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The thread: Two routes to a number — page 7

Essays 145 to 168 of the 228 on this theme, in the same order.
22.539343.078693.618034.157384.6967210⁻¹110¹10²stiffness damping βoverdamping marginβ* = 3.23607two routes to a boundaryclosed form β*3.2by certificate3.2difference4.7·10⁻¹³bisection steps44a factorisation that completesand a sine, agreeing to twelve digits The eigenvalue problem that is not linear

A class a longer chain takes away

Symmetry survives a bigger problem. Hyperbolicity does not. The damping that certifies a chain of seven masses is refused by a chain of eight, the damping the class demands grows like the length without bound, and the certificate has to be earned again at every size — which costs one Cholesky, and the alternative is a proof quietly inherited from a smaller problem.

darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.3eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it Eigenvalues, singular values, rank

The eigenvalues that are not there

For a normal matrix the resolvent norm is exactly one over the distance to the nearest eigenvalue, so a picture of it carries nothing the spectrum did not. Move one entry above the diagonal and the region a perturbation of 10⁻⁸ can put an eigenvalue into stops being a disc and reaches out past the unit circle, while every eigenvalue stays at 0.8.

the problem you posedA = H12b = A·(1, 2, …, 12)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 1.8·10⁻¹⁷the answer you wantedx = (1, 2, …, 12), exactlythe answer you gotx̂, wrong by 0.02 relativebackward error 1.8·10⁻¹⁷forward error 0.02κ = 1.8·10¹⁶κ · η = 0.33, and the measured forward error is 0.02.The algorithm is not at fault. The problem is.H12, LU with partial pivotingresidual and error differ Two errors, and whose fault they are

The fifth author

Four authors of a wrong answer have been named already and each is a statement about one computation. The fifth is not: it is what separates two computations that are both correct, it is a backward error of measurable size, and no residual, bound or condition number contains it.

10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸10⁻⁷10⁻⁶|h|, the curvature of H along the weak directionσ at which the count is first wrong10⁻⁴10⁻³10⁻²10⁻¹110¹10²no interchangesLDLᵀ of Krank tolerancethrough Z: right to 10⁻¹⁶10 × 4, eight draws per curvatureh = -1·10⁻⁴: LDLᵀ first wrong at σ1.8·10⁻¹¹h = -0.01: LDLᵀ first wrong at σ1.8·10⁻¹⁰h = -1: LDLᵀ first wrong at σ1.4·10⁻⁹h = -100: LDLᵀ first wrong at σ1.8·10⁻⁸the pair's small eigenvalue is σ²/|h|the count loses it long before the rank does The matrix a constraint makes

A constraint the count stops seeing

Let one constraint drift towards being a combination of the others and the inertia of the saddle-point matrix keeps its promise only while σ²/|h| can be resolved — σ the constraint's smallest singular value, h the curvature along the direction it barely constrains. At h = −1 the count stops seeing the constraint at σ = 1.4·10⁻⁹, six decades before any rank test would drop it, and below that it reports a genuine minimum as a saddle on three to six draws in eight. No shift of H brings the constraint back: the correction loop shifts a problem that needed nothing by as much as 2,620. A perturbation of the constraint block does not bring it back either — it decides, at σ = √(|h|δ).

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.

0123401020304050z, the effective noise ratiotrials mirrored, %1 thin: 1 × 1 model2 thin: 2 × 2 model3 thin: 3 × 3 modelper cent mirrored at z = 0.7, 600 trialsn = 3, 1 thin, 20 points9.7n = 5, 1 thin, 20 points13n = 10, 1 thin, 40 points12n = 5, 2 thin, 20 points23n = 10, 2 thin, 40 points24n = 8, 3 thin, 40 points32points: measured in n dimensionslines: a k × k determinant, no n in it Orthogonality, measured

A mirror decided in the thin directions

In n dimensions the nearest orthogonal matrix to a noisy alignment is still sometimes a reflection, and the rate at which it is does not depend on n. Three, five and ten dimensions with one thin direction mirror alike; two thin directions mirror like each other in five dimensions and in ten. The rate is the chance that a k × k matrix built from the k thin directions has a negative determinant — 21.7 per cent at a noise ratio of 0.7 for k = 2, measured at 23.0 — and it is well above k independent coin flips. With two or more thin directions the determinant correction still fires, and it no longer rescues the rotation: the answer is eleven noise-widths off whether or not it was mirrored.

