congruence-bars
At its defaults it draws how much two runs of the same fit agree about the factors, from 6 starting points each. Every run here reaches its target to the rounding level — 3·10⁻¹³, 1.16·10⁻¹¹ and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.0207. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.1089 and its factors mean nothing on their own.
congruence-bars is one function in lib/figures/cpals.js —
alternating least squares — an iteration that walks out of the set it is searching. 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 run here reaches its target to the rounding level — 3·10⁻¹³, 1.16·10⁻¹¹ and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.0207. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.1089 and its factors mean nothing on their own.
seeds: 6
The arguments are the ones A factorisation that is unique for once 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 run here reaches its target to the rounding level — 3·10⁻¹³, 1.16·10⁻¹¹ and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.0207. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.1089 and its factors mean nothing on their own.
seeds: 10
The arguments are the ones A factorisation that is unique for once 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 run here reaches its target to the rounding level — 3·10⁻¹³, 1.16·10⁻¹¹ and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.0207. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.0432 and its factors mean nothing on their own.
seeds: 3
The arguments are the ones A factorisation that is unique for once 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 run here reaches its target to the rounding level — 1.55·10⁻¹³, 3.59·10⁻¹² and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.1450. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.1089 and its factors mean nothing on their own.
seeds: 4
The arguments are the ones A factorisation that is unique for once 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 run here reaches its target to the rounding level — 3·10⁻¹³, 3.59·10⁻¹² and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.1026. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.1089 and its factors mean nothing on their own.
seeds: 8
The arguments are the ones A factorisation that is unique for once 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 run here reaches its target to the rounding level — 3·10⁻¹³, 1.16·10⁻¹¹ and 2.39·10⁻¹⁵ at worst — so all three are successful factorisations. The bar is the worst agreement between any two of them about the *factors*, matched over permutations and scalings, which is exactly the freedom the uniqueness theorem allows. Kruskal's condition k_A + k_B + k_C ≥ 2r + 2 holds for the first (9 ≥ 8) and fails for the second (6 < 8), and the bars are 1.0000 and 0.0207. The matrix is the comparison the whole thing rests on: AB = (AM)(M⁻¹B) for every invertible M, so its runs agree to 0.0432 and its factors mean nothing on their own.
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.
5 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.
a number of starting points
and only the first is recovered the same way twice
matmul shapes agree
one case satisfies Kruskal's condition and one does not
while every matrix run factorises
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 146
of 287 generators —
131 print a residual and
15 are exempt with a published reason;
141 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 that is unique for once
A rank-r factorisation of a matrix is never unique — AB is (AM)(M⁻¹B) for any invertible M, so no factor means anything on its own. For three indices a checkable condition on the factors' k-ranks makes the decomposition unique up to permuting and scaling the terms, and it holds generically.
When the index is a tupleA nearest point that is not there
Eckart and Young guarantee that a matrix has a best rank-k approximation and that the truncated SVD is it. For three indices the guarantee is false in the strongest available way — there are tensors whose distance to the rank-two set is zero and which no rank-two tensor equals.
Two errors, and whose fault they areA tensor that cannot be decomposed
Every member of a certain sequence is exactly a sum of two rank-one terms, and both terms are written down in closed form. A three-hundred-sweep fit from a random start does not find them, and gets further away as the sequence goes on — because the decomposition has a condition number of its own, and it is 2n².
When the index is a tupleAn iteration that walks out of the set
Every sweep of alternating least squares is the exact minimiser of its own subproblem, so the objective can only fall. What it cannot do is converge, when the target's nearest rank-r point is not in the rank-r set — and a plateau at a small residual looks identical to slow convergence unless the size of the terms is plotted beside it.
When the index is a tupleThe orthogonality that cannot be diagonal
A matrix decomposition hands over orthonormal factors and a diagonal middle at once. For three indices the two come apart, and there is no arrangement that has both — so the question stops being which decomposition to use and becomes which of the two properties the computation needs.