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The thread: Claims that can fail — page 7

Essays 145 to 168 of the 278 on this theme, in the same order.
10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³r½ recorded during one factorisationnoise that halves the growthT 6, h 0.3: 75T 9, h 0.3: 910T 12, h 0.3: 11000T 15, h 0.3: 1.3·10⁵T 12, h 0.25: 11000T 12, h 0.2: 11000T 9, h 0.15: 910measured ÷ predicted, shooting matricessmallest ratio0.92largest ratio2.4Wilkinson: recorded r½0Wilkinson: noise that halves it10⁻¹⁶one factorisation, one extra comparison a stepdashed: a factor of three either way Elimination, and the swap

A margin the factorisation records

Partial pivoting finds the second-largest candidate in every column it scans, and throws it away. Keep it, divide the gap by the growth reached at that step, and take the smallest over the steps before half the final growth has arrived. That one number, recorded by the factorisation that is already running, predicts the dense noise that halves the growth to within a factor of 0.92 to 2.4 on shooting matrices whose growth runs from 75 to 1.3·10⁵ and whose fragility spans three and a half decades — and on Wilkinson's matrix it is exactly zero.

111315171921232510⁻¹110¹sketch width l, for rank 10‖(I − QQᵀ)A‖ ÷ σ₁₁GaussianHadamardone nonzero a rowthree nonzeros a rowcoherent, geometric 0.8, width 20, twenty seedsGaussian: median 0.41, worst0.78Hadamard: median 0.37, worst0.94one nonzero a row: median 3.3, worst7.1three nonzeros a row: median 0.41, worst0.99solid: median · dashed: worst of twentybelow one: better than the best rank-10 error Randomised, and the guarantee that changes kind

A sketch that finds the columns it can see

A sparse sketch with one nonzero in each row is twenty times cheaper to apply than a Gaussian one, and on a matrix whose important directions are spread across its columns it finds the same range: a median error of 0.45 against 0.43. Put the same ten directions into ten particular columns and it is eight times worse — 3.33 against 0.41, with a worst draw of 7.1 — because two important columns hashed to one bucket are one direction. Three nonzeros a row repair it at a sixth of the Gaussian's cost, and a randomised Hadamard transform never had the problem.

010203040run, by bend then tolerancesteps beyond the better fixed degreestraightbend 10⁻⁵bend 10⁻⁴bend 0.001bend 0.01bend 0.1● second difference ■ residual at both ○ wrong fixed degreethirty runs: steps beyond the better degreesecond difference: runs matching the better degree, of 3026second difference: worst excess, steps5residual at both: runs matching the better degree, of 3029residual at both: extra residuals, in steps4.5wrong fixed degree: worst excess, steps35zero is the better degree, known in advanceeach group of five is one bend When the problem arrives again

The degree the history chooses

A sequence of solves can start each member from the line through its last two answers or the parabola through its last three, and which is better depends on how much its path bends — which a code does not know. Over thirty runs of bend and tolerance, always taking the line costs 456 inner steps more than the better choice; always taking the parabola costs 33. A free rule reading the stored answers closes that to 11. A rule that evaluates the residual at both starts picks the better one on 29 runs of 30, and pays 135 steps for the evaluations.

edge of the numerical rangespectrum, -3.49zero37.9-0.986-3.496.59unstable one-sided models, of nineteen placements051015-3-2-1.5-1-0.500.511.522.53cond 1.4·10⁶cond 112t, the rescaling exponent — the k-th state is multiplied by γ to the power tk; t = 1 symmetrisesorder 1order 2order 3order 6the transfer function is the same at every tonly the inner product changes Reduction, and what a model is for

A certificate written in coordinates

Every one-sided reduced model of the convection–diffusion system was stable, because its poles cannot leave the numerical range and the range sat in the left half plane. Rescale the state by a diagonal matrix and nothing about the system changes — not its eigenvalues, not its transfer function, not one pole of any two-sided model — but the numerical range moves. At Péclet 40 a rescaling with condition number 260 pushes its edge to +6.6 and ten placements of nineteen return unstable one-sided models. Projecting in the inner product of a Lyapunov solution puts the certificate back in any coordinates.

