The small m-mark dial
Rotation k0
Small-dial reading0
k − m read·
Lesson 10 · Capstone · 55–70 minutes
The Half-Modulus Unification
The halfway mark m/2 has turned up again and again: as Lesson 02's half-dial jump, Lesson 05's skipped mark, Lesson 06's +1 bonus, and T(m) in Lesson 08. In this capstone you will see that these are one thing. A single block of rotations turns every reading by the same amount, and on an even dial that amount is exactly half a turn. You will prove it, use it to explain the bonus, and see precisely what it does and does not imply.
Where this lesson is headed
T(k + m) ≡ T(k) + m/2 (mod m), for m even
In words: one more block of m rotations turns every reading half a dial. The half turn swaps each mark x with x + m/2. On a power-of-two dial it swaps the doubly hit mark 0 with the skipped mark m/2, which is exactly where the bonus comes from. For odd m the block turns the dial by 0.
m: marks on the small dial (Lesson 06: m = N/gcd(N, g))
d = gcd(N, g): how many blocks of m fit in N rotations
u = g/d: the growth on the small dial (shares no factor with m)
A: marks reached in the 1st block; B(m) = size of A
τ(x) = x + m/2: the half turn
J(x) = x + T(m): what one block does to a reading
01. Watch the 2nd block turn the dial
Start with Lesson 06's bonus clock, N = 8 and g = 2. Its small dial has m = 4 marks and there are d = 2 blocks. The 1st block skips mark 2 and hits 0 twice. Turn into the 2nd block and compare each reading with the one exactly m rotations earlier. (Shortcut: Space turns one rotation, ← undoes one.)
Guided experiments
Do these in order. "Set this up" chooses m, d and u and turns the clock for you.
The same number, twice
Question 1 · the bonus
Is the halfway mark 2 already in A, the 1st-block set?
Question 2 · the block jump
J(x) = x + T(m) is what one block of m rotations does to a reading x (for the growth part). Which reading does it send to 0?
Compare dials (d = 2, u = 1)
Click a row to load it.
| m | Type | B(m) | Added by 2nd block | J sends to 0 |
|---|
02. One half turn, several consequences
Each step has the general idea on the left and your dial from Stage 01 on the right. Almost every step is a result you proved in an earlier lesson, now put side by side.
1
A block adds T(m) to every reading
Replace k by k + m in k(k − 1)/2 and subtract:
T(k + m) − T(k) = km + T(m)
km is whole trips on the m-mark dial, so T(k + m) ≡ T(k) + T(m). Every reading moves by the same amount, T(m), whatever k is (Lesson 02).
2
What T(m) reads
m odd: (m − 1)/2 is whole, so T(m) = m × (m − 1)/2 reads 0.
m even, m = 2h: T(m) = h(2h − 1) = mh − h, which reads −h, and −h + 2h = h. So T(m) reads m/2.
T(m) ≡ 0 (m odd), T(m) ≡ m/2 (m even)
This is Lesson 02's jump and Lesson 08's T(m) ≡ m/2 in one line.
3
The half turn and its pairs
For even m, call τ(x) = x + m/2 the half turn. Doing it twice turns the dial m/2 + m/2 = m, a full trip, so τ(τ(x)) = x. The marks fall into pairs {x, x + m/2}, and τ swaps each pair.
Because of this, the 3rd block equals the 1st, the 4th equals the 2nd, and so on: readings repeat every 2m rotations (Lesson 02).
4
The growth u does not change the half turn
On the small dial the reading is u·T(k), so a block adds u × m/2. Since u shares no factor with the even m, u is odd, and
u·(m/2) − m/2 = m·(u − 1)/2
is whole trips. So u·(m/2) reads m/2: still a half turn.
5
The bonus is what the half turn adds
Let A be the 1st-block marks. With d ≥ 2 blocks, the rotations reach A ∪ τ(A), so
C(N, g) = B(m) + (marks in τ(A) but not in A)
m = 2, 4, 8, …: A is every mark except m/2 (Lesson 05). Then τ(A) is every mark except 0, and the half turn adds exactly m/2: +1.
other even m: m/2 is already in A, and then τ(A) = A (the no-bonus lemma, proved in Lesson 07): +0.
odd m: the block turns by 0: +0. d = 1: there is no 2nd block: +0. m = 1: one mark, nothing to add.
Check a count
6
The block jump has exactly one zero
J(x) = x + T(m) turns the dial by a fixed amount, so it hits every mark once; exactly one reading goes to 0:
J(x) ≡ 0 ⇔ x ≡ −T(m)
For even m that is x = m/2 (since −m/2 and m/2 are the same mark); for odd m it is x = 0. So the reading that one block sends to 0 is the halfway mark, the same mark that is skipped on a power-of-two dial.
