The collection

Every essay — page 12

Essays 265 to 288 of 567, in the same order.

Iterating, instead of factorising

At scale nobody factorises, and the reason is not accuracy — it is that the factors of a sparse matrix are not sparse. What replaces elimination is a sequence of approximations, and the question changes shape: not what the residual of the factorisation is, but how fast the error falls and whether the thing you can measure tracks the thing you cannot. On the model problem every rate is known in closed form before anything runs.

-213284300.250.50.751rotation of the anisotropy (degrees)factor / couplingusableconvergence factoraxis couplingdiagonal couplingthe standard answer, and the anglefactor at 0°0.19factor at 45°0.8axis ÷ diagonal coupling at 45°2the hierarchy reads the matrixand the matrix lost the direction

How much direction there was to lose

At 45° the nine-point stencil hands smoothed aggregation the same wrong hierarchy at every anisotropy — six strong neighbours per interior point, 121 aggregates, the identical partition from ε = 10⁻⁴ to 0.099.

6 figures · Anisotropy, essay 5
10⁻¹100.20.40.60.8strength threshold θresidual reduction per cycleθ = εsemi-coarseningkept whole rowsfull coarseningone parameter, two methodsbest factor above ε0.053best factor below ε0.1the ratio across the switch231×31 anisotropic operatora switch, not a dial

The switch does not know which side is better

The strength threshold moves the coarsening from full to semi at θ = ε exactly, at every anisotropy.

6 figures · Algebraic multigrid, essay 4
10⁻³10⁻²10⁻¹1024681012size of the negative eigenvalue, −λproducts before the test firesharder to find, and milder8 spectra, n = 50products at the largest λ3products at the smallest10smallest share of λ recovered0.14largest0.34the one that hidesis the one that matters least

A proof that does not ask how large the matrix is

Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n.

7 figures · Breakdown, essay 4
10³10⁶10⁹10¹²10¹⁵10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1largest iterate on the way, ‖x‖difference between the two residualsthe bound, linear in ‖x‖what it does, slope 0.46a bound of one, a walk of a halffitted slope0.46the bound's slope1smallest gap measured3.9·10⁻¹⁵largest gap measured1.5·10⁻¹⁰share of the bound, near end9.1·10⁻⁴share of the bound, far end3.4·10⁻⁹the bound is rightand loose by a square root

A walk needs a length

The gap between the two residuals grows as the square root of something, and a square root needs a length.

6 figures · Residual gap, essay 3
22.32.62.910⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised832storage, second-order416Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations

The answer that arrives when the space runs out

A second-order Krylov recurrence holds vectors of length n for a problem with 2n eigenvalues, so it is exact at n steps where the linearised route needs 2n.

6 figures · Krylov, essay 5
110¹00.30.60.91.2trust-region radius Δshare of the exact model decreasethe exact subproblem7 radii, n = 60share at the smallest radius0.95share at the largest0.3products, at most8radii stopped by the curvature4a few products against an eigendecompositionand most of the decrease

The certificate that arrives soonest is worth least

The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth.

7 figures · Trust-region, essay 1

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.

the matrix, lower triangle408 entriesits Cholesky factor1739 entries · 1331 created→‖A − LLᵀ‖/‖A‖1.4·10⁻¹⁶fill, symbolic1331fill, numeric1331n = 144 · density 3.2% · bandwidth 12same matrix, renumberedthe answer is identical to rounding

The factor is not sparse

A sparse matrix has a factor that is not sparse, and the gap between them is the entire reason iterative methods exist.

5 figures · Fill, essay 1
natural1739reverse Cuthill–McKee1354minimum degree1026nested dissection1413matrix: 408 entries · dense factor: 10440bandwidth 12 · 4.26× the matrixbandwidth 12 · 3.32× the matrixbandwidth 123 · 2.51× the matrixbandwidth 108 · 3.46× the matrixn = 144, five-point stencilevery ordering fills in; none avoids it

The order decides the memory

Four elimination orderings on one matrix give factors of 1,739, 1,354, 1,413 and 1,026 entries.