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 Sparsity, and what elimination costs

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.

027548110813510⁻⁶10⁻⁴10⁻²110²10⁴10⁶power‖Aᵏ‖Kreiss constant 6760e · n · K‖Aᵏ‖ρᵏtwo routes to one peakspectral radius0.8peak of ‖Aᵏ‖2·10⁴Kreiss constant6757e · n · K1.1·10⁵everything here decays in the endand one of these curves says how much first Eigenvalues, singular values, rank

A spectral radius that grows first

ρ(A) below one guarantees that the powers of A go to zero and says nothing about what they do on the way. Here they rise by a factor of twenty thousand before turning over, and the peak is bracketed above and below by a constant computed from the resolvent norms outside the unit circle — two routes to one number, one through the plane and one through the powers.

noise that halves the growthshooting, partial pivoting0.0018shooting, τ = 0.10.56Wilkinson, τ = 0.10.1800.10.20.30.40.50.60.70.80.9110⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1one minus the thresholdstored noise that halves the growthshooting, T = 12, h = 0.3Wilkinson, n = 24the same growth at every thresholdheld by a choice noise must push further to reverse Elimination, and the swap

A threshold that holds the growth still

Partial pivoting's large growth on the shooting matrix rests on a margin of 5.6·10⁻³, and Wilkinson's rests on exact ties, which a perturbation at rounding breaks. A sparse code pivots with a threshold instead, taking the sparsest row among candidates within a factor τ of the largest. At every threshold from 1 to 0.1 both matrices grow by exactly as much as under partial pivoting. But the growth is now held by row counts, which no perturbation of the values can reverse: at τ = 0.1 it survives noise on the stored entries up to 0.56 on the shooting matrix, three hundred times more, and 0.18 on Wilkinson's, where partial pivoting's goes at 10⁻¹⁶. The margin that predicts it is the chosen candidate's height above the threshold line, and it needs no division by the growth.

κ(A) = 10⁸, 152 saddlespassed, ordinary63passed, on wrong counts57passed, always510⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹10⁻⁴10⁻²110²10⁴10⁶10⁸curvature needed, μshift applied ÷ μordinary schedulenull space on wrong countsnull space alwaysbelow the dashed line a saddle was certifiedthe arbitrated shift sits at twice the need The matrix a constraint makes

A loop that asks the null space why

An inertia-correction loop sees only an integer, and two different faults produce the same wrong one: curvature that needs a shift, and a constraint too weak for the count to see. One QR of the constraint matrix on a wrong count tells them apart — it shifts none of the 22 weak-constraint minima the ordinary loop shifted by up to 2,621 — and its reduced eigenvalue gives the shift a saddle needs in one step, twice the need exactly, where the schedule overshoots by up to 17,783 times. But at κ(A) = 10⁸ the loop still certifies 57 saddles of 152, because a false certificate is a count that read right, and a check made only on wrong counts never sees it. Asking every time leaves five, all shallower than 2·10⁻⁸.

each bar is a percentage — of the sample, or of the true condition numberexactly right86.0%inside 10%91.0%inside a factor of 291.0%worst in the sample, ×10035.7%the constructed matrix, ×1007.7%usually exactexact share0.86worst of the sample0.36the constructed matrix0.077a routine that is right most of the timeand never wrong in the safe direction Two errors, and whose fault they are

The tail a sample never reaches

Hager's estimator is exactly right on four random matrices in five, and that share is stable — between 80.5 and 87.5 per cent across nine sizes. The worst underestimate is not stable at all: it falls every time more matrices are drawn, from 0.746 at sixty to 0.377 at four hundred, and the matrix built to defeat the estimator sits five times below anything four hundred draws found.

