Concept

Bit length — where it appears

How many binary digits an integer occupies, which in exact arithmetic is the unit every cost is measured in. An operation on a forty-thousand-bit integer is not one operation, so a count of arithmetic operations does not price an exact algorithm and a count weighted by operand length does.

Named by 12 essays across one field — each of them below, with the objects they name alongside it.

3456789101112110¹10²10³10⁴10⁵10⁶nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn12answer40Hadamard bound52fraction-free40reduced rationals39unreduced1.4·10⁶the error is zero on every curvethe cost is the length of the numbers

An answer with no error in it

An integer matrix eliminated over the rationals rounds nothing, so the forward error is zero, the residual is the zero vector, and the identity this site is built on has no terms left. The cost does not vanish with the error. It moves into the length of the numbers, where three correct routes differ by four orders of magnitude.

exact · Exact cost
1-2-6-7-474-26-12-5-8-524-394151435A, the matrix as given1-2-6-7-401840552700112217207003943011710017279153after 2 fraction-free steps3 × 3 minors25 divisions, all exactthe intermediates are minorswhich is why the divisions come out whole

Every intermediate is a minor

Fraction-free elimination divides by the previous pivot at every step and the division is always exact. Not usually, not for these entries — always, because the number being divided is a determinant with that pivot as a factor, which is a theorem and is checked here against the minors themselves.

exact · Fraction-free
3456789101110²10³nbitsthe answerthe question · det Abits in, bits outn11question484answer729det A33widest part37the answer is the floorand no route writes it down more cheaply

The answer is longer than the question

An exact solution of an integer system is a vector of fractions, each of them a ratio of two determinants. So the output carries 2n long integers where the input carried n² short ones, and no algorithm can write it down more cheaply — the length of the answer is a floor under every exact solver rather than a property of one.

exact · Exact cost
34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

How many primes the answer needs

Work modulo a word-sized prime and no intermediate can exceed twenty-six bits, whatever the matrix does. The catch is that the answer must be reassembled from several such computations, and the number of them has to be fixed before the first one runs — by a theorem about how large a determinant can be, not by trying more until it settles.

exact · Modular lift
2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

A fraction recovered from one remainder

A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

exact · Modular lift
10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110⁻⁷10⁻⁵10⁻³10⁻¹10¹relative error in the datarelative error in the answerthe answer's errorthe data's errornothing here is the arithmetic'sdata error10⁻¹²answer error10·10⁻⁴amplification10·10⁸rounding committed0residual0the residual is exactly zeroand the answer is wrong anyway

An exact answer to a measured problem

The residual is the zero vector, nothing was rounded at any step, and the answer is wrong in its first digit. Data accurate to fourteen places, an exact solve of the system it defines, and an error of 10⁻⁵ — because conditioning was never a statement about arithmetic and removing the arithmetic error removes none of it.

exact · Exact cost
10×10, ±6determinant, bits27Hadamard's bound36median slack10024680510152025303540elimination stepbitsHadamard, on the minorsthe intermediatesevery intermediate is a minor of the originalso a bound on the minors bounds the run

A bound on every intermediate at once

Fraction-free elimination's intermediates are minors of the original, which is a theorem about exactness. It is also a bound: Hadamard's inequality applies to every minor, so one inequality bounds the whole run before it starts. The bound on the k-th step is the one on (k+1)×(k+1) minors, not the one on the whole matrix — and on a 10×10 with entries in ±6 the difference is ten bits, with the run reaching 2.7 bits a step against the bound's 3.4.

exact · Fraction-free
20 matricespeak, every rule29the determinant's own length29area, smallest ÷ natural0.9602468051015202530elimination stepbits, longest entrythe determinant: 29 bitsswap only on a zerolargest pivotsmallest nonzero pivotthree orders, three sequences of minorsand one last entry they all arrive at

Three orders and one last entry

Over the integers there is no stability to pivot for, so a fraction-free elimination swaps rows only when the pivot is zero. Choosing a pivot for length instead does change the sequence of minors — the smallest-nonzero rule makes seven exchanges where the natural order makes none and keeps the profile two bits lower through the middle. It cannot change the peak. The last entry of the elimination is the determinant, and the determinant does not know what order it was computed in.

exact · Fraction-free
largest pivot, area1.037largest pivot, cost1.293smallest pivot, area0.969smallest pivot, cost0.704shortest row, original, area0.966shortest row, original, cost0.841shortest row, current, area0.964shortest row, current, cost0.760over the natural order: below one is cheaperthe rules agree on area and part on cost

The 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.

exact · Fraction-free
1.001.251.501.752.002.252.50cost ÷ the least any row order reachesnatural order1.733 · worst 2.41 · least on 0smallest pivot1.199 · worst 1.61 · least on 0cheapest step1.187 · worst 1.57 · least on 1two pivots ahead1.025 · worst 1.87 · least on 8three pivots ahead1.000 · worst 1.27 · least on 13smallest, then next1.217 · worst 1.89 · least on 0two ahead, estimated1.160 · worst 1.87 · least on 2bar: median · whisker: worst of twenty-fourthree pivots ahead finds the least on thirteen

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.

exact · Fraction-free
1.01.21.41.61.82.0median member's cost over the least any row order reachesindependent entriesa shared zero patternshared row scalesone matrix, two entries redrawnnatural order1.60the median order1.72another member's best1.54best over the other eleven1.46rows by their norms1.49smallest pivot entry1.24natural order1.65the median order1.72another member's best1.59best over the other eleven1.36rows by their norms1.35smallest pivot entry1.08natural order1.88the median order1.86another member's best1.13best over the other eleven1.16rows by their norms1.05smallest pivot entry1.01natural order1.28the median order1.51another member's best1.16best over the other eleven1.07rows by their norms1.42smallest pivot entry1.14dashed: the least cost of all 40,320 ordersthe rule wins wherever the rows show why

An order found once knows what the rows know

Searching for a fraction-free elimination's least-cost row order costs forty to two hundred and sixty times the elimination, so it can only pay if one order serves a whole family. The prediction was that it cannot: the best order depends on which rows happen to have short entries, so a reused order should do no better than the natural one. On independent entries and on a shared zero pattern that holds — another member's best order is an ordinary order, at the middle of the 40,320. On a family that shares its row scales it fails completely: another member's order costs a median 1.13 of the least where the natural order costs 1.88. But on every family, including that one, the rule that just takes the smallest pivot entry is cheaper than any reused or trained order — 1.24, 1.08 and 1.01. What an order carries from one matrix to the next is what the family shares — and only when that is a particular matrix, not a visible structure, does an order trained on the family beat the rule.

exact · Fraction-free
10³10⁵10⁷10⁹10¹¹size nbit operations48163264modular Hermitemodular Smithplain Hermiteplain Smithschoolbook bit operations, independent of the machinebounded is not cheaper

The swell was one-sided

The Smith form can be computed modulo the determinant, and it was predicted to return the right invariant factors, to stay within twice the determinant's width, and to cost about what the modular Hermite form costs instead of the far larger count of the plain Smith routine. The first three hold. The comparison does not: the plain Smith routine never needed a far larger count. With column operations as well as row operations, its smallest-entry pivot is a unit after a step or two, every entry it holds is a minor of the matrix, and nothing it writes is wider than the determinant — at every size to sixty-four, at a fiftieth of the modular route's work. The swell the Hermite form suffered belonged to doing elimination with one hand.

exact · Normal forms

Named alongside it

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

Exact arithmeticFraction-free eliminationDeterminantHadamard boundExact ground truthMinorSylvesters identityModular arithmeticPermutationBackward errorChinese remainder theoremCondition number

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