Generator

threshold-stops

One function in the tolerance library, called 5 times across 1 essay. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 8 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 where the arnoldi recurrence stops, against the tolerance it calls a zero, n = 27. The spectrum has 3 coarse clusters of width 10^-4 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁵ to 5.48·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

threshold-stops is one function in lib/figures/tolerance.js — tolerances — the zeros nobody meets and everybody writes. 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-4 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁵ to 5.48·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest5.5·10⁻¹⁷residual left, loosest1.5·10⁻⁵both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-4 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁵ to 5.48·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

logEta: -4

The arguments are the ones Deciding that a zero has arrived 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-4 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁵ to 5.48·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest5.5·10⁻¹⁷residual left, loosest1.5·10⁻⁵both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-4 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁵ to 5.48·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

logEta: -8

The arguments are the ones Deciding that a zero has arrived 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-8 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁹ to 2.92·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest2.9·10⁻¹⁷residual left, loosest1.5·10⁻⁹both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-8 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁹ to 2.92·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

logEta: -2

The arguments are the ones Deciding that a zero has arrived 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-2 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 0.00152 to 2.52·10⁻¹⁶. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest2.5·10⁻¹⁶residual left, loosest0.0015both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-2 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 0.00152 to 2.52·10⁻¹⁶. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

logEta: -6

The arguments are the ones Deciding that a zero has arrived 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-6 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁷ to 2.85·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest2.8·10⁻¹⁷residual left, loosest1.5·10⁻⁷both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-6 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁷ to 2.85·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

logEta: -3

The arguments are the ones Deciding that a zero has arrived 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.

Where the Arnoldi recurrence stops, against the tolerance it calls a zero, n = 27The spectrum has 3 coarse clusters of width 10^-3 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁴ to 5.26·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²0612182430tolerance called a zerostep the recurrence stops atsubspace of 3subspace of 9subspace of 27too late: dividing by roundingtoo early: an answer thrown awayone matrix, three answerssteps at the tightest tolerance27at the loosest3residual left, tightest5.3·10⁻¹⁷residual left, loosest1.5·10⁻⁴both directions are failuresand the constant decides which

The spectrum has 3 coarse clusters of width 10^-3 and 9 fine ones inside them, so this one matrix has three near-invariant subspaces and all three are real. The tolerance decides which the recurrence stops at: it stops at 3 at the loose end, at 9 in the middle, and at 27 at the tight end. The residual it leaves runs from 1.51·10⁻⁴ to 5.26·10⁻¹⁷. Neither end is a mistake and neither is a trade: stopping too early throws away an answer that was available, and stopping too late means dividing by rounding error.

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.

8 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 coarse cluster width inside the range the sweep separates

a power of ten rather than an exponent literal

a size the two levels of clustering divide evenly

a tighter tolerance runs further

an Arnoldi recurrence started from a vector that is not zero

and a looser one leaves a residual the tighter had removed

each of which is one of the subspaces the matrix genuinely has

one matrix, three different answers

Against the rule

It draws a decomposition and prints its residual. It calls arnoldiThreshold, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 113 of 219 generators — 98 print a residual and 15 are exempt with a published reason; 106 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.

The whole library · All essays · What must fail