## Answers

A simple proof that on the two-by-three rectangle osberg cannot be folded to spell borges (or vice versa) is to note that in each case the fold requires that two pairs of cells touching only at their corners would have to be brought together in the final

Figure 33

Neale's "Sheep and Goats" problem packet. It is evident that no fold can put a pair of such cells together.

The square puzzle with the faces and prison windows is solved from the starting position shown. Fold the top row back and down, the left column toward you and right, the bottom row back and up. Fold the right packet of three cells back and tuck it into the pocket. A face is now behind bars on each side of the final packet. The central face of the square cannot be put behind bars because its cell is diagonally adjacent to each of the window cells.

Space prevents my giving solutions for the eight pseudonyms of Beelzebub, but Beelzebub itself can be obtained as follows. Starting with the layout shown, fold the bottom row toward you and up to cover BBE. Fold the left column toward you and right to cover ZU. Fold the top row toward you and down, but reverse the crease between L and Z so that LZ goes between B and B on the left and the upper E goes on top of the lower E. You now have a rectangle of two squares. On the left, from the top down, the cells are BLZBUB, on the right EEE. The final move is difficult. Fold the right panel (EEE) toward you and left. The three E's are tucked so that the middle E goes between Z and B, and the other two E's together go between B and L. Once you grasp what is required it is easier to combine this awkward move with the previous one. The result is a tightly locked packet that spells Beelzebub. The solution is unique. If the cells of the original "map" are numbered 1 through 9, the final packet is 463129785.

To find the 5-face of the tetraflexagon, start with face 1 on the top and 2 on the bottom. Mountain-fold in half vertically, left and right panels going back, so that if you were to open the Hexagon at the center crease you would see the 4-face. Instead of opening it, however, move left the lower inside square packet (with 4 and 3 on its outsides) and move right the upper square packet (also with 4 and 3 on its outsides). Insert your fingers and open the flexagon into a cubical tube open at the top and bottom. Collapse the tube the other way. This creates a new tetraflexagon structure that can be flexed to show faces 1, 3 and 5.

A similar maneuver creates a structure that shows faces 2, 4 and 6. Go back to the original structure that shows faces 1,2, 3 and 4 and repeat the same moves as before except that you begin with the 2-face uppermost and the 1-face on the underside.

Figure 34 shows how to separate the sheep from the goats:

(1) Start with the two-color square folded as shown.

(2) Fold in half along the horizontal diagonal by folding the

Figure 34

Figure 34

Solution to the "Sheep and Goats" problem

Solution to the "Sheep and Goats" problem bottom corner up to make a "hat" with a white triangle at the lower left-hand corner.

(3) Open the hat's base and continue opening until you can flatten the hat to make the small square shown.

(4) Insert a left finger into the pocket on the right of the upper face of drawing No. 3. Pull upward and flatten as shown.

(5) Turn the paper over sideways and repeat the previous move on the other side. The result is a rectangle with a white triangle in the upper right-hand corner.

(6) Open the rectangle into a cubical tube open at the top and bottom. Collapse the tube the other way to make a rectangle again, except that now it is colored as shown.

(7) Insert your right thumb into the pocket on the left of drawing No. 6, lift up the flap and flatten it as shown.

(8) Turn the paper over sideways and do the same on the other side. You should now have a small square, black on both sides.

(9) Reach into the square from above, open it and flatten to make an inverted hat, black on both sides.

(10) Open the hat by separating its bottom points and flatten the large square that results. It will be the same size as the square you started with, but now it is all white on one side and all black on the other—all sheep and all goats.

Repeating the same sequence of moves will mix the sheep and the goats again. With practice the folds can be done so rapidly that you can hold the square out of sight under a table for just a few moments and produce the change almost as if by magic.

ADDENDUM!

The "little law" that Henry Ernest Dudeney hinted about in connection with his map-fold problem has probably been rediscovered. Mark B. Wells of the Los Alamos Scientific Laboratory used a computer to confirm that the 2x3 map has 10 folds for each cell on top. The program also found that the order-3 square has 152 folds for each cell on top. In his 1971 paper Lunnoim proved that for any rectangular map every cyclic permutation of every possible fold is also a possible fold. Thus it is necessary to determine only the folds for one cell on top because the cyclic permutations of these folds give all the other folds. For example, since 123654789 is a possible fold, so also are 236547891, 365478912 and so on. It is a strange law because the folds for cyclic permutations differ wildly. It is not yet known whether the law applies to all polyomino-shaped maps or to maps with equilateral triangles as cells.

In his 1971 paper Lunnon used an ingenious diagram based on two perpendicular slices through the center of the final packet. He was able to write a simple backtrack program for x-hy-y maps, extend the problem to higher dimensions and discover several remarkable theorems. For example, the edges of one cross section always diagram x linear maps of y cells each, and the edges of the other cross section diagram y linear maps of x cells each.

The 2 X 3, 2 x 4, 2 x 5, 2 x 6, and 3x4 maps have respectively 60, 320, 1980, 10512, and 15552 folds. The order-3 square has 1368 folds, the order-4 has 300608, the order-5 has 186086600. In all cases the number of folds is the same for each cell on top, as required by cyclic law. The order-2 cube, folded through the fourth dimension, has 96 folds. The order-3 cube has 85109616. Many other results are tabulated by Lunnon in his 197 1 paper, but a nonrecursive formula for even planar maps remains elusive.