6 figures · Ordering, essay 1
the matrix43 entriestip eliminated first253 entriestip eliminated last43 entries‖A − LLᵀ‖/‖A‖, tip first1.4·10⁻¹⁶‖A − LLᵀ‖/‖A‖, tip last0dense factor is n(n+1)/2 = 253 · sparse factor is 2n − 1 = 43one row swapped to the endnothing numerical chose between them

Two ends of the same arrow

One matrix, one row moved from the front of the elimination order to the back, and the factor goes from completely dense to no fill at all.

6 figures · Fill, essay 2
the matrix105 entriescorner first — sparsest227 entries, growth 1.9·10¹¹largest first — safe242 entries, growth 1.19the middle factor is the smaller one, and its answer has no correct digitsboth factorisations reproduce the matrix‖PA − LU‖/‖A‖, sparsest3.8·10⁻¹⁷‖PA − LU‖/‖A‖, pivoted5.4·10⁻¹⁷forward error, sparsest3·10⁻⁵forward error, pivoted4.8·10⁻¹⁶red marks are entries elimination createdthe fill argument and the stability argument disagree

Structure and stability stop being separable

The sparsest variable to eliminate on this matrix has a diagonal entry of 10⁻¹².

7 figures · Sparse pivoting, essay 1
10⁻³10⁻²10⁻¹1110¹10²10³10⁴pivot threshold τgrowth factor · entries in L+U, ÷ entries in Agrowthfillthe library default‖PA − LU‖/‖A‖ at τ = 0.12.9·10⁻¹⁶growth at τ = 0.138entries in L+U at τ = 0.1372one knob, two measurements, opposite directionsand the default is most of both

A threshold between fill and growth

One number decides how small a pivot an elimination will accept.

7 figures · Sparse pivoting, essay 2
the bound211no pivoting127τ = 0.00198τ = 0.003100τ = 0.01106τ = 0.03111τ = 0.1116τ = 0.3117τ = 1138entries in Ua bound, and its slackthe bound, from the graph alone211the worst that occurs138loose by1.5no arithmetic was done to compute the bound — only the pattern of AᵀA and its elimination graphGeorge and Ng: U fits inside chol(AᵀA), whatever the swapsprovable, cheap, and loose

What the symbolic phase can only bound

Without pivoting, the fill can be computed from the graph and the count is exact — 233 predicted, 233 measured.

5 figures · Sparse pivoting, essay 3
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

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.

8 figures · Sparse pivoting, essay 4
0481216202402468101214unknowns on the separatorcolumns above 10⁻⁸the same matrix, renumberedin the separator's own orderthe ordering the geometry hands overseparator 73separator 236renumbered, largest11the block, largest11share of the square stored0.52the fill is totaland it is not independent

The fill that is not independent

Eliminate both halves of a grid and what is left on the separator is 100 per cent nonzero — the sparsity field's result, unchanged.

8 figures · Fill, essay 3
natural113 predicted · 113 countedminimum-degree63 predicted · 63 countedreverse Cuthill–McKee63 predicted · 63 countedthe shaded entries are fill: zeros of K that the factorisation makes nonzeroallocated before the numbersnatural113minimum-degree63reverse-cuthill-mckee63predicted minus counted0the symbolic phase decides the memoryand nothing later is allowed to argue

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.

8 figures · Sparse pivoting, essay 5
02468101210⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹singular value, in orderσ ⁄ σ₁eight digitsrenumberedin the separator's orderthe same cliff, from the other endσ₂ ⁄ σ₁0.062σ₄ ⁄ σ₁1.3·10⁻⁴σ₆ ⁄ σ₁1.5·10⁻⁸renumbered σ₄ ⁄ σ₁0.63rank at 10⁻⁸6the same entriesin two orders

The cliff behind the count

The fill's rank is an integer between three and six across every separator two dense half-eliminations can afford, and this field has already recorded that a handful of such integers cannot carry a law.

6 figures · Fill, essay 4
6101418222610³10⁴10⁵grid side karithmeticND, totalMD, totalMD, critical pathND, critical pathdashed: the time on unbounded processorsthe same time, bought with more work

An ordering that buys processors, not time

Nested dissection loses to minimum degree on fill and on total work at every grid either measurement could draw.

