Generator

schulz-ball

One function in the polar library, called 7 times across 7 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 4 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 σ ↦ σ(3 − σ²)/2, the map newton–schulz applies to every singular value, started at 1.6. The cubic and the diagonal, with the iteration from 1.6 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.6 the limit is 1.000000.

schulz-ball is one function in lib/figures/polar.js — the polar factor — the nearest orthogonal matrix, and two ways to it without an svd. 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.

σ ↦ σ(3 − σ²)/2, the map Newton–Schulz applies to every singular value, started at 1.6The cubic and the diagonal, with the iteration from 1.6 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.6 the limit is 1.000000.the whole convergence theory of an n×n iteration is this cubic√3start 1.610−1fixed points 0, 1, −1start1.6after one step0.35limit1the boundary, √31.7outside the basin it still convergesto an orthogonal matrix that is not the answer

The cubic and the diagonal, with the iteration from 1.6 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.6 the limit is 1.000000.

alpha: 2.24

The arguments are the ones An answer that changes with the seed 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.

σ ↦ σ(3 − σ²)/2, the map Newton–Schulz applies to every singular value, started at 2.24The cubic and the diagonal, with the iteration from 2.24 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 2.24 the limit is not finite.the whole convergence theory of an n×n iteration is this cubic√3start 2.2410−1fixed points 0, 1, −1start2.2after one step-2.3limit10³⁰⁸the boundary, √31.7outside the basin it still convergesto an orthogonal matrix that is not the answer

The cubic and the diagonal, with the iteration from 2.24 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 2.24 the limit is not finite.

alpha: 1.74

The arguments are the ones An eigenvalue that arrives twice 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.

σ ↦ σ(3 − σ²)/2, the map Newton–Schulz applies to every singular value, started at 1.74The cubic and the diagonal, with the iteration from 1.74 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.74 the limit is -1.000000.the whole convergence theory of an n×n iteration is this cubic√3start 1.7410−1fixed points 0, 1, −1start1.7after one step-0.024limit-1the boundary, √31.7outside the basin it still convergesto an orthogonal matrix that is not the answer

The cubic and the diagonal, with the iteration from 1.74 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.74 the limit is -1.000000.

alpha: 1.6

The arguments are the ones An iteration that only multiplies 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.

σ ↦ σ(3 − σ²)/2, the map Newton–Schulz applies to every singular value, started at 1.6The cubic and the diagonal, with the iteration from 1.6 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.6 the limit is 1.000000.the whole convergence theory of an n×n iteration is this cubic√3start 1.610−1fixed points 0, 1, −1start1.6after one step0.35limit1the boundary, √31.7outside the basin it still convergesto an orthogonal matrix that is not the answer

The cubic and the diagonal, with the iteration from 1.6 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.6 the limit is 1.000000.

alpha: 1.7

The arguments are the ones Where the box is cut 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.

σ ↦ σ(3 − σ²)/2, the map Newton–Schulz applies to every singular value, started at 1.7The cubic and the diagonal, with the iteration from 1.7 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.7 the limit is 1.000000.the whole convergence theory of an n×n iteration is this cubic√3start 1.710−1fixed points 0, 1, −1start1.7after one step0.094limit1the boundary, √31.7outside the basin it still convergesto an orthogonal matrix that is not the answer

The cubic and the diagonal, with the iteration from 1.7 drawn as a cobweb. The fixed points are 0, 1 and −1, and zero is repelling — its slope there is 3/2. Every starting value in (0, √3) is carried to 1, so the matrix iteration converges exactly when every singular value is inside that interval; √3 = 1.7320508 is sent to exactly zero, and past it the sequence leaves. From 1.7 the limit is 1.000000.

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.

4 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 starting value the cubic keeps finite for a few steps

and every starting value above it goes somewhere else

and every starting value below √3 is carried to one

the cubic sends √3 to exactly zero

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 92 of 198 generators — 77 print a residual and 15 are exempt with a published reason; 106 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.

Methods that were designed apart

An answer that changes with the seed

A randomised rank-k solve is a truncation computed in a random subspace, and it reaches the same floor as the deterministic ones. What it does not do is return the same answer twice — a factor of 1.84 across four seeds at rank 8, and 1.02 at the rank where the method is best.

Eigenvalues, singular values, rank

An eigenvalue that arrives twice

A matrix with forty distinct eigenvalues, handed to Lanczos for eighty steps, returns twenty-five extra copies of thirteen of them — the largest arriving five times. Every copy is accurate to 1.9·10⁻⁸ relative. No arithmetic error was made, nothing overflowed, and a caller counting eigenvalues gets the wrong multiplicity from a computation in which no individual number is wrong.

Orthogonality, measured

An iteration that only multiplies

Newton's iteration for the polar factor needs an inverse every step. Newton–Schulz needs only matrix products — nothing that reads an entry, nothing that pivots — and it converges if and only if every singular value is below √3. At 1.73205 it converges and at 1.73206 it returns an orthogonal matrix that is not the answer, with a residual of 5·10⁻¹⁶ and nothing to say so.

Randomised, and the guarantee that changes kind

Counting what cannot be looked at

The trace is n additions and one of the most expensive quantities in the subject to estimate, because the matrices whose trace is wanted are never stored. Hutchinson's estimator is unbiased with one line of algebra — and its variance depends on which random vector is used, by a factor that is a property of the matrix, and on a diagonal matrix one choice is exact from the first probe and the other is not.

Orthogonality, measured

Orthogonal is a number

"Q is orthogonal" is a claim about a measurable quantity, ‖QᵀQ − I‖, and on the eight-by-eight Hilbert matrix two standard algorithms return 10⁻¹⁵ and 1 for it. The one that returns 1 still reconstructs the matrix perfectly, which is why nothing warns you.

Orthogonality, measured

The nearest orthogonal matrix

Every field that has to clean up a drifted rotation reaches for QR, and QR does not answer the question. The nearest orthogonal matrix is the orthogonal factor of the polar decomposition — nearer by about a tenth, and, more to the point, the same matrix whatever order the columns were written in. QR's answer changes completely.

The arithmetic underneath

Where the box is cut

A branch-and-bound with an interval operator settles a whole square — two roots proved unique, forty-two regions proved empty, nothing left undecided, in 87 evaluations. Move the roots so one lands on the first bisection and it proves nothing at all, at any depth. Cutting at 0.485 instead of 0.5 finds both, in a quarter of the work.

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