Generator

Three exact eliminations of the same random matrix, and the widest number each one forms

One function in the integer library, called 40 times across 10 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 6 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws three exact eliminations of the same random matrix, and the widest number each one forms. Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 11 the answer itself is 33 bits and Hadamard's bound allows 47. Fraction-free elimination never forms a number wider than 35 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 35 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 341287 bits — 9750 times the width of the fraction-free route and 10300 times the width of the answer. All three are correct.

bit-swell is one function in lib/figures/integer.js — exact arithmetic — no residual to print, and a cost measured in the length of the numbers. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

Three exact eliminations of the same random matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 11 the answer itself is 33 bits and Hadamard's bound allows 47. Fraction-free elimination never forms a number wider than 35 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 35 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 341287 bits — 9750 times the width of the fraction-free route and 10300 times the width of the answer. All three are correct.34567891011110¹10²10³10⁴10⁵nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn11answer33Hadamard bound47fraction-free35reduced rationals35unreduced3.4·10⁵the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 11 the answer itself is 33 bits and Hadamard's bound allows 47. Fraction-free elimination never forms a number wider than 35 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 35 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 341287 bits — 9750 times the width of the fraction-free route and 10300 times the width of the answer. All three are correct.

family: "random", upTo: 11

The arguments are the ones A basis that describes its lattice badly passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Three exact eliminations of the same random matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 11 the answer itself is 33 bits and Hadamard's bound allows 47. Fraction-free elimination never forms a number wider than 35 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 35 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 341287 bits — 9750 times the width of the fraction-free route and 10300 times the width of the answer. All three are correct.34567891011110¹10²10³10⁴10⁵nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn11answer33Hadamard bound47fraction-free35reduced rationals35unreduced3.4·10⁵the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 11 the answer itself is 33 bits and Hadamard's bound allows 47. Fraction-free elimination never forms a number wider than 35 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 35 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 341287 bits — 9750 times the width of the fraction-free route and 10300 times the width of the answer. All three are correct.

family: "random", upTo: 10

The arguments are the ones A basis that describes its lattice badly passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Three exact eliminations of the same random matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 10 the answer itself is 37 bits and Hadamard's bound allows 42. Fraction-free elimination never forms a number wider than 37 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 36 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 30683 bits — 829 times the width of the fraction-free route and 829 times the width of the answer. All three are correct.345678910110¹10²10³10⁴nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn10answer37Hadamard bound42fraction-free37reduced rationals36unreduced3.1·10⁴the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 10 the answer itself is 37 bits and Hadamard's bound allows 42. Fraction-free elimination never forms a number wider than 37 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 36 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 30683 bits — 829 times the width of the fraction-free route and 829 times the width of the answer. All three are correct.

family: "wide", upTo: 9

The arguments are the ones A fraction recovered from one remainder passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Three exact eliminations of the same wide matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 9 the answer itself is 91 bits and Hadamard's bound allows 98. Fraction-free elimination never forms a number wider than 91 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 89 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 182755 bits — 2010 times the width of the fraction-free route and 2010 times the width of the answer. All three are correct.3456789110¹10²10³10⁴10⁵nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn9answer91Hadamard bound98fraction-free91reduced rationals89unreduced1.8·10⁵the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 9 the answer itself is 91 bits and Hadamard's bound allows 98. Fraction-free elimination never forms a number wider than 91 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 89 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 182755 bits — 2010 times the width of the fraction-free route and 2010 times the width of the answer. All three are correct.

family: "random", upTo: 9

The arguments are the ones A prime that divides the answer passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Three exact eliminations of the same random matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 9 the answer itself is 25 bits and Hadamard's bound allows 37. Fraction-free elimination never forms a number wider than 25 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 25 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 5947 bits — 238 times the width of the fraction-free route and 238 times the width of the answer. All three are correct.3456789110¹10²10³10⁴nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn9answer25Hadamard bound37fraction-free25reduced rationals25unreduced5947the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 9 the answer itself is 25 bits and Hadamard's bound allows 37. Fraction-free elimination never forms a number wider than 25 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 25 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 5947 bits — 238 times the width of the fraction-free route and 238 times the width of the answer. All three are correct.

family: "wide", upTo: 10

The arguments are the ones A prime that divides the answer passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

Three exact eliminations of the same wide matrix, and the widest number each one formsEvery route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 10 the answer itself is 103 bits and Hadamard's bound allows 110. Fraction-free elimination never forms a number wider than 103 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 103 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 754790 bits — 7330 times the width of the fraction-free route and 7330 times the width of the answer. All three are correct.345678910110¹10²10³10⁴10⁵10⁶nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn10answer103Hadamard bound110fraction-free103reduced rationals103unreduced7.5·10⁵the error is zero on every curvethe cost is the length of the numbers

Every route here returns the same determinant exactly, so there is no error to plot and the y-axis is the length of the intermediates in bits. At n = 10 the answer itself is 103 bits and Hadamard's bound allows 110. Fraction-free elimination never forms a number wider than 103 bits, because each of its intermediates is a minor of the original matrix. Rational elimination that reduces every fraction to lowest terms reaches 103 bits and pays a gcd on every arithmetic operation to stay there. Rational elimination that does not reduce reaches 754790 bits — 7330 times the width of the fraction-free route and 7330 times the width of the answer. All three are correct.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

6 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a family the library builds

a size the exact routes are run to

at least two sizes to draw

Bareiss's widest intermediate is inside Hadamard's bound on a minor

every Bareiss division is exact

fraction-free and rational elimination return the same integer determinant

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 214 of 397 generators — 194 print a residual and 20 are exempt with a published reason; 183 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

Exact arithmetic, and what it costs instead

A basis that describes its lattice badly

The same set of points has infinitely many bases, they are all correct, and they are not equally useful. One measurement separates them — the product of the vectors' lengths over the lattice determinant — and the determinant is the invariant the reduction may not change, which is what makes the reduction checkable.

Exact arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

A prime that divides the answer

A modular elimination reports a singular matrix and is telling the truth — over the field with p elements the matrix is singular. Over the rationals it is not. Nothing in the residue distinguishes the two cases, no quantity is small enough to be suspicious, and the wrong answer is a correct computation of a different question.

Exact arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

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 arithmetic, and what it costs instead

The rank depends on the ring

A floating-point rank is a decision about a threshold. Remove the arithmetic error entirely and the threshold goes away — and the answer still is not a property of the array of numbers, because one integer matrix has rank six over the rationals, five modulo three and four modulo two, with nothing rounded and nothing decided.

Exact arithmetic, and what it costs instead

What a determinant does not determine

Two integer matrices can have the same determinant, the same rank and the same size, and define genuinely different maps. What separates them is a list of integers each dividing the next — computed here twice, once by unimodular elimination and once from the gcds of every minor, which share no algorithm at all.

The whole library · All essays · What must fail