5 figures · Ordering, essay 3
least possible76minimum degree76worst of 300 tie-breaks81reverse Cuthill–McKee85natural order9920 unknownstie-breaks at the least0.44minimum degree ÷ least1every order searched, by the set already eliminatedthe greedy rule found the minimum

The least fill there is

Finding the elimination order with the least fill is NP-hard, and that is a statement about the hardest graph and the largest size.

5 figures · Ordering, essay 2
all 2^16 subsets searchedboth minima, one order0.97minimum degree optimal0.82worst fill overshoot1.1worst work overshoot1.111.021.041.0611.041.081.121.16fill ÷ the least filloperations ÷ the leastovershoot squaredon the diagonal: the two overshoots are equalon the upper curve: the arithmetic overshoot is the square

Two minima that are one minimum

The order that decides the memory found the operation count behaving like the square of the fill, which leaves room for an order with slightly more fill but a shorter heaviest column to do less arithmetic.

5 figures · Ordering, essay 4
total operationswork, no dissection10⁵work, full dissection1.5·10⁵work, one level1.3·10⁵on unbounded processorscritical path, no dissection4.1·10⁴least, at depth 43.4·10⁴critical path, one level5.4·10⁴10⁵switching depthoperations01234568total operationscritical pathdepth 0 is minimum degree, depth 8 is nested dissectiondepth 1 is worse than both

The depth that is worse than both ends

Nested dissection to a chosen depth and minimum degree below it is the ordering codes ship, and sweeping the depth was supposed to find a setting that keeps most of dissection's parallelism for most of minimum degree's work.

5 figures · Ordering, essay 5
10⁻³10⁻²10⁻¹1110¹10²10³10⁴pivot threshold τgrowth factor · entries in L + U ÷ entries in Agrowth, rows onlygrowth, rows and columnsfill ratio, rows onlyfill ratio, rows and columns8×8 grid, 64 unknowns, 288 entriesτ = 0.001: rows only — 741 entries; growth2209τ = 0.001: rows and columns — 640 entries; growth2.5τ = 0.1: rows only — 875 entries; growth38τ = 0.1: rows and columns — 640 entries; growth2.5τ = 1: rows only — 986 entries; growth1.2τ = 1: rows and columns — 659 entries; growth1.2dashed: a fixed column · solid: the column chosen toothe same threshold rule in both

The column that was never fixed

Every threshold-pivoting measurement so far chose the pivot row in a fixed column, and the routine's own description said that choosing the column as well would change the constants and not the argument.

6 figures · Sparse pivoting, essay 6
80 matrices, τ = 0.1, ties to the largestno column search, median fill244one column136every column109share of the gain at one column0.804896144192240columns the search may look atmedian fill123468121624allno column search at allmedian fillmedian growthworst growth — see the captionthe fill is the quantity the width buysthe worst case is the draw

How few columns the search needs

A full row-and-column pivot search is quadratic in the active submatrix at every step, and no library performs one.

6 figures · Sparse pivoting, essay 7
a 28 × 28 saddle-point matrixnatural order, entries113sparsest order, entries72two-by-two pivots, loosest test0and at the strictest3023466992115the constant in the pivot testentries in the factor0.10.30.640.80.95133natural ordersparsest ordergrowth, naturalgrowth, sparsestnumbers above the points: two-by-two pivots takenboth curves favour the sparser order

The freedom a symmetric factorisation does not have

Permuting rows and columns together leaves no column to choose, so the conflict between the sparsest pivot and the sound one should be worse rather than better.

6 figures · Sparse pivoting, essay 8
rows and columns, from the coordinateslevel structures, from the graph aloneseparator, level 1separator, level 2separator, level 3same vertex count in the first cutdifferent pieces under it

The halves were the price

One level of nested dissection followed by minimum degree was worse than both no dissection and full dissection on a grid, and the explanation offered was the separator's dense block sitting on every path.

5 figures · Ordering, essay 6