Cluster tree — where it appears
Named by 11 essays across one field — each of them below, with the objects they name alongside it.
Which pairs are allowed to be small
A hierarchical representation is a partition of the matrix into blocks, and the rule that produces it reads four numbers per pair of index clusters and not one entry of the matrix. On a 256-square it yields 112 blocks, 66 of them stored as two thin factors, none of rank above five.
The test that costs what it saves
The partition that refuses to compress a touching pair keeps every rank at five while the other lets them climb from nine to thirteen. It also stores more numbers at every size measured — 67,968 against 61,440 at n = 512 — and which of those two facts matters is a question about how large the problem is going to get.
The same matrix, numbered twice
One symmetric permutation. The condition number is 24.3948 either way to eight digits and the Frobenius norm is 6.13996414·10³ either way to twelve. The partition that stored 27,008 numbers now finds no admissible pair anywhere and stores all 65,536, and the format that compresses regardless stores 118,208.
The offset that moved the slope
The accuracy is supposed to lift a storage curve and leave its growth alone. Measured at seven tolerances, the strong partition adds 10.9 numbers per unknown per doubling at two digits and 36.5 at twelve — the growth rate more than triples, so ten decades of accuracy cost 33 per cent more storage at 64 unknowns and 118 per cent at 512.
The partition that does not move
The storage curve's growth rate more than triples between two digits and twelve, and the earlier measurement said so without saying what moves it. There are two candidates and the measurement separates them decisively: the partition is identical at every tolerance — 250 blocks, 156 of them low-rank, 94 dense — and the mean rank rises by exactly one per two decades of accuracy. The slope is 5.11 numbers per unknown per doubling per unit of rank, with a worst residual of 0.03 over six tolerances.
Two knobs on one number
The tolerance raises every block's rank and leaves the partition alone. The leaf size does the opposite — it changes how many blocks there are and does not move a single rank. Both act on the same storage, and only one of them has a best setting: at eight digits the least storage is at a leaf of 16 and the largest leaf tried costs 59 per cent more. The best leaf rises with the accuracy, from 4 at two digits to 16 at twelve, and the influence runs only that way.
A second objective that is the first one doubled
Counting the operations a hierarchical product does was supposed to give the leaf size a second optimum, somewhere other than where the storage puts it. The count is 160,128, 139,904, 135,936, 155,136 and 216,064 against stored totals of 80,064, 69,952, 67,968, 77,568 and 108,032 — twice each, at every leaf, exactly. The two objectives cannot disagree, and what separates the leaf that stores least from the leaf that runs fastest is a charge of about 139 operations for reaching a block at all.
A geometry setting that is a second accuracy
The tolerance raises every rank and moves no block; the leaf moves every block and no rank. The constant that decides how far apart two clusters must be before their block may be compressed moves both — 592 blocks at mean rank 3.25 at one end and 94 at mean rank 8.94 at the other — and the storage falls the whole way while the error against the dense matrix rises by a factor of 3.5, at a tolerance that never changed. It has no best setting, and it is not a third knob on the storage.
A prediction that arrives a decade late
A rougher kernel raises every rank, a higher rank raises the side at which compressing a block stops paying, so the leaf that stores least should rise. Across four kernels whose largest rank runs 5, 5, 9 and 10, it is 16 on three of them and a tie between 8 and 16 on the fourth. The prediction is right in sign and out by a decade of tolerance, and the reason is a ceiling: an oscillation multiplies the rank by exactly two — 3 and 6, 4 and 8, 5 and 10, 6 and 12, 7 and 14 — at any frequency, because a cosine of a difference is a rank-two function.
A quarter of the leaf
The leaf that stores a hierarchical matrix most cheaply had been read off a table of nine points, with the suggestion that it depends on the mean rank alone. Across six kernels at five tolerances — thirty cells — it does, once the mean rank is read in the right place. Plotted against the mean rank at a fixed leaf of 16, two cells a tenth of a rank apart want leaves of 8 and 16. Plotted against the rank of the blocks a halving would create, one rule — halve the leaf while that rank is below a quarter of the leaf — predicts the best leaf on 28 of the 30, and the two it misses are the two whose rank sits within three per cent of the threshold. The fractional power the last measurement expected to be rough is smoother than 1/r.
One build tells the leaf
The rule that picks a hierarchical matrix's cheapest leaf — halve it while the rank at half the leaf is below a quarter of the leaf — reads ranks that exist only after the matrix has been built at every candidate leaf. Fed instead the rank a code can guess from the tolerance alone, one rank per two decades, it names the best leaf on twenty-three cells of thirty, and on the rank-one kernel it cannot see it stores up to seventy-two per cent too much. Doubling the guess for an oscillation, as earlier measurements suggested, makes it worse: eighteen of thirty. Building once, at a single leaf, and holding that build's mean rank fixed names the best leaf on twenty-nine — one more than the rule that builds at all five — and never stores three per cent more than the least.
Named alongside it
The objects these essays reach for when they reach for this one.
Hierarchical matrixAdmissibilityNumerical rankOff-diagonal rankLow-rank approximationToleranceStorageTruncationAsymptotic analysisOptimal-complexityWeak admissibilitySeparation ratio