What residual replacement costs and what it buys, at three periods
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.
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: -12
The arguments are the ones A walk needs a length passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
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 4.01·10⁻²¹ and the answer's stalls at 6.24·10⁻¹¹. Replacing every 5 steps costs 30 extra matrix–vector products on top of 150 — 20 per cent — and brings the answer's residual to 7.17·10⁻¹⁷, with the two residuals then agreeing to a factor of 2.2.
logEps: -14
The arguments are the ones Stable once, and three thousand times passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
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.
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.
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 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.
A walk needs a length
The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.
When the problem arrives againStable 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 factorisingThe 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.