The linear map-fold function, as Lunnon calls it, is the limit approached by the ratio between adjacent values of the number of possible folds for a 1 Xn strip. It is very close to 3.5. In his unpublished 1981 paper Lunnon narrows the upper and lower bounds to 3.3868 and 3.9821.

In 1981 Harmony Books in the United States, and Pan Books in England brought out a large paperback book called Folding Frenzy. It contains six 3 X 3 squares, with red and green patterns on both sides, and five pages partially die-cut. Without removing any pages, there are nine puzzles to solve by folding the squares. The puzzles are credited to Jeremy Cox.

In describing one of his map-fold puzzles (Modern Puzzles,

No. 214) Dudeney mentioned a curious property of map folds that is not at all obvious until you think about it carefully. It applies not only to rectangular maps, but also to maps in the shape of any polyomino; that is, a shape formed by joining unit squares at their edges. Assume that any such map is red on one side, white on the other. No matter how it is folded into a 1 x 1 packet, the colors on the tops of each cell will alternate regardless of which side of the packet is up. If the cells of the map are colored like a checkerboard, with each cell the same color on both sides, the final packet (after any sort of folding) will have leaves that alternate colors. If the checkerboard coloring is such that each cell is red on one side, white on the other, all cells in the folded packet will have their red sides facing one way, their white sides facing the other way.

It occurred to me in 1971 that the parity principles involved here could be the basis for a variety of magic tricks. One appeared under the title "Paradox Papers" in Karl Fulves' magic periodical, The Pallbearers Review. It goes like this: Fold a sheet of paper twice in each direction so that the creases make 16 cells. It is a good plan to fold the paper each way along every crease to make refolding easier later on.

Assume in your mind that the cells are checkerboard colored black and red, with red at the top left corner. Five red playing cards are taken from a deck and someone selects one of them. With a red pencil jot the names of the five cards in five cells, using abbreviations such as 4D and QH. Tell your audience that you are taking cells at random, but actually you must put the name of the chosen card on one of the "black" cells, and the other four names on "red" cells.

Have another card chosen, this time from a set of five black cards. Turn the sheet over, side for side, and jot the names of the five black cards on cells, again apparently at random. Use a black pencil. Put the chosen card on a "red" cell, the others on "black" cells.

Ask someone to fold the sheet any way he likes to make a lxl packet. With a pair of shears, trim around the four sides of the packet. Deal the 16 pieces on the table. Five names will be seen, all the same color except for one—the chosen card of the other color. Turn over the 16 pieces. The same will be true of the other sides.

Gene Nielsen, in the May 1972 issue of the same journal, suggested the following variant. Pencil X's and O's on all the cells, alternating them checkerboard fashion. Turn over the sheet horizontally, and put exactly the same pattern on the other side. Spectators will not realize that each cell has an

X on one side, a O on the other. Someone folds the sheet randomly into a packet. Pretend you are using PK to influence the folding so that it will produce a startling result. Trim the sides of the packet and spread the pieces on the table. All X's face one way, all O's face the other way.

Swami, a magic periodical published in Calcutta by Sam Dalai, printed my "Paper Fold Prediction" in its July 1973 issue. Start by numbering the cells of a 3x3 sheet from 1 through 9, taking the cells in the usual way from left to right and top down. Put the digits on one side of the paper only. After someone folds the sheet randomly, trim the sides of the packet and spread the pieces. Add all the numbers showing. Reverse the pieces and add the digits on the other sides. The two sums will be different. Explain that by randomly folding the sheet, the nine digits are randomly split into two sets. Clearly there is no way to know in advance what either sum will be when the pieces are spread.

Repeat the same procedure, but this time use a 4 x 4 square with cells numbered 1 through 16. The sheet is randomly folded and the edges trimmed. Before spreading the pieces, hold them to your forehead and announce that the sum will be 68. Put down the packet, either side uppermost, and spread the pieces. The numbers showing will total 68. Discard the pieces before anyone discovers that the sum on the reverse sides also is 68.

The trick works because if the original square has an odd number of cells, the sums on the two sides will not be equal. (On the 3x3 they will be 20 and 25.) However, if the square has an even number of cells, the sum is a constant equal to (n2 + n)/4 where n is the highest number. You can now repeat the trick with a 5x5 square, but instead of predicting a sum, predict that the difference between the sums on the two sides of the pieces will be 13.

The principle applies to cells numbered with other sequences. For example, hand a wall calendar to someone and ask him to tear out the page for the month of his birth. He then cuts from the page any 4x4 square of numbers. The sheet is folded, the packet trimmed, the pieces spread, and the visible numbers added. The sum will be equal to four times the sum of the sheet's lowest and highest numbers. You can predict this as soon as you see the square that has been cut, or you can divine the number later by ESP.

Some other suggestions. Allow a spectator to write any digit he likes in each cell of a sheet of any size, writing left to right and top down. As he writes, keep a running total in your head by subtracting the second number from the first, adding the third, subtracting the fourth, and so on. The running total is likely to fluctuate between plus and minus. The number you end with, whether plus or minus, will be the difference between the two sums after the sheet is folded, trimmed, and the pieces spread.

Magic squares lend themselves to prediction tricks of a similar nature. For example, suppose a 4 x 4 map bears the numbers of a magic square. After folding, trim only on two opposite sides of the packet. This will produce four strips. Have someone select one of the four. The other three are destroyed. You can predict the sum of the numbers on the selected strip because it will be the magic square's constant. Of course you do not tell the audience that the numbers form a magic square.

Many of these tricks adapt easily to nonsquare sheets, such as a 3 x 4. The underlying principles deserve further exploration by mathematical magic buffs.

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