A triptych, that is, a landscape sheet divided into three equal parts by two vertical folds, in principle can be folded, as we saw last week, in 8 different ways: in each fold we can cover the front or back of the sheet, so there are 4 possibilities (AA, AR, RA, RR) starting with one fold and 4 starting with the other, 8 in total; but only in principle, since two of the folds give rise to the same configurations obtained with two others (can you determine which ones they are?), so in reality there are only 6 different folds. If mental visualization is not your thing, I suggest that you make a triptych by folding a sheet of paper, and number the sides from 1 to 6 (or better A1, A2, A3 on the front and R1, R2, R3 on the back. ) you will be able to spend a time that is as entertaining as it is instructive studying the different folding possibilities.
Similarly, the possible different folds of a “quatriptych” (a landscape sheet divided into four equal parts by three vertical folds) are not 24 (2 x 2 x 2 = 8 possibilities starting with each of the 3 folds: 3 x 8 = 24), but only 16. If the triptych seemed too easy, try to find the 16 different folds of the “quatriptych” (or what amounts to the same thing, determine which ones are repeated).
The stamp strip problem
The apparently simple problem of folding a “polyptych”, that is, a landscape sheet with only vertical folds, is often called the “stamp strip problem”, in which it is a question of performing a complete folding, that is, of Place them all under one of them, forming a compact pile. In the trivial cases of 0 and 1 folds, there are, respectively, 1 and 2 different configurations, and, as we have seen, there are respectively 6 and 16 possibilities for 2 and 3 folds. The sequence continues to grow rapidly:
1, 2, 6, 16, 50, 144, 462, 1392, 4536…
Don’t try to look for a guideline: there is no formula that, for a strip of n stamps, gives the number of possible folds as a function of n. In 1968, John E. Koehler, who was the first to calculate the number of folds for long strips (he found the value 16,864,984 for n = 16), showed that the number of possible folds of a strip of n stamps is equal to the number of different ways of joining n points of a circle by means of ropes of two alternating colors without cutting ropes of the same color; but, as far as I know, this interesting equivalence did not contribute to facilitating the calculation of said number.
Borges deconstructed by Nabokov
Moving on from the triptychs and stamp strips to the maps themselves, in the (seemingly) simple case of the map with only two vertical folds and one horizontal fold, raised last week, the possibilities are 6 x 8 = 48 (can you explain because).
While you investigate the possibilities of folding the 2 x 3 map (for which I suggest you start with the 2 x 2), you can try to solve a puzzle inspired by the novel by Vladimir Nabokov Ada or the ardorin which Borges appears camouflaged behind the Osberg anagram as an apocryphal author of The little gypsy and, therefore, the indirect cause of Lucette’s suicide (an ironic and somewhat malevolent “tribute” that inspired Umberto Eco when it came to turning Borges into Fray Jorge de The name of the rose).
After producing two vertical folds and one horizontal fold on a sheet of paper, write the OSBERG letters in the six resulting boxes in an orderly manner, as indicated in the figure, and then try to fold the sheet so that the letters in the successive boxes are stacked together. form, from top to bottom, the word BORGES.
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#Borges #deconstructed