Concept

Unit roundoff — where it appears

The largest relative error a single rounding can introduce, which sets the accuracy every claim about a format is stated in. It is about 1.1·10⁻¹⁶ in double precision, and a tolerance set below it stops selecting anything and starts counting rounding error.

Named by 39 essays across 13 fields — each of them below, with the objects they name alongside it.

110¹10⁻¹¹10⁻⁸10⁻⁵pieces the vector was divided intodistance from the exact sum, relativethe published boundκ · uone vector, one algorithmdistinct answers21runs26spread, in ulps2.3·10⁷κ of the sum10⁸bound ÷ worst error2.6·10⁴nobody chose pand no answer is the answer

The same program, twice

One vector of 4,096 numbers, one summation algorithm, one precision, twenty-six runs — and twenty-one different answers. Nothing in the program chose between them, every one of them satisfies the textbook bound, and the exactly rounded answer is not among them.

machine · Reduction order
[½, 1)[1, 2)[2, 4)0.5124gap 0.125gap 0.25 — twice as wide8 values per octavespacing doubles at each power of two

What a float can hold

The representable numbers are not a fine fuzz spread evenly over the line. They are evenly spaced inside each power-of-two interval and twice as far apart in the next one up, and almost everything else in this subject is a consequence of that one fact.

arithmetic · Floating-point
00.250.50.75110⁻¹110¹10²10³fraction of the method's own rangerelative errorfloor 0.141truncation KTikhonov λCGLS steprandomised rankfour methods, one floortruncation K0.14Tikhonov λ0.14CGLS step0.14randomised rank0.14four knobs from four fieldsand one obstruction underneath them

Four knobs and one floor

A truncation, a Tikhonov parameter, a step count and a randomised rank, on one problem with an answer that is known. Their best errors are 0.1445, 0.1406, 0.1426 and 0.1449 — a spread of 3% across four methods that share no arithmetic.

combination · Parameter choice
10²10⁴10⁶10⁸10¹⁰10¹²10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹κ of the sumrelative sizethe boundκ · umeasured spreadtwo curves and one constantspread ÷ κu, low0.25spread ÷ κu, high0.26bound ÷ spread, low7932decades swept10the spread is computablethe bound cannot see the order

A bound every answer satisfies

The classical bound on a summation error is correct, it covers all twenty-six answers one vector produced, and it is 7,932 times larger than the difference between them. A statement true of every ordering cannot say which ordering you got.

machine · Reduction order
21018263442500102030405060708090100110120130140150160170180190200210220230240250260270280290300310320330340significand bitsiterationspreconditioner roundedarithmetic roundedno preconditionersame bits, different casualtyerror, 3-bit preconditioner8.8·10⁻¹³error, 3-bit arithmetic0.16‖A − LLᵀ‖/‖A‖ of the factor0.083a direction may be roundeda measurement may not

The part of a solver that may be rounded

A preconditioner computed and applied with a three-bit significand still returns thirteen correct digits — it costs seventeen extra iterations and nothing else. Round the working arithmetic instead and the step count barely moves while the answer loses exactly the digits the format dropped.

combination · Mixed-precision
1357911131510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1index kσₖ ÷ σ₁σ₁√utwo factorstheir productthe normal equations, againroutes agree to k =6product floors at9.7·10⁻¹⁰σ₁√u2.3·10⁻⁹κ(P)κ(Q)1.8·10³⁶√ of it1.3·10¹⁸do not form the productthe σ below the line are the bound

The product nobody had to form

The Hankel singular values are the square roots of the eigenvalues of PQ. Form that product and half of them stop existing, at a floor this site can predict from one number — and the fix is the one the least-squares field has had since its first essay, arriving in a place with no least-squares problem in it.

reduction · Balanced truncation
051210241536204810⁻¹⁷10⁻¹⁴terms consumederror accumulated so farone accumulator8 piecesthe steps are the partial sumspeak partial sum32the answer1.6·10⁻⁵error, one piece7.6·10⁻¹⁴error, 8 pieces2.4·10⁻¹⁴mean at p = 648.1·10⁻¹⁵the walk sets the sizeand nothing sets the value

Where the disagreement comes from

The error of a reduction is a walk whose step length is the spacing of the running total, not of the answer. That one sentence predicts the size of the disagreement to a factor of two, explains why dividing the work makes it smaller, and explains why the value cannot be predicted at all.

machine · Reduction order
10²10³10⁴10⁵10⁶10⁷10⁸10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²condition number‖QᵀQ − I‖one passsweeptwicetreewhat the second pass removesfitted slope, one pass2fitted slope, two passes0.96rounds, two passes6rounds, the sweep24one pass squares the condition numberand two do not

