Field

The arithmetic underneath

The representable numbers are spaced, and the spacing doubles at every power of two. Everything else follows from that: why subtracting two close numbers destroys the part you wanted, why the order of a sum changes its value, and why the mantissa is the one parameter on this site worth putting on a slider.
[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.

10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹xrelative error of the computed value(1 − cos x)/x², as written2 sin²(x/2)/x²no digits left at allbinary64 throughoutone function, two spellings · zero below 1.5·10⁻⁸

Cancellation takes the answer, not a digit

Subtracting two nearly equal numbers is exact. That is what makes it dangerous — the subtraction introduces no error at all, it exposes error the operands were already carrying, and the exposure can consume every significant figure at once.

10¹10²10³10⁴10⁵10⁶10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹number of terms addedrelative error against the exact sumin orderin a treecompensatedbinary32 · terms are 1/icompensated: 3·10⁻⁸

The order they are added in

Addition is associative in the algebra and is not associative in the arithmetic. The same million numbers, added in a different order, give answers that differ in the third significant figure — and the fix is not a wider float, it is a different order.

012345610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹refinement step‖x − x*‖ / ‖x*‖a full double-precision solveresidual in24-bitresidual indoubleone argument apartκ·u of the factorisation6·10⁻⁴double residual, final3.2·10⁻¹³same-precision, final1.3·10⁻⁴30×30, κ = 10⁴, same factors in both runsidentical cost

Buying the accuracy back

Factorise in single precision, then correct the answer using residuals computed in double, and the result is what a full double-precision solve would have given. Compute those residuals in single instead and the identical algorithm, at identical cost, recovers nothing.

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.

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.

-47-37-27-17-7313233301234log₁₀ of the vector's normfp1611 bitsbfloat168 bitstf3211 bitsbinary3224 bitspale: the format's range · blue: √(Σ(xᵢ/m)²)·m · red: √(Σxᵢ²)fp16 and tf32 have the same eleven significand bitsand their bars do not overlap

A norm that overflows before it is a norm

The vector of sixteen thousands has a Euclidean norm of 4,000, which fp16 represents exactly. Written as the square root of the sum of squares it returns infinity, because squaring doubles the exponent — and the expression costs half the format's range on the one computation every iterative method performs at every step.

10⁻⁷10⁻⁵10⁻³10⁻¹⁰10⁻⁸10⁻⁶magnitudespacing to the next numberthe smallest normalgradualflush to zerosmallest normal6.1·10⁻⁵smallest subnormal6·10⁻⁸octaves of subnormals10pairs that lie under FTZ10the spacing stops halving and stays putwhich is what makes x − y = 0 mean x = y

The numbers below the smallest one

Below the smallest normal number the spacing stops halving and stays put, all the way to zero. That is what gradual underflow is, and the thing it buys is the sentence every algorithm assumes without being told — x minus y is zero only when x equals y.

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.

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.

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.

half a stepthe 32 values of one block, in the order they arrivestep 0.0625one scale, thirty-two valuesoctaves inside the block1.9entries rounded to zero0worst error over its bound131 levels either side of zerothe largest entry chose the step

One exponent for thirty-two numbers

Share the exponent across a block and the cost per value drops from eight bits to 6.25, and the accuracy improves — up to about three octaves of spread inside a block. Past that a single outlier deletes the thirty-one values beside it, and the 2-norm barely notices.

entries rounded to zero, of 320block, as given310block, sorted by size22E4M3, either order0median entry's relative errorblock, as given1block, sorted0.00321E4M30.022the same numbers, three waysdeleted, as given310deleted, sorted22deleted, per-element08-bit significands, blocks of 32sorting is free and changes no value

A bit buys an octave

The outlier a block survives is exactly two raised to its significand width — 8 at three bits, 32 at five, 128 at seven, 512 at nine. Each extra bit doubles the range the block tolerates and halves the ordinary entry's error. Reordering the same numbers buys every octave at once and costs nothing.

10⁻⁹10⁻⁶10⁻³110³10⁶13212937κ · usignificand bitsκu = 1a bound was provedthe method refusedit never returns a wrong boundlargest κu with a proof0.45smallest κu without one0.89cases refused, of the grid13a refusal is not a wide bound — it is no bound at alland it is the only failure mode here

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.

11.522.511.52xyexactly one roota verdict, not a bound‖I − C F′(X)‖0.28width of X0.8width of K(X)0.23strictly inside is a proofand overlapping is nothing at all

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.

grey: proved empty · filled: proved to contain exactly one roota covered squareboxes proved empty42boxes proved unique2undecided0operator evaluations87every rectangle carries a proofand the two crosshairs are where the roots are

Where the box is cut

A branch-and-bound with an interval operator settles a whole square — two roots proved unique, forty-two regions proved empty, nothing left undecided, in 87 evaluations. Move the roots so one lands on the first bisection and it proves nothing at all, at any depth. Cutting at 0.485 instead of 0.5 finds both, in a quarter of the work.

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

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.

02468101201change of units, by exponent10⁰10¹10²10³10⁴10⁶10¹⁰10¹⁶10¹⁹10²⁰10⁴⁰10¹⁵⁰as writtenafter scalinga range questionlargest finite value3.4·10³⁸predicted boundary γ1.8·10¹⁹last γ that forms10¹⁹stops where scaling fails0not a poor answerno answer at all

The units that overflow before the answer does

A change of variable that is exact in the algebra requires γ² times a matrix to be a number the format can hold. In binary64 that is a bound nobody meets by accident. In binary32 it arrives at 10¹⁹ and in fp16 at 256, and past it there is no answer rather than a poor one.

036912151810¹10²10³rotationswidththe enclosurethe seta rotation is an isometrymeasured growth a step1.4√2, from the geometry1.4enclosure ÷ set after 201024no rounding error is responsible for any of thisa higher precision does not touch it

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.

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