°
presets:
first ring (stays)
second ring (slides: crossings θ apart)
centre line ÷ 2 = cos(θ/2), the amplitude of |0⟩
crossing line ÷ 2 = sin(θ/2), the amplitude of |1⟩
Every cosine, sine and tangent on the ring side of this page comes from Neal's Net trig engine, the ring's own anchors and recurrence, never the browser's trigonometric library. The library appears only in the textbook column, whose job is to check the ring against the standard. Engine against the library over 1,441 angles: largest difference 6.8e-11.
The setting. A gate angle θ is one placement of the rings with the crossings θ apart. The centres then stand 2cos(θ/2) apart and the crossings 2sin(θ/2) apart, and those two half-distances are the entries of every rotation gate: Rz has them as real and imaginary parts on its diagonal, Ry has them on and off the diagonal, Rx has cos and −i·sin, and the two-qubit Rzz and XX gates carry the same two numbers. The gate angle sits at ring position θ/12 (the half angle, six degrees a position); angles that are multiples of 12° sit on the top ring, every other angle on the ladder of finer rings, which carry the same sixty digits.
The certificate. The digits under the two crossings, at positions k and 60 − k, are equal for odd k and sum to ten for even k. At the cardinals (multiples of fifteen) both are 0 and the pair (F(n), F(n+1)) names which cardinal: (0,1), (0,7), (0,9), (0,3) at 0, 15, 30, 45.
The fold. Born's rule, P(0) = cos²(θ/2), is the half-angle law cos²(θ/2) = (1 + cos θ)/2: the amplitude is read at position k and the expectation value ⟨Z⟩ = cos θ at position 2k, the expectation ring, drawn on the right.
The turns. Merging two gates adds positions, with the landing digit fixed by the addition law. A gradient is the quarter turn: ±15 positions on the expectation ring, digit ×7 forward and ×3 back. The spinor's −1 after 360° is the half turn, digits summing to ten; 720° is the lap. The Pauli phases 1, i, −1, −i are the locked square n, n+15, n+30, n+45 with digits 1, 7, 9, 3 (from n = 1).
the setting
θ = 396°, position 33 of the amplitude ring (θ/12 = 33; level 0)
certificate (the digits under the crossings)
odd position 33: digits 8 = 8 (equal) ok
readings: centre line ÷ 2, crossing line ÷ 2
ring: cos(θ/2) = -0.9510565163 sin(θ/2) = -0.3090169944
textbook: cos(θ/2) = -0.9510565163 sin(θ/2) = -0.3090169944
difference 0, 4.4e-15
the gate Rz(θ) read off the setting
[ -0.9511 + 0.3090i 0.0000 ]
[ 0.0000 -0.9511 − 0.3090i ]
largest difference from the textbook matrix: 4.4e-15
Born’s rule, the fold: amplitudes at k, expectations at 2k
P(0) = (centre ÷ 2)² = 0.90450850 P(1) = (crossing ÷ 2)² = 0.09549150
fold: (1 + cos θ)/2 = (1 + 0.809017)/2 = 0.90450850 → same as P(0): diff 0
⟨Z⟩ = cos θ read at 2k = 66.00: ring 0.80901699, textbook 0.80901699; digit at k: 8, at 2k: 8
merge: Rz(θ)·Rz(48°) = Rz(444°), adding positions 33.00 + 4.00
largest entry difference between the product and Rz(444°): 6.5e-15
addition law on the digits: 33 + 4 = 37: D(37) = 7; law D(33)·D(5) + D(32)·D(4) = 8·5 + 9·3 ≡ 7 ✓
the spinor: 360° is the half turn (−1), 720° the lap
Rz(θ+360°) + Rz(θ) → largest entry 0 (so Rz(θ+360°) = −Rz(θ))
Rz(θ+720°) − Rz(θ) → largest entry 0
positions k = 33, k+30 = 3: digits 8 + 2 = 10 → sum to ten ✓; k+60 is the digit 8 again
the gradient: the quarter turn on the expectation ring
E(θ) = ⟨Z⟩ = cos θ. [E(θ+90°) − E(θ−90°)]/2 = [-0.587785 − 0.587785]/2 = -0.58778525
exact derivative −sin θ = -0.58778525 (ring), -0.58778525 (textbook); differences 0, 2.4e-15
digits: D(2k) = 8, D(2k+15) = 6 = 7·8 mod 10 ✓, D(2k−15) = 4 = 3·8 mod 10 ✓
the Pauli phases: the quarter-turn orbit, a certified Pauli frame
locked square 1, 16, 31, 46: digits 1, 7, 9, 3 = D(1)·(1, 7, 9, 3) mod 10 ✓
XY = i^1·Z → phase exponent 1, carried at position 16 with digit 7 = 1·7^1 mod 10 ✓
YZ = i^1·X → phase exponent 1, carried at position 16 with digit 7 = 1·7^1 mod 10 ✓
ZX = i^1·Y → phase exponent 1, carried at position 16 with digit 7 = 1·7^1 mod 10 ✓
The instrument in four pictures the ground rules under the cards
Everything in the cards below uses four facts. Each is stated once here, with its picture, in the words of the standard formalism beside the ring's. The glossary of words and symbols is the drawer at the end.
1. The law of the settinggeometry
Two rings of radius one. The second slides along the line of centres until they cross with the crossings θ apart.
Why the lengths are the gate's entries: each crossing lies on both rings, so it is at distance 1 from both centres. The crossings sit at positions k = θ/12 and 60 − k, θ apart on the ring, so the ring setting for a gate of angle θ is the half angle θ/2: the crossing sits θ/2 above the line of centres, so the centres stand 2cos(θ/2) apart and the two crossings 2sin(θ/2) apart. Those are the entries of Rz(θ), Ry(θ) and Rx(θ). A position is 6° (sixty on the ring), the half angle θ/2 sits at position θ/12. The instrument and its law are the paper Two Sliding Rings: A Self-Checking Instrument that Reads Sine, Cosine and Tangent at One Setting, with its own calculator page.
2. Why Fibonacci, why sixtythe net
The marks on the rings are the last digits of the Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 3 (13), 1 (21), 4 (34), 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1.
