The thread: Identical algebra, different arithmetic — page 4
An inner product with no fixed sign
‖QᵀQ − I‖ is how this site turns "orthogonal" into a number, and across ten partitionings it moves by 2.4%. The entries it is built from are not so lucky: 125 of the 1,128 off-diagonal pairs take both signs, and one of them takes five different values including zero.
Least squares, and the road not to takeThe 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.
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.
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.
Orthogonality, measuredWhat the appended block inherits
Modified Gram–Schmidt on [A b] solves least squares as well as Householder, although its Q is not orthogonal. A block code appends b as one more block. Block modified Gram–Schmidt inherits the rescue at every placement of the ill-conditioning: at κ = 10⁸ the appended block gives 6.9·10⁻¹⁰ where the same Q through Qᵀb gives 8.9·10⁻³. Block classical Gram–Schmidt gets the same wrong answer both ways, to the last bit. And the variant whose Q is orthogonal to 10⁻¹⁵ — two passes with Cholesky QR inside — is a hundred thousand times worse than Householder when the ill-conditioning is inside the blocks, because its R is wrong.
Where the flop count stopped predicting the timeThe leaf that sits on the edge
A recursive elimination's base case was found to be a block size in disguise, with a cliff where the blocked elimination's is, and the choice read as a trade: the processor wants wide leaves, the cache wants narrow ones. Measured at every width rather than at the halvings of 96, there is no trade inside the edge. Leaves exactly √M − 2 wide are the cheapest the recursion can have in words as well as calls — 0.81, 0.80 and 0.84 of the pure recursion's traffic at 64, 144 and 256 words — and one column wider moves 1.77 to 3.15 times it. And a matrix of 100 columns, which halves unevenly, meets the cliff in two steps rather than one.
Structure, and the solver that cannot see itA zero no twist can step around
Solved through its circulant wrap with the small capacitance system factorised stably, a banded matrix was found to lose only what the cancellation in the correction costs, and a twist of half a step kept that cancellation near one. That holds only while the cancellation is large. With it kept small, the corrected solve's error is a quarter of the wrap's condition number times u on every case measured — and on a symbol with a zero of fourth order, the square of the Laplacian, no twist can keep the wrap well conditioned, because the nearest sample any twist can reach sees (π/n)⁴. At n = 64 the corrected solve is 34 times less accurate than elimination, where the Laplacian's is four.
Exact arithmetic, and what it costs insteadThe pivot is in every product
A fraction-free elimination's intermediates are minors, and Hadamard bounds a minor by the lengths of the rows it is made of — so the rule that picks the smallest pivot entry looked like a proxy for a rule that picks the shortest row. Measured, the two keep the bit-length profile equally low: 0.969 and 0.966 of the natural order's area. They part on what the arithmetic costs. Counting every multiplication and division at the product of its operands' lengths, the smallest-pivot rule costs 0.70 of the natural order and the shortest-row rule 0.84, because the pivot multiplies every entry of the step and divides every entry of the next. Three per cent of area is thirty per cent of arithmetic.
Reduction, and what a model is forThe smaller cell downstream
Nodal values on a convection–diffusion mesh graded towards its outflow layer put the numerical range in the right half plane, and Galerkin reduced models then fail. A code that grades towards the layer usually also upwinds, and upwinding adds diffusion. It does not give the certificate back: at Péclet 100 and a grading of a thousand the range's edge falls from 8.94 to 6.78 and stays positive, and 18 reduced models fail where 23 did. Each row of the upwind stencil's symmetric part sums to half the difference between the inverse widths of the cell after the node and the cell before it, so wherever the cells shrink downstream the dissipation upwinding adds is outweighed by the coupling to the smaller cell next door.
Structure, and the solver that cannot see itOne step past the zero
The squared Laplacian, solved through its twisted circulant and a rank-four correction, lost 6 to 34 times more accuracy than elimination, because its symbol's fourth-order zero keeps the wrap as ill conditioned as the matrix whatever the twist. One step of iterative refinement — a residual formed with the band itself, a second solve by the same route — takes it to 0.59, 0.68, 0.73 and 1.40 times elimination's error at n = 16, 32, 64 and 128, and the second and third steps go no lower. The loss the zero imposed was the solver's, and a solver's loss is what refinement removes.
Iterating, instead of factorisingOne sequence and two recurrences
CGLS and LSQR compute the same iterates — the minimiser over a space is unique, so there is nothing to choose between them in the algebra. At κ = 10⁶ they cost 42 steps and 47. At κ = 10¹⁰ they cost 110 and 209, across four seeds, and the quantity that separates them is the orthogonality of a basis neither of them keeps.
