The thread: Measured, not assumed — page 17

Least squares, and the road not to take

A multiplier is a force

A third constraint nearly parallel to the first made the multipliers of a constrained fit rise in exact proportion to κ(B), which looked like the conditioning measured a second, dearer way. It was not. Give the third constraint a datum that asks for nothing new and, at the same κ(B) = 4.6·10¹², the multipliers are eighteen thousand times smaller; give it a strain δ and they are 0.0133 δ/ε², a force on a lever of length ε. What they measure is what the constraint asks. What they do not measure is the error of the best route, which sits at the same level whether the constraint asks for nothing or for a displacement of 3·10⁹.

Methods that were designed apart

A tail from Tikhonov and a corner from truncation

Below half the answer, conjugate gradients beat Tikhonov at a matched count of directions by up to 69 per cent, and the proposed measurement was the sharpness p of a roll-off between Tikhonov and truncation that matches the iteration there. Any p from 2 to 4 closes the gap to a tenth; truncation, the family's limit, reopens it to a third. But no p describes the iteration. Its filter has Tikhonov's slope exactly in its tail and a local sharpness of 2 to 5 on its shoulder, and a single fitted p is a compromise that drifts from 2.3 at the first step to 1.7 at the answer.

The matrix a constraint makes

An augmentation read in the smallest eigenvalue

An augmented Lagrangian preconditioner's least work sat at γ = 1, 10⁴ and 1 on three systems, and two of the three optima sat where the smallest generalised eigenvalue was about 0.06 — which suggested a rule written in ν rather than γ. On twenty-four systems whose norms are all one, the least-work γ spans seven decades and ν at the optimum spans a factor of thirty, so there is no one ν. There is still a rule: take the smallest γ at which ν reaches 0.03, and the median system pays 7 per cent over its least work and the worst 29, where the best single γ pays 61. Its price is digits — twelve times the best forward error on the median system, 4,561 times on the worst.

Where the flop count stopped predicting the time

Leaves cut to the edge on purpose

A recursive elimination's cheapest leaf is exactly the square root of M, less 2, columns wide, and the rule drawn from it was to set the base case there and let the halvings put the leaves at or below it. On thirty-two sizes from 96 to 127 columns, halving to that base case moves more words than the pure recursion on sixteen of them at 144 words of fast memory, because the halvings stop at six and seven columns, not ten. Cut every dimension at a multiple of the edge instead and every size keeps the saving: 0.79 to 0.86 of the pure recursion's words, against halving's 0.91 to 1.07, with the same arithmetic and the same pivots. What it cannot make full is the one leftover leaf, and a leftover of one column is where it loses.

Structure, and the solver that cannot see it

The staircase a separable kernel builds

A block-circulant preconditioner took 18, 21, 22, 21, 19 and 18 steps on a 10 × 10 Toeplitz-block-Toeplitz system as the correlation rose from 0.5 to 0.98 and the unpreconditioned count rose from 43 to 178 — the parameter that makes the problem hard was the one the solver did not notice. That kernel factorises. The isotropic kernel with the same correlation along each axis does not, and on it the preconditioned count climbs 17, 22, 27, 29, 33, 36, while the preconditioner buys a factor of 1.4 where it bought ten. The reason is the spectrum's shape: a product of two one-dimensional spectra is a staircase of ten treads, and a kernel that is not a product gives a ramp of thirty-eight.

Reduction, and what a model is for

A mode that rings is counted twice

Balanced truncation's bound, twice the sum of the discarded Hankel singular values, was attained at every order on the heat model — a formula for the error rather than a bound on it. On six lightly damped oscillators it is loose at every order but the last, by factors between 1.3 and 5, at every damping from 0.7 to 0.01. The reason is visible in the singular values: a mode that rings contributes two of them, nearly equal, and removing the mode costs about what removing one would. The last mode's bound over its error is 2.02, 2.00 as the damping falls to 0.03 and 0.01. The equality was a property of modes with one singular value each.

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.

Sparsity, and what elimination costs

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.

