Generator

tolerance-agreement

One function in the race library, called 6 times across 3 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 asking for more accuracy buys accuracy, and buys no agreement at all. Conjugate gradients on a 200 × 200 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 1.5·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.34, 1.48, 1.71, 1.17, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.

tolerance-agreement is one function in lib/figures/race.js — what a reduction decides when the decision is discrete — a step count, a rank, and a test's tolerance. 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.

Asking for more accuracy buys accuracy, and buys no agreement at allConjugate gradients on a 200 × 200 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 1.5·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.34, 1.48, 1.71, 1.17, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.-12-10-8-610⁻¹³10⁻¹¹10⁻⁹10⁻⁷log₁₀ of the residual tolerance asked forforward error of the answer1.34×1.48×1.71×1.17×the band does not closetolerances swept4runs at each7accuracy gained1.5·10⁶ratio at 10⁻⁶1.3ratio at 10⁻¹²1.2the bars falland they keep their height

Conjugate gradients on a 200 × 200 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 1.5·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.34, 1.48, 1.71, 1.17, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.

n: 200

The arguments are the ones A stopping test is a race 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.

Asking for more accuracy buys accuracy, and buys no agreement at allConjugate gradients on a 200 × 200 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 1.5·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.34, 1.48, 1.71, 1.17, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.-12-10-8-610⁻¹³10⁻¹¹10⁻⁹10⁻⁷log₁₀ of the residual tolerance asked forforward error of the answer1.34×1.48×1.71×1.17×the band does not closetolerances swept4runs at each7accuracy gained1.5·10⁶ratio at 10⁻⁶1.3ratio at 10⁻¹²1.2the bars falland they keep their height

Conjugate gradients on a 200 × 200 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 1.5·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.34, 1.48, 1.71, 1.17, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.

n: 80

The arguments are the ones The tolerance that buys no agreement 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.

Asking for more accuracy buys accuracy, and buys no agreement at allConjugate gradients on a 80 × 80 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 5.4·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.04, 4.83, 14.83, 1.36, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.-12-10-8-610⁻¹³10⁻¹¹10⁻⁹10⁻⁷log₁₀ of the residual tolerance asked forforward error of the answer1.04×4.83×14.83×1.36×the band does not closetolerances swept4runs at each7accuracy gained5.4·10⁶ratio at 10⁻⁶1ratio at 10⁻¹²1.4the bars falland they keep their height

Conjugate gradients on a 80 × 80 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 5.4·10⁶ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.04, 4.83, 14.83, 1.36, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.

n: 300

The arguments are the ones The tolerance that buys no agreement 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.

Asking for more accuracy buys accuracy, and buys no agreement at allConjugate gradients on a 300 × 300 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 8.7·10⁵ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.27, 1.36, 1.11, 1.38, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.-12-10-8-610⁻¹³10⁻¹¹10⁻⁹10⁻⁷log₁₀ of the residual tolerance asked forforward error of the answer1.27×1.36×1.11×1.38×the band does not closetolerances swept4runs at each7accuracy gained8.7·10⁵ratio at 10⁻⁶1.3ratio at 10⁻¹²1.4the bars falland they keep their height

Conjugate gradients on a 300 × 300 matrix with κ = 10⁴, solved at four tolerances, seven partition counts each. The bar at each tolerance spans the smallest and largest forward error the seven runs produced. The bars fall by 8.7·10⁵ across the sweep, which is the tolerance doing exactly what it is for. The ratio between the top and bottom of each bar is 1.27, 1.36, 1.11, 1.38, which does not fall with the tolerance. So the disagreement between machines is not a residue of an insufficiently converged answer that a tighter tolerance would remove; it is proportional to whatever accuracy was reached, and the runs stay a fixed factor apart all the way down.

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 4 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 problem the sweep can solve twenty-eight times

and the disagreement between runs does not close

every tolerance shows a spread of iteration counts

LU is for square matrices

matmul shapes agree

the accuracy improves by orders across the sweep

there is a disagreement to speak of at all

Against the rule

It draws a decomposition and prints its residual. It calls toleranceSweep, 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 197 of 363 generators — 179 print a residual and 18 are exempt with a published reason; 166 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