The two facts the ring rests on: the last digits repeat with period sixty (the Pisano period of 10), so sixty positions carry them all; and the digits pair off in sums of ten across the two cardinal axes, F(n) + F(60−n) ≡ 0 at even n and F(n) + F(30−n) ≡ 0 at odd n (mod 10). Draw each pair as a chord and every chord is level or upright: that grid of twenty-eight chords is Neal's Net, and its twin, joining the equal digits, completes the fourteen rectangles. The four cardinals 0, 15, 30, 45 carry the digit 0 and sit on no chord. The net, this grid and its reading as a checksum, is Strassner's discovery, stated and verified in Neal's Net, Version 3 (the paper), with its own calculator page.
3. The three digit lawsthe turns
Every turn of the ring carries its digit with it by a fixed rule, and every sum of positions has a predicted digit.
Written out, all mod 10: quarter turn F(n+15) ≡ 7·F(n); half turn F(n+30) ≡ 9·F(n) ≡ −F(n); three-quarter turn F(n+45) ≡ 3·F(n); and the addition law F(a+b) ≡ F(a)·F(b+1) + F(a−1)·F(b). Since 7² = 49 ≡ −1, the digit 7 is the ring's i, and the four digits d, 7d, 9d, 3d at n, n+15, n+30, n+45 are d times 1, i, −1, −i. That is the locked square, and it is what the Pauli frame rides.
In plain words: start at any position on the ring and read its digit. Move a quarter of the way round, clockwise, fifteen positions. The digit you land on is the digit you started with times 7, last digit kept. Move half way round, thirty positions, and the digit you land on is the starting digit times 9, last digit kept, which is the same as 10 minus the starting digit. Move three quarters of the way round and it is the starting digit times 3. In the picture the start is position 1 with the digit 1, so the four corners read 1, 7, 9, 3: the dim square joins them, and that square sits on the ring at every starting position, with its four digits always d, 7d, 9d, 3d. The same rule read twice is why 7 behaves as the ring's i: a quarter turn and another quarter turn is a half turn, and 7 × 7 = 49 ends in 9, which is −1 in last digits.
4. The certificate, and what it catchesthe guard
A quantum computer's classical half holds angles as positions: the virtual-Z phase accumulator of every qubit, the compiler's merged rotation, the Pauli frame. Beside each position the ring stores its digit pair (F(n), F(n+1)). Corrupt the position and the recomputed pair no longer matches the stored one.
corrupt the stored position 10 (the gate Rz(120°)):
position 10 holds Rz(120°) with its pair (5,9). Flip a bit, or write any other position, and read it back.
Coverage, counted here on every case: arbitrary corruption of a position, all 3540 cases: one digit catches 90.40%, the pair 100.00%, a mod-3 residue 67.80%. Single-bit upsets of the six-bit register, all 360: one digit 97.78%, the pair 100.00%, and a mod-3 residue also 100% (no power of two is a multiple of three). The pair is one to one with the sixty positions, which is why nothing escapes it; the register card below extends this to 32-bit phase words with three rings and to the ladder.
▼ Click to expand the glossary: the words and symbols used on this page ▼▲ Click to collapse the glossary ▲
the symbols
θ (theta) the gate angle, in degrees: the angle a rotation gate turns a qubit through. Rz(θ), Ry(θ), Rx(θ) are the three rotation gates.
n a position on the ring: one of the sixty marks, numbered 0 to 59, six degrees apart, 0 at the left and counting clockwise.
F(n) the digit written at position n: the last digit of the n-th Fibonacci number. F(6) = 8 because the sixth Fibonacci number is 8; F(7) = 3 because the seventh is 13.
k the position of a gate: k = θ/12. A gate of angle θ turns the amplitudes by the half angle θ/2, and a position is 6 degrees, so θ/2 ÷ 6 = θ/12. Rz(72°) sits at k = 6; Rz(120°) at k = 10.
2k twice that position: the full angle θ, read at θ/6. Rz(72°) reads its expectation values at 2k = 12.
d any digit 0 to 9; 7d means seven times that digit, last digit kept.
(F(n), F(n+1)) the pair: the digit at a position and the digit at the next one. The sixty pairs are all different, so a pair identifies its position.
|0⟩, |1⟩ the two basis states of a qubit; cos(θ/2) and sin(θ/2) are their amplitudes after Ry(θ) on |0⟩.
P(0), P(1) the probabilities of reading 0 or 1: the squares of those amplitudes.
⟨Z⟩, ⟨X⟩ expectation values: the average of the ±1 outcomes of measuring Z or X many times. For the state above ⟨Z⟩ = cos θ and ⟨X⟩ = sin θ.
≡ (mod 10) equal in the last digit: 55 ≡ 5.
the words
the setting one placement of the two rings. The value is a length read off it: the centre line is 2cos(θ/2), the crossing line 2sin(θ/2).
the fold turning a product into a sum by the half-angle identity cos²(θ/2) = (1 + cos θ)/2. The amplitude is read at position k, the expectation value at position 2k.
the turn moving a position around the ring. Its digit follows by the laws of picture 3: a quarter turn multiplies the digit by 7, a half turn by 9, three quarters by 3.
amplitude ring the ring read at position k = θ/12. Its two half-lengths are the gate's entries and the qubit's amplitudes.
expectation ring the ring read at position 2k = θ/6, the full angle, where ⟨Z⟩ = cos θ and ⟨X⟩ = sin θ.
cardinal the positions 0, 15, 30 and 45, on the two axes. Each carries the digit 0 and no chord; the pair there names which cardinal it is: (0,1), (0,7), (0,9), (0,3). On a finer ring, any multiple of fifteen.
the ladder the finer rings for angles between the sixty marks. Level 1 has 3,600 positions of 0.1°, level 2 has 216,000, level L has 60·60^L, and every level carries the same sixty digits in the same order.
rung one position on a finer ring. The T gate, Rz(45°), is rung 225 of the 3,600 on level 1.
the vesica the setting with the centres one radius apart (the vesica piscis of geometry): crossings 120° apart, position 10, half-lengths ½ and √3/2. The quarter-turn vesica has its crossings 90° apart, position 7.5.
the net the twenty-eight chords joining each digit to the digit that makes ten with it: 60−n at an even position (an upright chord), 30−n at an odd one (a level chord). Neal's Net.
the twin the twenty-eight chords joining equal digits, the other mirror partner of each position. Net and twin together cut the ring into fourteen rectangles.
the locked square the four positions n, n+15, n+30, n+45 and their digits d, 7d, 9d, 3d, which are d times 1, i, −1, −i. The Pauli frame's phases ride it.
gnomon reading an angle read as a tangent on an upright line standing at position 30. The optimizer finds its best angle this way, where tan θ* = C/B.
the certificate the digit test on a setting: the two digits under the crossings are equal at an odd position and sum to ten at an even one, and the stored pair (F(n), F(n+1)) is recomputed on every read of a register.