Regularisation, and the answer that is chosen

More samples take the floor and leave the dip

GCV's catastrophic misses on fine grids were blamed on squareness: on an n × n system the residual and n − t both reach zero as λ does, and their ratio can dip there. With more samples than unknowns neither reaches zero, and the prediction was that the dip would be gone by construction. Half of it is. The minimum at the floor of the scale, 21 draws in 240 on square systems, is gone at every ratio. The interior dip is not — 28, 20, 13, 8 and 4 draws at one to four samples per unknown — and at four per unknown one draw still misses the oracle by 877 times.

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.

When the index is a tuple

The rank that stops being typical

A random 2 × 2 × 2 tensor has rank two with probability π/4 and rank three otherwise, and the sentence has no analogue for matrices. It is the first of a family. An n × n × 2 tensor is a pencil of two slices, and it has rank n exactly when the pencil's eigenvalues are all real, n + 1 otherwise. Over draws, the share of rank n is 0.786 at n = 2, 0.500 at 3, 0.264 at 4, 0.039 at 6, 0.004 at 8 and none of 4,000 at 10 — falling like e^(−0.087n²) — because the mean number of real eigenvalues grows only like the square root of n, to Edelman, Kostlan and Shub's closed form within two per cent. Both ranks stay typical in theory; in practice the lower one disappears.

Orthogonality, measured

The residual the appended block cannot remove

Appending b as one more block made block modified Gram–Schmidt solve least squares as well as Householder, ten million times better than the same Q through Qᵀb at κ = 10⁸ — on problems with no residual. Give b a component outside the range and every stable route's error rises with it, while Qᵀb's, already at κ²u, does not move. The appended block's advantage then falls as one over the residual: 5,400 at a relative residual of 10⁻⁴, 54 at a per cent, none at one. It never falls behind Householder by more than a factor of four. What the residual decides is whether the extra block is worth its synchronisations, and the answer is yes up to a residual of about a per cent.

Elimination, and the swap

A trigger finer than the growth

Threshold pivoting cannot remove the shooting matrix's growth of 1.1·10⁴, because at τ = 1 it is partial pivoting. The proposal was a factorisation that notices growth on a step and searches as rook pivoting does after it, paying for the stricter search only where growth is made. But no step makes much: the largest one-step rise is exactly the growing mode's rate over one shooting interval, 1.284 at a step of 0.3, and 1.133 at 0.15. A trigger finer than that rate holds the growth at 2.00 for about two thirds of rook's extra search; one coarser never fires. And a trigger fine enough searches as rook on a quarter of an ordinary random matrix's steps.

The matrix a constraint makes

The right-hand side that hides the saddle

Refinement against an unregularised saddle-point matrix gives the second-order verdict the pivot signs cannot, from step 62 on the shallowest saddle — for one right-hand side. The verdict depends on how much of the growing direction the right-hand side contains, and it can contain none. Each decade removed delays the verdict by a fixed number of steps, 147 on the shallowest saddle, set by the growth against the minimum's slowest contraction rather than by the growth alone, which predicted 1,611. Rounding stops the hiding at about 2,700 steps — but by then refinement has solved the system to a residual of 2.6 times ten to the minus thirteen, and any stopping test has already accepted it. A random kick to the starting point costs nothing and finds it.

Exact arithmetic, and what it costs instead

What reading the next pivot buys

In a fraction-free elimination every pivot is paid for twice — it multiplies every entry of its own step and divides every entry of the next — and the rule that picks the smallest pivot entry left a median 20 per cent above the least arithmetic any row order reaches. A rule that charges two steps ahead brings the median matrix to within 2.5 per cent, and on one matrix of twenty-four costs 1.87 times the least, worse than the smallest pivot ever does. Charging three steps ahead finds the least of all 40,320 orders on thirteen matrices and is never more than 27 per cent above it. And none of it pays: choosing that way costs forty to two hundred and sixty times the elimination it chooses.