10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁴10⁻¹10¹²10²⁵10³⁸10⁵¹10⁶⁴δ, the gap between consecutive eigenvaluesrelative error in eᴬan answer with no correct digitsV f(Λ) V⁻¹scaling and squaring‖A(δ) − A₀‖exact eigenvalues throughoutκ(V) at the smallest δ3.3·10⁸²eigen route2.9·10⁶⁵scaling and squaring4.1·10⁻¹²distance to the limit7·10⁻¹²the eigenvalues are the diagonaland they are exact at every stop Eigenvalues, singular values, rank

A function of a matrix is not a function of its entries

Everybody learns that f(A) means diagonalise, apply f to the eigenvalues, undiagonalise. That is a definition, not a method. On a matrix seven picometres from a defective one — with exact eigenvalues and eigenvectors from a closed form — the definition returns an answer wrong by sixty-five orders of magnitude, and a method that never mentions an eigenvalue returns the right one.

Re λIm λwhat survives the arrowsvertices18arcs18worst row sum0worst column sum0largest |Im λ|0.98asymmetry1the null vector is still exactand nothing else about the spectrum is real The matrix that is a graph

A Laplacian that is not symmetric

Point the edges and the matrix stops being symmetric. Its row sums are still exactly zero, so the null vector survives; everything built on the quadratic form does not, and the eigenvalues need a plane rather than a line. Asymmetry permits that and does not force it, which the smallest case here demonstrates by being asymmetric and real.

n = 512, ε = 10⁻⁸1/r, least at leaf16log r, least at leaf8cos(40r)/r, least at leaf16cos(120r)/r, least at leaf1604794141188235leaf sizestored per unknown481632641/rlog rcos(40r)/rcos(120r)/rthe rank doubles and the optimum does not movewhich the next figure explains Neither sparse nor dense

A prediction that arrives a decade late

A rougher kernel raises every rank, a higher rank raises the side at which compressing a block stops paying, so the leaf that stores least should rise. Across four kernels whose largest rank runs 5, 5, 9 and 10, it is 16 on three of them and a tie between 8 and 16 on the fourth. The prediction is right in sign and out by a decade of tolerance, and the reason is a ceiling: an oscillation multiplies the rank by exactly two — 3 and 6, 4 and 8, 5 and 10, 6 and 12, 7 and 14 — at any frequency, because a cosine of a difference is a rank-two function.

96 × 64, κ = 10⁸blocks of eight, final1.4·10⁻¹⁴their root-sum-square1.2·10⁻¹⁴one at a time, final2.9·10⁻¹⁴081624324048566410⁻¹⁵10⁻¹⁴10⁻¹³columns factorised‖QᵀQ − I‖sum of the blocks' ownone reflector at a timeblocks of eightroot-sum-squareeach block is one factor, each reflector anotherthe departure grows with the square root of the count Orthogonality, measured

Eight blocks and sixty-four reflections

One block of sixteen reflectors, assembled as I − Y T Yᵀ, departed from orthogonality five times as far as a single reflection, and the question was whether a factorisation of many blocks multiplies that factor. It does not. On 96 × 64 matrices eight blocks of eight end at 1.42·10⁻¹⁴ — within a fifth of the root-sum-square of their own departures, and less than half their sum — while the same factorisation taken one reflector at a time ends at 2.91·10⁻¹⁴. A block departs more than a reflector, and there are an eighth as many of them. And a nearly dependent column that swells the triangle's entries to 10²⁶ costs the product nothing.

median ratioone nonzero, t = 03.5one nonzero, t = 0.011.210⁻⁴10⁻³10⁻²10⁻¹11mixture t (t = 0 at the left edge)error ÷ σ₁₁one nonzero a rowheavy columns reservedthree nonzeros a rowGaussianthe left edge is exact coherencethe failure fades over four decades of mixing Randomised, and the guarantee that changes kind

