The smaller cluster sets the rank

The kernel with nothing to compress held a block’s geometry at the separation ratio every compressible measurement in this field had used, $q = \tfrac12$, and scaled it up. The smooth kernel $1/r$ needed six columns at every scale. The oscillatory kernel $\cos(\kappa r)/r$ needed 12, 16, 22, 33 and 53, growing like the square root of the number of wavelengths across the picture, with no size at which it stopped. It also showed the block to be a function of $\kappa L$ alone: different wavenumbers and sizes with the same product gave the same integer rank.

Every block in that measurement was square in the physical sense — two segments of the same length $L$, a distance $2L$ apart. A hierarchical partition is mostly made of such blocks, because its tree pairs clusters of the same level. But the essay’s conclusion, that the rank is set by the size of the picture, has a hidden variable once the two clusters can differ: whose size. If the rank belongs to the larger cluster, an unequal block is as expensive as the square block on its long side. If it belongs to the smaller, an unequal block is as cheap as the square block on its short side, and a partition that pairs small clusters with large ones — which this field’s partitions never do — would escape the growth.

A grid of shapes at one separation #

The construction keeps everything the earlier measurement held and frees the one thing it did not. A target segment of length $a$ and a source segment of length $b$ lie parallel and centred on each other, a distance $D = a + b$ apart, so the separation ratio $q = (a/2 + b/2)/D$ is exactly one half for every pair of lengths. The kernel is $\cos(40 r)/r$, a wavelength of 0.157, sampled at twelve points per wavelength and never fewer than forty points a segment; the rank is the number of singular values above $10^{-8}$ of the largest, which is the reading a rank that is a number of digits established and rank is a decision qualified. Both lengths run over $\tfrac18, \tfrac14, \tfrac12, 1, 2, 4$ — from under one wavelength to twenty-five.

Down the diagonal the grid repeats the earlier essay: square blocks of side $\tfrac18$ to 4 need 7, 9, 12, 16, 22 and 33 columns. Off the diagonal it answers the question. Along the bottom row the source is fixed at four and the target shrinks: 33, 26, 18, 13, 10, 8. Along the last column the same, by symmetry — a block and its transpose have the same singular values, and the grid agrees with itself to within one column wherever the two are compared. The rank follows the smaller cluster. A target an eighth long costs eight columns against a source of any length measured here, which is what the smallest square block costs, and against the longest source it is a quarter of what the square block of that source costs.

The same thirty-six shapes with the smooth kernel $1/r$ in place of the wave are the control, and they show the question has no content there. Every shape needs between four and six columns; the square blocks need six and the most lopsided four, a two-column difference that is the whole of the shape effect for a kernel with no length of its own. A smooth block’s rank is decided by the separation ratio, which is the same everywhere on this grid, so it has nothing left to depend on. For the wave kernel the separation ratio is also the same everywhere, and the rank ranges from seven to thirty-three: everything that moves it is a length measured in wavelengths, and the grid is the test of which length.

Lengthening one side stops mattering #

Read along a row, the rank climbs while the source is shorter than the target and then stops. With the target at one half, sources of an eighth, a quarter, a half, one, two and four need 8, 9, 12, 13, 13 and 13 columns; once the source is twice the target, doubling it again adds nothing. Turn the dial: a target of a quarter levels off at 10 or 11 and a target of one at 18, while a target of two is still climbing, at 26, when the source reaches four — the longest measured, and only twice the target. Each row’s plateau sits between the square block of the target’s side and the square block of twice that side, while the square blocks’ own line keeps climbing, because both of their sides grow.

That is the whole practical content of the measurement in one sentence. The cost of an oscillatory block is set by its shorter side, to within a factor of two in length, and a source can be made as long as the problem likes without the rank noticing. The earlier essay’s growth was real for the blocks it measured, and those blocks were exactly the ones in which the shorter side grows too.

What the rank is a function of #

Holding one side and moving the other is one way to vary a block; there is a single number that the whole grid, and every other shape at this separation, turns out to depend on almost alone. Call it

$$F = \frac{\kappa\, a\, b}{2\pi D},$$

the product of the two lengths divided by the distance, counted in wavelengths. For the square blocks it is $\kappa L/4\pi$, a quarter of the wavelength count the earlier essay used, so its collapse in $\kappa L$ is this collapse restricted to squares. For a short target against a long source, $D$ grows with $b$ and $F$ tends to $\kappa a/2\pi$ — the number of wavelengths across the short side — which is why the rows level off.