Iterating, instead of factorising

Exact along one axis

The tuned diffusion makes the answer exact at every node, and in two dimensions it holds at exactly one flow angle. Five degrees off the grid the relative error goes from 1.2·10⁻¹⁴ to 6.9, and by twenty degrees the scheme is worse than the upwinding it was built to improve on.

Eigenvalues, singular values, rank

The vector was what was wanted

Nobody who computes a matrix exponential wants the matrix. They want eᴬᵗb — one vector, the state of a system at a later time. Twenty matrix–vector products get it to sixteen digits on a hundred-by-hundred problem, without ever forming a hundred-by-hundred exponential, and the exponential that does get computed is twenty by twenty.

The eigenvalue problem that is not linear

Where a contour's budget should go

A contour method's ceiling is the number of probes times the number of moments, and the moments are free in solves while the probes are not. Four ways of reaching one ceiling come out four orders apart, the ordering is not monotone, and what separates the best two is not the usual draw but the unlucky one.

The answer that depends on the machine

The reading that never moves

Across thirty runs — five grids from 36 to 196 unknowns, six working precisions from 53 significand bits down to 8 — the residual conjugate gradients stops on stays between 1.10·10⁻¹³ and 9.95·10⁻¹³. Over the same thirty runs the error of the answer spans a factor of 2.39·10¹¹, and the step count more than doubles. The one number the run publishes is the only one that responds to neither axis.

When the problem arrives again

The step the two rows owe

A sliding least-squares window can start each step's correction from a fresh solve or from the answer it already has. The answer it has is three orders worse on noisy data, because the exact answer moves by most of itself in a step. The proposal was a start that moves too: the previous answer plus the change the entering and leaving rows imply, two triangular solves from the factor the window keeps. Its start lands exactly where one correction of the carried answer lands — the update is that correction, computed from two rows instead of twenty-four — and one correction after it ends 2.9 to 440 times below the fresh start at κ(A) = 3·10⁶, and level with the carried answer when the data agree exactly.

The arithmetic underneath

The well on the far side of the band

Gradual underflow was said to buy a predicate and not an answer, because a quantity that has decayed into the subnormal range is already lost. A quantity that passes through the band on its way somewhere else is not. The stationary distribution of a two-well chain, computed in half precision across a barrier whose top is two to the minus sixteen of the near well, keeps its far well's probability of 0.2454 to three digits with subnormals and returns exactly zero without them. The normwise backward error calls both answers exact.

Two errors, and whose fault they are

Balanced is not symmetric

The Sylvester–Kac matrix is made of small integers, so a double holds it exactly, and its eigenvalues are the integers from −(n − 1) to n − 1 in steps of two. Every digit an eigensolver loses on it is therefore the solver's own, and it loses them at exactly the rate first-order perturbation theory predicts: the median error is half the prediction across 1,568 eigenvalues. Balancing, the preprocessing libraries apply for this kind of matrix, divides every condition number by about sixty and leaves their growth untouched, and at order 112 the unbalanced solver returns eighteen complex eigenvalues for a spectrum of integers.

The matrix that is a graph

A finer invariant that hears less

The Laplacian spectrum fixes a graph's number of spanning trees and not the group those trees form, so the Smith normal form of the grounded Laplacian is a strictly finer integer invariant — and on the six-vertex pair the spectrum cannot separate, it does: ℤ₁₂ against ℤ₂ ⊕ ℤ₆. Over all 1,022 Laplacian-cospectral pairs of connected graphs on eight vertices it separates 435. The adjacency polynomial, a second spectrum rather than a finer one, separates all of them.

Neither sparse nor dense

The smaller cluster sets the rank

An oscillatory kernel block between two equal clusters needs a rank that grows without limit as they grow. Make the clusters unequal at the same separation ratio and the rank stops following the larger one: a target an eighth long against a source of four needs 8 columns where the square block of side four needs 33, and a target a third long against a source 130 wavelengths long needs 11. What decides the rank is the product of the two lengths over their distance — the Fresnel number, the count optics gives for the waves two apertures can exchange.