swamp-trace
At its defaults it draws alternating least squares on a tensor with a rank-two answer and on one without, over 20,000 sweeps. The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00141 and has not finished, while the rising curve is its largest term: 4.698 to 10.15, a factor of 2.16. Fitted over 199 points, the error falls as the 2.03 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
swamp-trace 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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00141 and has not finished, while the rising curve is its largest term: 4.698 to 10.15, a factor of 2.16. Fitted over 199 points, the error falls as the 2.03 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
sweeps: 20000
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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00141 and has not finished, while the rising curve is its largest term: 4.698 to 10.15, a factor of 2.16. Fitted over 199 points, the error falls as the 2.03 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
sweeps: 200
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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00616 and has not finished, while the rising curve is its largest term: 4.698 to 4.918, a factor of 1.05. Fitted over 149 points, the error falls as the 2.07 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
sweeps: 5000
The arguments are the ones A nearest point that is not there 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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00266 and has not finished, while the rising curve is its largest term: 4.698 to 7.42, a factor of 1.58. Fitted over 197 points, the error falls as the 2.04 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
sweeps: 1000
The arguments are the ones An iteration that walks out of the set 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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00469 and has not finished, while the rising curve is its largest term: 4.698 to 5.618, a factor of 1.20. Fitted over 189 points, the error falls as the 2.06 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
sweeps: 40000
The arguments are the ones An iteration that walks out of the set 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.
The steeply falling curve is a rank-three fit to a tensor built from rank-three factors: it reaches 9.74·10⁻¹⁵ in 58 sweeps and its largest term does not move — 12.202 at sweep twenty and 12.202 at the end. The other is a rank-two fit to the border-rank tensor, whose error goes from 0.00677 to 0.00101 and has not finished, while the rising curve is its largest term: 4.698 to 11.99, a factor of 2.55. Fitted over 199 points, the error falls as the 2.02 power of the term size — a better path than the explicit sequence's 1/n, and still one that leaves the set rather than converging inside it.
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 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 run long enough to see the plateau
and the benign run's terms do not move
matmul shapes agree
the swamp's error keeps falling
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.
When the index is a tupleA rank that is not a property of the tensor
The same eight real numbers have rank three over the reals and rank two over the complexes, and a random 2 × 2 × 2 tensor has rank two with probability exactly π/4. Neither sentence has an analogue for matrices, where the rank is one number and a random matrix has the largest one.
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.