Doing it twice

Cholesky QR squares the condition number — a fitted slope of 1.95 in κ against the Householder sweep's 1.00. Run the identical routine a second time on the Q it returned and the slope is 0.93, the orthogonality is at or below the sweep's at every κ, and the price is one more all-reduce.

cost · Communication
distinct answersone accumulator303eight pieces72compensated119pre-rounded1exact1worst error 1.17·10⁻⁸worst error 4.61·10⁻⁹worst error 4.96·10⁻¹⁰worst error 4.91·10⁻⁶worst error 0bitwise, or not at allpermutations400one accumulator303pre-rounded1its error4.9·10⁻⁶compensated error5·10⁻¹⁰accuracy and agreement are different propertiesand the accurate one is not the agreed one

The sum that cannot be wrong

Snap every addend to a common multiple before adding, and every partial sum is exact — so the order stops mattering, by construction rather than by luck. Four hundred permutations return one value where an ordinary reduction returns three hundred and three.

machine · Reproducible summation
10¹10³10⁵10⁷10⁹10¹¹110¹10²10³10⁴condition number of the matrixgrowth factorbound 2^11partial pivotingCholeskyno pivot to gain fromCholesky growth, every κ1Cholesky interchanges0partial pivoting, worst9the bound, 2^112048both eliminations reach the same growthand only one of them had to swap to get there

A factorisation with nothing to pivot for

Cholesky's growth factor is not bounded by one. It is equal to one, at every size and every condition number, and the two-line reason is why the algorithm needs no pivoting at all — not "usually gets away without it". Its only failure is the square root of a non-positive number, which is exactly the test for definiteness, and in floating point that test moves with the precision.

elimination · Cholesky
10¹10²10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰110⁴10⁸10¹²10¹⁶condition number κ(A)× short of a double solvebf16fp16fp32bf16fp16fp32the reference is soundreference solve, worst backward error10⁻¹⁶bfloat16 threshold κ256fp32 threshold κ1.7·10⁷eight refinement steps, residual always in doubleflat at 1 means it reached double

Where the hardware went

bfloat16 carries eight mantissa bits, which puts its refinement threshold at a condition number of 256. That is not an exotic matrix. It is an ordinary one, and past it the method still improves the answer by a factor of four hundred while getting nowhere near a usable one.

arithmetic · Mixed-precision
3456789101100.250.50.751bits in the exponent fieldeach curve as a fraction of its own maximumbfloat16fp16rangeprecisionto 617 decadesto 3.9 digitswhat the split buysbfloat16: largest number3.4·10³⁸fp16: largest number6.6·10⁴a bit of exponent doubles the rangea bit of significand adds a third of a digit

The other half of a format

fp16 and tf32 have the same eleven significand bits and their largest numbers are 65,504 and 3.4·10³⁸. For two phases this site simulated the significand alone, so it was obliged to report them as the same format — which is a claim, and a false one.

arithmetic · Floating-point
012345610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹round‖AΩ‖ ⁄ ‖A‖ at that roundredrawn each roundhalf redrawnkept: the sketch is now the zero matrixthe randomness is spentkept, round 12.6kept, round 29·10⁻¹⁵redrawn, round 23.1kept, final error0.28redrawn, final error0.11a random matrix used twiceis not random the second time

The sketch that is spent

Every other object a sequence carries has a shelf life. A random sketch has one use. Deflate what its first round found and apply it again, and it returns the zero matrix — 9.0·10⁻¹⁵ where the first round saw 2.62 — because the input has been made orthogonal to the very draw the guarantee is over.

randomised · Sketching
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

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.

elimination · Determinant
10⁻³10⁻¹10¹10³01magnitude448NaN — no ∞0.0156 — the smallest normaldrawn from the format's own rulespositive finite values126largest finite value448worst round-trip error0the subnormals are the evenly spaced ticks at the lefteverything a byte can be

Eight bits, and a format that breaks the rules

E4M3 reuses the exponent code IEEE reserves for infinities, so it reaches 448 where the same bits under IEEE's rules would reach 240 — and has no infinity left to signal an overflow with. The same computation is a NaN on one conforming device and 448 on another.

arithmetic · Floating-point
01234510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1correction steprelative errorLU route: η = 2.2·10⁻¹⁷LU route: forward 2.8·10⁻⁴forward errorbackward errorwhat a correction buysη before refinement4.5·10⁻⁵η after four steps2.8·10⁻¹⁷forward, unchanged7.1·10⁻⁴cost of a step, flops1800the residual is repairableand the accuracy floor is the problem's

The gap refinement can close

Multiplying by a computed inverse is not backward stable, and refinement at the working precision repairs it. That much is settled. The claim beside it — that the forward error does not move — was read at one conditioning and four corrections too late. Swept over ten, it moves at every one, and it lands on the LU route's own number after a single correction.

