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The thread: Claims that can fail — page 3

Essays 49 to 72 of the 213 on this theme, in the same order.
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 Sparsity, and what elimination costs

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.

2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one Exact arithmetic, and what it costs instead

A fraction recovered from one remainder

A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1relative size of the entrywise perturbationrelative forward errorκ∞ · εcond(A,x) · εmeasuredboth bounds holdκ∞(A)1.9·10⁸cond(A, x)4.8ratio of the bounds4·10⁷both curves above the data are boundsand only one of them is a measurement Two errors, and whose fault they are

A condition number scaling cannot move

Skeel's componentwise condition number is invariant under any row scaling — exactly, before any norm is taken, because two diagonal factors cancel entry by entry. It is never larger than the normwise one and can be arbitrarily smaller, and the ratio between them is a diagnostic for which kind of ill-conditioning a matrix has.

081624324048566410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹index kmagnitudethe floor: k = 43best truncation: k = 31σₖ|uₖᵀb| exact|uₖᵀb| with noise|uₖᵀe|, smoothedcorrelation ρ = 0.9noise energy, first 28 directions0.98tilt of the noise floor, decades1.1the crossing43the best truncation31the floor tilts, and is still a floorρ = 0.9 Regularisation, and the answer that is chosen

Noise that spares the answer and fools the rules

Make each noise sample remember the last one, keep its size fixed, and the best answer available gets slightly better — 0.1056 to 0.1010 — because slow noise hides in the directions where dividing by σ costs nothing. The Picard crossing still lands two dozen indices past the best truncation. What breaks is the rules. Generalised cross-validation more than doubles the best error on 14 draws of 48 instead of 3, the discrepancy principle's typical cost triples, and the two miss in opposite directions. Whitening by the covariance takes GCV back to 3.

00.250.50.751-1-0.500.51mode frequency θ / πdamping factor per sweeppredictedmeasured±0.333a coarse grid seestwo routes to one factorsmoothing factor, scanned0.33smoothing factor, closed form0.33worst mode disagreement4.4·10⁻¹⁶63 interior points, one sweepthe left-hand end is what the coarse grid is for Iterating, instead of factorising

The error smoothing cannot reach

One weighted Jacobi sweep multiplies every mode of the error by a number, and the number is a sine. Half the modes are cut by three or better, and the other half come back at 0.999 — which is not a failure of the method but the fact the whole of multigrid is built on.

-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6.4·10⁵growth × γ0.64the ordering is free to chooseand not free of consequence The matrix a constraint makes

The regularisation that legalises every order

Perturb a saddle-point matrix's two blocks in opposite directions and it acquires a factorisation with a diagonal D under every symmetric permutation — not under a good one, under all of them. Five hundred random orderings, five hundred successes, and a growth factor that spans six orders across them.

D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero Elimination, and the swap

When symmetry is not enough

The matrix [[0, 1], [1, 0]] is symmetric, nonsingular and perfectly conditioned, and there is no diagonal entry to pivot on. Every factorisation restricted to symmetric interchanges and one-by-one pivots fails on it, at any depth of searching, because every entry it could search is zero. The repair is to take two variables at once.

-47-37-27-17-7313233301234log₁₀ of the vector's normfp1611 bitsbfloat168 bitstf3211 bitsbinary3224 bitspale: the format's range · blue: √(Σ(xᵢ/m)²)·m · red: √(Σxᵢ²)fp16 and tf32 have the same eleven significand bitsand their bars do not overlap The arithmetic underneath

A norm that overflows before it is a norm

The vector of sixteen thousands has a Euclidean norm of 4,000, which fp16 represents exactly. Written as the square root of the sum of squares it returns infinity, because squaring doubles the exponent — and the expression costs half the format's range on the one computation every iterative method performs at every step.

