The thread: Whose fault is it — page 7
Half the conditions and a certificate
A two-sided interpolatory reduction of a stable convection–diffusion system came back unstable at four placements of nineteen. The one-sided reduction built from the same kind of solves came back stable at all 152 placements measured across eight Péclet numbers, and not by luck: every pole of a Galerkin model lies inside the numerical range of the operator, whose edge here is exactly the diffusion term's largest eigenvalue, −0.986 at Péclet 10. The price is the slope conditions, and spent as six one-sided shifts instead of three two-sided ones it beats the stable two-sided model at seven placements of ten.
Sparsity, and what elimination costsThe 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.
Methods that were designed apartA count that marks the edge and not the pace
The number of directions a truncated preconditioner divides is counted for free when it is built, and it was proposed as a stand-in for the stride it buys. On three blurs it is not one — the runs leave the answer at strides of 5.71, 3.44 and 2.41. What the count does predict is the edge: on all three blurs, at two noise levels, a run stops landing on the answer's path within five per cent of the point where the count reaches the answer's own effective dimension. And the halved cutoff rule, measured on one blur, crosses that line on the narrowest.
Iterating, instead of factorisingThe stencil that is not symmetric
Past a cell Péclet number of exactly one — measured by bisection at 1.0000000000000002 — the central-difference solution of a convection–diffusion problem oscillates from point to point and leaves the interval the equation guarantees, at 16 of 31 grid points. It is the exact solution of its own linear system, to 4.6·10⁻¹⁸. No solver was involved.
Two errors, and whose fault they areThe 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.
Where the flop count stopped predicting the timeWiden the beam where the ranking is right
A beam that is wide at its first pairings and narrow at its last looks like the right shape, since the early commitments are the damaging ones. Measured, it is worse than no beam: at nine tensors a width of 16, 8, 4, 2 and then 1 prices 742 pairings and has a median of 1.263 against width 1's 1.062 for 120. The reverse shape — width 1 early, doubling to 16 at the end — prices 78 pairings on seven tensors and finds the exhaustive order on 37 of 60 networks, against 30 for a constant width of 4 at 161. A beam should be wide where cost so far is nearly the whole cost.
Randomised, and the guarantee that changes kindA 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.
Reduction, and what a model is forA 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.
Sparsity, and what elimination costsHow 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.
Methods that were designed apartThe 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.
Regularisation, and the answer that is chosenA corner the penalty can afford
Every smooth reading of the deconvolution's grid needed about forty points and then stopped improving, and the step was the suspect. Give the step one coefficient of its own and forty-eight points reach an error of 0.0070 at 0.1% noise, against 0.118 for the best smooth reading on ninety-six — the step was most of the error. But the same step given two coefficients recovers half as well, and given a doubled node at each edge it recovers worse than no breakpoint at all, while representing the signal to 0.07%. What decides is what the penalty is charged for the corner, and whether the data can say where it is.
Exact arithmetic, and what it costs insteadThe room a relation has to stand out
A lattice search for an integer relation returns its shortest vector, and the proposal was to return the gap to the next one as well, so a caller could tell a relation from an accident. Measured, the gap is a certificate with a budget: the digits the numbers really have, shared among all but one of them, less the size of the relation. A found relation's gap sits half a digit under that budget, accidents stay near zero, and a one-digit gap vouches for 81 of 96 relations among three numbers and for 1 of 35 among six. The test numbers the proposal came from turned out to have relations of their own.
Structure, and the solver that cannot see itThe correction lost to its own two-by-two solve
Solving a band matrix through a circulant and a small correction was measured losing the answer to 10⁻⁴ where elimination kept it to 10⁻¹⁴, and a wrap that landed one sample on a zero was measured costing four orders more than one that landed a pair. Both measurements solved the two-by-two correction system by Cramer's rule. Solved with a row interchange, the corrected solve on the same matrix loses 2.9·10⁻¹⁰ — the cancellation, and nothing multiplied onto it — and the single landing costs what the pair costs. The loss was in the determinant.
When the index is a tupleThe rank a sweep can vouch for
To decide a tensor's rank, fit it at one rank after another and stop where something changes. Three things can be read off each rank's fits — the error, how nearly parallel the terms are, and whether fits from different starting points agree — and over six planted rank-three tensors at four noise levels they decide it 24, 12 and 15 times out of 24; the ratio of consecutive errors, which needs nothing, decides it 24 times. The error is right every time and has to be told the noise level. The collinearity is a coin toss. Agreement among starts is never wrong and, under noise, cannot decide on nine of the twenty-four — and the rank past the answer, where every decision is made, costs ten times the answer.
