Origami & Folding
what one sheet can do
Series

Map folding — the series

9 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 1 × 4161 × 5501 × 61442 × 282 × 3602 × 43203 × 31,3684 × 4300,608filled — counted here, by exhaustive search over stacking ordersopen — Lunnon's published count, quoted rather than computed

    The oldest open problem

    In how many ways can a map be folded? The question needs no notation to state, the answer is a small integer for small maps, and after sixty years there is still no formula — only a list of numbers, each one found by searching every possibility.

    part 1 · complexity
  2. 2468101222.22.42.62.833.23.4stampsratio to the term beforeodd terms, from aboveeven terms, from belowfilled: computed here, to 9 stamps · hollow: 10 and 11 and 12, computed once and quoted4,536 foldings at 9 stamps

    Where the exponent comes from

    The number of ways a strip of stamps folds grows exponentially, and the base of the exponential is a number nobody has proved exists. The ratio of one term to the last climbs past three and is still climbing where the computation stops — which is the only structural handle anybody has on the sequence.

    part 2 · complexity
  3. 1 × 4161 × 5501 × 61442 × 282 × 3602 × 43203 × 31,3684 × 4300,608filled — counted here, by exhaustive search over stacking ordersopen — Lunnon's published count, quoted rather than computed

    Two directions that will not separate

    A map has rows and columns, and a strip of stamps is a map with one row. The obvious hope is that the two-dimensional count is built from the one-dimensional one — fold the rows, then fold the columns. It is not: a two-by-three map folds 60 ways against a product of 12, and the discrepancy grows from a factor of two to a factor of thirty-eight over the counts anybody has.

    part 3 · complexity
  4. how many ways each map foldsa strip of five50of 120a plus120= 5! — every stackinga tee120= 5! — every stackinga two-by-three60of 720a two-by-three, one gone40of 120one corner gone848of 40320the middle gone8016of 40320the full square1368of 362880

    The map that is not a rectangle

    Take one square out of a three-by-three map and the number of ways it folds does not go down by an eighth. It goes up — to 848 if the square came from a corner, and to 8,016 if it came from the middle. Two maps of eight squares in the same box, differing by nearly a factor of ten, and no function of the box tells them apart.

    part 4 · complexity
  5. the pale bar is the published count, the dark one the objectsneither operation ever fixes a folding; doing both sometimes does, and that is why it is not a quarter2 stamps2 labelled · 1 objects · 2 fixed by doing both3 stamps6 labelled · 2 objects · 2 fixed by doing both4 stamps16 labelled · 5 objects · 4 fixed by doing both5 stamps50 labelled · 14 objects · 6 fixed by doing both6 stamps144 labelled · 38 objects · 8 fixed by doing both7 stamps462 labelled · 120 objects · 18 fixed by doing both8 stamps1392 labelled · 353 objects · 20 fixed by doing both

    The count counts labels

    One, two, six, sixteen, fifty, a hundred and forty-four: the oldest sequence in the subject counts foldings of a strip of numbered stamps. A folded strip of blank paper has no first stamp and no top side, and neither of those operations ever leaves a folding alone — so the count of objects is 1, 2, 5, 14, 38, 120, and it is not the count over four.

    part 5 · complexity
  6. the bar is the number of foldings, on a logarithmic scaleboth routes give the number printed; a disagreement anywhere would be a defect in one of them2 × 122 letterings · 1 creases3 × 164 letterings · 2 creases4 × 1168 letterings · 3 creases5 × 15016 letterings · 4 creases6 × 114432 letterings · 5 creases2 × 288 letterings · 4 creases3 × 26032 letterings · 7 creases4 × 2320128 letterings · 10 creases3 × 31,368256 letterings · 12 creasesa strip of stamps is the one-row case, and the classical sequence 2, 6, 16, 50, 144 is the top of the table

    The map counted from the layers

    The classical map-folding counts are computed from a rule that never places a panel: work out which edge of the folded square each fold wraps around, and refuse the orderings that interleave two folds at one edge. Place the panels instead and order them by the general non-crossing rules, and the same numbers come out — 2, 6, 16, 50, 144, 8, 60, 320, 1368 — on nine sizes, by machinery that shares no line of code with the first.

    part 6 · complexity
  7. the bar is how many letterings pass every condition at every vertexand the note is how many of those close a loop in the arcs2 by 122 panels · 2 letterings pass every vertex · 0 close a loop3 by 143 panels · 4 letterings pass every vertex · 0 close a loop4 by 184 panels · 8 letterings pass every vertex · 0 close a loop5 by 1165 panels · 16 letterings pass every vertex · 0 close a loop2 by 284 panels · 8 letterings pass every vertex · 0 close a loop3 by 2326 panels · 32 letterings pass every vertex · 0 close a loop4 by 21288 panels · 128 letterings pass every vertex · 0 close a loop3 by 32569 panels · 256 letterings pass every vertex · 4 close a loopa map's difficulty is not here — it is in the rules about which panels may lie between which

    The test that never fires on a map

    The cheapest refusal this collection has reads a crease list once and reports that no arrangement of the layers exists. Enumerate every labelling of every map from two panels to nine and it fires on four of the four hundred and fifty-four — all four on the largest map, none at all below it. On the oldest open problem in the subject, the cheap test has essentially nothing to say.

    part 7 · complexity
  8. the period cell of the gridone period, with its neighbours round it1 interior vertices in the cell4 crease pieces drawnperiod 1.000 × 1.000one square, because a grid repeats at every linethe cell is a rectangle of ordinary paper until somebody says its edges are one edge

    A map with no edges

    Counting the ways a rectangular map folds is the oldest open problem in the subject, and every version of it assumes the map has an edge. Join the map's opposite edges and the question changes shape: half the sizes have no folded state at all, and the ones that do have no bottom layer to count from.

    part 8 · complexity
  9. the pieces that are one panelleft and right edges identified — 9 pieces, 6 panels9 pieces on the drawing6 panels on the sheet10 creases, 4 verticeskeeps the sidetwo pieces of one shade are one piece of paper, a cell apart

    The tube a map makes

    Join one pair of a map's edges and the result is a tube — a real object, foldable in the hand, and neither the strip's problem nor the torus's. It has one loop that cannot be shrunk instead of two, it keeps its bottom layer because it keeps half its rim, and half its sizes are refused by a parity the flat map does not have.

    part 9 · complexity

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