Generator

replacement-repair

One function in the resgap library, called 5 times across 4 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 6 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 what residual replacement costs and what it buys, at three periods. Each row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 6.92·10⁻²¹ and the answer's stalls at 5.07·10⁻¹⁰. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 8.31·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.8.

replacement-repair is one function in lib/figures/resgap.js — residual gap — the number a method reports against the residual of its answer. 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.

What residual replacement costs and what it buys, at three periodsEach row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 6.92·10⁻²¹ and the answer's stalls at 5.07·10⁻¹⁰. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 8.31·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.8.never replace5.07·10⁻¹⁰replace every 251.5·10⁻¹⁶replace every 101.34·10⁻¹⁶replace every 58.31·10⁻¹⁷reported 6.92·10⁻²¹ · 0 extra productsreported 5.3·10⁻¹⁷ · 6 extra productsreported 1.08·10⁻¹⁶ · 15 extra productsreported 2.95·10⁻¹⁷ · 30 extra productsthe residual of the answer the run returnsone line, at three pricesnever: the answer's residual5.1·10⁻¹⁰never: what it reported6.9·10⁻²¹every 5: the answer's8.3·10⁻¹⁷every 5: what it reported3·10⁻¹⁷extra products for that30ask the matrix againand the recurrence forgets what it did

Each row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 6.92·10⁻²¹ and the answer's stalls at 5.07·10⁻¹⁰. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 8.31·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.8.

logEps: -14

The arguments are the ones Stable once, and three thousand times 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.

What residual replacement costs and what it buys, at three periodsEach row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 6.92·10⁻²¹ and the answer's stalls at 5.07·10⁻¹⁰. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 8.31·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.8.never replace5.07·10⁻¹⁰replace every 251.5·10⁻¹⁶replace every 101.34·10⁻¹⁶replace every 58.31·10⁻¹⁷reported 6.92·10⁻²¹ · 0 extra productsreported 5.3·10⁻¹⁷ · 6 extra productsreported 1.08·10⁻¹⁶ · 15 extra productsreported 2.95·10⁻¹⁷ · 30 extra productsthe residual of the answer the run returnsone line, at three pricesnever: the answer's residual5.1·10⁻¹⁰never: what it reported6.9·10⁻²¹every 5: the answer's8.3·10⁻¹⁷every 5: what it reported3·10⁻¹⁷extra products for that30ask the matrix againand the recurrence forgets what it did

Each row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 6.92·10⁻²¹ and the answer's stalls at 5.07·10⁻¹⁰. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 8.31·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.8.

logEps: -8

The arguments are the ones The residual the method reports 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.

What residual replacement costs and what it buys, at three periodsEach row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 7.91·10⁻²¹ and the answer's stalls at 5.9·10⁻¹³. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 9.15·10⁻¹⁷, with the two residuals then agreeing to a factor of 1.5.never replace5.9·10⁻¹³replace every 251.28·10⁻¹⁶replace every 101.17·10⁻¹⁶replace every 59.15·10⁻¹⁷reported 7.91·10⁻²¹ · 0 extra productsreported 4.9·10⁻¹⁷ · 6 extra productsreported 9.07·10⁻¹⁷ · 15 extra productsreported 6.04·10⁻¹⁷ · 30 extra productsthe residual of the answer the run returnsone line, at three pricesnever: the answer's residual5.9·10⁻¹³never: what it reported7.9·10⁻²¹every 5: the answer's9.1·10⁻¹⁷every 5: what it reported6·10⁻¹⁷extra products for that30ask the matrix againand the recurrence forgets what it did

Each row is the same conjugate gradient run with the recomputed residual assigned back into the recurrence every k steps. Without it, the reported residual reaches 7.91·10⁻²¹ and the answer's stalls at 5.9·10⁻¹³. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 9.15·10⁻¹⁷, with the two residuals then agreeing to a factor of 1.5.

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.

6 distinct claims across 3 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 size the whole run is affordable at

an eigenvalue small enough for there to be a gap to repair

an eigenvalue small enough to make the iterates large

and the answer itself is orders of magnitude better, not merely better described

and with it every five steps they agree to within a small factor

without replacement the two residuals are orders of magnitude apart

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 113 of 238 generators — 98 print a residual and 15 are exempt with a published reason; 125 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.

When the problem arrives again

Stable once, and three thousand times

A sliding window adds a row and removes one at every step and never looks at the data again. No single step of it amplifies by more than 2.72, no downdate fails, and after three thousand steps the triangular factor in memory is 3.9·10⁻¹⁴ from the matrix it is supposed to be a factor of — six hundred times growth from a per-step bound that says nothing about chains.

Iterating, instead of factorising

The number that is re-derived

GMRES prints a residual it never computes from its answer either. On the matrix that sends a conjugate gradient recurrence 7.3·10¹⁰ wrong, and on two others chosen to be worse, its number is never more than a factor of 2.86 out — while the basis it is computed from has lost orthogonality entirely. The disease is not iterative methods, and it is not floating point.

Iterating, instead of factorising

The residual the method reports

Conjugate gradients prints a relative residual of 6.9·10⁻²¹. The unit roundoff is 1.1·10⁻¹⁶, so that is not a small residual and not a large one — it is not a residual. The vector the method is holding at that step has ‖b − Ax‖/‖b‖ = 5.1·10⁻¹⁰, and nothing in the run says so.

The arithmetic underneath

Three walks and one bound

A left-to-right sum, a chain of three thousand rotations and a conjugate gradient residual recurrence share no arithmetic and no vocabulary. Each has a standard bound that is linear in whatever it accumulates against. All three come out at a half — 0.486, 0.554 and 0.507 — and nothing is rescaled.

The whole library · All essays · What must fail