A table of jumps: put J on a grid, entry (x, m) = x + T(m) on m marks. Along x = n², m = 2n the entry is n(n + 1) mod 2n, which reads 0 exactly when n is odd.
| x | J(x) | zero? |
|---|
03. Prove the unification, and its limits
Every claim keeps its conditions. In particular, the half turn exists on every even dial, but it only produces a bonus on powers of two.
Proof workshop
Complete proof
1. The engine. T(k + m) = T(k) + km + T(m), so on m marks T(k + m) ≡ T(k) + T(m). For even m, T(m) ≡ m/2, so each block acts on readings by the half turn τ(x) = x + m/2.
2. Growth. Write d = gcd(N, g), m = N/d, u = g/d (Lesson 06). As u shares no factor with the even m, u is odd and u·(m/2) ≡ m/2. So the 2nd block of the small-dial clock reaches τ(A), where A = {u·T(k) : 0 ≤ k < m} on m marks.
3. Counting. For d ≥ 2 at least two blocks occur, and since τ undoes itself later blocks add nothing further, so C(N, g) = B(m) + (size of τ(A) minus A). If m = 2r ≥ 2, Lesson 05 gives A = every mark except m/2, so τ(A) adds exactly one mark. For other even m, the no-bonus lemma (Lesson 07) gives τ(A) = A. For odd m the shift is 0. If d = 1 no 2nd block occurs; if m = 1 there is nothing to add.
4. The zero of the jump. J(x) = x + T(m) is 0 exactly when x ≡ −T(m), which for even m is m/2. So the mark added in the power-of-two case and the zero of the block jump are the same mark. This does not say every even dial has a bonus: the half turn exists on every even dial, but it adds a mark only when A is missing m/2.
5. The jump table. Entry (n², 2n) is n² + T(2n) ≡ n(n + 1) (mod 2n), which is a multiple of 2n exactly when n + 1 is even, that is, n odd. ∎
Edge cases · don't overgeneralize
| Case | Added | Why |
|---|---|---|
| m = 8, d = 2 | +1 | A skips 4, and the 2nd block happens. |
| m = 8, d = 1 | +0 | A skips 4, but there is no 2nd block. |
| m = 12, d = 2 | +0 | A already contains 6, so τ(A) = A. |
| m = 1, d = 3 | +0 | A 1-mark dial has nothing to add. |
| m = 7, d = 2 | +0 | T(7) reads 0: no turn at all. |
| jump table, n = 4 | reads 4 | The (n², 2n) entry is not always 0. |
What rests on what
The block shift and the zero of the jump are pure algebra, proved here. The full bonus rule also uses Lesson 05 (powers of two) and Lesson 07 (the no-bonus lemma). The computer check is evidence, never a replacement for these proofs.
Computer check
Checks the block shift, that u keeps the half turn, the full bonus rule against direct listing, the unique zero of the jump, and the jump-table entries, all over stated ranges.
Not run yet.
Final capstone · four answers, one mark
Take m = 8, d = 2, u = 1, so N = 16 and g = 2. Work out each answer from the formulas, then check with the clock.
The course in one page
| Lesson | Main idea |
|---|---|
| 01 · The Growing Clock | Total distance S(k) = kN + gT(k), a quadratic. |
| 02 · Distance as a remainder | Read S(k) on an M-mark dial; one block jumps by 0 or M/2, so readings repeat within 2M. |
| 03 · The missing remainders | Mark a is reachable exactly when 8a + 1 is a square. |
| 04 · Counting the reachable marks | Rotations pair up as k and 1 − k: (p + 1)/2 marks on a prime dial. |
| 05 · Powers of two | The 1st block skips only m/2; the 2nd block fills it in. |
| 06 · The +1 bonus | Shrink the dial by gcd(N, g); a bonus exactly when d > 1 and m is a power of two. |
| 07 · Composite dials | CRT splits a dial into prime-power pieces; counts multiply; the no-bonus lemma. |
| 08 · Symmetry | Every clock on a prime dial has one mirror, c ≡ (g − 2N)(2g)⁻¹. |
| 09 · Tables of clocks | Four tables square to zero because a few sums read 0 when p > 3. |
| 10 · This lesson | One block is a half turn: the skipped mark, the bonus and the jump's zero are the same mark m/2. |
Core course complete
One identity, several views
For even m: T(m) mod m = m/2 = the half turn of one block = the reading the jump sends to 0.
On a power-of-two dial the same mark is the one the 1st block skips. That is the unification. It does not claim that unrelated results, like the zero tables of Lesson 09, are the same theorem.
Where to go next: replace T(k) with other polygonal numbers (Bonus lesson 11), or study the full group of symmetries of these sequences.