The readings, arranged in a circle
Reading at c − r
Reading at c + r
Pairs that match
Lesson 08 · Symmetry · 50–65 minutes
Symmetry & Palindromic Sequences
On the 5-mark dial the simplest clock reads 0, 1, 3, 1, 0: the same forwards and backwards. That is no accident. In Lessons 03 to 07 the mirror pairs k and 1 − k kept doing work. Here you will find the mirror for any growing clock on a prime dial, compute where it sits without listing readings, and see the same mirror in tables of clocks.
Where this lesson is headed
mirror center c ≡ (g − 2N) × (2g)−1 (mod p)
In words: the readings of a growing clock on a p-mark dial read the same going either way around one special rotation c, and c can be computed from N and g. Setup: a clock with starting circumference N and growth g, read on a dial with p marks (p an odd prime, as in Lesson 02 the dial is chosen freely). The reading after k rotations is S(k) = kN + gT(k) mod p.
N: starting circumference (only its reading on the dial matters)
g: growth per rotation, not a multiple of p
p: number of dial marks, an odd prime
c: the mirror center, a rotation count
Mirror: S(c − r) and S(c + r) read the same for every r
01. Find the mirror
Start with the simplest clock (N = 1, g = 1) on a 5-mark dial. The circle shows rotations 0 to p − 1 in order, each with its reading inside. Since readings repeat every p rotations, rotation p is rotation 0 again, so the list really is a circle. Guess a center c and test whether the readings match in pairs c − r and c + r.
Guided experiments
Do these in order. "Set this up" sets the clock; you move the center.
The same mirror in a table of clocks
Stack clocks in a table: row i is the clock with starting circumference N = i and growth 1, so entry (i, k) is ik + T(k) read on p marks. Pick an anti-diagonal, the cells with i + k reading a. Its entries, row by row, form a sequence.
Change a. Does the mirror center of the anti-diagonal move?
A second table: growth rows
Now row g is the clock with growth g on the special dial N = p (Lesson 02, step 5), so entry (g, k) is gT(k) read on p marks.
02. Compute the mirror center
Each step has the general idea on the left and your clock from Stage 01 on the right.
1
The readings come from a quadratic
From Lesson 01, step 5:
S(k) = kN + gT(k) = g2k² + (N − g2)k
A parabola is symmetric about its vertex. On a dial the same thing happens, but the "vertex" is a rotation count found by dial arithmetic. A mirror at c means S(c − r) and S(c + r) read the same for every r.
2
Compare two readings the same distance from c
Expand T(c ± r) = (c ± r)(c ± r − 1)/2 and subtract:
T(c + r) − T(c − r) = r(2c − 1)
The straight part adds N·(c + r) − N·(c − r) = 2rN, so
S(c + r) − S(c − r) = r · [2N + g(2c − 1)]
For every pair to match, the bracket must read 0 on the dial. (With r = 1 it has to; and if it does, every r works.)
3
Solve for c
2N + g(2c − 1) ≡ 0 ⇔ 2g · c ≡ g − 2N (mod p)
p is an odd prime and g is not a multiple of p, so 2g shares no factor with p and can be undone (Lesson 02, step 4). Undoing it gives exactly one center:
c ≡ (g − 2N) × (2g)−1 (mod p)
(2g)−1 is the number that multiplies 2g to read 1 on the dial.
Try it by hand
For p = 13, N = 2, g = 3: 2g = 6 and g − 2N = −1, which reads 12. Find 6−1 on 13 marks, then c.
4
Check both directions
If there is a mirror, the r = 1 pair matches, which forces the equation of step 3, so c must be the value found there.
If c is that value, the bracket reads 0, so r × bracket reads 0 for every r: every pair matches, not just the ones we tested.
5
Why anti-diagonals all share one center
On the anti-diagonal i + k ≡ a, write the row as ℓ and the column as a − ℓ. Then the entry is
ℓ(a − ℓ) + T(a − ℓ) ≡ T(a) − T(ℓ) (mod p)
(expand T(a − ℓ): the aℓ terms cancel). So every anti-diagonal is a constant minus the same sequence T(ℓ). T has its mirror at c = (p + 1)/2, the lone rotation from Lesson 04, since then 2c − 1 = p reads 0. The center does not depend on a.
6
A mirror sum: partners add up to a square
A different kind of mirror: add the readings at k and m − k instead of comparing them. Expanding,
T(k) + T(m − k) = T(m) + k² − mk
On an m-mark dial mk is whole trips, so
T(k) + T(m − k) ≡ T(m) + k² (mod m)
On an odd prime dial T(p) reads 0, so the two add up to the square k². On an even dial T(m) reads m/2, the halfway mark from Lesson 05, which will come back in Lesson 10.
03. Prove the three symmetry facts
All three come from one kind of calculation: comparing a quadratic at two points. Check the key steps, then read the full proofs.
Proof workshop
Complete proofs
- Every clock has one mirror. For S(x) = Nx + gT(x), subtraction gives S(c + r) − S(c − r) = r[2N + g(2c − 1)]. If every pair matches, take r = 1 to get 2gc ≡ g − 2N (mod p). Since 2g shares no factor with p, this has the single solution c ≡ (g − 2N)(2g)−1. Conversely, for that c the bracket reads 0, so every pair matches. Because readings repeat every p rotations, rotation counts can be taken around the circle.
- Anti-diagonals. On i + k ≡ a set i = ℓ and k ≡ a − ℓ. Expanding, ℓ(a − ℓ) + T(a − ℓ) = T(a) − T(ℓ). For h = (p + 1)/2, T(h + r) − T(h − r) = r(2h − 1) = rp, a multiple of p. So the anti-diagonal reads the same at h − r and h + r for every r, whatever a is.
- Growth rows. T(p + 1 − k) = T(k) + p × (whole number) reads the same as T(k), since this is the mirror pair k ↔ 1 − k shifted by p. Multiplying by g keeps it. So each row of the growth table, from column 2 to column p − 1, reads the same both ways, and the full row is a mirror around (p + 1)/2.
- Mirror sum. T(k) + T(m − k) = T(m) + k² − mk, and mk is whole trips on an m-mark dial. For the whole clock the straight parts add to kN + (m − k)N = mN, also whole trips, so S(k) + S(m − k) ≡ g[T(m) + k²] (mod m). On an odd prime dial T(p) reads 0, leaving gk². ∎
The mirror sum and the mirror match are different statements: one adds the readings at k and m − k, the other compares readings at c − r and c + r.
Computer check
For nine odd prime dials, tests every N and every allowed g by comparing readings directly. Also checks every anti-diagonal, every growth row, and the mirror sum on every dial up to 80. Many matches are strong evidence, but only the proofs cover every case.
Not run yet.
Exit problem · compute, don't guess
Find the two mirror centers with the formula, and one table entry with T(a) − T(ℓ).