The forty tools, and one from 2025, computed 41 tools: 14 settings, 10 folds, 17 turns, and the register · 42 of 42 cards check
Each card runs one tool of quantum computing in your browser on a small instance, the way the paper's appendix does on explicit matrices, and prints the ring's reading beside the textbook value. Seven cards are about the one gate you set in the instrument above, and they recompute as you move its angle: the rotation gates, Born's rule, the variational energy, the uncertainty, the entanglement of the state, gate composition and the gradient. The other thirty-five are complete worked problems, each on its own fixed example, and they run once when the page loads. The last of the turns is a measurement published in 2025, after the paper's forty, read as the same moves. Each card's corner says which of the three moves the tool is and whether every check on it passed. Where a card draws random trials it uses the random seed in the box below, 42 by default; change the seed and every such card reruns with new trials, and the checks pass for any seed.
The setting — a value read off one placement (14)
Rotation gates Rz, Ry, Rx, Rzz, XXa setting ✓ all checks pass: ring = textbook
Turn one qubit, or a pair, by an angle. Standard: a 2 × 2 or 4 × 4 matrix of cos(θ/2) and sin(θ/2).
On the ring: one setting at crossing separation θ; the two half-lengths are every entry. Check: the crossing digits.
1. place the crossings 396° apart (position 33.00)
2. centre line ÷ 2 = -0.95105652 = cos(θ/2)
3. crossing line ÷ 2 = -0.30901699 = sin(θ/2)
Rz(θ) from the two lengths vs textbook: diff 4.4e-15
Ry(θ) from the two lengths vs textbook: diff 4.4e-15
Rx(θ) from the two lengths vs textbook: diff 4.4e-15
Rzz(θ) from the two lengths vs textbook: diff 4.4e-15
XX(θ) from the two lengths vs textbook: diff 4.4e-15
certificate: odd position 33: digits 8 = 8 (equal)
Any single-qubit gatea setting ✓ all checks pass: ring = textbook
Any change to one qubit. Standard: U = Rz(β)·Ry(γ)·Rz(δ), a product of three matrices (Nielsen and Chuang 4.1).
On the ring: three settings; |U₀₀| = cos(γ/2) and |U₀₁| = sin(γ/2) are the lengths of the middle setting, and the phase of U₀₀ is −(β + δ)/2, the two z-positions added by the addition law.
1. three settings: positions 6, 10, 4
2. |U₀₀| − cos(γ/2) = 2.3e-15, |U₀₁| − sin(γ/2) = 3.9e-15
3. phase of U₀₀ against e^(−i(β+δ)/2) read at position 10: diff 7.1e-15
4. addition law: 6 + 4 = 10: D(10) = 5; law D(6)·D(5) + D(5)·D(4) = 8·5 + 5·3 ≡ 5 ✓
U†U = I to 1.4e-14
Hadamard, CZ, CNOTa setting ✓ all checks pass: ring = textbook
The fixed gates every circuit is built from. Standard: fixed matrices.
On the ring: H is a quarter turn and a half turn, H = Ry(90°)Rz(180°); CZ is three quarter turns, Rz(90°)⊗Rz(90°)·Rzz(−90°); CNOT is CZ dressed by H on the target. Positions 7.5 and 15 of the amplitude ring, cardinals of their level.
H from the two turns: diff (up to a global phase) 5.0e-11
CZ from three quarter turns: diff 1.4e-10
CNOT from that CZ dressed by H on the target: diff 7.1e-11
cardinals: D(15) = 0, pair (0,7); rung 450 → D(30) = 0, pair (0,9)
Noise channelsa setting ✓ all checks pass: ring = textbook
How a qubit degrades. Standard: Kraus operators on a density matrix (Nielsen and Chuang ch. 8).
On the ring: the same setting on smaller rings of radius r. Depolarizing shrinks both rings; dephasing shrinks the crossing share only; amplitude damping shrinks it by √(1−γ) and pulls the centre toward |0⟩. Check: Var X + Var Y + Var Z = 3 − r².
depolarizing p=0.3: (x,z) 0.7660,0.6428 → 0.5362,0.4500 = (1−p)·(x,z); r = 0.7000
dephasing λ=0.3: x → (1−2λ)x = 0.3064, z kept 0.6428
damping γ=0.3: x → √(1−γ)x = 0.6409, z → γ+(1−γ)z = 0.7500
total uncertainty 3 − r²: pure 2.0000, depolarized 2.5100, dephased 2.4929, damped 2.0268
Magic statesa setting ✓ all checks pass: ring = textbook
The fuel for T gates, which Clifford circuits cannot make. Standard: the stabilizer octahedron |x| + |y| + |z| ≤ 1 (Bravyi and Kitaev 2005).
On the ring: the square on the four cardinals inside the ring, with the inscribed ring at r = 1/√2 = cos 45°, the quarter-turn setting’s half-length. Depolarizing walks the radius inward: the H-type state leaves magic at p = 1 − 1/√2, the T-type at p = 1 − 1/√3.
1/√2 read as the half-length of the 90° setting: 0.70710678
H-type state (45°) leaves magic at p = 1 − 1/√2 = 0.2929
T-type state (cube diagonal) at p = 1 − 1/√3 = 0.4226
both tested just inside and just outside the square
Error mitigation, benchmarking, readouta setting ✓ all checks pass: ring = textbook
Undo or measure the noise. Standard: zero-noise extrapolation, randomized benchmarking, readout calibration.
