrap is.

7 figures · Circulant, essay 3
at σ = 10⁻⁸Cramer's rule1.2·10⁻⁴pivoted2.9·10⁻¹⁰elimination on T1.1·10⁻¹⁴cancellation × u1.7·10⁻¹⁰-10-9-8-7-6-5-4-3-210⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1σ, as a power of tenforward errorCramer's rulepivotedcancellation × uelimination on Tleft: σ small, the wrap nearly singularthe pivoted solve follows the cancellation

The correction lost to its own two-by-two solve

Solving a band matrix through a circulant and a small correction was measured losing the answer to 10⁻⁴ where elimination kept it to 10⁻¹⁴, and a wrap that landed one sample on a zero was measured costing four orders more than one that landed a pair.

6 figures · Circulant, essay 4
n = 64, twist of half a stepnearest sample, Laplacian0.0024nearest sample, squared5.8·10⁻⁶κ(wrap), squared2.8·10⁶10⁻³10⁻²10⁻¹110⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹10¹θsymbolπ/nLaplacian, θ² near 0its square, θ⁴ near 0no twist puts a sample further than π/n from θ = 0and the symbol's order decides what it sees there

A zero no twist can step around

Solved through its circulant wrap with the small capacitance system factorised stably, a banded matrix was found to lose only what the cancellation in the correction costs, and a twist of half a step kept that cancellation near one.

5 figures · Circulant, essay 5