1357911131510⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1k, where h = 1 − 10⁻ᵏrelative error of 1 − hsubtractwith a correction stepcorrection kept as a pairfrom the reflectors40 × 6k = 12, subtracting1.1·10⁻⁴k = 12, with a correction1.1·10⁻⁴k = 12, from the reflectors3·10⁻¹⁵reflector operations900dashed: a unit of roundoff over the divisorthree routes sit on it and one does not Least squares, and the road not to take

The factor a sparse code keeps anyway

Every deletion diagnostic divides by one minus a leverage, and computing it as a subtraction loses a digit for every decade the leverage is from one. The route that does not subtract needs the orthogonal factor, which a sparse factorisation is supposed not to have. Three repairs that avoid it all fail at exactly a unit of roundoff over the divisor — and the fourth, which reaches the orthogonal factor through the Householder vectors a sparse code keeps in order to solve anything at all, returns the same bits as a stored factor in 900 operations.

n = 512, ε = 10⁻⁸operations over stored numbers2cheapest leaf, arithmetic alone16cheapest leaf at 0 a block16097194291388485leaf sizeoperations per unknown48163264the arithmetic aloneand the blocks it reacheslarge dot: the cheapest leaf at this chargethe lower curve is the storage curve doubled Neither sparse nor dense

A second objective that is the first one doubled

Counting the operations a hierarchical product does was supposed to give the leaf size a second optimum, somewhere other than where the storage puts it. The count is 160,128, 139,904, 135,936, 155,136 and 216,064 against stored totals of 80,064, 69,952, 67,968, 77,568 and 108,032 — twice each, at every leaf, exactly. The two objectives cannot disagree, and what separates the leaf that stores least from the leaf that runs fastest is a charge of about 139 operations for reaching a block at all.

falls through a half, % of the answerfast, truncated96exact, shifted105fast, shifted66the answer's own dimensionblur 2.5 points wide2300.250.50.7511.251.500.20.40.60.81the preconditioner's own dimension ÷ the answer'sshare of the run on the arcfast, truncatedexact, shiftedfast, shifteddashed: as much dimension as the answer carrieshorizontal: half the run on the arc Methods that were designed apart

The shift had an edge, and the approximation moved it

A fast-transform preconditioner made invertible by a shift was recorded as never reaching the unpreconditioned floor, and predicted to sit off the answer's path at every shift. At a large shift it sits on the path and reaches the floor to a tenth of a per cent. It has an edge like the truncated one — but on the exact operator that edge is where the shift's own effective dimension reaches the answer's, 1.02 to 1.05 of it on six problems, and on the fast approximation it arrives at 0.49 to 0.77. The difference is sixteen samples at the ends of the signal, where the approximation is wrong and a shift divides the error by α.

00.10.20.30.40.50.60.70.80.9110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹decay rate of the spectrumrelative spread of the rank-k errorthe seedthe machinetwo kinds of variationseeds drawn8partitionings7seed spread0.31machine spread9.8·10⁻¹⁶ratio3.1·10¹⁴one of these is recordedand it is the large one The answer that depends on the machine

The variation that comes with a seed

A randomised low-rank approximation's error moves by 31% between draws and by 10⁻¹⁵ between partitionings of one draw. In the one field whose answer already comes as a band, the machine is inside the width of the line — and it is still there.

a triangle and a square, joinedK₂,₃ with a pendant edge1234560123456indexeigenvaluethe same, and not the samevertices each6edges each7spanning trees12spectra differ by5.3·10⁻¹⁵highest degree, left4highest degree, right3one has a trianglethe other is bipartite The matrix that is a graph

The spectrum is not the graph

Two graphs on six vertices with the same Laplacian characteristic polynomial — as integer polynomials, not to fourteen digits. One contains a triangle; the other is bipartite. Every method in this field that reads only the spectrum is answering about the class.

10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²-1-0.500.51relative backward error acceptedfraction of the work not doneQR deflation criterionincomplete Cholesky droplow-rank truncationthree fields, no shared arithmeticslope, deflation criterion0.039slope, drop tolerance0.051slope, rank truncation0.23widest apart, as a ratio5.7a tolerance is an offerand the three offers are one offer Two errors, and whose fault they are

A tolerance is priced by the problem

Three tolerances from three fields sit on one pair of axes and agree to within a factor of 5.74. That factor is the ratio of the two curves that cannot move. Change the only problem in the comparison and the third curve's fitted slope swings from 0.188 to 0.040 while the printed spread does not shift by a digit.

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.

110¹10²10³110¹10²10³the accumulating quantity, relative to its first valuethe error, relative to its first valuethe bounds: slope 1what all three do: slope ½three mechanisms, one exponenta left-to-right sum0.49a chain of rotations0.55a residual recurrence0.51every bound's slope1spread of the three0.067a bound is a sum of the roundingsand the roundings have signs The arithmetic underneath

Three walks and one bound

A left-to-right sum, a chain of three thousand rotations and a conjugate gradient residual recurrence share no arithmetic and no vocabulary. Each has a standard bound that is linear in whatever it accumulates against. All three come out at a half — 0.486, 0.554 and 0.507 — and nothing is rescaled.