On the ring: readings of the radius: extrapolate two radii to r = 1, the decay r^m per Clifford, the read radius 1 − 2ε. Check: two radii return the clean value exactly.
⟨Z⟩ at r = 0.9 and 0.8: 0.578509, 0.514230 → line to r = 1: 0.642788; clean 0.642788; diff 0
benchmarking: ⟨Z⟩ after m depolarizing gates = 0.9^m·⟨Z⟩: 0.5785, 0.5207, 0.4686, 0.4217, 0.3796 ✓
readout flip ε = 0.03: read ⟨Z⟩ = (1 − 2ε)·⟨Z⟩ = 0.604220 ✓
Entropya setting ✓ all checks pass: ring = textbook
How mixed a qubit is. Standard: the von Neumann entropy of the density matrix.
On the ring: a function of the radius alone: the eigenvalues are (1 ± r)/2 whatever the direction, so the same radius reads the same entropy at every setting.
direction 20°: eigenvalues 0.850000, 0.150000 → S = 0.609840
direction 110°: eigenvalues 0.850000, 0.150000 → S = 0.609840
direction 250°: eigenvalues 0.850000, 0.150000 → S = 0.609840
one radius, one entropy: 0.609840 bits
Phase estimation, QFTa setting ✓ all checks pass: ring = textbook
Read a phase as bits. Standard: controlled rotations R_k = diag(1, e^(2πi/2^k)), angles 360°/2^k, and a binary fraction m/2ⁿ.
On the ring: exact on the ladder: level L has 60·60^L positions and carries 2^(2L+2), so an n-bit phase is an exact position at level ⌈n/2⌉ − 1 and lands on a cardinal (a multiple of fifteen), certified by its zero and pair. Check: n = 1 to 12, two bits per rung.
1-bit phase 1/2^1: exact at level 0 (⌈n/2⌉−1 = 0), rung 30, multiple of 15: yes, digit 0, pair (0,9)
2-bit phase 1/2^2: exact at level 0 (⌈n/2⌉−1 = 0), rung 15, multiple of 15: yes, digit 0, pair (0,7)
3-bit phase 1/2^3: exact at level 1 (⌈n/2⌉−1 = 1), rung 450, multiple of 15: yes, digit 0, pair (0,9)
4-bit phase 1/2^4: exact at level 1 (⌈n/2⌉−1 = 1), rung 225, multiple of 15: yes, digit 0, pair (0,3)
5-bit phase 1/2^5: exact at level 2 (⌈n/2⌉−1 = 2), rung 6750, multiple of 15: yes, digit 0, pair (0,9)
6-bit phase 1/2^6: exact at level 2 (⌈n/2⌉−1 = 2), rung 3375, multiple of 15: yes, digit 0, pair (0,7)
7-bit phase 1/2^7: exact at level 3 (⌈n/2⌉−1 = 3), rung 101250, multiple of 15: yes, digit 0, pair (0,9)
8-bit phase 1/2^8: exact at level 3 (⌈n/2⌉−1 = 3), rung 50625, multiple of 15: yes, digit 0, pair (0,3)
9-bit phase 1/2^9: exact at level 4 (⌈n/2⌉−1 = 4), rung 1518750, multiple of 15: yes, digit 0, pair (0,9)
10-bit phase 1/2^10: exact at level 4 (⌈n/2⌉−1 = 4), rung 759375, multiple of 15: yes, digit 0, pair (0,7)
11-bit phase 1/2^11: exact at level 5 (⌈n/2⌉−1 = 5), rung 22781250, multiple of 15: yes, digit 0, pair (0,9)
12-bit phase 1/2^12: exact at level 5 (⌈n/2⌉−1 = 5), rung 11390625, multiple of 15: yes, digit 0, pair (0,3)
Shor’s order findinga setting ✓ all checks pass: ring = textbook
Find the period that factors a number. Standard: phase estimation, then continued fractions.
On the ring: orders dividing 60 sit on the top ring (the polygons that fit); the register’s readings are the convergent wheel. Simulated here: the 8-bit order-finding register for 15 (base 7) and 21 (base 2).
orders: 15/7 → 4, 21/2 → 6, 33/2 → 10, 35/2 → 12, 77/2 → 30; all divide 60 ✓
15 base 7: readings 0, 64, 128, 192 with probability ¼ each, the four cardinals ✓
21 base 2: six most likely readings 0, 43, 85, 128, 171, 213 ✓
convergents: 43/256 → 6, 213/256 → 6, 171/256 → 3, 128/256 → 2 ✓; gcd(2³ ∓ 1, 21) = 3, 7 ✓
Stabilizer codesa setting ✓ all checks pass: ring = textbook
Catch an error without looking at the data. Standard: parity checks and syndromes ([[4,2,2]], [[5,1,3]], the surface code).
On the ring: the parity is the sum to zero, the syndrome is the odd one out, the net’s rectangle rule. Computed by Pauli commutation: every single-qubit error against every stabilizer.
[[4,2,2]]: 12 of 12 single-qubit errors flip a stabilizer ✓
3-qubit code: syndromes for X on qubit 0, 1, 2 = (−+), (−−), (+−) — three distinct, the odd one out ✓
[[5,1,3]]: 15 errors → 15 distinct nonzero syndromes, none silent, each used once ✓
distance-3 surface code: 27 errors → 23 syndromes, none silent, 4 shared pairs each a weight-two boundary stabilizer ✓
Linear-systems algorithm, HHLa setting ✓ all checks pass: ring = textbook
Solve A·x = b on the amplitudes. Standard: phase estimation, a controlled rotation, amplitude amplification (Harrow, Hassidim, Lloyd 2009).
