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6.BD. BRIDGE A MOAT WITH PLANKS
In the simplest case, one has a 3 x 3 moat with a 1 x 1 island in the centre. One wants to get to the island using two planks of length 1 or a bit less than 1. One plank is laid diagonally across the corner of the moat and the second plank is laid from the centre of the first plank to the corner of the island. If the width of the moat is D and the planks have length L, then the method works if 3L/2 > D2, i.e. L > 22 D/3 = .94281.. D. One really should account for the width of the planks, but it is not clear just how much overlap is required for stability. Depew is the only example I have seen to use boards of different lengths. With more planks, one can reach across an arbitrarily large moat, but the number of planks needed gets very large. In this situation, the case of a circular moat and island is a bit easier to solve.

The Magician's Own Book (UK version) version is quite different and quite erroneous.

Magician's Own Book (UK version). 1871. The puzzle bridge, p. 123. Stream 15 or 16 feet across, but none of the available planks is more than 6 feet long. He claims that one can use a four plank version of the three knives make a support problem (section 11.N) to make a bridge. However the diagram of the solution clearly has the planks nearly as long as the width of the stream. In theory, one could build such a bridge with planks slightly longer than half the width of the stream, but to get good angles (e.g. everything crossing at right angles or nearly so), one needs planks somewhat longer than 2/2 of the width. E.g. for a width of 16 ft, 12 ft planks would be adequate.

Mittenzwey. 1880. Prob. 298, pp. 54 & 105; 1895?: 330, pp. 58 & 106; 1917: 330, pp. 52 53 & 101. 4 m gap bridged with two boards of length 3¾ m. He only gives a diagram. In fact this doesn't work because the ratio of lengths is 15/16 = .9375

Lucas. RM2. 1883. Le fossé du champ carré. Bridge the gap with two planks whose length is exactly 1. Notes this works because 3/2 > 2.

Hoffmann. 1893. Chap. VII, no. 9, pp. 289 & 295. Matchstick version = Hoffmann-Hordern, p. 193.

Benson. 1904. The moat puzzle, p. 246. Same as Hoffmann, but the second plank is shown under the first!!

Dudeney. CP. 1907. No. 54: Bridging the ditch, pp. 83-85 & 204. Eight 9' planks to cross a 10' ditch where it makes a right angle.

Pearson. 1907. Part I, no. 34: Across the moat, pp. 122 & 186.

Blyth. Match-Stick Magic. 1921. Boy Scouts' bridge, p. 21. Ordinary version done with matchsticks.

J. C. Cannell. Modern Conjuring for Amateurs. C. Arthur Pearson, London, nd [1930s?]. Boy Scouts' bridge, pp. 68-69. As in Blyth.

Depew. Cokesbury Game Book. 1939. Crossing the moat, pp. 225-226. Square moat 20 feet wide to be crossed with boards of width 18 and 15 ft. In fact this doesn't work -- one needs L₁ + L₂/2 > D2.

"Zodiastar". Fun with Matches and Matchboxes. Op. cit. in 4.B.3. Late 1940s? The bridge, pp. 66-67 & 83. Matchstick version of the square moat & square island problem.

F. D. Burgoyne. Note 3106: An n plank problem. MG 48 (No. 366) (Dec 1964) 434 435. The island is a point in the centre of a 2 x 2 lake. Given n planks of length s, can you get to the island? He denotes the minimal length as s(n) and computes s(1) = 1, s(2) = 2 2/3, s(3) = .882858... and says s() = 2/2, [but I believe it is 0, i.e. one can get across an arbitrarily large moat with a fixed length of plank].

Jonathan Always. Puzzling You Again. Tandem, London, 1969.

Prob. 10: A damsel in distress, pp. 15 & 70 71. Use two planks of length L to reach a point in the centre of a circular moat of radius R. He finds one needs L²  4R²/5.

Prob. 11: Perseus to the rescue again, pp. 15 16 & 71 72. Same with five planks. The solution uses only four and needs L²  2R²/3.

C. V. G.[?] Howe? Mathematical Pie 75 (Summer 1975) 590 & 76 (Autumn 1975) 603. How big a square hole can be covered with planks of unit length? Answer says there is no limit, but the height of the pile increases with the side of the square.

Highlights for Children (Columbus, Ohio). Hidden Picture Favorites and Other Fun. 1981. Brain Buster 4, pp. 12 & 32: Plink, plank, kerplunk? Two children arrive at a straight(!) stream 4m wide with two planks 3m long. Solution: extend one plank about 1¼ m over the stream and one child stands on the land end. The second child carries the other plank over the stream and extends it to the other side and crosses. He then pulls the plank so it extends about 1¼ m over the stream. The first child now extends her plank out to rest on the second plank and crosses, pulling up her plank and taking it with her. I theory this technique will work if L > ⅔D, but one needs some overlap space, and the children may not have the same weight.

Richard I. Hess. Puzzles from Around the World. The author, 1997. (This is a collection of 117 puzzles which he published in Logigram, the newsletter of Logicon, in 1984-1994, drawn from many sources. With solutions.) Prob. 64. Usual problem with D = 10, but he says the board have width W = 1 and so one use the diagonal of the board in place of L. In my introduction, we saw that the standard version leads to L²  8D²/9, so Hess's version leads to L²  8D²/9 - W² and we can get across with slightly shorter planks, but we have to tread very carefully!