Numerical Linear Algebra
where the algebra stops being the arithmetic
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The truncation error of a smooth tensor against the rank kept, between the two bounds the theorem gives

One function in the hosvd library, called 12 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 17 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 the truncation error of a smooth tensor against the rank kept, between the two bounds the theorem gives. The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.688, 0.717, 0.879, 0.933 … 0.999982, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7320 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

tucker-error is one function in lib/figures/hosvd.js — the higher-order svd — d matrix decompositions, and a core that cannot be diagonal. 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 truncation error of a smooth tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.688, 0.717, 0.879, 0.933 … 0.999982, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7320 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.024681010⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnssmooth: pinned to the upper boundrank 10 error1.1·10⁻¹¹its upper bound1.1·10⁻¹¹the lower bound6.3·10⁻¹²error ⁄ bound1error ⁄ lower1.7inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.688, 0.717, 0.879, 0.933 … 0.999982, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7320 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

family: "smooth"

The arguments are the ones A decomposition made only of SVDs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The truncation error of a smooth tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.688, 0.717, 0.879, 0.933 … 0.999982, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7320 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.024681010⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnssmooth: pinned to the upper boundrank 10 error1.1·10⁻¹¹its upper bound1.1·10⁻¹¹the lower bound6.3·10⁻¹²error ⁄ bound1error ⁄ lower1.7inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.688, 0.717, 0.879, 0.933 … 0.999982, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7320 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

family: "hilbert"

The arguments are the ones A decomposition made only of SVDs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The truncation error of a hilbert tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.689, 0.829, 0.906, 0.950 … 0.999812, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7317 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.024681010⁻¹²10⁻⁹10⁻⁶10⁻³1rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnshilbert: pinned to the upper boundrank 10 error1.4·10⁻¹²its upper bound1.4·10⁻¹²the lower bound8.3·10⁻¹³error ⁄ bound1error ⁄ lower1.7inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.689, 0.829, 0.906, 0.950 … 0.999812, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.7317 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

family: "smooth", d: 2

The arguments are the ones A decomposition made only of SVDs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The truncation error of a smooth tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √2 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.707, 0.707, 0.707, 0.707 … 0.707107, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.0002 against √2 = 1.4142. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.024681010⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnssmooth: pinned to the upper boundrank 8 error2.9·10⁻¹⁰its upper bound4.1·10⁻¹⁰the lower bound2.9·10⁻¹⁰error ⁄ bound0.71error ⁄ lower1inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √2 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.707, 0.707, 0.707, 0.707 … 0.707107, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.0002 against √2 = 1.4142. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

family: "smooth", d: 4, n: 8

The arguments are the ones A decomposition made only of SVDs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The truncation error of a smooth tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √4 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.690, 0.738, 0.955, 0.963 … 0.999709, so by rank 8 the bound is attained to five decimals and the ratio to the lower bound is 1.0000 against √4 = 2.0000. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.0246810⁻¹⁷10⁻¹³10⁻⁹10⁻⁵10⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnssmooth: pinned to the upper boundrank 6 error1.2·10⁻⁶its upper bound1.2·10⁻⁶the lower bound6.2·10⁻⁷error ⁄ bound1error ⁄ lower2inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √4 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.690, 0.738, 0.955, 0.963 … 0.999709, so by rank 8 the bound is attained to five decimals and the ratio to the lower bound is 1.0000 against √4 = 2.0000. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

family: "noise"

The arguments are the ones A decomposition made only of SVDs passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The truncation error of a noise tensor against the rank kept, between the two bounds the theorem givesThe middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.618, 0.648, 0.687, 0.725 … 0.927736, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.5839 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.024681010⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnsnoise: pinned to the upper boundrank 10 error0.5its upper bound0.54the lower bound0.31error ⁄ bound0.93error ⁄ lower1.6inside the boundand sitting on it

The middle curve is the measured error of the projection; the upper dashed one is √(Σₖ tail_k²), which the theorem says it cannot exceed, and the lower one is maxₖ tail_k, which the best possible error cannot fall below. They are a factor of √3 apart. The measurement is that the projection sits on the upper one, and not between them: the ratio of error to bound runs 0.618, 0.648, 0.687, 0.725 … 0.927736, so by rank 10 the bound is attained to five decimals and the ratio to the lower bound is 1.5839 against √3 = 1.7321. That reads as a bad result and is not one — what it says is that the lower bound is weak, which only a second measurement can establish.

What it checked while drawing

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks 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.

17 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.

the error at rank 1 is inside its bound — checked 8 times

a dimension the dense unfoldings are affordable at

a family this site defines

a side the tensor fits at

and inside √d of the lower bound

hilbert is not reproduced exactly at every rank drawn

matmul shapes agree

noise is not reproduced exactly at every rank drawn

smooth is not reproduced exactly at every rank drawn

wave is not reproduced exactly at every rank drawn

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 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 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 decomposition made only of SVDs

Everything the definition of tensor rank loses comes back if the SVD's algorithm is carried across instead of its definition — take the leading left singular subspace of every unfolding and project onto all of them. It exists, it costs d matrix decompositions, and its error is within √d of the best there is.

 When the index is a tuple

The 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.

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