forcing-u
At its defaults it draws total inner iterations for a whole inexact newton solve, against the constant forcing term. Every point is a complete solve of the same problem to the same outer tolerance of 10^-10. At η = 10⁻¹⁴ it takes 9358 conjugate gradient iterations across 9 Newton steps; at η = 0.1 it takes 980 across 10. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 1009 iterations and reaches 3.37·10⁻¹¹.
forcing-u is one function in lib/figures/sequence.js —
sequences — a solve inside an outer loop, and the accuracy the loop throws away. 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.
Every point is a complete solve of the same problem to the same outer tolerance of 10^-10. At η = 10⁻¹⁴ it takes 9358 conjugate gradient iterations across 9 Newton steps; at η = 0.1 it takes 980 across 10. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 1009 iterations and reaches 3.37·10⁻¹¹.
logTol: -10
The arguments are the ones A factorisation kept past its date 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.
Every point is a complete solve of the same problem to the same outer tolerance of 10^-10. At η = 10⁻¹⁴ it takes 9358 conjugate gradient iterations across 9 Newton steps; at η = 0.1 it takes 980 across 10. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 1009 iterations and reaches 3.37·10⁻¹¹.
logTol: -12
The arguments are the ones A tolerance that reads its own residual 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.
Every point is a complete solve of the same problem to the same outer tolerance of 10^-12. At η = 10⁻¹⁴ it takes 9358 conjugate gradient iterations across 9 Newton steps; at η = 0.1 it takes 1256 across 12. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 1206 iterations and reaches 4.8·10⁻¹³.
logTol: -8
The arguments are the ones A tolerance that reads its own residual 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.
Every point is a complete solve of the same problem to the same outer tolerance of 10^-8. At η = 10⁻¹⁴ it takes 8230 conjugate gradient iterations across 8 Newton steps; at η = 0.1 it takes 730 across 8. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 822 iterations and reaches 4.75·10⁻⁹.
logTol: -6
The arguments are the ones The accuracy that is thrown away 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.
Every point is a complete solve of the same problem to the same outer tolerance of 10^-6. At η = 10⁻¹⁴ it takes 8230 conjugate gradient iterations across 8 Newton steps; at η = 0.1 it takes 459 across 6. The left arm is oversolving and the right arm is too many outer steps. The open circle is the adaptive rule, which is not a constant: it costs 593 iterations and reaches 4.47·10⁻⁷.
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.
8 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 conditioning the SPD construction can hold
a size the dense factorisations are affordable at
an outer tolerance inside the range the sweep is stable over
an outer tolerance that is a relative residual
and the rule that reads the residual costs a fraction of it
enough constants converge to draw a curve
matmul shapes agree
the tightest constant costs several times the cheapest
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.
A factorisation kept past its date
One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.
When the problem arrives againA tolerance that reads its own residual
The cheapest constant forcing term costs 980 inner iterations and arrives with a hundred times the forward error of the dearest, which costs 9,358. A rule that sets each step's tolerance from the ratio of the last two residuals costs 1,009 and arrives with neither problem — and it is not a constant, so it does not appear on the curve the constants are compared on.
When the problem arrives againThe accuracy that is thrown away
A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.
When the problem arrives againWhat a rebuild is worth
One sequence, one drift, one preconditioner — and six different right answers, because the cheapest rebuild period depends on what a rebuild cost to build and on nothing else. The optimum walks from every member to every twelfth as the setup gets dearer, and the free rule that reads the iteration count beats it in the middle of that range and loses at both ends.