Plotted against $F$, the thirty-six shapes of the grid fall on one rising band, and no block with a larger $F$ than another by more than a tenth ever needs fewer columns. Blocks built deliberately at fixed $F$ and stretched sit just under the square blocks at the same $F$. The square blocks run 8, 10, 13, 18, 25 and 39 at $F$ of a half, one, two, four, eight and sixteen, a growth exponent of 0.53 between $F = 2$ and $16$ — the earlier essay’s square root, recovered in the new variable.

Where $F$ comes from can be read off the kernel. Across the block the path length $r$ from a target point to a source point varies, and the phase $\kappa r$ varies with it; to first order in the offsets along the two segments, the part of $r$ that couples a target offset to a source offset is their product over the distance, $xy/D$. The largest such cross term over the block is $ab/4D$, and in wavelengths that is $F/2$. A separable approximation has to follow that cross phase, and the number of oscillations it makes across the block is what sets how many terms it needs.

The number has a name outside linear algebra. Two apertures of widths $a$ and $b$ a distance $D$ apart, exchanging waves of wavelength $\lambda$, can carry about $ab/\lambda D$ independent beams between them: it is the Fresnel number of the pair, and in optics it counts the degrees of freedom of the field one aperture can put on the other. With $\lambda = 2\pi/\kappa$ it is exactly $F$. The singular vectors of a kernel block are those beams — the pairs of patterns on the two segments that couple most strongly — and the rank at $10^{-8}$ is how many of them are needed before the rest are too weak to matter. The kernel here is a cosine, which is a wave going each way, so the count applies to each direction.

Why the long side adds nothing #

The optics reading also says why the rows of the grid level off, and the argument is short enough to give in full. Stand on the short segment and look at the long one. A field on the short segment can only distinguish directions that differ by about a wavelength over its own length — an angular resolution of $\lambda/a$. The long segment fills an angle of about $b/D$ as seen from the short one. The number of distinguishable directions under which the short segment sees the long one is the angle divided by the resolution, $ab/\lambda D$, which is the Fresnel number again, reached from one end.

Now lengthen the source while keeping the separation ratio. The distance $D = a + b$ grows with $b$, so the angle $b/D$ does not grow without limit: it approaches a fixed value, the angle a long segment subtends from a point a comparable distance away. The resolution $\lambda/a$ has not changed. So the count of distinguishable directions stops growing, and with it the rank — at about the number of wavelengths across the short side, which is where the rows of the grid flatten. The source is longer, and every additional stretch of it is seen through directions the short segment has already resolved.

The square blocks have no such ceiling, because both factors grow: a larger target resolves finer directions and a larger source fills the same angle with more of them to resolve. That is the earlier essay’s growth, and it is why square blocks, and only square blocks, cannot be escaped by making the problem larger. A block nobody can call sparse called a smooth kernel’s block a small number of patterns that matter; for a wave kernel the patterns are directions, and their number is the one the two apertures can exchange.

Stretching at a fixed Fresnel number #

If the rank were exactly a function of $F$, stretching a block at fixed $F$ would change nothing. It changes a little, always in the same direction: at $F = 8$ the square block needs 25 columns, the 1:4 block 23 and the 1:16 block 21; at $F = 2$, 13, 12, 11 and 11 out to 1:64; at $F = 16$, 39 and 33. From square to 1:16 the rank falls by one to four columns at every $F$ measured, never rises. So the square block is the worst shape at a given Fresnel number, which is the useful direction for the error to go — a rank budget set by the square blocks of a partition is an upper bound for every other shape with the same $F$.

The extreme case makes the point without a plot. At $F = 2$ and a ratio of sixty-four, the target is 0.32 long and the source 20.4, which is 130 wavelengths. That block, 40 by 1,560 samples, has rank 11. The square block at the same $F$, 0.63 on a side, has rank 13.

The singular values show where the difference lives. Both spectra have the same shape — a plateau of values near one, then a cliff — and the cliff is where the Fresnel count of well-coupled beams runs out. The 1:16 block, 1.34 by 21.4, has its cliff four values earlier than the square block of side 2.51, although it is sixteen times longer on one side and has eight and a half times as many samples on that side. The long side adds length and does not add directions: every extra point on it sees the short side through the same narrow set of beams.

