Generator

The determinant's length, and the bound that decides how many primes to use before any of them is chosen

One function in the integer library, called 32 times across 10 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 7 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 the determinant's length, and the bound that decides how many primes to use before any of them is chosen. Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 10 the random family needs 42 bits, which is 2 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

prime-budget 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.

Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 10 the random family needs 42 bits, which is 2 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

`upTo: 9`

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.

Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 9 the random family needs 37 bits, which is 2 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

`upTo: 11`

Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 11 the random family needs 47 bits, which is 2 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

`upTo: 10`

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.

`upTo: 12`

Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 12 the random family needs 52 bits, which is 3 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

`upTo: 8`

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.

Solid: Hadamard's bound on |det A|, the product of the row 2-norms, in bits. Dotted: the determinant's actual length. The bound is what a modular determinant budgets against — at n = 8 the random family needs 32 bits, which is 2 word-sized primes at 26 bits each, and that count is fixed before the first residue is computed. On a Sylvester Hadamard matrix of order 8 the bound is 13 bits and the determinant has 13, so the bound is attained rather than merely satisfied. On the unit triangular family the determinant is 1 at every size while the bound grows, which is the gap the budget pays for: a matrix can have a determinant of one and rows that are long.

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.

7 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 range of sizes

a Sylvester Hadamard order, a power of two

and is attained on a Hadamard matrix, so it is not loose by construction

enough primes for Hadamard's bound on this determinant

every Bareiss division is exact

Hadamard's bound holds on every family drawn

the assertion refuses a counterexample

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.

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