10⁻²10⁻¹110¹10²10³10⁴10⁻⁴10⁻³10⁻²10⁻¹1backward error, in units of ushare of systems above ituCramereliminationCramer, control24-bit arithmeticworst Cramer, in u395worst elimination, in u1.2control, worst Cramer4.9κ of the worst system3.4·10⁶one derivation, two computationsand only one of them is stable Elimination, and the swap

A rule that is correct and unusable

Cramer's rule gives every component of the solution in closed form, in terms of determinants, and it is a theorem. On two-by-two systems whose rows are nearly parallel it returns an answer with a backward error of 458 units of roundoff where elimination returns 1.3 — on a matrix whose condition number is 32,000 and which elimination solved perfectly.

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 Sparsity, and what elimination costs

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.

numbers stored, 256 × 256clustered, strong27,008clustered, weak24,064the dense matrix65,536shuffled, strong65,536shuffled, weak118,208the same matrix, twiceκ, clustered24κ, shuffled24Frobenius norm, clustered6140Frobenius norm, shuffled6140shuffled weak ⁄ dense1.8the compressibility is in the numberingand the numbering is not in the matrix Neither sparse nor dense

The same matrix, numbered twice

One symmetric permutation. The condition number is 24.3948 either way to eight digits and the Frobenius norm is 6.13996414·10³ either way to twelve. The partition that stored 27,008 numbers now finds no admissible pair anywhere and stores all 65,536, and the format that compresses regardless stores 118,208.

0.46-1.11.5-0.470.2-0.181.90.690.42-0.160.028-0.0160-1.50.160.880.099-0.1100.680.762.30.930.0480-0.250.110.952.11.500001.52.3after reflector 2the highlighted entries are the bulge — the only thing that is not Hessenbergentries below the subdiagonal3reflectors used2reflectors in a whole step5the shifts are never formed — only their sum and productand both of those are real Eigenvalues, singular values, rank

Two shifts that are never formed

The double shift is defined as a factorisation of (A − μI)(A − μ̄I), which nobody computes. What is computed is the first column of that product — three numbers — and the bulge those three numbers create, pushed down the subdiagonal by n − 2 reflectors until it falls off the bottom.

-2-101234-5-3-1135real partimaginary part4 insidea countable spectruminside the contour4drawn12existing∞worst branch residual1.6·10⁻¹⁵there is no last eigenvalueso the question has to change The eigenvalue problem that is not linear

A problem with infinitely many eigenvalues

Let the matrix depend on λ through something that is not a polynomial and three things stop being true at once. There is no linearisation, there is no characteristic polynomial, and "compute the spectrum" is not a request that can be granted — the only finite question is how many eigenvalues are inside this circle.

02468101210⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²binary64 operations per elementrelative errorruntime orderfixed partitionpre-rounded, one passpre-rounded, two passescompensated (Kahan)exact accumulationwhat determinism costsreproducible policies3cheapest, ops/element3its error0.0024two passes10⁻¹⁷ordinary, 8 pieces3.6·10⁻⁷nothing is cheap and exactthe middle of the figure is the answer The answer that depends on the machine

Accuracy and agreement are different properties

The most accurate policy on this site's summation figure returns 119 different answers, and the one that returns a single answer is four orders less accurate. Neither property implies the other, and the vocabulary has one word for both.

ℚ6𝔽24𝔽35𝔽56𝔽76𝔽116𝔽136𝔽1016𝔽655376rank, by the ring the entries are read inthe rank of one matrixover ℚ6over 𝔽24over 𝔽35over 𝔽56over 𝔽76nothing is rounded hereand the answer still is not a property of the matrix Exact arithmetic, and what it costs instead

The rank depends on the ring

A floating-point rank is a decision about a threshold. Remove the arithmetic error entirely and the threshold goes away — and the answer still is not a property of the array of numbers, because one integer matrix has rank six over the rationals, five modulo three and four modulo two, with nothing rounded and nothing decided.