On the ring: A = 1.5 I + 0.5 X has eigenvalues 2 and 1; at a quarter turn per unit the eigenphases are ½ and ¼ of a turn, positions 30 and 15, exact on the top ring. The inversion rotations Ry(2·asin(C/λ)) are the settings 60° and 180°, whose crossing half-lengths are C/λ. Check: A⁻¹b to fidelity one on 20 random b.
b #1: fidelity of the ancilla-1 branch to A⁻¹b = 1.000000000000, its probability = C²|A⁻¹b|² ✓
b #2: fidelity of the ancilla-1 branch to A⁻¹b = 1.000000000000, its probability = C²|A⁻¹b|² ✓
b #3: fidelity of the ancilla-1 branch to A⁻¹b = 1.000000000000, its probability = C²|A⁻¹b|² ✓
eigenphases 2/4 and 1/4 of a turn: positions 30 and 15, digits 0, 0 (cardinals)
all 20 random b: fidelity 1 and the stated probability ✓
Classical shadowsa setting ✓ all checks pass: ring = textbook
Learn a state from few random measurements. Standard: the inverse of the measurement channel, 3|o⟩⟨o| − I (Huang, Kueng, Preskill 2020).
On the ring: the radius times three: the Bloch vector of the outcome state scaled by three. Check: the Born-weighted average of the snapshots over the three bases returns 30 random mixed qubits exactly.
30 random mixed qubits (random angle, random radius 0.2 to 1)
largest entry difference between the averaged snapshots and the state: 0
Probabilistic error cancellationa setting ✓ all checks pass: ring = textbook
Undo the noise by a quasi-probability. Standard: the inverse channel sampled with signed weights (Temme, Bravyi, Gambetta 2017).
On the ring: the radius divided by 1 − p, weights 1/(1−p) and −p/(1−p), sampling cost γ = (1+p)/(1−p). Check: 30 random states restored exactly at p = 0.1 and 0.3.
p = 0.1: weights 1.1111 and -0.1111, cost γ = 1.2222
p = 0.3: weights 1.4286 and -0.4286, cost γ = 1.8571
30 states × 2 strengths restored; largest difference 0
Lindblad equationa setting ✓ all checks pass: ring = textbook
How a qubit decays in time. Standard: a master equation on the density matrix (Lindblad 1976).
On the ring: the setting turning at ω while the radius shrinks: the crossing share decays on T₂ with 1/T₂ = γ₁/2 + 2γ_φ and the centre relaxes on T₁ = 1/γ₁. Check: Runge–Kutta against the closed form to 1e-7.
t = 0.5: RK4 (x,y,z) = 0.29448, 0.75329, 0.40464; closed form differs by 2.3e-10
t = 1.0: RK4 (x,y,z) = -0.08840, 0.69051, 0.46129; closed form differs by 4.5e-10
t = 1.5: RK4 (x,y,z) = -0.35171, 0.48509, 0.51256; closed form differs by 6.0e-10
t = 2.0: RK4 (x,y,z) = -0.46583, 0.22128, 0.55894; closed form differs by 7.0e-10
The fold — a product turned into a sum (10)
Born’s rulea fold ✓ all checks pass: ring = textbook
Turn an amplitude into a probability. Standard: square the amplitude.
On the ring: cos²(θ/2) = (1 + cos θ)/2: the amplitude is read at position k and the expectation value at 2k. Check: the squared half-length at k equals half of one plus the centre reading at 2k, and the digits at k and 2k certify both settings.
P(0) = (cos(θ/2))² = 0.90450850; fold (1 + cos θ)/2 = 0.90450850; diff 0
P(1) = (sin(θ/2))² = 0.09549150; fold (1 − cos θ)/2 = 0.09549150; diff 3.2e-15
off-diagonal ⟨X⟩: 2cos(θ/2)sin(θ/2) = 0.58778525 = sin θ read at 2k = 0.58778525
digits: at k = 33 → 8 and 8; at 2k = 6 → 8 and 2
Variational energya fold ✓ all checks pass: ring = textbook
Score a trial state. Standard: a sum of measured Pauli averages.
On the ring: E(θ) = A + B cos θ + C sin θ, a fold of the setting. On the run H = Z₀ + ½Z₁ + (√3/4)(X₀X₁ + Y₀Y₁) with the Givens trial state, E(θ) = cos(θ + 60°): the ring read ten positions on. Check on the matrices, and ⟨XX⟩ = ⟨YY⟩ on the data.
1. the trial state G(θ)|01⟩ = cos(θ/2)|01⟩ − sin(θ/2)|10⟩ at θ = 396°
2. ⟨ψ|H|ψ⟩ on the 4×4 matrices = -0.10452846
3. the ring read at position 2k + 10 = 16.00: cos(θ+60°) = -0.10452846; diff 3.0e-15
4. data checksum ⟨XX⟩ = -0.587785, ⟨YY⟩ = -0.587785: equal ✓
the ground state: position 20 (vesica), E = −1 = −√(¼ + ¾), digits at 20, 35, 5 = 5, 5, 5
Uncertainty, shot noisea fold ✓ all checks pass: ring = textbook
How sure a reading is. Standard: variances and 1/N statistics.
On the ring: the other share: for a qubit at Bloch angle θ, Var Z = sin²θ and Var X = cos²θ, the two shares swapped, adding to one at every setting; for any pure qubit Var X + Var Y + Var Z = 2. Shot noise on ⟨Z⟩ is the crossing share over N.
Ry(θ)|0⟩ at θ = 396°: ⟨X⟩ = 0.587785, ⟨Z⟩ = 0.809017
Var Z = 0.34549150 = sin²θ = 0.34549150; Var X = 0.65450850 = cos²θ = 0.65450850
Var X + Var Y + Var Z = 2.00000000 (two for every pure qubit)
shot noise on ⟨Z⟩ at N = 1000: 0.00034549 (the crossing share over the shots)
Symmetry-verification statisticsa fold ✓ all checks pass: ring = textbook
Decide when a check has failed. Standard: a confidence threshold.
On the ring: the threshold from the shares: at the ground state ⟨XX⟩ = ⟨YY⟩ = −√3/2, each share ¼, so the standard error of ⟨XX⟩ − ⟨YY⟩ is √(½/N). Flag beyond five standard errors. Sampled here: 2,000 clean trials and 2,000 with a bit flip at N = 100.
standard error from the shares: √(¼/N + ¼/N) = 0.07071; threshold 0.3536
clean: 0 false alarms in 2,000
bit flip: gap √3 = 24.5 standard errors; 2000 of 2,000 flagged
State verificationa fold ✓ all checks pass: ring = textbook
Confirm a state was made right. Standard: tomography or fidelity estimation.
