Generator

tucker-storage

One function in the hosvd library, called 9 times across 5 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 3 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 numbers stored against the number of indices, at n = 20 and rank 4: the tensor, its core, and a train. The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 4 it is 1.1·10¹², smaller than the tensor by 9.54·10¹³ and still unstorable. A train is (d − 2)nr² + 2nr, which is 5,920 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

tucker-storage 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.

Numbers stored against the number of indices, at n = 20 and rank 4: the tensor, its core, and a trainThe tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 4 it is 1.1·10¹², smaller than the tensor by 9.54·10¹³ and still unstorable. A train is (d − 2)nr² + 2nr, which is 5,920 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core1.1·10¹²train5920core ⁄ train1.9·10⁸entries ⁄ core9.5·10¹³the definition is repairedthe size is not

The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 4 it is 1.1·10¹², smaller than the tensor by 9.54·10¹³ and still unstorable. A train is (d − 2)nr² + 2nr, which is 5,920 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

r: 4

The arguments are the ones A decomposition made only of SVDs 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.

Numbers stored against the number of indices, at n = 20 and rank 4: the tensor, its core, and a trainThe tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 4 it is 1.1·10¹², smaller than the tensor by 9.54·10¹³ and still unstorable. A train is (d − 2)nr² + 2nr, which is 5,920 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core1.1·10¹²train5920core ⁄ train1.9·10⁸entries ⁄ core9.5·10¹³the definition is repairedthe size is not

The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 4 it is 1.1·10¹², smaller than the tensor by 9.54·10¹³ and still unstorable. A train is (d − 2)nr² + 2nr, which is 5,920 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

r: 8

The arguments are the ones A decomposition made only of SVDs 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.

Numbers stored against the number of indices, at n = 20 and rank 8: the tensor, its core, and a trainThe tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 8 it is 1.15·10¹⁸, smaller than the tensor by 9.09·10⁷ and still unstorable. A train is (d − 2)nr² + 2nr, which is 23,360 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core1.2·10¹⁸train2.3·10⁴core ⁄ train4.9·10¹³entries ⁄ core9.1·10⁷the definition is repairedthe size is not

The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 8 it is 1.15·10¹⁸, smaller than the tensor by 9.09·10⁷ and still unstorable. A train is (d − 2)nr² + 2nr, which is 23,360 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

r: 3

The arguments are the ones A decomposition made only of SVDs 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.

Numbers stored against the number of indices, at n = 20 and rank 3: the tensor, its core, and a trainThe tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 3 it is 3.49·10⁹, smaller than the tensor by 3.01·10¹⁶ and still unstorable. A train is (d − 2)nr² + 2nr, which is 3,360 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core3.5·10⁹train3360core ⁄ train10⁶entries ⁄ core3·10¹⁶the definition is repairedthe size is not

The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 3 it is 3.49·10⁹, smaller than the tensor by 3.01·10¹⁶ and still unstorable. A train is (d − 2)nr² + 2nr, which is 3,360 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

r: 2

The arguments are the ones A decomposition made only of SVDs 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.

Numbers stored against the number of indices, at n = 20 and rank 2: the tensor, its core, and a trainThe tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 2 it is 1.05·10⁶, smaller than the tensor by 9.99·10¹⁹ and still unstorable. A train is (d − 2)nr² + 2nr, which is 1,520 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core10⁶train1520core ⁄ train690entries ⁄ core10·10¹⁹the definition is repairedthe size is not

The tensor is n^d, which at d = 20 is 1.05·10²⁶. A Tucker representation of it is r^d + d·n·r — the core is still exponential in d, so at rank 2 it is 1.05·10⁶, smaller than the tensor by 9.99·10¹⁹ and still unstorable. A train is (d − 2)nr² + 2nr, which is 1,520 — linear in d. The three lines are the field's whole argument: fixing the definition of the decomposition does not fix the size of what it returns, and the second fix is the same projection cut in a different place.

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.

3 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.

a rank the storage is counted at

and at two indices they are the same object

the core outgrows the train

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 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

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

An index that is a pair

A discretisation on a two-dimensional grid of n points a side has n² unknowns and a matrix with n⁴ entries — 10⁸ at n = 100. What that matrix is instead is two Kronecker products of an n × n matrix, which is 2n² numbers, and nothing has been approximated: assembling it was the mistake.

When the index is a tuple

The format that does not notice the dimension

A Tucker core is r^d numbers, so the format that repaired the definition still cannot go past five indices. Cutting between the indices rather than across them gives d − 1 ranks instead of d, storage linear in the number of indices, and a family whose ranks are two everywhere by an addition formula.

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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