Interactive companion · page 2

Digital shifts and the question of Dick, Goda and Suzuki

Dick, Goda and Suzuki asked for digital nets that are well separated at every size, and they allow a digital shift: a random constant added to every digit. This page shows what a shift does to the nets of the paper, why most of the separation bounds survive every shift, and which constructions are fragile.

1

Adding digits without carrying

A digital shift adds a fixed digit to each digit of each coordinate, modulo q, and throws the carries away. In base 5, the coordinate 0.34 shifted by 0.21 becomes 0.00, not 0.10 as ordinary addition would give. Each coordinate gets its own shift digits, one per digit position.

Without carries, a shift sends every elementary box to another box of the same shape: the first d digits of a coordinate are only relabelled. So a net with one point in every box keeps that property, whatever the shift. A digital shift is the simplest of Owen's nested scrambles, which all preserve the net property (Owen, 1995). It is the digital counterpart of the random shift of Cranley and Patterson. Click a point to see its digits; click a shift digit to raise it by one.

In the paper: def:dshiftlem:shiftshifted graph formsrem:lowdisc
Net
Shift digits (underlined: the digit after the cell digits)
2

Why the bounds survive every shift

The separation proofs never look at a single point. They look at two points in nearby cells and at the change of each digit between them. That change is an integer between −q and q, and its remainder modulo q is fixed by the cell step alone. Only two integers have that remainder: σ and σ − q, the second one when the digit wraps around. The proofs already take the worse of the two.

A shift adds the same constant to a digit at both points. The remainder of the change stays the same; the shift only decides at which cells the digit wraps, so which of the two values occurs. Since the bound covers both values, it holds for every shift. In the paper's terms, a shifted graph form has offsets Φn + γ with a constant γ, and every proof uses only differences, in which γ cancels.

Try it on the net 𝒫2,q of page 1: shift its eighteen digits however you like, and the exact separation stays above the bound c(q) of the theorem. The table follows the closest pair of the shifted net digit by digit.

In the paper: lem:shiftprop:shiftthm:mainrem:ppshiftper-prime netscor:allmshift
5
Shift digits (digit 3 is the geometric digit)
Each dot is the exact separation of one shifted net. Click a dot to load its shift.
3

Column nets: three ways a shift can act

In the column net 𝒞q of page 1 all three coordinates read the same second and third digit, so every point sits on the diagonal of its cell. A shift with equal digits in the three coordinates at these two positions keeps that structure and leaves the minimum distance exactly as it was. If only the second digits agree, the distance can change, but δN1/3 stays above 1 − (7q − 3)/(√3 q2), close to the cube value 1.

If the second digits differ, the three coordinates no longer wrap at the same cells. The ring shows, for one step direction, the values of the shared digit at which each coordinate wraps: its wrap window. A shift turns each window by that coordinate's own shift digit. Where one coordinate wraps alone, the step along its axis shrinks, and the value falls to about 1/√3. A lower bound still holds for every shift, so the family stays well separated, only with a smaller constant.

In the paper: prop:colfreeprop:colshiftrem:shiftsharedwrap windowscor:allmshiftthm:col
Shift digits (first, second, third)
Step along
4

A fragile lock

The net 𝒲p of page 3 beats the cube value 1 because its wraps are locked: z reads a digit ζ, and x and y both read 2ζ mod p, so z wraps only where x and y wrap too. A shift of the second digits by (t, t, t′) gives x and y the digit 2ζ + t − 2t′ after a translation of the cells. If t ≡ 2t′, this is the same lock. If t ≡ 2t′ + 1, the digit wraps one step earlier, which allows the same wrap patterns, so the proven bound still holds (for p ≥ 53).

Every other shift lets z wrap alone at some cells, and there the step along ez shrinks to about (4ε, 4ε, 2ε). The map on the left tries every shift of the second digits for a small prime. That only these two lines keep the lock is a computed observation, not a theorem.

In the paper: rem:wshiftprop:wnetlem:wwrapfig:wnetpanel (c)

Columns: the shift t of the second digit of x and y. Rows: the shift t′ of z. Colour: the exact δN1/3 of the shifted net, from pale (near 0) to full blue (near 1). Click a square to select that shift.

0
0
5

Bases two and three: every shift at once

In bases 2 and 3 the nets of the paper come from a lift that adds three digits at a time, and their bounds come from finite certificates instead of the lattice picture. For these, the worst shift can be computed exactly. Take two points whose digits differ by d. A nonzero digit of weight bk changes the coordinate by dkbk, or by (dk − b)bk when that digit wraps, and a shift can choose the wraps freely, digit by digit and coordinate by coordinate. The worst case μ(d) is the distance from d to the nearest sum of the numbers bk+1, one for each wrapped digit.

So the smallest separation over all shifts is a minimum over the differences of points of Σ μ(Δc)2, a minimum weight of a code. It drops to almost nothing when some difference reads 0…01…1 in every coordinate: a shift can put both points on a carry chain, such as 0.00111 and 0.01000 in base 2: their digits differ by 0.01111, their values by one unit in the last place. These carry chains are why the every-shift certificates of the paper vanish at some resolutions and appear at others.

In the paper: lem:shiftcertrem:shiftparitythm:base23part (3)tab:base23

The worst wrap of one digit difference

Base
Digit difference d, most significant first (click to change)

The worst shift of a whole net