The Weight of a Carry

The collision program begins with a simple digit question. In long division, how often do two digits a fixed distance apart agree? As the arithmetic inputs vary, the number of agreements changes.

These counts lead to a finite carry table. An exact rearrangement turns its centered entries into samples of a continuous wave made of rising, resetting ramps. Collision capacity is the average squared size of that wave. It measures the strength of the response underlying the table.

The capacity has a useful description in terms of remainder clocks. As the size of the calculation increases, a period-two clock reads $0,1,0,1,\ldots$. A period-three clock reads $0,1,2,0,1,2,\ldots$. There is a clock for every later period too. These periods describe the remainder counters, rather than the length of a repeating decimal.

Each clock receives a positive weight telling us how much it contributes. Their weighted readings, multiplied by $2/3$, give the shortfall of the capacity below its leading growth. The weights add to one. In this sense they divide one unit of mass among the clocks. Period two receives a quarter of that unit and period three a sixth.

Many other assignments would still add to one. The question is why the digit calculation forces these particular shares.

This paper derives them from ordinary addition and the rows supplied by the digit rule. It also shows how much the capacity retains. Within a specified regular class, recording one capacity at every resolution is enough to recover the entire primitive boundary that produced it.

A carry on a five-position clock

Number the positions zero through four. Adding three to three takes us past five and leaves a remainder of one. There has been one carry. Doubling every position gives a small table.

Residue	0	1	2	3	4
Doubled residue, reduced modulo 5	0	2	4	1	3
Carry	0	0	0	1	1

The carry has mean $2/5$ and variance $6/25$. Divide the ordinary residue by five and subtract its mean. That centered row has variance $2/25$. The carry variance is exactly three times as large.

There is a structural reason for the three. On the whole addition square, choose two residues independently and uniformly. Each residue and their reduced sum have the same uniform distribution. Any two of these three quantities are independent. After centering, the variance of the first plus the second minus the reduced sum is therefore the sum of three equal variances.

The collision calculation samples a shared phase instead of two independent residues. Its rows are $aA$, $bA$ and $(a+b)A$ modulo the same integer. When all three multipliers permute the residues, the carry has the same mean as on the addition square. A carry with only the values zero and one has its distribution fixed by that mean.

Expanding the resulting variance identity gives

$$\langle U_a,U_b\rangle =\langle U_a,U_{a+b}\rangle +\langle U_{a+b},U_b\rangle.$$

The angle brackets measure how two centered rows vary together. The parent pairing divides exactly into two child pairings. In the small table above, $2/25$ divides into $1/25+1/25$.

The shared-phase rows are dependent. Their covariance is the quantity being conserved.

The choice of measurement

The argument uses squared size. The paper also asks how much freedom there is in that choice.

Suppose the cost of a centered value depends only on that value. Use one continuous, even rule $\Phi(t)$ at every modulus and every amplitude, with zero cost at zero. Average these individual costs to measure a row. Require the pairing formed from the costs of a sum and a difference to obey the carry-refinement law.

Then the rule must be

$$\Phi(t)=ct^2.$$

Tests at growing odd moduli force doubling a value to multiply its cost by four. A calculation modulo five forces tripling to multiply the cost by nine. Continuity and those two independent scaling relations determine the whole rule. One variance calibration fixes its scale.

The assumptions matter. A measurement that first combines the entire row and then squares its variance can also pass the carry tests. Locality is what excludes that alternative. The theorem identifies quadratic measurement within the stated class.

The rows supplied by digits

Addition leaves the choice of rows open. The digit condition supplies it.

In base five, the two-digit words with equal digits are 00, 11, 22, 33 and 44. Their centered carry table admits an exact reordering of phases. After that reordering, its response is twice the sum of the first four centered sawtooth rows.

Each sawtooth rises steadily and resets at an integer. The same straightening works in every integer base and at every two-point lag. The lag is the separation between the two digit positions being compared. In base $B$ at lag $\ell$, the consecutive frequency rows end at $N=B^\ell-1$.

Their labels contain information a scalar energy cannot show directly. A reduced frequency pair first appears at a definite cutoff. Higher cutoffs admit exact common-scale copies of that pair. Counting those copies fixes its period address and contribution.

There is one small but necessary correction. The ordinary centered residue row and the sampled sawtooth differ at the endpoint. Dropping that term leaves a refinement defect of $1/25$ at modulus five. The proof restores it before identifying the affine boundary pairing. This keeps the finite table and the continuous capacity properly connected.

Splitting a boundary

For coprime positive coordinates $(a,b)$, the affine pairing gives boundary mass $1/(2ab)$. The pair $(2,3)$ has mass $1/12$. Its children $(2,5)$ and $(5,3)$ have masses $1/20$ and $1/30$.

There is a literal interval behind this calculation. The neighboring fractions $1/2$ and $2/3$ are separated by $1/6$. Insert their mediant, obtained by adding numerators and denominators, to get $3/5$. The two new gaps have lengths $1/10$ and $1/15$. Each boundary mass is half its interval length.

The source labels force the period address $a+b$. Dividing the boundary mass by that period gives the kernel

$$K(a,b)=\frac1{2ab(a+b)}.$$

The fractions assigned to the clocks now have a source. Period two has the single pair $(1,1)$, which contributes $1/4$. Period three has $(1,2)$ and $(2,1)$, each contributing $1/12$. Together they give $1/6$.

The same calculation assigns every later period its share. Refining the primitive tree proves that the total is one. Combining the weighted, centered remainder ages gives the global potential in mean square.

This also gives a uniqueness test at the boundary. Among symmetric kernels of degree minus three with a continuous normalized angular profile, conservation under this same subdivision forces the displayed reciprocal form. Unit mass fixes its constant.

Recovering the boundary

The forward calculation reduces many pairs to one capacity at each resolution. It is reasonable to expect that reduction to lose the source. Under the paper’s regularity assumptions, the complete sequence retains the primitive boundary ledger.

Take differences between successive capacities. Divisor inversion separates the period weights. The full set of weights then recovers the symmetric angular profile through grid averages and Fourier inversion. This last step requires three continuous derivatives. The complete global potential determines the same data in the fixed phase coordinates.

The first four capacities are $1/3$, $1$, $5/3$ and $23/9$. From those four numbers, the inverse recovers the masses $1/4$, $1/6$ and $1/12$ at periods two, three and four. Each additional resolution exposes another period. The complete sequence recovers the boundary, although any finite record leaves room for a different one.

A finite prefix is different. The paper constructs smooth positive boundary rules that agree with collision through any prescribed finite cutoff and differ later.

Exact recovery also has a cost. The paper gives a signed arithmetic readout that recognizes the collision mass through the prime-power coefficients of the von Mangoldt function. At a growing first changed address, its sensitivity to a potential perturbation has sharp order $n^2\log n$. A precise inverse need not turn a small average error into a small recovered coefficient.

The quarter at period two is fixed by the local calculation. So are the sixth at period three and every weight after it. Together they determine a global state whose complete capacity sequence allows the primitive boundary to be recovered. The passage from a finite carry to that state can be followed in both directions. That is what makes these capacities more than a measure of how much collision occurs.