One period of the square twist tessellation, with its edges joined
no-rim is one function. Everything below came out of it during this
build, at arguments taken from the essays rather than invented for this page — so a figure
here is the same figure a reader meets in an essay, and when the generator changes, this
page changes with it.
At its defaults
view: "bottom"
view: "counts", kinds: [square, triangular, hexagonal], sizes: [1, 2, 3]
view: "cell", kind: "square", cells: 2
What it checked while it drew
Collected by running this generator with a listener on the assertions, not written here. The count is how many separate times this build put that claim to the test.
- its 16 vertices are all interior, because a sheet with no edge has nowhere for a vertex to be otherwise ×7
- 9 glued cells over 3 tilings, every one with V − E + F = 0, which is a torus and is checked rather than assumed ×4
- folding the tessellation flat multiplies the plane's lattice by 0.410373441 and turns it by 36.62°, the same on all 3 tilings to eight figures ×4
- across 10 patches on 4 tilings, every panel with nothing below it touches the paper's edge, and none is in the interior ×3
- on 4 of 5 tilings the collection's own cycle test exhausts the 2×2 glued cell — a proof that no lettering of it is consistent — while the test that reads each arc's lattice step finds one ×3
- the 2×2 square cell is drawn with 40 crease pieces and has 32 creases once its opposite edges are one edge, and 25 drawn panels which are 16 panels of the sheet ×3
- the lettering the collection's test called impossible passes every vertex condition and forces no loop on all 8 ordinary clipped patches it was written onto, up to 924 creases and 432 vertices ×3
- it is the same along the cell as across it, which is what makes it a similarity rather than a squash, and every pair of identified panels agrees to 4e-14 of a sheet width ×2
- the 1×1 hexagonal cell is drawn with 34 crease pieces and has 24 creases once its opposite edges are one edge, and 23 drawn panels which are 12 panels of the sheet ×2
- the 1×1 triangular cell is drawn with 34 crease pieces and has 24 creases once its opposite edges are one edge, and 23 drawn panels which are 12 panels of the sheet ×2
- the 2×2 square cell's loops are ruled out in 2 directions rather than one, because a single direction has to be positive on every loop at once and these are not all one way ×2
- an exhausted search is a proof and not a budget, which is what makes the disagreement worth drawing ×1
- checkPattern and forcedCycle are the instruments a sheet of paper with an edge is checked with, and neither knows where these letters came from ×1
- each round removes the arcs a zero-summing walk could not use, and what is left splits into smaller pieces ×1
- the 1×1 elongated cell is drawn with 52 crease pieces and has 40 creases once its opposite edges are one edge, and 33 drawn panels which are 20 panels of the sheet ×1
- the 1×1 rhombille cell is drawn with 60 crease pieces and has 48 creases once its opposite edges are one edge, and 37 drawn panels which are 24 panels of the sheet ×1
- the 1×1 triangular cell's loops are ruled out in 3 directions rather than one, because a single direction has to be positive on every loop at once and these are not all one way ×1
- the count of them grows with the rim rather than with the sheet: 1 of 25, 2 of 81, 3 of 169 ×1
- the count of them grows with the rim rather than with the sheet: 3 of 69, 5 of 233 ×1
- the three counts come from three separate identifications — panels, creases and vertices — so Euler's number is a test and not a restatement ×1
Where it is called
Changing this generator changes every figure on this list, which is what makes the list worth publishing rather than keeping in a check script.
A bottom layer on half a rim
The bottom of a folded stack lives at the paper's edge, which is why a sheet with no edge has an order with no least element. A cylinder has half a rim, so it has a bottom — and the count of panels that could be it falls with the rim, which makes the claim a measurement rather than a boundary case.
A loop that goes somewhere
Every crease says which of its two panels lies above the other, and a loop in those statements is a proof that the pattern has no flat folded state. On a sheet with no edge that sentence is false. The loops of a periodic pattern carry a lattice step each, and a loop that ends one cell to the right is not a contradiction — it is a stack of paper with no bottom layer.
A sheet with no edge
A twist tessellation repeats, so a rectangle of it is a description of the whole plane rather than a piece of paper. Joining the rectangle's opposite sides makes that explicit and produces an object every gate in this collection can read: twenty-five drawn panels become sixteen, forty crease pieces become thirty-two, sixteen vertices are all interior, and the three counts add to nothing.
A test imported without its hypothesis
The rule that a loop in a folded sheet's layer relations proves the pattern cannot fold arrives from the layer-ordering literature, where the sheet is a disc and the panels are finitely many. This collection took the rule and not the sentence that says which sheets it is about, then applied it for years to patterns whose whole interest is that they repeat.
An order with no least element
Enumerating every way a folded pattern can be stacked works by building upward from a panel with nothing below it. The smallest square twist patch has exactly one such stacking and takes eleven thousand steps to find it. The pattern that patch was cut from has no panel with nothing below it at all, so the enumeration has nothing to start from — and the sheet is perfectly well stacked anyway.
Folding it flat is one similarity
Where a cell of paper goes when a twist tessellation collapses is a scale and a turn: multiply the plane's lattice by 0.410373441 and rotate it by 36.62°. That is the answer on the square grid, the triangular grid, the honeycomb, the elongated triangular tiling and the rhombille alike, agreeing to eight figures — while the collection's other answer to how much smaller it gets gives those five tilings five different numbers.
