After I shared my thoughts about Martin Gardner’s aha! Insight earlier this month, Tikitu de Jager wrote me with some interesting clarifications and corrections about this unusual book.

Though I never personally encountered it, a reprint of this 1978 work does indeed exist as a hardcover volume, published in 2006 by the Mathematical Association of America. It collects aha! Insight with its 1982 followup, aha! Gotcha — the existence of which also came as news to me.

The friend I mentioned in the earlier post, who recalled the math joke I based my encoder toy on as originating from aha! Insight, ends up half-right after all: the story about the alien who can record anything in a single, tiny mark actually appeared in the sequel volume. I have updated the toy’s afterword appropriately.

And finally, the Canadian artist behind both books’ delightful, scribbly artwork is named Jim Glen — any further information about whom I cannot find on the web, alas.

Updated to add: Joe Cabrera tells me that Ray Salmon illustrated the latter book, though carefully imitating Jim Glen’s style.

My own copy of aha! Insight arrived last week, purchased before I shared my entry (and therefore before I learned about the more recent and more complete reprint). Happily, I can report it in excellent shape, wearing the intervening decades rather better than the much-traveled and wrinkly copy I had borrowed from the public library. I find the pages white and supple and the covers clean, at least once I had carefully peeled away the University of Minnesota used-book sticker from the cheerful Glen-bird’s face.

A situation described on page 104 bothered me when I read this in May, and now that I can open back up to it, it bothers me still.

The author describes a notional game where two players sit facing one another, each with a card affixed to their forehead, such that they can see their opponent’s card but not their own. Each card has a positive integer written on it. The two integers are consecutive, and the players know this.

The rules of the game: Starting with either one, the players — let’s name them Andy and Beth — take turns answering the yes-or-no question: Do you know what your own number is? The players both hail from the tribe of infallibly rational and honest logic-puzzle people, so the first to utter “yes” wins the game.

Gardner asserts that, for the higher of the two numbers n, the winner will say “yes” on either turn n or turn (n - 1). I thought about this for a while and I just couldn’t see it, at least not beyond the lowest-numbered cases. So, for our mutual entertainment and enlightenment, dear reader, I thought I’d puzzle it out longhand in this blog post.

The degenerate case of (1, 2) is simple enough. If the first player — let’s make that Andy — sees 1, he says “yes”: he can only possibly have 2 on his head, according to the game’s rules, and so wins immediately. If he sees 2, he thinks: Well, I could have either 1 or 3. I don’t know yet, and so must say “no”. Beth, of course, doesn’t care what he says, because she sees 1, and the earlier logic applies. Done in one or two turns. Fine.

(2, 3) gets more interesting. If Andy sees 2, then he says “no”, for the same reasons as before. His opponent sees 3, but doesn’t know where she stands — 2 or 4? — so she says “no” too. And then the jig is up; Andy knows that Beth would have said “yes” if she saw a 1, due to the situation described in the previous case. But she didn’t, and he must therefore wear 3. Game over in three turns.

If instead Andy saw 3, he again must say “no” at the start. Beth sees 2, and she knows from Andy’s ignorance that she’s not 1, so she must be 3, and so she nails it in two turns.

So far, so good. However, I start losing the plot at the (3, 4) case, because the series starts feeling untethered from the start of the number line at 1. Let’s work it out.

Andy sees 3, says “no” as before. Beth sees 4, so she knows she wears either 3 or 5. Lacking clues, she says “no” too. Andy knows that if Beth saw 2 she’d have said “yes”, per the previous case, and deduces that he must wear 4 instead. Done in three turns.

Can we have Beth win in four? Well: Andy sees 4 now. “No.” That leaves Beth, seeing 3, with 2 or 4 on herself — meaning that she has no more idea which it is than she did at the start of the (2, 3) case. So: “No.” Back to Andy, who knows that he’s wearing 3 or 5, but still has no way of knowing which. “No.” And that’s all Beth needs, because his not saying “yes” means he didn’t recapitulate our earlier (2, 3) where Beth saw the 3 on Andy’s head. That means Beth can’t possibly be wearing the 2, leaving only 4. And she does indeed win in four turns.

Argh! I’m very sorry, but I insist on one more round of this. (4, 5):

Andy sees 4 again. “No.” Beth sees 5. “No.” This doesn’t tell Andy anything; Beth’s still following the previous script! “No.” Beth, still lacking information, goes off that script with “no”, which tells Andy everything he needs to know — specifically, that he can’t possibly wear 3, or Beth would have said “yes”, per the earlier case. That leaves him to say “yes” instead, and finish the game in five turns.

Which means Beth in four again, if we switch the cards, by Gardner’s rule? Because now Andy sees 5. “No.” Beth sees 4, and so falls into her script from the first part of our (3, 4) case: “No.” Andy goes off that script with “no”, as he must. This tells Beth that she doesn’t wear 3, so she says “yes”, taking the game in four.

Okay! Well, I see how this works now. Our theoretically perfect players, on sight of their opponent’s card, follow a mental pattern something like this:

  • If I see 1, then I wear 2. I’ll win on my first turn.

  • If I see a number larger than 1, then I’ll say “no” on my first turn (because I really don’t know which of two possible numbers I wear). On subsequent turns, unless I have reason to say otherwise (as defined below), I’ll still say “no”.

  • I know the turn on which my opponent would say “yes” if I wear the number less than their own, based on a following an accretive script that builds one step at a time all the way back to the (1, 2) case. If they do not say “yes” on that turn, then I know I wear the number greater than theirs instead, and I will say “yes”.

Gosh, that seems unsatisfying, somehow. I feel like I’ve missed a lesson, that I have merely codified the external pattern I’ve observed, rather than arrived at something deeper. And that leaves me feeling even denser, though I’ve managed to convince myself of the accuracy of Gardner’s stipulation.

It does amuse me to think that the case of (99, 100) would play out the way it must, though — the players doomed to bat “no” back and forth dozens of times before either needs to start paying attention.

While I hadn’t heard of this frustrating puzzler before this year, it brings to mind two related classics of my earlier acquaintence. Reading it in aha! Insight immediately made me think of the (English-language-specific) gem of Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo. And to my surprise, working out the numbers-game conundrum longhand just now made me think of the unexpected hanging paradox, a nasty little mindbender which I learned about from an earlier (and eponymous) Gardner book.

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