Generator

The preconditioned spectrum on four grids (ρ = 0.9)

One function in the bttb library, called 14 times across 3 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 22 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 the preconditioned spectrum on four grids (ρ = 0.9). Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.

cluster-thinning is one function in lib/figures/bttb.js — block toeplitz — the second dimension, where the cluster thins. 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.

The preconditioned spectrum on four grids (ρ = 0.9)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.357911110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.

rho: 0.92

The arguments are the ones Four orders of conditioning, and four steps passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The preconditioned spectrum on four grids (ρ = 0.92)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.357911110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.

rho: 0.85

The arguments are the ones Four orders of conditioning, and four steps passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The preconditioned spectrum on four grids (ρ = 0.85)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.357911110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.

rho: 0.95

The arguments are the ones Four orders of conditioning, and four steps passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The preconditioned spectrum on four grids (ρ = 0.95)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.357911110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17 while the number of unknowns goes 16, 36, 64, 100 — so the share clustered falls from 56% to 17%.

rho: 0.9, grids: [8, 9, 10, 11, 12]

The arguments are the ones Four orders of conditioning, and four steps passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The preconditioned spectrum on four grids (ρ = 0.9)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 13, 15, 17, 19, 21 while the number of unknowns goes 64, 81, 100, 121, 144 — so the share clustered falls from 20% to 15%.791113110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 8: inside of 6413m = 9: inside of 8115m = 10: inside of 10017m = 11: inside of 12119m = 12: inside of 14421the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 13, 15, 17, 19, 21 while the number of unknowns goes 64, 81, 100, 121, 144 — so the share clustered falls from 20% to 15%.

rho: 0.98, grids: [4, 6, 8, 10, 12]

The arguments are the ones Four orders of conditioning, and four steps passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The preconditioned spectrum on four grids (ρ = 0.98)Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17, 21 while the number of unknowns goes 16, 36, 64, 100, 144 — so the share clustered falls from 56% to 15%.35791113110¹grid side meigenvalue of C⁻¹Awithin ½ of onethe cluster grows like the sidem = 4: inside of 169m = 6: inside of 3611m = 8: inside of 6413m = 10: inside of 10017m = 12: inside of 14421the cluster grows with the sideand the spectrum with the area

Every eigenvalue of C⁻¹A, drawn as a point above the grid it belongs to, on a logarithmic vertical axis. The shaded band is within half a unit of one. The number of eigenvalues inside it goes 9, 11, 13, 17, 21 while the number of unknowns goes 16, 36, 64, 100, 144 — so the share clustered falls from 56% to 15%.

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.

22 distinct claims across 6 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.

so the share inside falls at m = 6 — checked 6 times

the cluster grows at m = 6 — checked 6 times

a correlation the family is conditioned at

a correlation the sweep draws

a grid the dense eigensolve is affordable on

a p-norm kernel between the separable and the isotropic

a two-dimensional kernel this file builds

and grows like the side rather than the area

Jacobi needs a symmetric matrix

matmul shapes agree

the similarity transform is symmetric

the two-dimensional operator is positive definite

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.

Structure, and the solver that cannot see it

Four orders of conditioning, and four steps

On a 10×10 grid the two-dimensional kernel's condition number runs from 62 at ρ = 0.5 to 818,561 at ρ = 0.98. The preconditioned step count over the same range runs 18, 21, 21, 22, 21, 19, 18, and the count of eigenvalues the preconditioner actually brings within half a unit of one does not move at all — it is 9, 11, 13, 17 at every correlation the figure will draw.

Structure, and the solver that cannot see it

The staircase a separable kernel builds

A block-circulant preconditioner took 18, 21, 22, 21, 19 and 18 steps on a 10 × 10 Toeplitz-block-Toeplitz system as the correlation rose from 0.5 to 0.98 and the unpreconditioned count rose from 43 to 178 — the parameter that makes the problem hard was the one the solver did not notice. That kernel factorises. The isotropic kernel with the same correlation along each axis does not, and on it the preconditioned count climbs 17, 22, 27, 29, 33, 36, while the preconditioner buys a factor of 1.4 where it bought ten. The reason is the spectrum's shape: a product of two one-dimensional spectra is a staircase of ten treads, and a kernel that is not a product gives a ramp of thirty-eight.

Structure, and the solver that cannot see it

Two dimensions, and the cluster that thins

The same kernel, the same averaging, the same transform — applied along two axes instead of one. In one dimension the preconditioned step count is 7, 10, 10, 10; on square grids with the same unknown counts it is 10, 18, 20, 21, and the share of the spectrum near one falls from 56% to 17%.

The whole library · All essays · What must fail