Generator

decomposition-kappa

One function in the cpals library, called 11 times across 5 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 33 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 along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achieves. Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 4.5·10⁻¹³. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 128. Started at the answer instead, the same code stays within 1.4·10⁻¹⁰ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

decomposition-kappa 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.

Along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achievesEvery A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 4.5·10⁻¹³. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 128. Started at the answer instead, the same code stays within 1.4·10⁻¹⁰ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.110¹10²10⁻¹²10⁻⁸10⁻⁴110⁴ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 1283.3·10⁴its closed form3.3·10⁴cosine of the terms1from a random start0.012from the answer1.4·10⁻¹⁰the answer existsand cannot be found

Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 4.5·10⁻¹³. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 128. Started at the answer instead, the same code stays within 1.4·10⁻¹⁰ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

upTo: 7

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.

Along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achievesEvery A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 4.5·10⁻¹³. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 128. Started at the answer instead, the same code stays within 1.4·10⁻¹⁰ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.110¹10²10⁻¹²10⁻⁸10⁻⁴110⁴ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 1283.3·10⁴its closed form3.3·10⁴cosine of the terms1from a random start0.012from the answer1.4·10⁻¹⁰the answer existsand cannot be found

Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 4.5·10⁻¹³. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 128. Started at the answer instead, the same code stays within 1.4·10⁻¹⁰ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

upTo: 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.

Along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achievesEvery A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 1.1·10⁻¹¹. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 1024. Started at the answer instead, the same code stays within 9.2·10⁻⁹ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.110¹10²10³10⁻¹²10⁻⁸10⁻⁴110⁴ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 10242.1·10⁶its closed form2.1·10⁶cosine of the terms1from a random start0.012from the answer9.2·10⁻⁹the answer existsand cannot be found

Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 1.1·10⁻¹¹. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0118 at n = 1024. Started at the answer instead, the same code stays within 9.2·10⁻⁹ of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

upTo: 5

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.

Along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achievesEvery A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 8·10⁻¹⁵. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0117 at n = 32. Started at the answer instead, the same code stays within 8.9·10⁻¹² of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.110¹10⁻¹²10⁻⁹10⁻⁶10⁻³110³ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 322049its closed form2049cosine of the terms1from a random start0.012from the answer8.9·10⁻¹²the answer existsand cannot be found

Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 8·10⁻¹⁵. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0117 at n = 32. Started at the answer instead, the same code stays within 8.9·10⁻¹² of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

upTo: 3

The arguments are the ones A tensor that cannot be decomposed 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.

Along a sequence of exactly rank-two tensors: the step's condition number, and what a fit from a random start achievesEvery A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 2.2·10⁻¹⁶. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0103 at n = 8. Started at the answer instead, the same code stays within 1.5·10⁻¹² of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.110¹10⁻¹²10⁻⁹10⁻⁶10⁻³1ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 8129its closed form129cosine of the terms0.99from a random start0.01from the answer5.6·10⁻¹³the answer existsand cannot be found

Every A_n on this sequence *is* a rank-two tensor and its two rank-one terms are written down, so nothing here is about existence. The rising curve is the condition number of the r × r system each alternating sweep solves, which on this sequence has a closed form in n whose asymptote is 2n² — the marks are measured and the dashed line is that closed form, agreeing to 2.2·10⁻¹⁶. The lower marks are what a three-hundred-sweep fit from a random start returns: 2.1·10⁻⁴ at n = 2 rising to 0.0103 at n = 8. Started at the answer instead, the same code stays within 1.5·10⁻¹² of it at every n — not the rounding level, because the drift from an exact start is itself about κ times the unit roundoff, but nine orders below what a random start reaches. That is the control that says the failure is the conditioning and not the implementation. A tensor away from the boundary conditions its step at 6.03.

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.

33 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.

and the step's condition number at n = 2 — asserted 10 times

started at the answer, n = 2 stays near it — asserted 10 times

the two terms' cosine at n = 2 — asserted 10 times

a range of n the sequence is followed over

and far nearer than a random start gets

matmul shapes agree

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.

When the index is a tuple

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 tuple

A 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 tuple

A 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 are

A 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 tuple

An 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.

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