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Nerd World Diversions–Puzzles and Problems

[Warning: This is a post from Nerd World.]

Over the years, I’ve come to make a distinction between two kinds of diversion: puzzles and problems.  To me, a puzzle is an intellectually stimulating activity whose only goal is the completion of the activity.  Naturally, many of these diversions are actually called puzzles–for example, crossword puzzles or jigsaw puzzles.  Others don’t use the word “puzzle”, but certainly are–the current favorite is Sudoku.

But, even within this category, there are some activities that are more engaging than others.  I read that Mary Leakey, one of the anthropologists of the Olduvai Gorge project, spent her time as a child solving jigsaw puzzles with all the pieces turned over so the picture doesn’t show.  She claimed that this was an early indication of her astounding ability to piece together fossil hominid skulls from hundreds of pieces.

I’m not nearly so visual.  But I do love words, so one of my favorites is cryptic crosswords.  It’s a type of crossword, but one where each clue has to be solved, rather than deducing a synonym to fill in the blanks.  It’s perhaps easiest to explain using an example.  Here’s a clue:

He’s about fifty and lives in the room upstairs, but is strong and healthy (8)

The number at the end tells the length of the word.  Each clue will have a definition in it someplace, and another way of deriving the word from some kind of wordplay.  It might be hidden, or an anagram, or a pun, or–in this case–a combination of pieces.  One trick is that numbers often turn into Roman numerals.  So, fifty is L.  Next: “He’s about fifty”, so we put the letters of HE around L, we have HLE so far.  “The room upstairs” is the ATTIC, so we put what we have so far inside it: AT(HLE)TIC and get ATHLETIC–strong and healthy.  If you want to try one, they’re published monthly in The Atlantic and Harper’s Magazine.

The distinction that I’m trying to make with the diversions that I call “problems” is that they’re much more open-ended.  And they can lead to interesting discoveries.  Let me take an example. 

Suppose you have a chain where the lengths of the links are the successive numbers starting with 1.  Suppose you have four links: 1″, 2″, 3″, 4″.  Connect the two ends.  Now, the question is: Can you place this loop of chain over two posts, so that the links don’t bend in the middle. Like this: 0==========0.

 In our example, we can.  2+3=4+1–the posts are 5″ apart.  What happens when we try a chain with five links?  Well, 1+2+3+4+5 = 15, so we can’t divide that evenly in half.   If we continue trying longer and longer chains, we find that 5 and 6 don’t work; 7 and 8 do; 9 and 10 don’t; 11 and 12 do.  At this point we see a pattern.  Here’s where it turns into a problem [rather than a puzzle].  What’s the underlying explanation of why this works? The place to start is the algebraic formula for the total length of the chain.  The sum 1+2+3+….+n [up to some number, n] is equal to n (n+1)/2.  So, if n is a multiple of 4, then when we divide by 2, there is still another factor of 2 for us to use to divide the chain in half.  [And this works when (n+1) is a multiple of 4 also.]  It’s a bit harder to trace through what happens when n is even, but not a multiple of 4 [like 6, for example].  This is one of the situations where the length of the chain is an odd number, so we can’t divide it in half.

Actually, the original version of the problem had three posts, arranged in an equilateral triangle and asked what was the shortest chain that could be looped around the three posts without bending any links.  To just find one example is a “puzzle” in my lexicon, but to find some formula or explanation that gives all, or most, of them–that’s a “problem”.  And this one leads in some interesting directions, which I’ll postpone for the future.

[If you’re trying to figure it out, one thing that has to be true is that the length of the chain has to be evenly divisible by 3–that’s not enough, but it rules out some cases.]

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