rank-share
At its defaults it draws the share of random 2 × 2 × 2 tensors with real rank two, against the number drawn, and π/4. Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 19,953 draws 15,705 have rank two — a share of 0.7871 against the exact value π/4 = 0.7854, inside 0.6 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
rank-share is one function in lib/figures/borderrank.js —
border rank — a nearest point that is not there, and a rank that is a sign. 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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 19,953 draws 15,705 have rank two — a share of 0.7871 against the exact value π/4 = 0.7854, inside 0.6 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
logDraws: 4.3
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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 19,953 draws 15,705 have rank two — a share of 0.7871 against the exact value π/4 = 0.7854, inside 0.6 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
logDraws: 3.3
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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 1,995 draws 1,589 have rank two — a share of 0.7965 against the exact value π/4 = 0.7854, inside 1.2 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
logDraws: 2
The arguments are the ones A rank that is not a property of the tensor 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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 100 draws 73 have rank two — a share of 0.7300 against the exact value π/4 = 0.7854, inside 1.3 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
logDraws: 4.7
The arguments are the ones A rank that is not a property of the tensor 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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 50,119 draws 39,500 have rank two — a share of 0.7881 against the exact value π/4 = 0.7854, inside 1.5 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
logDraws: 4
The arguments are the ones A rank that is not a property of the tensor 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.
Each draw is eight independent standard normal entries and its rank is decided exactly, by the sign of the hyperdeterminant, with no iteration involved. Over 10,000 draws 7,865 have rank two — a share of 0.7865 against the exact value π/4 = 0.7854, inside 0.3 standard errors. The band is ±2 of them. A random matrix has one typical rank; this is the picture of a random object having two, each with a probability that is a number rather than an experiment.
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.
2 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 draws the run can afford
the measured share is π/4 within four standard errors
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.
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.