elimination · Inversion
10²10³10⁴10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴terms summedrelative error+∞, −∞ 1.01zero 1.00stochastic 0.50nearest 0.47√n against n, fittednearest, fitted exponent0.47stochastic, fitted exponent0.5toward +∞, fitted exponent1twelve seeds averaged at each sizethe slope is the bias, not the precision

The direction the error leans

The size of one rounding error is set by the precision. How ten thousand of them combine is set by something else entirely — the rounding mode — and the fitted exponents are 0.47 for round-to-nearest and 1.01 for round-toward-infinity, on identical data at identical precision.

arithmetic · Rounding
02505007501000250275300325350375additionsrunning totalround to nearest: nothing arrivesexactstochasticnearesta thousand additionshalf an ulp at 2561moves, round to nearest0moves, stochastic46relative error, nearest0.28relative error, stochastic0.0228 significand bits, unbounded exponenta flat line is not a small error

A coin flip that fixes the average

Add 0.1 to 256 a thousand times at eight significand bits and the answer is 256. Not approximately — the total never moves, not once, and no error bound says so. Round up one time in twenty instead of never, and it arrives at 348 against a true 356.

arithmetic · Rounding
condition numbers, on a logarithmic scaleκ(P)5.66·10¹⁸κ(Q)5.71·10¹⁸√(κ(P)κ(Q)) — the SVD route5.68·10¹⁸κ(P)κ(Q) — the product route3.23·10³⁷1/u4.5·10¹⁵the mean, or the productand only one of them fits

A condition number that is not the model's

κ(P)κ(Q) is quoted as the reason one route to the Hankel singular values fails, and it is. It is also read as a measure of how reducible a model is, and it is not — the norm of the Gramian does not move at all as the McMillan degree runs from four to fourteen, and the condition number wanders over a factor of thirty-five with no trend.

reduction · Gramian conditioning
1357911131510⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1k, where h = 1 − 10⁻ᵏrelative error of 1 − h1 − ‖R⁻ᵀa‖²1 − ‖row of Q₁‖²‖row of Q₂‖²40 × 6k = 12, Cholesky route1.1·10⁻⁴k = 12, thin QR route4.4·10⁻⁴complement, worst k3·10⁻¹⁵one minus a sum of squaresor the sum of the other squares

One minus a leverage is a subtraction

Every deletion diagnostic divides by 1 − h, and computing it as one minus a computed leverage loses digits in proportion to 1/(1 − h), however accurate the leverage. The complementary block of a QR factor gives the same number as a sum of squares and loses κ(A)·u instead: every digit on a well-conditioned design, and half the digits the subtraction loses on a design whose far point is what made 1 − h small.

leastsquares · Leverage
4914192429343900.20.40.60.81decimal digits of the numbers the lattice is givenshare of trials recovering the relationa double has about this many digitsnumbers that are exactthe same numbers, measureda window with two edgescoefficients up to30exact, at 12 digits1exact, at 401doubles, at 121doubles, at 250doubles, at 400.13one edge is the relationand the other is the data

A relation among digits that were not there

A lattice finds the exact integer combination of several numbers that vanishes, and it needs enough digits of them to do it. Give it more digits than the numbers have and it finds a relation anyway — among the rounding, with the same confidence and no warning. Recovery runs 8 of 8 at twelve digits and 1 of 8 at forty.

exact · Lattice reduction
-12-10-8-610⁻¹³10⁻¹¹10⁻⁹10⁻⁷log₁₀ of the residual tolerance asked forforward error of the answer1.34×1.48×1.71×1.17×the band does not closetolerances swept4runs at each7accuracy gained1.5·10⁶ratio at 10⁻⁶1.3ratio at 10⁻¹²1.2the bars falland they keep their height

The tolerance that buys no agreement

Ask for four more orders of accuracy and you get them — the answers improve by a factor of 1.5 million. The ratio between the best and the worst run is 1.34, 1.48, 1.71 and 1.17 across the same sweep. The band falls and it does not close.

machine · Stopping test
38434853586368110¹10²10³10⁴e, where the skew is 2ᵉ + 12345orthogonality defect of the basis returneda double holds k exactlyan orthogonal basis — and the determinant, unchanged throughoutthe invariant sees none of itfloat, at 2^521float, at 2^531.4float, at 2^701.2·10⁴residue at 2^701.2·10⁴exact, everywhere1determinant, always1every step was unimodularand the answer is 12,345 times worse

The knob and the rounding

The Lovász parameter is the number a lattice reduction is specified by, and moving it from 0.50 to 0.99 strengthens the proved bound from 16.00 to 1.83, costs 91 per cent more steps, and returns a basis with the same orthogonality defect. The one rounding nobody writes down decides everything: above 2⁵³ the reduction returns a basis 12,345 times worse than it should, with the determinant invariant equal to one throughout.

