Series

Map folding — the series

12 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 of a folded state 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
  10. the four letterings of a three-by-three map the loop test refusesdash-dot: mountain; dashed: valley; each passes all four vertices and still has no layer orderone in sixty-four of the letterings a three-by-three map allows

    A map refuses in small pieces

    The one-pass refusal — a circle in the arcs a map's letters force — fires on four of a three-by-three map's 256 letterings and on none below it, and the question was how it grows. It grows by windows. Every lettering it refuses on a four-by-three or five-by-three map contains one of those four refused three-by-three patterns somewhere inside it, so the share refused is almost exactly one minus 63⁄64 to the power of the map's three-by-three windows. Four by four is the first map with a refusal no window explains, 64 of its 2,112. The rise is smooth, not sharp: a quarter at six by six, near half at eight, seven tenths at ten.

    part 10 · complexity
  11. the circle behind a refusal, on the smallest maps that need each onedash-dot: mountain; dashed: valley; heavy loop: the circle, each step to a panel the letters put above the last3 by 38 creases4 by 412 creases5 by 414 creases5 by 516 creases

    The long circles go round the middle

    A four-by-four map has 64 letterings the one-pass refusal catches with no three-by-three pinwheel anywhere in them, and they are not larger pinwheels. They are four shapes, sixteen versions of each, none unchanged by any turn or mirror, and every one is a circle of twelve creases running round the middle of the map. Read the same way on larger maps, every refusal is a circle round some rectangle of panels, and it needs the whole of that rectangle: one in 64 three-by-three rectangles closes one, one in 512 four-by-fours, and rectangles three cells across close nothing a pinwheel does not. Two window sizes explain all but 0.07 per cent of a twenty-by-twenty map's letterings.

    part 11 · complexity
  12. what refuses a lettering of a tube 5 panels longone bar to a circumference; the whole bar is the share refused0%25%50%75%100%2 round49.0%4 round72.9%6 round59.5%8 round52.6%10 round52.7%12 round54.7%a three-by-three windowa four-by-four windowa circle round the tubea longer circle on the sheet1,500 letterings drawn at random for each tube; each is filed under the first test that refuses it

    A tube refuses round itself

    On a flat map the cheap refusal — a circle in the arcs a lettering forces — is almost entirely local: small windows of the map explain all but a sliver of it. Glue the map's left edge to its right and the windows still refuse at exactly the rate their number predicts, counted across the seam. But a new refusal appears that no window can contain, a circle that goes once round the tube. It refuses half of every lettering of a tube two panels round, over half of one four round, a quarter at six, and it falls by about half with every two panels of circumference while the tube's length hardly matters.

    part 12 · complexity

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