What changes
when “same”
changes?

Place the endpoints at (a, 0) and (b cos γ, b sin γ). Their squared distance is (a − b cos γ)² + (b sin γ)². Expanding gives c² = a² + b² − 2ab cos γ. At 90°, cos γ = 0. The Pythagorean theorem follows exactly.

At the middle of each edge the radius is cos(π/n), so the maximum radial deficit is 1 − cos(π/n). It is positive for every finite n and tends to zero as n grows. With fractional tolerance ε, the first eligible n is ceil(π / arccos(1 − ε)), with n ≥ 3. This is a deliberately chosen measurement criterion, not a model of human vision.

A polygon vertex begins at angle 0. At zero rotation, all S rays land on vertices whenever S divides n. The sampled radii are then 1 even though each edge has a smaller radius between its endpoints. At n = S + 1 some rays meet edge interiors. The polygon improves geometrically while this particular test starts detecting it.

A pass means every sampled radius agrees with 1 to within 10⁻⁹. This numerical threshold is fixed and independent of the tolerance instrument above. The sweep tests only sides 3–48; it makes no claim about a final passing threshold. This ray detector is a checkable model of sampling alignment, not the document’s more difficult pixel-raster conjecture.

The image generator knows the shapes. The classifier receives only pixel arrays and class labels. Training computes a mean image for each class; prediction chooses the smaller mean squared pixel distance. Gray pixels in the prototypes are averages of binary training pixels.

Training varies polygon orientation and both classes’ radius. The test polygon has radius 0.78 of the image half-width. This is a small nearest-centroid learner, not a neural network or an LLM. A prediction demonstrates the behavior of this dataset and classifier; it does not establish a universal limit on machine learning.

Define R(θ) = 1 + t[0.22 cos(3θ) + 0.12 sin(5θ)], with 0 ≤ t ≤ 1. Since R ≥ 0.66, the radial map (r, θ) ↦ (rR(θ), θ) has a continuous inverse obtained by dividing the radius by R(θ). Both maps send the origin to the origin and are continuous there.

The hole option removes a centered disk of radius 0.28. The source has one boundary component; the target has two. They are not homeomorphic as planar regions. Probe pairing is disabled because the demonstrated whole-disk map no longer applies.

The current state is a permutation of the labels. The record is the list of states visited. A word followed by its inverse has the identity permutation, while its record still contains the outward and return journeys. “Clear sequence” deletes that record; appending an inverse preserves it. This is an ordinary action log, inspired by the notebook’s unerased frames, not a claim to recover a particular algebra from the drawings.

R rotates counterclockwise by one vertex; F reflects across the vertical axis through vertex A. On vertex positions modulo n these act as R(i) = i + 1 and F(i) = −i. Thus Rⁿ = identity, F² = identity, and FR = R⁻¹F (standard function-composition notation). The interface lists operations in the order you perform them.

These rotations and reflections form the dihedral group with 2n elements. The labeled vertices reveal differences that the outline alone hides. Reference: Wolfram’s dihedral-group documentation ↗

Our base shape is r(θ) = 1 + 0.22 cos(3θ) + 0.12 sin(5θ). Averaging all rotations uniformly removes the sine and cosine terms, leaving radius 1. With finitely many equally spaced rotations, a harmonic survives precisely when the number of rotations divides its frequency. Try two shapes, then three, then five.

Aligned copies keep the original irregularity. There is no defined “mean of all compact shapes” here: that would require a specified space of shapes, a distribution, an alignment convention, and an averaging operation. Comparing this averaging operation to an LLM is a philosophical analogy, not a theorem about how language models work.

The tape starts as 1ᵃ01ᵇ. Replacing the separator with 1 makes a + b + 1 marks. Erasing the rightmost mark leaves exactly a + b. This implementation takes a + b + 3 transitions. It also works when either input is zero; zero is an empty run.

This is one explicit unary encoding and one small program. It does not prove addition is inherently difficult on Turing machines, and the visible tape is a finite window into the model. Primary source: Turing’s 1936 paper ↗

Let H(x, y) be the scalar landscape and let J(a, b) = (−b, a). Along a stroke with velocity v = J∇H − μ∇H, the chain rule gives dH/dt = ∇H · v = −μ‖∇H‖². The perpendicular term contributes zero. At μ = 0 the stroke stays on its starting level. For μ > 0 its level decreases wherever the gradient is nonzero.

The displayed paths are traced by numerical integration, with distance rather than time used as the parameter. This changes speed, not direction or the sign of level change. The graph is a measured trace; the identity above is the proof.

A smooth autonomous field has unique local trajectories. Distinct trajectories cannot cross at a regular point: that would give the same initial point two different futures. Adding drift can make a stroke spiral, but does not remove uniqueness. A second field can cross the first because it is a different rule. This is the distinction between one flow and an overlay of drawings. MIT notes on uniqueness and trajectories ↗

Short strokes, line thickness, interpolation, and numerical integration can create apparent contacts in the rendered image. They do not change the exact-field statement.

The landscape is a sum of smooth, rotated Gaussian hills and wells. One group sets up surrounding swirls; another is arranged to suggest a head, shoulders, body, and facial features. The head handle translates its hills and facial wells. The shoulder handle shifts the shoulders and influences the neck, arms, and upper torso. Each swirl handle moves one background hill. Widths and amplitudes stay fixed. “Figure strength” weights that second group. The overlay adds a smooth oscillatory term to create a different field. The gradient and selected stroke are recalculated from the moved hills, so the level-change law applies at each fixed arrangement. While a handle is moving, the landscape itself changes; that is a sequence of new fields, not a single conserved trajectory.

The figure is a designed constraint, not something the computer discovered in random marks. Its recognizability is a question for the viewer. The experiment is about how local rules carry a global design—and which mathematical relationships survive when those rules change.

The strip has length 8. Each of its seven interior integer positions may have a cut, so there are exactly 2⁷ = 128 candidate dissections, including the uncut strip. The chosen naming rule assigns letters from left to right. The observer knows that rule and the total length, but is not told the number of regions.

“Letters only” merges consecutive copies of a letter and forgets lengths. “Visible run lengths” also measures the combined length of each letter run; a cut between two C regions stays hidden. “Every cut position” records the physical boundaries, even when the names agree. The count checks every candidate in this finite model.

The notebook’s uncertain repeated letter inspires the question. This strip experiment is not a proof of the document’s proposed reconstruction threshold for chord dissections: different objects and different observations would require a separate argument.

The shape is one disk minus its overlap with a second disk of the same radius r, whose center is d units away. Write u = d/r. Its area is A = r²[π − 2 arccos(u/2) + (u/2)√(4 − u²)]. The display uses a common scale. Rotating preserves area; changing the opening and compensating with the radius can preserve area too.

The token rule assigns C to every shape in this family. Area comparisons use the displayed value rounded to 0.01 unit²; direction is measured in degrees, with 360° identified with 0°. The full-parameter test compares opening, radius, and direction numerically. A pass certifies only the chosen record.

The two descriptions are available at once. This is an explicit representation choice, not an experiment measuring human or machine cognition. The notebook’s crescent suggests the comparison; claims about a special cognitive cost would need participants, tasks, and controls.