Free Letters to a Young Mathematician by Ian Stewart

to change (but not move) exactly one letter, and the result must be a valid word (as determined by, say, Webster’s).
    Solving this word puzzle isn’t particularly hard: for instance,
    SHIP
    SHOP
    SHOT
    SLOT
    SOOT
    LOOT
    LOOK
    LOCK
    DOCK
    There are plenty of other solutions. But I’m not after a solution as such, or even several: I’m interested in something that applies to every solution. Namely, at some stage, there must be a word that contains two vowels. Like SOOT (and LOOT and LOOK) in this particular answer. Here I mean exactly two vowels, no more and no less.
    To avoid objections, let me make it clear what “vowel” means here. One thorny problem is the letter Y. In YARD the letter Y is a consonant, but in WILY it is a vowel. Similarly, the W in CWMS acts as a vowel: “cwm” is Welsh, and refers to a geological formation for which there seems to be no English word, although “corrie” (Scottish) and “cirque” (French) are alternatives. We need to be very careful about letters that sometimes act as vowels but on other occasions are consonants. In fact, the safest way to avoid the kinds of words that all Scrabble players love is to throw away Webster’s and redefine “vowel” and “word” in a more limited sense. For the purposes of this discussion, a “vowel” will mean one of the letters A, E, I, O, U, and a “word” will be required to contain at least one of those five letters. Alternatively, we can require Y and W always to count as vowels, even when they are being used as consonants. What we can’t do, in this context, is allow letters to be sometimes vowels, sometimes consonants.
I’ll come back to that later.
    It’s not a question of what the correct convention is in linguistics; I’m setting up a temporary convention for a specific mathematical purpose. Sometimes in math the best way to make progress is to introduce simplifications, and that’s what I’m doing here. The simplifications are not assertions about the outside world: they are ways to restrict the domain of discourse, to keep it manageable. A more complicated analysis could probablyhandle the exceptional letters like Y too, but that would complicate the story too much for my present purpose.
    With that caveat, am I right? Is it true that every solution of the SHIP–DOCK puzzle includes a word (in the new, restricted sense) with exactly two vowels (in the new, restricted sense)?
    One way to investigate this is to look for other solutions, such as
    SHIP
    CHIP
    CHOP
    COOP
    COOT
    ROOT
    ROOK
    ROCK
    DOCK
    Here we find two vowels in COOP, COOT, ROOT, and ROOK. But even if a lot of individual solutions have two vowels somewhere, that doesn’t prove that they all have to. A proof is a logical argument that leaves no room for doubt.
    After a certain amount of experiment and thought, the “theorem” that I am proposing here starts to seem obvious. The more you think about how vowels can change their positions, the more obvious it becomes thatsomewhere along the way there must be exactly two vowels. But a feeling that something is “obvious” does not constitute a proof, and there’s some subtlety in the theorem because some four-letter words contain three vowels, for instance, OOZE.
    Yes, but . . . on the way to a three-vowel word, we surely have to pass through a two-vowel word? I agree, but that’s not a proof either, though it may help us find one. Why must we pass through a two-vowel word?
    A good way to find a proof here is to pay more attention to details. Keep your eye on where the vowels go. Initially, there is one vowel in the third position. At the end, we want one vowel in the second position. But—a simple but crucial insight—a vowel cannot change position in one step, because that would involve changing two letters. Let’s pin that particular thought down, logically, so that we can rely on it. Here’s one way to prove it. At some stage, a consonant in the second position has to change to a vowel, leaving all