On the ring: prepare twice, then the swap test: the ancilla reads 0 with probability (1 + F)/2, the half-angle fold with the fidelity as its cosine. A setting one position off has F = cos²(Δθ/2). Run here as the five-qubit circuit on the ground state.
identical P(0) = 1.000000 → F = 2P(0) − 1 = 1.000000; direct fidelity estimation 1.000000
bit flip X₀ P(0) = 0.500000 → F = 2P(0) − 1 = 0.000000; direct fidelity estimation 0.000000
phase flip Z₀ P(0) = 0.625000 → F = 2P(0) − 1 = 0.250000; direct fidelity estimation 0.250000
one position off, 126° P(0) = 0.998630 → F = 2P(0) − 1 = 0.997261; direct fidelity estimation 0.997261
one position off: F = cos² 3° = 0.997261 (the twin test resolves one ring step)
Entanglement, Bell testsa fold ✓ all checks pass: ring = textbook
How two qubits’ outcomes are linked. Standard: E(a, b) = −cos(a − b); CHSH at most 2 classical, 2√2 quantum.
On the ring: the ring at the difference of two positions; 2√2 is twice the centre line of the quarter-turn setting (position 7.5, centres √2 apart). Check on the matrices over a grid, and the product state stays at √2.
E(a,b) + cos(a − b) over a 20 × 20 grid: largest 5.6e-15
singlet CHSH |S| = 2.82842712; 2 × centre line of the 90° setting = 2.82842712; textbook 2√2 = 2.82842712
product state |00⟩: S = 1.41421356 = √2
Entanglement of a statea fold ✓ all checks pass: ring = textbook
How entangled a trial state is. Standard: Wootters’ concurrence |⟨ψ|σ_y⊗σ_y|ψ*⟩|.
On the ring: the fold: for G(θ)|01⟩ the concurrence is 2cos(θ/2)sin(θ/2) = |sin θ|, the crossing half-length at 2k. Zero at the cardinals (a product state), one at the quarter turn (a Bell state).
G(θ)|01⟩ at θ = 396°: concurrence on the matrices = 0.58778525
2 × cos(θ/2) × sin(θ/2) = 2 × -0.95106 × -0.30902 = 0.58778525
= |sin θ|, the crossing half-length at 2k: 0.58778525; diff 5.3e-15
GHZ states, Mermin testsa fold ✓ all checks pass: ring = textbook
How three qubits’ outcomes are linked. Standard: E(a, b, c) = cos(a + b + c); Mermin at most 2 classical, 4 quantum.
On the ring: the ring at the sum of three positions (two qubits read the difference, three read the sum, both by the addition law). The four Mermin settings are cardinals and quarter turns, and the value is the ring read at 0° four times.
E(a,b,c) − cos(a + b + c) over a 12 × 8 × 6 grid: largest 1.0e-14
E(30°, 60°, 90°) = -1.00000000 = cos 180°
Mermin combination = 4.00000000 (classical bound 2)
Cross-entropy benchmarkinga fold ✓ all checks pass: ring = textbook
Score a noisy machine against the ideal. Standard: F_XEB = 2ⁿ Σ p(x)q(x) − 1 over ideal and measured probabilities (Boixo et al. 2018).
On the ring: a fold of the readings: under depolarizing noise with fraction f the benchmark is exactly f times the ideal value, so it reads the radius. Check: 50 random two-qubit circuits of settings at three strengths.
50 circuits: ideal F_XEB = 4Σp² − 1 from 0.054 to 1.836
noisy F_XEB − f × ideal over 150 cases: largest 2.3e-11
Quantum volumea fold ✓ all checks pass: ring = textbook
One number for how big a circuit a machine runs well. Standard: the heavy-output probability over random circuits (Cross et al. 2019).
On the ring: a fold of the readings shrunk by the radius: HOP = f × ideal + (1 − f)/2 exactly, so the two-thirds threshold is a line on the radius. Check: 50 random two-qubit circuits of settings.
mean ideal heavy-output probability over 50 circuits: 0.8213
HOP(f) − [f × ideal + (1 − f)/2] over 100 cases: largest 0
the ⅔ threshold is crossed at f* = 0.5188 for these circuits
The turn — a shift of position with its digit (17)
Gate compositiona turn ✓ all checks pass: ring = textbook
Merge two rotations into one. Standard: add the angles in the compiler.
On the ring: add positions: Rz(a)Rz(b) = Rz(a + b), and the landing digit is fixed by the addition law F(a+b) ≡ F(a)F(b+1) + F(a−1)F(b) (mod 10), a certificate on the merge.
product Rz(a)·Rz(b) against Rz(a+b): largest entry difference 6.5e-15
addition law: 33 + 4 = 37: D(37) = 7; law D(33)·D(5) + D(32)·D(4) = 8·5 + 9·3 ≡ 7 ✓
Parameter-shift gradienta turn ✓ all checks pass: ring = textbook
Find which way to move an angle. Standard: E′(θ) = [E(θ + 90°) − E(θ − 90°)]/2 (Mitarai et al. 2018; Schuld et al. 2019).
On the ring: the quarter turn: ±15 positions on the expectation ring, digit ×7 forward and ×3 back, and 7 × 3 ≡ 1 says the two shifts undo each other. On the run E(θ) = cos(θ + 60°). Where a gate carries two frequencies, four eighth-turn shifts with weights y₁, y₂.
E(θ±90°) on the matrices: [-0.994522 − 0.994522]/2 = -0.99452190
exact derivative −sin(θ + 60°) from the ring: -0.99452190; diff 7.4e-12
digits: D(16) = 7 → ×7 = 9 = D(31) ✓; ×3 = 1 = D(1) ✓
two frequencies (state outside the number block): two shifts wrong at 59 of 60 positions; four eighth-turn shifts with y₁ = 0.7452, y₂ = -0.0381 right at all 60 ✓
Clifford phases S, Z, S† and Ta turn ✓ all checks pass: ring = textbook
The fixed phase gates every circuit uses. Standard: diagonal matrices with i, −1, −i, e^(iπ/4).