The leverage that did not move

A one-nonzero sketch fails on a matrix whose leading directions sit on ten particular columns, and coherence — the largest column leverage — is the statistic that names the failure. Turn the directions away from their columns by a hundredth of a radian and the sketch's median error falls from 3.47 to 1.23 times σ₁₁ while the coherence stays at 6.40 to three figures. Giving the heaviest columns buckets of their own repairs the rest, but only when it reserves more buckets than the rank: ten reserved leave 1.16, sixteen reach 0.36, below the Gaussian's 0.41.

start 1, errorreaches2.8·10⁻¹³stopped by δ = 0.010.0099stopped by δ = 0.0010.0098stopped by δ = 10⁻⁴2.6·10⁻¹³110¹10²10³10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²sweeprelative errorδ = 0.01δ = 0.001δ = 10⁻⁴δ = 10⁻⁵a plateau is a still errorand a still error is what the test asks for When the index is a tuple

A still error is not a settled one

A CP fit with one term too many on a noisy tensor never meets the usual stopping test, and its error has settled by sweep 150. A test that asks only whether the error has moved by less than δ of itself over ten sweeps stops those fits by sweep 25, and over twenty-four noisy tensors it names the same rank as the usual test every time, at δ = 10⁻² for 30,250 sweeps against 311,458. But δ = 10⁻² is the tolerance a rank decision states, and on a slow fit that converges it stops five starts of six on a plateau ten orders above the answer and names the wrong rank for four tensors of six. A tenth of the decision's tolerance keeps every decision on both families.

Cheeger's band, for the circulationcirculation · arcsvertices on the smaller sidewhich of the two the theorem is aboutλ₂ of 𝓛0.066λ₂/20.033√(2λ₂)0.36circulation Φ0.12arc conductance0.04their ratio3.1the inequality holdsabout a quantity nobody counts The matrix that is a graph

A conductance the arcs do not measure

Symmetrising a directed Laplacian with respect to its walk recovers everything the arrows took — a real spectrum, a sweep cut, a Cheeger inequality. What it does not recover is the quantity: the inequality bounds the probability that a step of the walk crosses the cut, which on one graph here is three times the weight of the arcs that do.

012345610⁻¹10²10⁵10⁸10¹¹log₁₀ γ, the change of unitscondition numberthe linearisationthe quadraticone problem, two amplifiersκ(quadratic), first6.6κ(quadratic), last6.6κ(linearisation), last10¹²how far the first moved1the problem is as well conditioned as everand the method is not The eigenvalue problem that is not linear

The number that moves when the problem does

Two quantities are offered as the condition number of one eigenvalue. One is unmoved to eight digits by a change of variable that is exact in both directions, and grows like the square root of the chain's length. The other is inflated by ten orders by that change of variable, and is ten times too large before anything has been done at all.

‖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 answer that depends on the machine

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.

036912151810¹10²10³rotationswidththe enclosurethe seta rotation is an isometrymeasured growth a step1.4√2, from the geometry1.4enclosure ÷ set after 201024no rounding error is responsible for any of thisa higher precision does not touch it The arithmetic underneath

Nine steps of pessimism

A proved bound is 8 to 26 times the error it bounds, at every precision from 16 to 40 significand bits. A carried interval is (√2)ᵐ times too wide after m re-enclosures. The two cross between eight and nine, so the method everybody warns against is the tighter of the two for a short computation.

10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁴10⁻¹10¹²10²⁵10³⁸10⁵¹10⁶⁴δ, the gap between consecutive eigenvaluesrelative error in eᴬan answer with no correct digitsV f(Λ) V⁻¹scaling and squaring‖A(δ) − A₀‖exact eigenvalues throughoutκ(V) at the smallest δ3.3·10⁸²eigen route2.9·10⁶⁵scaling and squaring4.1·10⁻¹²distance to the limit7·10⁻¹²the eigenvalues are the diagonaland they are exact at every stop Eigenvalues, singular values, rank

The series that has to be squared back

The Taylor series for the matrix exponential is not wrong — every term is computed correctly — and on Moler and Van Loan's two-by-two its largest term is 5.4 million times the answer it sums to. The method that replaces it scales the matrix down and squares the result back, and both halves of that sentence cost: too few squarings and the approximant is out of range, too many and each one doubles the rounding.

