Later in the letter, Carroll mentions that he "was told there is an American game involving a similar principle. I have never seen it ..." Whatever the truth, the game's origins faded into obscurity and the game itself long since became part of the multilingual folklore. It recently took a mathematical turn, under the name of the Ship-Dock Theorem. The name originates with a particular puzzle, that of getting from SHIP to DOCK. Here's a couple of solutions: SHIP, SLIP, SLOP, SLOT, SOOT, LOOT, LOOK, LOCK, DOCK [Stewart, p 41], and a shorter one SHIP, SHOP, CHOP, COOP, COOK, COCK, DOCK [Gale, pp 11-112].
Ship-Dock Theorem In any solution of the problem, there must be a word at least two of whose letters are vowels.
Well, as it stated, the theorem might be incorrect. For it implicitly relies on the assumption that every English word contains at least one vowel. Which is simply not true. One of the more exotic counterexamples is the word nth - of a mathematical origin - that probably managed to slip into the English vocabulary before the current epidemic of innumeracy. (As a supportive counterexample, Nathan Bowler offers the following sequence with one possible blemish to an exacting eye: SHIP, SKIP, SKIS, SKYS, SAYS, SACS, SACK, SOCK, DOCK.)
Three quarters of a century after appearance of Lewis Carroll's puzzle, on March 17, 1953 [Gazalé, p 25], Frank Gray, a research scientist at Bell Labs, filed patent no. 2632058, for the Gray code encoding vacuum tube. n-digit (binary) Gray code is a sequence of strings of n symbols "0" and "1" such that any two consecutive strings differ only in a single position. Compare this to representations of 7 (0111) and 8 (1000) in the binary positional system that differ in all 4 positions.
The Gray codes are widely used in industry to improve fidelity of telegraphic transmission, among other things. Martin Gardner [Knotted Doughnuts] popularized application of Gray codes to puzzle solving.
Mathematicians seem to have turned puzzle solving into a gainful occupation.
Here's one puzzle. In all, there are 2n binary strings of length n. If written one after another they will produce a cancatenated string of length n2n. This string is redundant in that there are numerous repetitions. For example, using the Gray ordering for n = 2, we get the string 00011110 in which "00" appears twice while "11" appears 3 times. How short may be a string to contain, as substrings, all binary strings of length n exactly once? The absolute minimum is 2n + n - 1. And, as was shown by N.G. de Bruijn in 1946, there are 2n-1 - n such arrangements [Stein, p 149]. One particular arrangement, known as the de Bruijn cycle is presented below [Knuth, v1, exercise 2.3.4.2-23, and Knuth, v2, exercise 3.2.2-17].
Note that there are always enough zeros at the tail of the string to identify them with all but one zero at the front. This explains the name de Bruijn cycle. Such circular arrangement contains exactly 2n digits.
This was a difficult puzzle. It required not only ingenuity, but also a lot of advanced knowledge.
Here is an easier one. In a triangle, mark 1/3 of each side counting from vertices in a certain order. Connect those points to the opposite vertices. Prove that the area of the middle triangle thus obtained is 1/7 that of the given triangle. This is a well-known puzzle with multiple solutions. One straightforward solution that makes use of Ceva's and VanObel's theorems generalizes to the case where we mark 1/N-th of each side. The illustration below draws on the knowledge derived from those two theorems.
(Note that for N = 2, we get what may be called a proof without words for the fact that the area of the triangle (shown in red) formed by the medians of a given triangle equals 3/4 of the area of the latter. Another such pww appeared in the April's issue (1999) of Mathematics Magazine. )
Recently I ran into a puzzle [Kanga] whose relevance to the above was hard to miss. Given a triangle, extend each side by its own length in one direction following a certain order. Connect the newly obtained points. Prove that the area of the big triangle is 7 times that of the given one.
Of course, we can add 1, 2, 3, or, in general, N lengths to each side. The triangles so obtained have areas 7, 19, 37, or, in general, 3N2 + 3N + 1 times greater than the given triangle. (This this #A003215 in the On-Line Encyclopedia of Integer Sequences.)
Arrange all integers in a hexagonal spiral pattern. In a moment, I'll have another chance to mention a spiral. For now, let's continue talking about numbers, the sequence 1, 7, 19, 37, ..., in particular.