The answer that depends on the machineThe 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.
Two errors, and whose fault they areA 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.
The eigenvalue problem that is not linearAn estimate that does not move
The tropical roots are said to miss the bottom of a spectrum by a factor growing like n². The estimate does not get worse. It changes by four per cent between four masses and sixty-four while the modulus it names falls by a factor of a hundred and sixty, and an estimate that is constant in the variable the answer depends on is a different defect from an inaccurate one.
Neither sparse nor denseA geometry setting that is a second accuracy
The tolerance raises every rank and moves no block; the leaf moves every block and no rank. The constant that decides how far apart two clusters must be before their block may be compressed moves both — 592 blocks at mean rank 3.25 at one end and 94 at mean rank 8.94 at the other — and the storage falls the whole way while the error against the dense matrix rises by a factor of 3.5, at a tolerance that never changed. It has no best setting, and it is not a third knob on the storage.
Regularisation, and the answer that is chosenThe data count their dimensions, not the step's
Every grid in the deconvolution essays was chosen with the answer in hand, and so was every λ. From the data alone, the discrepancy principle's worst draw is within 16 per cent of the oracle on every grid from 16 points to 96; generalised cross-validation is better on the median draw and, on grids of thirty points and more, has draws thousands of times worse. And the data can say how many dimensions they carry — about 20, 25 and 29 at three noise levels, one number once the grid exceeds it — but not how many more the step needs: the grid that count chooses is 14 to 19 per cent worse than forty points at the lower two.
When the index is a tupleA fit that has an answer and cannot stop
Fit a noisy rank-three tensor with four terms and the prediction was that the spare term would find real structure in the noise and stop pairing off with the others. It does not: the largest cosine between two fitted terms has a median of 0.977 at 1% noise against 0.975 without noise. What changes is the solver. Without noise the four-term fit stops in fifty sweeps; with noise it never stops: its error keeps moving by a part in a million a sweep for six thousand sweeps, while on most tensors its terms stand still. The extra term takes exactly its share of the noise, and the error is settled by sweep 150.
Randomised, and the guarantee that changes kindThe miss a normal table already priced
A trace estimator that stops when its own standard error reaches a target misses the target on about a third of draws, and the essay that measured it read the loose targets as the worst calibrated. Over 400 draws the loose target is the better covered — 76% at 10% against 64% at 3% — because the warm-up stops most of its runs with probes to spare. Where the criterion decides, the misses are a normal distribution's: a margin of c on the standard error buys what a normal table says, 94.8% at 1.96, and costs c² in probes, 3.86 times.
Least squares, and the road not to takeThe weight the factor met first
The route to one minus a leverage through the orthogonal factor was said to lose a digit for every decade of the condition number, whatever else it does. Put a weight on one row and it does not. With the heavy row first, the complement keeps every digit at κ(A) = 2.5·10⁹ while both subtractions return nothing. With the same row last it loses digits as the row's scale grows. And two heavy rows that leave κ(A) at 3.1 still lose six digits when the light rows come first. The law was about the order the factor met the rows, and the condition number had been standing in for it.
When the problem arrives againThe answer the last window left
A sliding window that corrects its least-squares answer at every step could start each correction from the previous step's corrected answer instead of from a fresh solve: the two windows share all but one row. On a stream with any noise in it, that start is three orders worse. The window's exact answer moves by 0.79 of itself in one step at κ(A) = 3·10⁶, a fresh seminormal solve is wrong by only 1.8·10⁻⁴, and one correction contracts either start by the same factor — so the fresh start ends at 2.2·10⁻⁸ and the carried one at 3.7·10⁻⁵. Only on data that agree exactly does carrying win.
Reduction, and what a model is forThe inner product the mesh already computed
A Galerkin reduced model is certified stable when the operator's numerical range sits in the left half plane, and the certificate belongs to the coordinates. The coordinates that broke it before were a constructed rescaling. Grade a convection–diffusion mesh towards its outflow boundary layer — the grading anyone resolving the layer would choose — and nodal values break it too: the range's edge is past zero at a ratio of ten and reaches +8.9 at a thousand, and at Péclet 100 seven reduced models come back unstable. Weight the projection by the cell sizes the discretisation already computed and every model at every grading is stable, with the range's edge back at −0.098.