Field

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. The entries elimination creates can be counted before any arithmetic runs, from the graph alone.

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. All four factorisations are exact, all four return the same answer, and the one with the better asymptotics is not the one that wins.

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. Both factorisations are exact to rounding, and nothing numerical chose between them.

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. Eliminating it produces the smaller factor, reproduces the matrix to 3.8·10⁻¹⁷ — better than pivoting does — and returns an answer wrong in the fifth digit.

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. At 0.001 the factor holds 172 entries and the matrix grows by 1,330; at 1 it holds 260 and grows by 1.2. The libraries ship 0.1, and the measurement says why.

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. With pivoting it is 233 predicted and 242 measured, and what survives is a bound that is right at every threshold and loose by 1.7 times at the largest grid drawn.

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. 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.

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. Its off-diagonal block is 11 by 12 and six columns describe it to eight digits. Renumber the separator and the same block needs all eleven.

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. On one family of matrices they do not — the ordering can be chosen for fill alone, and the fill the symbolic phase predicts is the fill the factorisation produces — exactly, not as a bound.

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. The singular values underneath are real numbers. They say the cliff's first step is 23.0 at a separator of eleven, 19.1 at fifteen and 16.2 at twenty-three — and that a control with no differential operator behind it gives 14,672.

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. Read along the elimination tree a parallel factorisation works on, it does not win back the time either: its critical path is within 28 per cent of minimum degree's at every size from 8 to 24 points a side, in both directions, and the tree heights and widest columns are nearly the same. What it wins is the ratio. Its total work over its critical path — the most a factorisation on unbounded processors can speed up by — grows from 2.7 to 4.5 while minimum degree's stays between 2.1 and 2.6.

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. On a graph of twenty vertices every one of the 20! orders can be searched at once, through the million sets of vertices already eliminated, and the least fill is a number. On the 4×4, 4×5 and 3×7 grids minimum degree finds it exactly. On eighty random sparse graphs of eighteen vertices it finds it on 53 and misses by at most 7.6 per cent, and on every one of the eighty some breaking of its ties finds it.

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. Searched exactly over every elimination order on forty graphs, that order does not exist: one order attains both minima on thirty-nine of forty, and on the fortieth the least-fill order's arithmetic is 1.0099 times the least. Minimum degree attains both on the same thirty-three graphs and neither on the same seven.

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. It does not exist: total work rises with the depth at every grid size, and one depth — the first — is worse than both extremes on work and on the critical path at all four sizes measured. One bisection buys nothing because there is no recursion under it to amortise the separator.

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. Measured, it changes the argument. On the 8×8 conflict grid the factor shrinks from 875 entries to 640 at the library default, and the growth factor that climbed to 2,209 as the threshold loosened stays at 2.54 at every threshold from 0.3 down to 0.001. What does most of the work is not the column but which of several equally cheap entries is taken — and on random sparse matrices, choosing the column without that makes the growth worse.

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. Looking at a single sparsest column takes the median fill from 244 to 136 where the full search reaches 109 — four fifths of the benefit for a linear scan — and that share is 79, 83, 80, 89 and 92 per cent across five thresholds. The worst growth appears to favour the narrow search by a factor of six, and on the next draw it favours the wide one by two.

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. On a saddle-point matrix whose constraint rows have no diagonal entry at all, it is not there: taking the sparsest available pivot holds 70 entries against the natural order's 113 and a growth of 1.28 against 1.83 — better on both currencies at once, at every setting of the pivot test. The two-by-two blocks that make it legal cost 1.33 entries apiece.

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. That predicted a sign: a separator found from the graph, rather than read off the coordinates, should make the penalty larger. It makes it smaller at every size — 44,116 against 53,508 at 24×24 — and the separator is not where the difference is: both first separators have 24 vertices and cost exactly 4,900. What differs is the path through the halves. Minimum degree takes 48,608 through a rectangular half of the 24×24 grid, longer than its 41,072 through the whole grid, and 39,216 through a triangular one.

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. A code computes its ordering once, from the pattern. That static minimum-degree order holds fewer entries than re-reading the degrees on fourteen of sixteen settings, growth under two throughout — and it does not defer. It spreads the constraint rows through the elimination, each after two thirds of its own variables. Scatter small pivots through a grid instead and the plan is refused on up to 25 of 64 steps; its saving falls from 31 per cent to 3.