What it costs, counted in numbers stored #

A rank is a count of columns, and what a format pays is the rank times the two sides’ sample counts. On the grid the two readings agree in direction and differ in size. The square block of side four stores $33 \times (306 + 306) = 20{,}196$ numbers against 93,636 dense, a saving of 4.6 times; the block of an eighth against four stores $8 \times (40 + 306) = 2{,}768$ against 12,240, a saving of 4.4 times. The lopsided block is not more compressible as a ratio — it is smaller in the first place, and its rank is low enough that its compressed form is dominated by the long side’s samples, one column’s worth of numbers per unit of rank. What the smaller cluster buys is that the long side’s cost enters linearly and not through the rank.

That is the sense in which where the format starts paying has to be re-read for a wave kernel. The crossover it computed assumes a rank that does not grow; a square block’s rank does, and a lopsided block’s rank, at fixed short side, does not. The format pays on the blocks whose short side stays short, which on a partition of equal halves means only the blocks near the leaves.

What this changes for a partition #

Which pairs are allowed to be small built the partition every hierarchical format in this field uses: split both clusters in half at every level, and compress a pair when its separation ratio is small enough. That rule only ever forms blocks whose two sides are the same length, and on a smooth kernel it does not matter, because a smooth block’s rank is set by $q$ alone — the size the rank does not notice measured it flat across an eight-fold change in sampling.

On an oscillatory kernel it matters twice. First, the square blocks are the worst shape at each Fresnel number, so the partition is choosing the most expensive way to cover each interaction. Second, and more usefully, the measurement says what a cheaper covering looks like: pair a short target with a long source, so that the product of the two lengths over their distance stays bounded while one side grows. A family of factorisations built on that principle exists — the butterfly, which at each level halves one side and doubles the other — and what is measured here is the property it relies on, checked at one separation on one kernel: holding $ab/D$ fixed holds the rank.

That is also where the test that costs what it saves comes back with a different sign. There, a stricter admissibility test bought lower ranks with more blocks. A test written in the Fresnel number rather than in the separation ratio — compress a pair when $\kappa ab/D$ is small, whatever $q$ is — would admit lopsided pairs that the separation test and the level structure never produce, and whether that covers a matrix in fewer numbers is a question about the whole partition, which this measurement does not answer.

What one separation and one kernel leave out #

Everything here is at $q = \tfrac12$, two parallel segments centred on each other, one kernel, one accuracy. A hierarchy with no grid behind it builds its clusters from scattered points, and a cluster of scattered points has no single length; the count would need the extent of each cluster perpendicular to the line joining them, which this measurement does not vary. Segments that are offset sideways see each other at an angle, and the Fresnel count then involves the projected lengths; segments in a plane or a volume have two-dimensional apertures, and the count becomes a product of two such numbers. The kernel is real, a standing wave, where the kernels of scattering are complex travelling waves; the real kernel’s rank is at most twice the complex one’s and probably close to it. And the accuracy $10^{-8}$ adds a margin of columns to the Fresnel count that grows as the accuracy tightens, which is the separable part of the earlier essays’ story riding on top of this one.

Still open: where the square root turns straight, and a partition built on F #

The large-F regime. The optics count predicts that the number of well-coupled beams grows in proportion to the Fresnel number, not as its square root. Divided by the square root of $F$, the square blocks’ ranks here are flat at about nine from $F = 2$ to $8$ and rise to 9.8 at $16$, and the earlier essay’s largest block, at $F$ of about 25, needed 53 — 10.5 times the square root. The prediction with a sign is that past $F \approx 30$ the rank becomes a straight line in $F$ with a slope near two, one beam for each direction the real kernel travels in, and that the square-root law was the regime in which the margin of the accuracy dominates.

A partition written in F. The measurement says a lopsided block is cheaper than a square one at the same Fresnel number and much cheaper than a square one of its long side. A partition that pairs clusters by $\kappa ab/D$ instead of by level and separation — with the storage summed over the whole matrix, dense blocks included — is the experiment that would say whether that saving survives being assembled, on the same segment geometry the earlier essays used.

Sideways offsets. Two segments that do not face each other see each other foreshortened. The prediction is that the rank follows the Fresnel number computed from the lengths projected perpendicular to the line between their centres, so that a long source seen end-on behaves like a short one — the directional statement the high-frequency methods build on, and one number to test it.