exact · Lattice reduction
8×8, κ = 10⁸direction at ε = 10⁻², ‖QᵀQ − I‖1.5·10⁻¹⁵scalar at ε = 10⁻², ‖QᵀQ − I‖0.065direction at ε = 10⁻², residual0.00510⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²relative perturbation εdeparture‖QᵀQ − I‖, scalar‖A − QR‖/‖A‖, scalar‖A − QR‖/‖A‖, direction‖QᵀQ − I‖, directionflat line: a different reflection is still a reflectionsloped line: a non-reflection is not

One number that has to be right

Householder's orthogonality was called structural: a reflection is built from a unit vector, so rounding the vector names a different reflection rather than a broken one. Tested by breaking it, the claim is narrower and sharper. Perturb every component of the reflector by a relative 10⁻², and ‖QᵀQ − I‖ stays at 1.5·10⁻¹⁵ while the factorisation moves to 5·10⁻³. Perturb the one stored scalar by the same amount and ‖QᵀQ − I‖ is 6.5·10⁻². The structure is one degree of freedom, and the departure is four times its relative error.

orthogonality · Householder

A bound that is proved

Every error statement on this site so far is a measurement of one run. Interval arithmetic makes a different kind of claim — the answer lies in this set, for this input, with no probability attached — and its failure mode is that it returns nothing at all. On a Hilbert system it proves a bound 23 times the error it bounds, and one size later it refuses.

arithmetic · Interval

A triangle where the scalar was

Every level-3 QR assembles a block of reflectors into Q = I − Y T Yᵀ, and T is computed by a recurrence whose inputs are its own previous columns. A block of sixteen carries 136 computed numbers where sixteen separate reflections carry sixteen. The orthogonality it produces is 3.9·10⁻¹⁵ against the single reflector's 7.8·10⁻¹⁶ — a factor of five for a hundred and thirty-six times as many things that have to be right.

orthogonality · Householder

Proving the answer is in the box

Every other method here computes a number and estimates how wrong it is. This one returns a verdict: there is exactly one solution in this box, or there is none, or — the honest third outcome — nothing can be said. Two of the three are proofs about infinitely many points from finitely many operations.

arithmetic · Interval

Three errors and one number

This site's identity has two factors and a division of blame between them. Two fields have now added a third party and a fourth, and only one of the four is a property of anything — the others are decisions, made before the arithmetic, reported by nothing.

error · Backward error

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.

leastsquares · Leverage

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.

arithmetic · Summation

The 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.

leastsquares · Leverage

One 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.

iterative · Krylov

The licence is not the boundary

Cholesky QR is licensed by κ²u ≪ 1, which reaches equality at κ = 9.5·10⁷ in double precision. At 10⁸ the factor it returns is already 0.37 away from orthogonal, and it goes on returning factors as far as 10¹³ — refusing at scattered condition numbers in between, at different ones for eight columns and for six.

machine · Algorithm selection

Nine steps of pessimism

A proved bound is 8 to 26 times the error it bounds, at every precision from 16 to 40 significand bits. A carried interval is (√2)ᵐ times too wide after m re-enclosures. The two cross between eight and nine, so the method everybody warns against is the tighter of the two for a short computation.

arithmetic · Interval

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.

machine · Stopping test

The residual the method reports

Conjugate gradients prints a relative residual of 6.9·10⁻²¹. The unit roundoff is 1.1·10⁻¹⁶, so that is not a small residual and not a large one — it is not a residual. The vector the method is holding at that step has ‖b − Ax‖/‖b‖ = 5.1·10⁻¹⁰, and nothing in the run says so.

iterative · Residual gap

A threshold the matrix does not set

Two numbers come out of a relative-accuracy comparison and they belong to different things. The size of the matrix moves the constant of the routes that never fail, by a factor of 2.7 between n = 4 and n = 10; it does not move the point where the route through BᵀB stops returning an answer, which sits between ten and eleven decades of grading at every size drawn.

spectra · Relative accuracy

The error the method already knows

Summing the exponential's Taylor series throws away a known number of digits, and the number is on the machine while the sum is being formed. The largest term divided by the answer, times the unit roundoff, tracks the relative error that comes out — to within a factor of nine, across fourteen orders of magnitude of it — and nothing reports it.

spectra · Matrix function

A walk needs a length

The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.

iterative · Residual gap

Named alongside it

The objects these essays reach for when they reach for this one.

Condition numberBackward errorError accumulationForward errorResidualCatastrophic cancellationConjugate gradientsExact ground truthNormal equationsHalf-precisionHouseholder reflectionQR factorisation

All concepts