penalty-null-space
At its defaults it draws two penalties on the same problem, against the offset in the signal. Best relative error for each penalty at four offsets. With no offset the two are within 7% of each other. At an offset of 10 the derivative penalty is 1.85 times better, because a constant lies in its null space and costs it nothing, while the norm penalty pays for the whole offset at every λ.
penalty-null-space is one function in lib/figures/krylovreg.js —
iterative regularisation — a step count as the parameter, and the filter it applies. 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.
Best relative error for each penalty at four offsets. With no offset the two are within 7% of each other. At an offset of 10 the derivative penalty is 1.85 times better, because a constant lies in its null space and costs it nothing, while the norm penalty pays for the whole offset at every λ.
n: 48
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.
Best relative error for each penalty at four offsets. With no offset the two are within 7% of each other. At an offset of 10 the derivative penalty is 1.85 times better, because a constant lies in its null space and costs it nothing, while the norm penalty pays for the whole offset at every λ.
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 2 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.
the derivative penalty's advantage grows at offset 2 — asserted 3 times
a range of offsets wide enough to show the trend
a size the λ sweep is affordable at
enough noise for the penalty to matter
with almost nothing between them at no offset
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 66
of 131 generators —
51 print a residual and
15 are exempt with a published reason;
65 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 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.
Least squares, and the road not to takeThe valley with no bottom
A degree-nine fit's coefficients can be moved by a third of their own size before the residual changes in the sixth significant figure. The arithmetic did not lose those digits. The data never contained them.
Regularisation, and the answer that is chosenWhen 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 chosenWhere 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.