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 6 times across 1 essay. 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.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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.345678910110¹10²nbitsthe budget and what it buysrandom, bound42random, actual37primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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: 11

The arguments are the ones How many primes the answer needs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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.34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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: 5

The arguments are the ones How many primes the answer needs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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 = 5 the random family needs 18 bits, which is 1 word-sized prime 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.345110¹10²nbitsthe budget and what it buysrandom, bound18random, actual15primes needed1Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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 = 5 the random family needs 18 bits, which is 1 word-sized prime 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: 13

The arguments are the ones How many primes the answer needs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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 = 13 the random family needs 56 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.345678910111213110¹10²nbitsthe budget and what it buysrandom, bound56random, actual44primes needed3Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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 = 13 the random family needs 56 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: 9

The arguments are the ones How many primes the answer needs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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.3456789110¹10²nbitsthe budget and what it buysrandom, bound37random, actual25primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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: 8

The arguments are the ones How many primes the answer needs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The determinant's length, and the bound that decides how many primes to use before any of them is chosenSolid: 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.345678110¹10²nbitsthe budget and what it buysrandom, bound32random, actual24primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

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 checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks 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 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 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.

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