Generator

hybrid-curves

One function in the hybrid library, called 7 times across 6 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 5 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 one subspace, solved and regularised, at 1.0% noise. Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1426 at step 20 and then climbs to 161 — a factor of 1127. The iterate with a penalty on the projected problem reaches 0.1429 and stays within 0.00% of it for the rest of the run.

hybrid-curves is one function in lib/figures/hybrid.js — hybrid regularisation — a penalty inside the subspace, and the rule that transfers to it. 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.

One subspace, solved and regularised, at 1.0% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1426 at step 20 and then climbs to 161 — a factor of 1127. The iterate with a penalty on the projected problem reaches 0.1429 and stays within 0.00% of it for the rest of the run.1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 20no penaltypenalty insidewhat stopping is worthbest without a penalty0.14and at step 40161best with one0.14and at step 400.14the same floor, reached twiceand only one run stays on it

Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1426 at step 20 and then climbs to 161 — a factor of 1127. The iterate with a penalty on the projected problem reaches 0.1429 and stays within 0.00% of it for the rest of the run.

noise: 0.005

The arguments are the ones A parameter chosen on a smaller problem 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.

One subspace, solved and regularised, at 0.50% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1221 at step 22 and then climbs to 80.4 — a factor of 658. The iterate with a penalty on the projected problem reaches 0.1331 and stays within 0.00% of it for the rest of the run.1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 22no penaltypenalty insidewhat stopping is worthbest without a penalty0.12and at step 4080best with one0.13and at step 400.13the same floor, reached twiceand only one run stays on it

Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1221 at step 22 and then climbs to 80.4 — a factor of 658. The iterate with a penalty on the projected problem reaches 0.1331 and stays within 0.00% of it for the rest of the run.

noise: 0.01

The arguments are the ones A parameter that counts steps 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.

One subspace, solved and regularised, at 1.0% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1426 at step 20 and then climbs to 161 — a factor of 1127. The iterate with a penalty on the projected problem reaches 0.1429 and stays within 0.00% of it for the rest of the run.1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 20no penaltypenalty insidewhat stopping is worthbest without a penalty0.14and at step 40161best with one0.14and at step 400.14the same floor, reached twiceand only one run stays on it

Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1426 at step 20 and then climbs to 161 — a factor of 1127. The iterate with a penalty on the projected problem reaches 0.1429 and stays within 0.00% of it for the rest of the run.

noise: 0.02

The arguments are the ones When the answer is a choice 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.

One subspace, solved and regularised, at 2.0% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1498 at step 10 and then climbs to 322 — a factor of 2146. The iterate with a penalty on the projected problem reaches 0.1500 and stays within 2.02% of it for the rest of the run.1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 10no penaltypenalty insidewhat stopping is worthbest without a penalty0.15and at step 40322best with one0.15and at step 400.15the same floor, reached twiceand only one run stays on it

Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1498 at step 10 and then climbs to 322 — a factor of 2146. The iterate with a penalty on the projected problem reaches 0.1500 and stays within 2.02% of it for the rest of the run.

noise: 0.05

The arguments are the ones Where the answer stops being in the data 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.

One subspace, solved and regularised, at 5.0% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1562 at step 5 and then climbs to 804 — a factor of 5147. The iterate with a penalty on the projected problem reaches 0.1572 and stays within 10.03% of it for the rest of the run.1591317212529333710⁻¹110¹10²10³bidiagonalisation stepsrelative errorleast without: 5no penaltypenalty insidewhat stopping is worthbest without a penalty0.16and at step 40804best with one0.16and at step 400.17the same floor, reached twiceand only one run stays on it

Two error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1562 at step 5 and then climbs to 804 — a factor of 5147. The iterate with a penalty on the projected problem reaches 0.1572 and stays within 10.03% of it for the rest of the run.

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.

5 distinct claims across 5 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 noise level small enough to be noise

a size the reference solve is affordable at

and the regularised one does not

enough steps for the unregularised run to turn

the unregularised run leaves its own best behind

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 70 of 151 generators — 55 print a residual and 15 are exempt with a published reason; 81 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.

Methods that were designed apart

A parameter chosen on a smaller problem

Inside a hybrid method the regularisation parameter is chosen on a 25×24 problem rather than a 64×64 one. The rule that reads a residual transfers exactly; the rule that reads a trace is biased by exactly two grid steps at twenty-four steps and one at forty, at every noise level from 10% to 0.1%.

Methods that were designed apart

A parameter that counts steps

The regularisation field's knob is a positive real number chosen by one of three rules. The iterative field's is an integer nobody called a knob — where to stop. On the same problem the best step is 20 and the best λ is 0.025, and they reach 0.1426 and 0.1406.

Two errors, and whose fault they are

A small residual is not a small error

Substituting the answer back and finding that it fits is the most natural check there is, and it verifies the wrong thing. A residual of 10⁻¹⁷ is entirely compatible with an answer whose second digit is wrong.

Methods that were designed apart

The step that stops mattering

Regularise the problem the iteration has built rather than the problem it was given, and the error curve stops turning. The unregularised run ends 1,127 times above its own best; the same run with a penalty inside it ends 1.000000000003 times above.

Regularisation, and the answer that is chosen

When the answer is a choice

A backward-stable least-squares solve of this problem returns an answer whose relative error is 5.5·10⁸. Nothing went wrong. The singular values decay exponentially with no gap anywhere in them, the data does not determine the answer, and something outside the data has to choose — which is the computation rather than a preliminary to it.

Regularisation, and the answer that is chosen

Where the answer stops being in the data

The Picard condition finds the index where a noisy right-hand side stops carrying signal, from the data alone, with no knowledge of the answer. It lands at 32 where the truncation that actually minimises the error is 28 — and at 45 where the best is 38. It overshoots at every noise level from 1% to 0.001%, and it overshoots for a reason.

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