On the ring: quarter, half and three-quarter turns on the expectation ring (digit ×7, ×9, ×3), and the ladder for T at 7.5. The gates’ own positions 15, 30, 45 are cardinals (digit 0, pairs (0,7), (0,9), (0,3)); the Pauli frame’s phases 1, i, −1, −i ride the locked square 1, 16, 31, 46 with digits 1, 7, 9, 3.
S from the quarter turn: diff 7.1e-11; Z from the half turn: 1.0e-14; S† from −90°: 7.1e-11; T from 45°: 7.1e-11
cardinals: D(15), D(30), D(45) = 0, 0, 0; pairs (0,7), (0,9), (0,3)
locked square 1, 16, 31, 46 → digits 1, 7, 9, 3 = 1·(1, 7, 9, 3): the phases 1, i, −1, −i
Optimizera turn ✓ all checks pass: ring = textbook
Find the best angle. Standard: gradient descent with a step size.
On the ring: three settings and a gnomon reading: E at θ and at θ ± 90° fix A, B, C exactly; the minimum sits where tan θ* = C/B, a tangent-line reading, and its value is A − √(B² + C²), a hypotenuse. On the run: readings at 0°, ±90° give θ* = 120°.
run: readings E(0°), E(90°), E(−90°) = 0.50000, -0.86603, 0.86603
θ* found on the ladder = 120.000000° (position 20.000 of the expectation ring); E_min = -1.00000000
20 random energies with random working angles: A, B, C recovered to 8.7e-15; each landed on its true minimum ✓
Convergencea turn ✓ all checks pass: ring = textbook
Know the search will settle. Standard: a convergence proof per optimizer.
On the ring: exact coordinate minima are monotone: a gnomon sweep never raises the energy and the energy has a floor, so it converges. Check: Ry(a)⊗Ry(b) then CNOT on Z, ZZ and XX terms from (10°, 200°); the energy never rises and reaches the minimum over 14,400 grid pairs.
energies: 0.3097 → -0.9460 → -1.1383 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524 → -1.2524
never rises ✓; final -1.25241747, at or below the 120 × 120 grid minimum -1.25241636
Grover’s searcha turn ✓ all checks pass: ring = textbook
Find a marked item faster. Standard: amplitude amplification, success sin²((2k+1)θ) with sin θ = √(M/N).
On the ring: a setting turned by 2θ per query; the success probability is the crossing share of the setting at (2k+1)θ. Two qubits, one marked: sin θ = ½, θ = 30°, position 5 (digit 5), one query lands at 90° exactly. Three qubits: θ on the ladder, checked at k = 1, 2, 3.
two qubits, one query: the marked state’s probability = 1.000000000000
three qubits: after 1, 2, 3 queries the matrices give 0.781250, 0.945312, 0.330078
the setting turned to 3θ, 5θ, 7θ (addition law from sin θ = 1/√8): 0.781250, 0.945312, 0.330078 ✓
Hamiltonian simulationa turn ✓ all checks pass: ring = textbook
Run a many-term interaction step by step. Standard: the Lie–Trotter product of the terms’ rotations.
On the ring: settings chained: commuting terms merge exactly by the addition law; anticommuting pairs carry the quarter-turn phase (XZ = −iY, the ×7 and ×3 of the locked square), and the Trotter error is theirs, falling as the square of the step.
merge Rz(40°)Rz(70°) against Rz(110°): diff 7.1e-11
XZ = i^3·Y: phase exponent 3, the three-quarter turn (digit at 1+45 = 3 = 3·1)
Trotter error at steps 0.1, 0.05, 0.025: 9.972e-3, 2.498e-3, 6.249e-4; ratios 3.992, 3.998 (4 = quartering)
exact − product at t = 0.1 against the leading term −i t² Y: remainder 3.34e-4 (t³/3 = 3.33e-4)
QAOAa turn ✓ all checks pass: ring = textbook
Approximate a combinatorial optimum. Standard: alternating cost and mixer layers (Farhi, Goldstone, Gutmann 2014).
On the ring: two kinds of setting alternating. For one edge the expected cut is exactly ½(1 + sin 4β sin γ); its maximum, the full cut, sits at γ = 90° (the cost layer at the quarter turn) and 2β = 45° (the mixer at the eighth turn).
expected cut against ½(1 + sin 4β sin γ) over 1,800 grid pairs: largest 9.9e-15
at γ = 90°, 2β = 45°: cut = 1.000000000000
Teleportationa turn ✓ all checks pass: ring = textbook
Send a state with two classical bits. Standard: a Bell measurement, then a Pauli correction (Bennett et al. 1993).
On the ring: the correction is two half turns, positions 0 or 30, read from the two bits with their digits, so it is certified before it is applied. Check: 20 random states, four outcomes each with probability ¼, all 80 restored.
bits (0,0): probability 0.2500, correction Z^0 X^0 = turns 0, 0 → fidelity 1.0000000000
bits (0,1): probability 0.2500, correction Z^0 X^1 = turns 0, 30 → fidelity 1.0000000000
bits (1,0): probability 0.2500, correction Z^1 X^0 = turns 30, 0 → fidelity 1.0000000000
bits (1,1): probability 0.2500, correction Z^1 X^1 = turns 30, 30 → fidelity 1.0000000000
20 states × 4 outcomes: all restored ✓
Superdense codinga turn ✓ all checks pass: ring = textbook
Send two bits with one qubit. Standard: a Pauli on half of a Bell pair (Bennett and Wiesner 1992).
On the ring: the same four turns run the other way: I, X, Z, XZ on one half give four orthogonal Bell states, and Bob’s Bell measurement reads the two bits.
overlaps |⟨Bell_i|Bell_j⟩|:
1.000 0.000 0.000 0.000
0.000 1.000 0.000 0.000
0.000 0.000 1.000 0.000
0.000 0.000 0.000 1.000
four orthogonal states ✓
Dynamical decouplinga turn ✓ all checks pass: ring = textbook
Cancel a drifting phase. Standard: the spin echo, X pulses in sequence (Hahn 1950; Viola and Lloyd 1998).
