That book was an essential part of my life growing up. I remember it fondly. The puzzles were hard but not cruel. They didn’t rely on weird and/or random interpretations of questions or premises like lower quality puzzles often do. It was instrumental in me experiencing the satisfaction that comes from intellectual effort. It may not be much of a stretch to say it significantly affected the path I took in terms of education and career.
“If Avtonom leaves Moskva on train at fyorty kilometers per the hour, and Ljuba is with train from Nizhny Novgorod at fifetyn kilometers per the hour, which is first to be mauled by bear?”
This book blew my mind when I was a kid. It is easily the best gift you could ever get a bright 9 or 10 year old (well, this and Practical Mental Magic)
Of the many awesome puzzles in it, the one I always first recall is #256: A Singular Trip
The puzzle itself is challenging and interesting, but the improvised plan within the story (to this day!) makes me smile at how clever / stupid it is.
For the latter puzzle shown:
Basic method: find the loops of numbers out of place (for instance: In space #1 is a 7, in space #7 is a 20, etc., making a loop of 1,7,20,16,11,2,24, and back to 1). There are 6 such loops. Swap them around so that each ends up in the right place.
Minimum moves: 19 (it takes x-1 moves to resolve a loop of x numbers; since there are 25 numbers and 6 loops, that’s 19 moves).
OK. if we randomly place the numbers in this grid what is the expected number of loops and thus minimum moves. The smallest is zero if you are lucky and they all end up in the correct spot. What is the highest? (1 loop 25 moves long I think) I don’t know the answer, but it is the way these mathematical puzzles go, something simple and then tell me something about the entire permutation space of the problem.
Next it would be for any square, and then rectangle NxM. I suspect it only matters how many numbers there are for the exchange.
Yes - e.g. when they are all out of place by 1 (1 is in space 2, 2 is in space 3…)
In Soviet Russia, book puzzles you!
I haven’t the foggiest if this is right, but:
It seems to me that there will be about one loop of length one, about one loop of length two, and so on until we fill the square (with some numbers being omitted). So, for instance, there are 25 numbers, which is between the triangle numbers of 21 and 28, so we’d expect 6-7 loops (which is what we get).
The triangle numbers are x =n(n+1)/2. I do not want to solve that equation for n, but, happily, it trends proportionally closer to n2 as n grows, so I’d say to expect about √(2n) loops for a grid with n squares. So the expected number of moves to solve a grid of size n would be n-√(2n).
Of course, I could totally be wrong in my guess of how many loops there are going to be, as I have no idea how what math you’d use to figure that out.
I totally controlled myself to not post that, but I am glad someone did! I said it to myself with the accent.
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