Eigenvalues, singular values, rankThe 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.
When the problem arrives againA fit wins where the steps were few
The line through the last two answers of a sequence of solves amplifies their stored error by √5, and the parabola through the last three by √19. A least-squares line through five amplifies it by 1.05 and a parabola through six by 1.79, and on a straight path both beat their interpolants at every tolerance: 12 inner steps against 17 at 10⁻¹⁴. On a bent path they lose, by exactly the ratio of their truncation constants, and they lose in the runs that cost a hundred steps rather than ten. Over thirty runs no fitted start beats the parabola through three, and the best of eight starts chosen per run saves 58 steps out of 1,212.
Where the flop count stopped predicting the timeLeaves 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.
Sparsity, and what elimination costsAn 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.
Orthogonality, measuredThe 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.
The answer that depends on the machineThe 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.
The arithmetic underneathThe 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.
Elimination, and the swapWhere the multipliers go
Bunch–Kaufman bounds the growth in D and not the entries of L, and the warning attached to that is that everything which later uses the factors inherits the size of L. On a matrix built to make those entries 1.2 over ε, they reach 1.2·10¹⁰ while the solve's backward error stays at 1.6·10⁻¹⁶, |L||D||Lᵀ| stays at 6.7 times ‖A‖, and a step of refinement changes nothing. The large multipliers are where the rule has put the matrix's ill-conditioning. A direction of negative curvature read from those factors finds 7·10⁻¹⁶ of the curvature that is there; the bounded rule's finds 13%.
Orthogonality, measuredThe factor nobody forms
A blocked Householder factorisation's orthogonal factor, multiplied out, departs from orthogonality half as far in blocks of sixteen as one reflector at a time, and that was read as blocking buying a factor of two. Libraries do not multiply it out. Applied to vectors through its stored blocks — which is how every caller uses it — the same factor departs by 3.1 to 4.0·10⁻¹⁵ at every block size from one to sixty-four, and stops growing after about twenty reflectors instead of adding them up. The factor of two was the price of forming the product, and a factor that is never formed never pays it.
Sparsity, and what elimination costsPieces ordered blind
One level of nested dissection followed by minimum degree lengthened the factorisation's critical path, and the blame moved from the separator to the halves: minimum degree's path through a 12 × 24 half was longer than through the whole 24 × 24 grid. A 12 × 24 grid on its own has a path of 12,913, a third of the square's. What made the half expensive is that it was ordered as if the separator were not there. Count the separator's vertices in every degree and number them last, and the depth-one path falls from 53,508 to 41,050 — minimum degree's own — while the pieces' own arithmetic does not change at all. At depth two the same ordering beats full dissection.
Where the flop count stopped predicting the timeThe short leaf belongs at the bottom
A recursive LU that cuts every dimension at a multiple of the cache's edge moves 14 to 21 per cent fewer words than the pure recursion, except where one leaf per dimension has to be short — and the one-column leftovers cost the most. Moving the short leaf to the top, where it becomes a thin strip of a large product, was predicted to cure that. It makes it worse on all twenty-nine sizes that have a leftover, by up to three hundredths. The measurement that matters more is the edge itself: one column under it the aligned cut keeps most of its saving, one column over it every size costs two and a half to three times the pure recursion. Halving to the same base case barely notices either mistake.
Eigenvalues, singular values, rankTwo matrices and one problem
Ax = λBx is what a finite element model, a structural vibration and a constrained optimisation actually produce, and it is not the one-matrix problem with a change of variables. Everybody is told not to form B⁻¹A because it is not symmetric. That is true, the departure from symmetry is about one, and it is not what decides the accuracy.
Structure, and the solver that cannot see itThe symbol that builds the staircase
A block-circulant preconditioner left the isotropic kernel's spectrum a ramp of thirty-eight clusters where the separable kernel's was a staircase of ten, and the proposal was a circulant sampled from the isotropic kernel's own symbol, to gather the ramp into treads of its own. It gathers the separable kernel instead: six clusters at every grid from 4 to 10 and every correlation, sixty-four of a hundred eigenvalues at exactly one, and eight conjugate-gradient steps at every side from 6 to 16 while Chan's circulant climbs from 18 to 28. On the isotropic kernel it shortens the ramp from 38 clusters to 30 and the step count by about a sixth, and both still grow with the grid. The treads come from a Kronecker product, not from a small difference: on both kernels the matrix and its symbol circulant differ in dozens of directions.