In this weekâ€™s puzzle, two players face off in an unusual tabletop game. Itâ€™s one you can play yourselfâ€”and, if you know the strategy, win every time.

### Sunday Puzzle #41: Dueling Cigars

Two people are seated at a square-topped table. One places an ordinary cigar (flat at one end, pointed at the other) on the table, then the other does the same, and so on alternately, a condition being that no cigar shall touch another. Which player should succeed in placing the last cigar, assuming that they each will play in the best possible manner? The size of the table top and the size of the cigar are not given, but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar, we will say that the table must not be less than 2 feet square and the cigar not more than 4Â˝ inches long. With those restrictions you may take any dimensions you like. Of course we assume that all the cigars are exactly alike in every respect. Should the first player, or the second player, win?

Weâ€™ll be back next week with the solutionâ€”and a new puzzle! Got a great brainteaser, original or otherwise, that youâ€™d like to see featured? E-mail me with your recommendations. (Be sure to include â€śSunday Puzzleâ€ť in the subject line.)

### SOLUTION To Sunday Puzzle #40: In Search of an Unusual Book

Last week I asked you to identify an unusual book in an unusual library. Said library contained no two books containing the same number of words, and a total number of books greater than the number of words in the largest book. The challenge was to deduce how many words one of the books contained and what the book is about.

Commenter DanRFarmer was the first person to provide a correct response. â€śOne book is empty and itâ€™s about nothing,â€ť he wrote. However, commenter Hawkingdo was the first to explain the reasoning behind the solution. They did so by simplifying the problem considerably, by imagining a library with only two books:

If there are two books the greatest number of words possible is 1 and the other must have zero.

Since the number of books is greater than the number of words in the largest book without repetition in number of words the number of books will always be one more than the number of words in the largest book and there will always by necessity be books with every number of words possible less than the largest book and there will always be a book with no words to increase the total number of books. This assumes (correctly) you canâ€™t have books with negative number of words.

2 books: 0 words and 1 word (singular/plural is (are?) weird .. how can you have multiple zeroes ... anyway).

3 books: 2 words, 1 word, 0 words.

n books: n-1 words, n-2 words ... 0 words.

The problem hinges on the pigeonhole principle, a straightforward concept with many applications. Esther described the principle in a post on io9 last year:

Imagine fifteen pigeonholes and sixteen pigeons. A storm comes along, and all of the pigeons take shelter inside the pigeonholes. They could be arranged any number of ways. For all we know, all sixteen pigeons could be inside one hole, and the rest of the holes could be empty. What we know for sure, no matter what, is that there is at least one hole that contains more than one pigeon.