Half a rim
A rectangle of tessellation cut out of the plane has four edges; glued into a torus it has none. Gluing one pair and leaving the other gives the middle of the scale — the same drawing, the same vertices, the same conditions asked of them, and exactly half the rim. What the rim costs turns out to be measurable per edge rather than only at the ends.
Pruning on proofs alone
A search that discards a branch it cannot prove wrong is not a search. Deciding whether a periodic pattern's layer relations really contradict themselves is far dearer than the disc's one-pass test, so the cheap test is asked first — it is sufficient, so it settles almost everything — and the expensive one runs only on what the cheap one rejects. Five of nine steps on a small cell, fifty thousand of fifty-seven on a large one.
The arc that arrived twice
Which of two panels a crease calls its near one is decided by the order a face walk happened to number them, and the mirrored record is the same relation. Except on one sheet, where it is not — and that sheet turned out to be the one whose folded state comes back the other way up, which is how a duplicate in a graph became a diagnosis.
The bottom layer is at the rim
A hundred and sixty-nine panels of folded tessellation, and three of them have nothing underneath. All three touch the paper's edge, and the same is true on every tiling at every size measured. Which panel is at the bottom of a stack turns out to be a fact about where the sheet was cut rather than about the pattern, and the pattern itself has no bottom at all.
The cost of asking the wrong sheet
A test written for a sheet with an edge, run on a sheet without one, does not fail. It exhausts — proving, at three, thirty-five and three thousand four hundred and fifty-five nodes, that no lettering exists — and the letterings it proved impossible fold, on the collection's own machinery, at every size they were tried at.
The cost of proving something false
A search closing its whole tree is the strongest result this collection can produce, and on a glued tessellation it produces one that is wrong. What it costs to reach is three steps at one period, thirty-five at four, three thousand four hundred and fifty-five at nine, and more than two hundred thousand at sixteen — growing far faster than the cost of finding the lettering it says does not exist.
The edge was not what made it hard
Five families of pattern searched at one step per panel and a tessellation patch did not, and the property left standing after four alternatives were killed was having a rim. Measured under a fixed letter order the patches cost between a half and two-thirds of a step per panel, at every tiling and every size — below the line rather than above it, and the rim is why.
The lettering that was proved impossible
A search closed its whole tree on a glued square tessellation and reported that no mountain-and-valley assignment of it is consistent. Written onto ordinary patches of one, four and nine periods and handed to the four vertex theorems and a folded sheet rebuilt from scratch, the assignment it says cannot exist passes every check, on four tilings, up to fifteen hundred creases.
The most decided vertex here
Sixteen ways to letter four creases; Maekawa allows eight; the big-little-big lemma allows four. A twist polygon's corner is one of the few vertices in this collection where the second cut applies, so it keeps four labellings where a grid, a leaf, a Miura and a crumple all keep eight — and the family the collection long called difficult turns out to be the one whose conditions decide the most.
The period nobody measured
Every repeating pattern in this collection has its drawn period recorded, because a drawing cannot be generated without one. Its folded period is recorded nowhere, and on one of the families measured the two differ by a factor of three — which means the number that has always been quoted is the wrong one for anything about the folded object.
The rim is four letters a cell
Cut a rectangle out of a tessellation and it asks exactly the vertices the tessellation asks, exactly the same questions. What it adds is four free letters for every period of edge — the creases the cut divides, which become two independently answerable creases instead of one. Eight letters on a two-period square, sixteen on a four-period one, and nothing else about the two objects differs at all.
The symmetry a gluing adds
A patch of a tessellation has whatever symmetry its outline allows — a few reflections, a rotation or two. Glue its edges and it acquires translations, and a lettering of the glued sheet has to be invariant under them. That is a much stronger requirement than a lettering of the patch, and it is why one answer covers every patch at once.
Two panels that are one panel
Paper cannot pass through paper, and every test for it compares pairs of panels. On a glued sheet two pieces of the drawing can be the same piece of paper — so a test that does not know the identification either reports a collision between a panel and itself, or misses one where the sheet meets itself round the loop.
What the rim was doing
One rectangle of a twist tessellation, cut out of the plane in the ordinary way, gives up a consistent lettering in forty-eight steps. Join its opposite edges so that no crease is divided and the same drawing, at the same vertices, under the same conditions, takes fifty-six thousand seven hundred and seventy-two. The edge of the paper was never the difficulty. It was the slack.
Where you cut hardly matters
Slide the same rectangle across one whole period of the same tessellation and every position gives a different patch: different creases divided, different half-panels round the edge, panel counts from forty-nine to sixty-one. The cost of lettering them runs from twenty-five steps to thirty-three. Whether a cut is made changes the answer by three orders of magnitude; where it falls changes it by a third.
Which pair is glued
A cell's two cylinders have the same Euler number, the same amount of rim and the same name. On a symmetric drawing they have identical counts of letters, panels and vertices — and searching them costs twenty-four nodes one way and eighty-five the other. Half the rim is a description of the topology and not of the object.
Every generator · The flat-folding field · The patterns a reader can fold