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 Sparsity, and what elimination costs

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.

each method's best, dimensionconjugate gradients24Tikhonov23truncated SVD21and the error thereconjugate gradients0.14Tikhonov0.14truncated SVD0.140510152025301effective dimensionrelative errorconjugate gradientsTikhonovtruncated SVDdashed vertical: the iteration's bestthe iteration drawn while its dimension still rises Methods that were designed apart

One arc, and what each filter pays to be on it

Conjugate gradients and Tikhonov stop at the same effective dimension, and that could have meant two curves crossing once or one curve. It is one curve over a stretch — on six problems, Tikhonov and truncation reach the iteration's error at the iteration's dimension to within 7.2 per cent from 0.7 of the answer to its top — and the two separate on either side. But the curve is shared by a trade, not by an identical answer: at the same dimension Tikhonov carries 22 to 42 per cent more noise than the iteration and up to five per cent less bias.

6 numbers, coefficients to 30relations, both agree34relations, they differ1accidents, both agree5accidents, they differ88510152025300123decimal digits the numbers are scaled togap, in digitsrelation, both agreerelation, they differaccident, both agreeaccident, they differhorizontal: a gap of one digitdashed vertical: the digits a double has Exact arithmetic, and what it costs instead

Two precisions guard the other edge

Run a lattice relation search at N digits and again at N/10, and accept its answer only if both runs return the same vector. Among six measured numbers, where the gap between the shortest and next vector vouches for one found relation in 35, the two runs agree on 34. Past a double's sixteen digits they never once agree on an accident, 0 of 561. They do agree on 62 accidents at fifteen digits or fewer — approximate relations that really are the shortest vector there — and the gap, which cannot see a relation among six numbers, can see those.

κ of the suma component of b − Ax8.09·10¹⁷ad − bc, near-degenerate3.6·10¹⁶qᵢᵀqⱼ, an orthogonality check3.26·10¹⁶zᵀAz, a trace probe95.7pᵀAp, a curvature73.7rᵀr, a residual norm1measured, not assumedhighest8.1·10¹⁷lowest1above 10¹⁰3terms128sums of squares are safeand nobody decides anything from one The answer that depends on the machine

Two machines, one certificate

Nothing a solver returns says which of its answers you got. Four things could be reported instead — the summation condition number, the partition count, an exactly accumulated residual and a directed-rounding interval — and each costs about one pass over data the routine already has in hand.

finite eigenvalues (degree of det Q)13at infinity (2n − degree)1at infinity, by the rank of M12n, if M were nonsingular14a degree, not a decisiondegree of det Q13at infinity1by the rank of M1singular-value gap∞the count is a degreeand the other route is a judgement The eigenvalue problem that is not linear

One mass removed, and one eigenvalue gone

A coordinate with no inertia reads like a coordinate that has been deleted, and a chain of eight masses with one of them removed would then be a chain of seven, with fourteen eigenvalues. It has fifteen. The massless coordinate is still there, still carrying a damper, and it contributes a first-order equation rather than none.

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.

sum with overshoot clippedwidest departure, 1.1 to 1.30.17widest departure, 0.3 to 0.50.690.30.50.70.91.11.30.7511.251.51.7522.252.5clipped sum ÷ the answer'serror ÷ the iteration'sTikhonovtruncated SVDshaded column: 0.7 to 1.0 of the answerthe axis decides the band above the top Methods that were designed apart

The overshoot was the lead

Past its best, conjugate gradients' error rose more slowly than Tikhonov's at the same effective dimension — on the narrow blur at 0.1% noise Tikhonov was 2.35 times worse at 1.3 of the answer — and the reading was that the iteration spends its dimension where the data has content. Count admitted directions instead of summing factors, so that a factor of 1.81 counts once, and the lead is gone: 0.96 on that problem, and within 0.17 of one on all six from 1.1 to 1.3. Tikhonov now carries less noise than the iteration there. Below half the answer, where the three filters also disagree, the count changes nothing.

7 tensors, 60 networksτ = 2, networks exact46constant 8, networks exact4210²0102030405060pairings pricednetworks exacttie-widened, τ 0 to 8constant width 1 to 8wide late, two ratesup and to the left is betterwidth spent where the ranking is uncertain Where the flop count stopped predicting the time

A near-tie is a factor of four

A beam over contraction orders spends its width by level — wide late was the best schedule, finding the exhaustive order on 37 of 60 seven-tensor networks for 78 pairings priced. The alternative was to widen only where the ranking is a near-tie. Measured, near-ties buy nothing: keeping every candidate within a quarter of the best changes nothing at all, because the pairing the exhaustive order wants is almost never tied with the ranking's first choice. It sits a factor of two to four further down. Widen to keep everything within a factor of three and the beam finds the exhaustive order on 46 networks for 150 pairings, and at nine tensors has a median of 1.007 where every schedule tried before had 1.056 or worse below 1,380 pairings.

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