polar: U₁P − U₂QR: Q₁P − Q₂-3.3·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶-1.9·10⁻¹⁶5.6·10⁻¹⁷-1.7·10⁻¹⁶6.7·10⁻¹⁶-8.9·10⁻¹⁶2.8·10⁻¹⁶-5.6·10⁻¹⁶-3.6·10⁻¹⁶6.1·10⁻¹⁶-8.3·10⁻¹⁶8.3·10⁻¹⁶-7.2·10⁻¹⁶-10·10⁻¹⁶2.2·10⁻¹⁶4.4·10⁻¹⁶1.7·10⁻¹⁶-4.4·10⁻¹⁶-5.6·10⁻¹⁷5.6·10⁻¹⁷0-7.2·10⁻¹⁶-4.4·10⁻¹⁶3.9·10⁻¹⁶5.6·10⁻¹⁷2.8·10⁻¹⁶-1.1·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶1.1·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶0.67-0.35-0.23-0.620.19-0.72-0.770.170.20.290.190.93-0.180.55-0.0290.06-0.460.11-0.17-0.61-0.230.38-0.55-0.380.750.870.661.1-0.36-0.530.42-0.670.180.180.52-0.3Frobenius norms, columns reordered‖U₁P − U₂‖2.8·10⁻¹⁵‖Q₁P − Q₂‖3‖Q‖, for scale2.4κ of the matrix100Frobenius distance between the two answers, on one scale0 to 4polar2.8·10⁻¹⁵QR3.048‖Q‖ = 2.449the column space did not moveand one of the two answers did Orthogonality, measured

A test with no answer in it

A caller with no reference answer can still ask whether a routine answered the right question: reverse the columns, run it again, compare. The polar factor's two answers agree to 10⁻¹⁵ at every conditioning drawn; a QR's differ by 2.353 on matrices whose own norm is 2.449. The test has a floor, and the floor is measurable too.

110¹10²10³10⁴10⁵10⁶error ÷ the oracle's, on the same drawten times the oracleover tendiscrepancy principletold ‖e‖0 of 1000unbiased risktold σ²15 of 1000cross-validationtold nothing57 of 1000quasi-optimalitytold nothing0 of 1000L-curvetold nothing0 of 1000blue: median · bar to the 90th percentile · line to the 99th · red: worstthe counts are the tail Regularisation, and the answer that is chosen

One draw in twenty

Sixteen draws gave generalised cross-validation a worst case of 12%. A thousand draws at each of five noise levels give it a second answer on four to six in every hundred, ten to seven million times worse than the oracle, while its median stays among the best of five rules. The quasi-optimality criterion, told nothing either, never costs more than 1.41 in five thousand draws. The share settles by a thousand draws, and letting the search look further down more than triples it.

-5-3-11350eigenvalueHZᵀHZK4 negative — Cholesky of H stops at row 30 negative — a minimum on the constraint(10, 4, 0) = In(ZᵀHZ) + (4, 4, 0)one factorisation, no Zpositive, LDLᵀ of K10negative, LDLᵀ of K4negative in H4negative in ZᵀHZ0the count follows the reduced Hessiannot the Hessian The matrix a constraint makes

A minimum the Hessian cannot see

A Hessian with four negative eigenvalues can sit at a constrained minimum, and a Cholesky of it stops at the third row. One symmetric indefinite factorisation of the saddle-point matrix settles the question anyway — ten positive pivots and four negative — without a basis for the null space ever being formed. The count is exact in the algebra and blind in floating point, in a band that grows like κ(A)²; the route through the null space is blind in one that grows like κ(A).

the estimator maximises this quantity over the columns it visitscolumn 1 ‹visited›12column 2 ‹the answer›114column 311.4column 411.4column 511.4column 611.4column 711.4column 811.4column 911.4column 1011.4column 1111.4column 1211.4estimate 12.0a walk that stopped earlythe estimate returned12the true 1-norm114columns visited1products with the matrix5the walk's own stopping test firedand every column it could see was smaller Two errors, and whose fault they are

An estimate that can be fooled

Nobody computes a condition number, because forming an inverse costs more than the solve did. Every library estimates it instead, from four or five products with a factorisation already in hand. The estimate is exactly right on four random matrices out of five — and there is a matrix, three distinct entries wide, on which it returns a twentieth of the truth.

