022 — The Colouring Book Has No Area
agylövés: Lysarith
mathematics, verification & prose: Claude Opus 5 (plain-chat)
build: Claude Opus 5 (CLI room)
2026-07-31
Colour in as much of it as you like. You will never fill it, and not because it is large. Three procedures with nothing in common — a pair of scissors, a question about odd and even, and a single bitwise test — arrive at the same set of cells, and none of them can explain the others. Then a fourth, driven by nothing but chance, is thrown at it and never lands in a hole.
SCISSORS · PARITY · BITMASK · CHANCEDEPTH 5 · 243 CELLS
The screen and the download share the same lattice. The SVG is a clean black-and-white A4 page: every surviving cell becomes an outlined triangle, so a pencil has real spaces to fill. It stops at a finite depth, because paper is finite — and that is the joke, not a shortcut. The page will print at depth 5.
Cut it yourself
Click any filled triangle. It splits into three, with a hole where its middle was. The counter beside it does the thing this piece is about: the number of triangles goes up, and the area goes down. Draw more, have less. You can deepen one corner twenty times and leave the rest alone — the reading stays exact, because it is a sum over the triangles that are actually there and not a formula about a depth you no longer have.
CLICK A TRIANGLE · IT BECOMES THREE
- triangles
- 1
- area, of the original
- 100%
The area is summed, not assumed. Every triangle contributes exactly 4−k of the original, where k is how many times its own line of descent has been cut. The total is computed over a common denominator in exact integer arithmetic, so the figure is not a rounded reading of a float — it is the number.
Why is this worth a colouring page?
Because a colouring page is a promise that something can be filled in, and this is the object for which that promise is false in an exact, provable way. At every round three quarters of what remains survives and one quarter is cut out. After one round you have 75% of the triangle; after twenty, less than one part in three hundred. The fraction is (3/4)ⁿ, and it goes to zero.
So the gasket has zero area — not "very little", zero — while its edges get longer at every round without limit. Infinite perimeter, no area at all. It is all boundary. And the holes are not gaps in the picture: the holes are the picture. Remove them and there is no gasket, only a triangle. Every attempt to complete it by adding more colour destroys the only thing that made it itself.
ROAD 01 · SCISSORSCut the middle out
Take a triangle. Remove the smaller one whose corners are the midpoints of the sides. Three are left. Do the same to each. Then to each of the nine. A child can follow it and nothing about it is subtle.
ROAD 02 · PARITYAsk only: odd or even?
Write out Pascal's triangle and throw away everything except whether each entry is odd. Colour the odd ones. The gasket appears — not something like it, the same cells. There is no shape in this procedure anywhere.
ROAD 03 · BITMASKOne condition, no arithmetic
Entry (r, c) is odd exactly when the binary digits of c are a subset of those of r. That is the whole test: r & c == c. It draws the gasket without ever mentioning a triangle.
What the formula says, when the cutting is even
If every triangle is cut at every round, the bookkeeping collapses to 3n of 4n and the table below is the whole story. The moment the cutting is uneven — which is what happens the instant a reader starts clicking — that formula is answering a question about a depth the object no longer has, and the panel above stops using it. The highlighted row is the depth on the main screen.
| depth | cells filled | of the whole | share |
|---|
3n of 4n. The table is finite; the object is not.
How to play with it
Switch roads and watch nothing happen
Move between scissors, parity and bitmask. The picture does not change — that is the entire result. Three procedures that share no code and no idea of each other produce the same set, cell for cell. The readout in the corner is the live comparison, not a caption: it compares the three sets against each other and reports the number of cells they disagree on. The last option in the list draws that disagreement — every cell one road has and another lacks, in red — which is why it looks like nothing happens there too. An empty red layer is the result; if it ever filled in, it would be the only thing worth looking at.
Throw chance at it
The fourth road has no plan. Drop a pencil anywhere inside, pick one of the three corners at random, move half the distance to it, mark where it lands, repeat forever. The marks do not scatter. They fall into the gasket and keep falling into it, and the counter of points landing outside stays at zero.
The sentence this piece refuses to write
The chaos game does not reproduce the gasket, and it cannot in finitely many points. That is the tempting sentence and it is false — the picture on the screen looks finished, so the strong claim feels true. What is tested is the weaker thing that is actually true: it never leaves. Randomness cannot find the holes, because the holes are exactly where the halving rule cannot reach.
Take it off the screen
The download is an A4 SVG, sharp at any printer resolution. Every surviving cell is an outlined triangle with real space inside it. Colour neighbouring cells, follow one of the three big copies down, alternate by row — or ignore all of it. You cannot cheat the page by scribbling over the white parts, because the white parts are not unfilled regions. They are places where the object is not.
What is on the screen, honestly
Everything drawn stops at a finite depth, and the object has no last step. Area zero is a statement about the limit set, not about any drawing — anything actually rendered has area, because pixels do. The check counts cells at a stated depth rather than pretending to reach the limit.
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