On the ring: a half turn about x carries the position to its antipode and the addition law then sums φ and −φ to zero: X·Rz(φ)·X·Rz(φ) = I up to a global phase at all sixty φ. The echo is the sum-to-ten law applied to a drift.
X Rz(φ) X Rz(φ) against I (up to phase) at all 60 grid angles: largest 0
digits of φ and −φ at position 8: 1 and 9, sum 10
Oracle algorithms, Deutsch–Jozsa and Bernstein–Vazirania turn ✓ all checks pass: ring = textbook
Read a hidden property in one query. Standard: Hadamards, a sign oracle, Hadamards.
On the ring: quarter and half turns around half turns: the Hadamards are turns, the oracle’s signs are half turns, and the final interference is the fold landing on one cardinal, the secret string read with probability one. Check: all eight three-bit strings.
s = 000: P(read s) = 1.0000000000
s = 001: P(read s) = 1.0000000000
s = 010: P(read s) = 1.0000000000
s = 011: P(read s) = 1.0000000000
all eight strings read with probability one ✓
Quantum signal processing, QSVTa turn ✓ all checks pass: ring = textbook
Shape a polynomial of a matrix. Standard: a rotation chain with classical phases (Low and Chuang 2017; Gilyén et al. 2019).
On the ring: settings chained; with every phase zero the chain of d identical settings is the addition law d times, ⟨0|W(x)^d|0⟩ = cos(d·arccos x) = T_d(x), the Chebyshev polynomial. The signal x = cos 6n is the centre half-length at position n, so no arccos is needed: W is the setting at n.
29 signals x = cos 6n (n = 1 … 29), degrees d = 1 … 6: ⟨0|W^d|0⟩ against T_d(x) by the recurrence and against the ring at d·n
largest difference 3.2e-14
Quantum walksa turn ✓ all checks pass: ring = textbook
Spread over a graph faster than a random walk. Standard: a coin operator and a shift operator (Aharonov et al. 2001; Kempe 2003).
On the ring: the coin is the quarter turn and the half turn (the Hadamard) and the shift is a turn of positions. On a cycle of sixty sites every momentum block has |sin ω| = |sin k|/√2, the quarter-turn setting read at each k; after twenty steps the spread is 9.29 sites against the classical √20 = 4.47.
probabilities sum to 1.000000000000
standard deviation after 20 steps: 9.2910 sites; classical √20 = 4.4721
120 eigenvalues on the unit circle with |sin ω| = |sin k|/√2 ✓
Solovay–Kitaev wordsa turn ✓ all checks pass: ring = textbook
Write any rotation as a word in H and T. Standard: a recursive search over gate words (Dawson and Nielsen 2006).
On the ring: a chain of quarter, half and eighth turns toward a setting on the ladder. Every word of H and T to length 20 (no two H in a row): the best approximation to U3(50°, 70°, 110°) falls from 0.31 at length 4 to 0.062 at length 20, never rising.
words at length 20 (no HH): 17711 (the Fibonacci count)
best distance at length 4, 8, 12, 16, 20: 0.3061, 0.1035, 0.1035, 0.0818, 0.0624
Measurement-based computationa turn ✓ all checks pass: ring = textbook
Compute by measuring a cluster state. Standard: measurement angles with Pauli byproducts (Raussendorf and Briegel 2001).
On the ring: a setting, the fold and a turn: on a two-qubit cluster, measuring the first qubit at angle α leaves the second as X^s·H·Rz(α)|ψ⟩, each outcome with probability ½, the byproduct X^s the half turn of the frame. Check: 20 random states and angles, both outcomes.
20 random states and angles, both outcomes: probability ½ and X^s H Rz(α)|ψ⟩ to fidelity 1 ✓
W-state measurement (Park, Hofmann, Okamoto and Takeuchi, 2025)a turn ✓ all checks pass: ring = textbook
Tell which of the N W states a set of N qubits is in: the Bell measurement generalised to N qubits. Standard: a discrete Fourier transform over the N photon modes, then photon counting; the sum of the output-mode indices, K = Σ k_j (mod N), names the W state (Science Advances 11, eadx4180, 2025).
On the ring: a turn: the cyclic shift of the qubits is a turn of positions, and its eigenvalue e^(2πiK/N) is a vertex of the N-gon on the ring, the triangle 0, 20, 40 for three qubits. The readout K = Σ k_j (mod N) is the addition law landing on that vertex, and the vertex’s digit pair certifies the K register. N = 2 is the Bell measurement, positions 0 and 30.
1. the three W states on the 8-dimensional matrices: orthonormal ✓; each an eigenstate of the cyclic shift with eigenvalue e^(2πiK/3) ✓ (ring phase vs textbook: diff 5.2e-15)
2. the optical measurement simulated, 18 detector patterns per input: probabilities sum to 1 ✓; every pattern with nonzero probability has k_V + k_H + k_H ≡ K (mod 3) ✓ (18 nonzero patterns over the three inputs)
3. the K register on the ring: vertices 0, 20, 40 carry the pairs (0,1), (5,6), (5,1), all different ✓: the vertex names itself
4. N = 2 is the Bell measurement: K = 0 at position 0, K = 1 at position 30, the half turn ✓
5. N = 4, 5, 6: the W states turn on the square, the pentagon and the hexagon of the ring ✓
And the register: the guard on the phase word
The 32-bit phase worda register ✓ all checks pass: ring = textbook
The classical accumulator that implements Rz on hardware, 2³² to the turn. Standard: no check, or a mod-3 residue.
On the ring: three positions on rings of 60, 61 and 59 marks run in lockstep: a change escapes only if it is a multiple of all three, 215,940. Every single-, double- and triple-bit upset is caught, exhaustively; the ladder guard (six base-60 digits with their pairs) is one to one with the word.
on one 60-ring the pair (F(n), F(n+1)) catches every corruption of a position (picture 4); the word guard below extends that to 32-bit words
every single-bit change escapes no ring ✓
double-bit changes: 1924 (1,924); 196 escape the 60-ring alone (196); 0 escape all three
triple-bit changes: 34518 (34,518); 0 escape all three
2,000 random pairs of words: the sum’s position from the positions and the carry ✓