567891010¹10²10³10⁴10⁵10⁶log₂ nnumbers the construction touchedentries, n²products with the operatoran operator, applied a few hundred timesproducts at n = 512256entries at n = 5122.6·10⁵per doubling48relative compression error4·10⁻⁷excess over the compression7.3no entry of the matrixwas ever read Randomised, and the guarantee that changes kind

Built from products alone

A 512-square hierarchical representation, at a relative error of 4·10⁻⁷, from 256 applications of an operator that is never assembled. The compression route reads 262,144 entries; this one reads none, and pays for it with a factor of seven against the representation the entries would have given.

567891010⁴10⁵10⁶10⁷10⁸10⁹log₂ nmultiplicationsdense factorisation, n³⁄3the recursion, countedwhere the format starts payingratio at n = 641.5ratio at n = 5120.16exponent, first doubling2.1exponent, last doubling1.7backward error1.4·10⁻¹⁰cheaper is a sizenot a property Where the flop count stopped predicting the time

Where the format starts paying

A hierarchical solve costs 1.48 times a dense factorisation at 64 unknowns and 0.16 times it at 512. The crossover is between 64 and 128, it walks right when the accuracy is tightened, and the exponent between consecutive sizes is 2.13, 1.93, 1.74 — falling towards one and never arriving.

10¹10¹.³10¹.⁶10¹.⁹⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰¹10².²00.250.50.751grid points nresidual reduction per stepJacobiGauss–SeidelV-cycleV-cycle spread, 8× in size0.003Jacobi at n = 1270.99work exponent, fitted0.079the dashed curve is cos(πh), Jacobi's closed formthe flat line is the whole method Iterating, instead of factorising

A rate that does not notice the size

The V-cycle reduces the residual by a factor of ten a cycle at fifteen points and at a hundred and twenty-seven. Jacobi on the same four problems goes from 0.981 to 0.9978, climbing towards one. One of those is a constant and the other is an exponent, and that is the whole distinction the field turns on.

10⁻⁷10⁻⁵10⁻³10⁻¹⁰10⁻⁸10⁻⁶magnitudespacing to the next numberthe smallest normalgradualflush to zerosmallest normal6.1·10⁻⁵smallest subnormal6·10⁻⁸octaves of subnormals10pairs that lie under FTZ10the spacing stops halving and stays putwhich is what makes x − y = 0 mean x = y The arithmetic underneath

The numbers below the smallest one

Below the smallest normal number the spacing stops halving and stays put, all the way to zero. That is what gradual underflow is, and the thing it buys is the sentence every algorithm assumes without being told — x minus y is zero only when x equals y.

chain · cheapest1.51·10⁴chain · greedy5.06·10⁴chain · dearest5.13·10⁸train-inner · cheapest3344train-inner · greedy1.51·10⁴train-inner · dearest6.58·10⁹als-step · cheapest1.05·10⁵als-step · greedy1.13·10⁵als-step · dearest1.13·10⁵multiply-adds, on a logarithmic scaleone value, many priceschain, best ⁄ worst3.4·10⁴train, best ⁄ worst2·10⁶als step, best ⁄ worst1.1worst greedy excess4.5no answer changesand the price does Where the flop count stopped predicting the time

The order the products are taken in

The sparsity field's first essay says the elimination order decides the memory. This is the same sentence about arithmetic: a contraction of several tensors over shared indices has one value and many evaluation orders, and on the inner product of two trains they differ by a factor of two million.

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