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. A 12 × 24 grid on its own has a path of 12,913, a third of the square's. What made the half expensive is that it was ordered as if the separator were not there. Count the separator's vertices in every degree and number them last, and the depth-one path falls from 53,508 to 41,050 — minimum degree's own — while the pieces' own arithmetic does not change at all. At depth two the same ordering beats full dissection.

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. Redraw the plan for the remaining rows only when the pivoting criterion rejects the planned pivot, and it keeps 85 to 100 per cent at every share up to three in ten, reading a tenth to a quarter of the rows the step-by-step rule reads. Each redrawn plan heads off refusals that would have followed it: 3 re-plans where the fixed plan was refused 7 times, 9 where it was refused 17. Where the plan was never refused, redrawing costs nothing — and with half the diagonal small, no plan of any kind beats the natural order.

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. Gather the same small pivots into one region and the prediction was that the fixed plan would do worse, because a whole region's ranking is wrong at once. It does better. On 8 × 8 and 10 × 10 grids the fixed plan is refused 0 to 6 times with the small pivots clustered, against 7 to 19 scattered, and at three in ten it saves 10 and 23 per cent of the natural order's factor where scattered it saved 3 and 9. The reason is in the Bunch–Kaufman test: a small pivot whose neighbour is also small cannot be passed over for that neighbour, so the test pairs the two into a block that keeps the planned row. Both plans still save something with half the diagonal small. And the step-by-step sparsest rule, best of all when the small pivots are scattered, falls behind the redrawn plan in a cluster.

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. Put the small pivots along lines, each with two small neighbours, and the prediction of one refusal for every three small entries holds — between 0.25 and 0.42 on every share and grid. The rest of the reading does not. From a fifth of the diagonal up, the three layouts add within a factor of 1.6 of the same number of entries to the fixed plan's factor while their refusal counts differ by three to seven times. What moved the earlier savings was the baseline: the natural order's factor changes with the layout too, by 13 per cent at three in ten, and on the 10 × 10 grid the lines 'save' 19 per cent where the scatter saves 9 — with fixed-plan factors of 907 and 898 entries.

-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. Turning the lines to run along columns was predicted to bring the natural order below 1,000 entries, leave the plan within 3 per cent of 907, and drop the saving to the scatter's. It misses on all three: 1,008, 977 and 3.1 per cent. What the natural order paid for on rows was its own interchanges. Offered by position, as a dense code offers it, every Bunch–Kaufman interchange moves a row out of sequence, and on rows that costs 125 entries at three in ten and 191 at half — the most of any layout at every share. Kept to the grid's index, the natural order barely notices the layout, and the row lines' saving halves to 9.0 per cent.

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. On every separator the dissections here produce — the grid's rows, its level-set diagonals, and the level sets of three irregular graphs of 300 vertices — twenty-four shuffles of the separator's order give the same factor to the last entry. The reason is one line of graph theory: once the pieces are eliminated, a separator whose vertices all touch one connected piece is a clique, and a clique's order changes nothing. The order matters only on a separator that is not a clique, and the one built here to be that — two grid rows thick — loses to a separator of one row under every order tried. The separator's size is the decision; its order is not one.

0102030405060radius 0.11graph 1cliquegraph 2cliquegraph 3cliquegraph 4cliquegraph 5cliquegraph 6cliquegraph 7cliquegraph 8cliqueradius 0.13graph 1cliquegraph 2cliquegraph 3cliquegraph 4cliquegraph 5cliquegraph 6cliquegraph 7cliquegraph 8cliqueradius 0.16graph 1cliquegraph 2cliquegraph 3cliquegraph 4cliquegraph 5cliquegraph 6cliquegraph 7cliquegraph 8cliqueseparator vertices; dark: touching only one sideevery one a clique, at every densityspare vertices, no broken clique

The spare vertices lean one way

Every separator a dissection drew on grids and sparse geometric graphs was a clique once its pieces were eliminated, so the order inside it changed nothing; only a separator two rows thick, built by hand, was not. On geometric graphs at twice the density, level-set separators were predicted to come out several vertices thick where the level crosses a dense patch, with vertices touching only one side — failing the clique condition, making their order matter by more than five per cent, and paying more for trimming than for any order. On twenty-four graphs at three densities the spare vertices are there: up to thirteen in one separator. Not one separator fails the clique condition, because on every graph the spare vertices all touch the same piece, and two vertices that share a piece are joined through it. No shuffle of any separator moves the fill or the work by a single entry. Trimming the spare vertices into the piece they touch shortens the critical path by up to 48 per cent — and denser graphs have fewer of them, not more.

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