LAPACK's walk, exact83.0% · 5.2 productsLAPACK's walk, worst × 10037.7% · 5.2 productsblock of one, exact84.0% · 4.4 productsblock of one, worst × 10037.7% · 4.4 productsblock of two, exact96.5% · 8.5 productsblock of two, worst × 10059.6% · 8.5 productsblock of four, exact100.0% · 16.8 productsblock of four, worst × 100100.0% · 16.8 productseach bar a percentage — of the sample, or of the truthtwo random vectors close most of the tail Two errors, and whose fault they are

Two columns see what one walk cannot

The condition estimator every library ships walks from the all-ones vector, and a matrix whose largest column cancels against that vector hides from it: at n = 24 it reports five per cent of the truth. The block estimator behind MATLAB's condest walks with two vectors, the second random. On the same matrix at three sizes it is exact on every one of twenty seeds. On four hundred random 8 × 8 matrices it is exact on 96.5 per cent where the single walk is exact on 83.0, and its worst case, 0.596, is reached in the first fifty draws and not lowered by the next 1,550. The single walk's worst was still falling at 1,600. Four vectors are exact on all 400.

k < s/4cells agreeing28cells3010⁻³10⁻²10⁻¹11distance from the threshold, relativepredicted leaf ÷ bestred: the rule's two missesboth within three per cent of the break-even Neither sparse nor dense

A quarter of the leaf

The leaf that stores a hierarchical matrix most cheaply had been read off a table of nine points, with the suggestion that it depends on the mean rank alone. Across six kernels at five tolerances — thirty cells — it does, once the mean rank is read in the right place. Plotted against the mean rank at a fixed leaf of 16, two cells a tenth of a rank apart want leaves of 8 and 16. Plotted against the rank of the blocks a halving would create, one rule — halve the leaf while that rank is below a quarter of the leaf — predicts the best leaf on 28 of the 30, and the two it misses are the two whose rank sits within three per cent of the threshold. The fractional power the last measurement expected to be rough is smoother than 1/r.

points of coveragegain at c = 1, fast decays, best5.3worst0.750.811.21.41.61.822.22.42.6-30-20-10010products ÷ the sequential rule'scoverage gained, pointsdecay 0.8decay 0.9decay 0.97dashed: no gaina few points, bought with a fifth to double the probes Randomised, and the guarantee that changes kind

A spread measured on probes it does not average

A trace estimator that stops on its own standard error misses its target a few points more often than a normal table says, because the runs that stop earliest are the ones that underestimated their noise. Spend a pilot of probes only on the spread, fix the number of probes to average in advance, and the selection is gone: with Student's margin the two-stage rule covers 64.8 to 73.3 per cent at one standard error where the table says 68.3. With the normal's margin and a pilot of four it covers 58.5. At one standard error it recovers one to five points for a fifth to two fifths more probes; at 1.96 there was nothing to recover, and the guarantee costs a tenth to double.

pooled over every settingaccidents, one digit apart62accidents, three digits apart5relations lost on the way109110¹10²10³digits between the two searchesruns123relations keptaccidents acceptedlogarithmic vertical axiseach digit apart divides the accidents by three or four Exact arithmetic, and what it costs instead

The digits between the two searches

A lattice relation search run at N digits and again at N/10 never agrees with itself on a relation among the rounding, but it does agree on 62 approximate relations at fifteen digits or fewer — accidents that really are the shortest vector there, and that dropping a digit was said to be unable to dislodge, since a combination that cancels to D digits cancels to D − 1. It cancels, but it stops being the shortest. Two digits apart the accepted accidents fall to 17 and three digits apart to 5, while the relations kept fall from 711 of 755 to 652 and 602 — every one of the losses at fifteen digits or fewer. Past a double's sixteen digits the separation costs nothing and buys nothing.

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