Can a chess knight visit every square on a board with 4 rows by a series of successive moves?

No, it can’t, and Hungarian mathematics prodigy Louis Pósa proved this while still in his early teens. Suppose that such a tour is possible. Then, on any board bearing the standard checkerboard pattern, the knight will land alternately on white and black squares. But now imagine that the board’s top and bottom rows have been colored red and the middle two rows are blue. Now a knight on any red square must jump to a blue square, and because the board has an equal number of red and blue squares, any knight on a blue square must jump to a red one (if it visits two blue squares in a row, it won’t be able to make up for this later by visiting two red ones in a row). So the knight’s tour on any 4 × n board must alternate strictly between red and blue squares.

“But this is impossible,” notes Colorado College mathematician John J. Watkins. “The same knight’s tour alternated between white and black squares in the one coloring, and between red and blue squares in the other coloring, which would mean the two color patterns must be the same, which of course they aren’t. Isn’t that a clever proof, especially for a teenager to discover?”

(John J. Watkins, Across the Board, 2004. Pósa’s proof is given more rigorously here, and it’s also presented in Ross Honsberger’s 1973 book Mathematical Gems.)

UPDATE: It’s important to note that it’s a “closed” knight’s tour that’s impossible — that’s one that ends where it began. An open tour, which can end anywhere, is possible — it breaks Pósa’s proof because it need not alternate strictly between red and blue squares. Thanks to reader Marjan Milanović for pointing this out.

A late, beautiful contribution by Lee Sallows:

See “Extra Magic” Realized.

(Thanks, Lee!)

This is interesting — in order for a simple arch to stand, its shape when inverted must fit into the limits described by a hanging chain (a catenary).

Spanish Catalan architect Antoni Gaudí followed this principle in designing some of his buildings — he created inverted models and let the shapes of hanging chains or weighted strings determine the shapes of the arches.

In the early 1940s a curious question began to circulate among the members of the Princeton physics department. An ordinary lawn sprinkler like the one shown here would turn clockwise (in the direction of the long arrow) as its jets ejected water (short arrows). If you reversed this — that is, if you submerged the sprinkler in a tank of water and induced the jets to suck in the fluid — would the sprinkler turn in the opposite direction?

The problem is associated with Richard Feynman, who was a grad student at the time (and who destroyed a glass container in the university’s cyclotron laboratory trying to find the answer).

In fact Ernst Mach had first asked the question in an 1883 textbook. The answer, briefly, is no: The submerged sprinkler doesn’t turn counterclockwise because counterbalancing forces at the back of the nozzle result in no net torque. Experiments tend to bear this out, although in some cases the sprinkler turns slightly counterclockwise, perhaps due to the formation of a vortex within the sprinkler body.

I don’t know whether this is contrived or whether a student offered it on an actual exam — Ed Barbeau presented “this little beauty of a howler” in the January 2002 College Mathematics Journal, citing Ross Honsberger of the University of Waterloo in Ontario.

In his Harmonices Mundi of 1619, Johannes Kepler wrote, “The heavenly motions are nothing but a continuous song for several voices, to be perceived by the intellect, not by the ear; a music which, through discordant tensions, through syncopations and cadenzas as it were, progresses toward certain pre-designed six-voiced cadences, and thereby sets landmarks in the immeasurable flow of time.” In 1979 Yale geologist John Rodgers and musician Willie Ruff scaled up the frequencies of the planetary orbits into the range of human hearing so that Kepler’s “harmony of the world” could become audible:

Mercury, as the innermost planet, is the fastest and the highest pitched. It has a very eccentric orbit (as planets go), which it traverses in 88 days; its song is therefore a fast whistle, going from the E above the piano (e′′′′′) down more than an octave to about C# (c#′′′′) and back, in a little over a second. Venus and Earth, in contrast, have nearly circular orbits. Venus’s range is only about a quarter tone, near the E next above the treble staff (e′′′); Earth’s is about a half tone, from G (g′′) to G# at the top of that staff. … Next out from Earth is Mars, again with an eccentric orbit … it ranges from the C above middle C (c′′) down to about F# (f#′) and back, in nearly 10 seconds. The distance from Mars to Jupiter is much greater than that between the inner planets … and Jupiter’s song is much deeper, in the baritone or bass, and much slower. It covers a minor third, from D to B (D to B₁) just below the bass staff. Still farther out and still lower is Saturn, only a little more than a deep growl, in which a good ear can sometimes hear the individual vibrations. Its range is a major third, from B to G (B₂ to G₂), the B at the top being just an octave below the B at the bottom of Jupiter’s range. Thus the two planets together define a major triad, and it may well have been this concord … that made Kepler certain he had cracked the code and discovered the secret of the celestial harmony.

(The outer planets, discovered after Kepler’s time, are represented here with rhythmic beats.) “The Earth sings Mi, Fa, Mi,” Kepler wrote. “You may infer even from the syllables that in this our home misery and famine hold sway.”

(John Rodgers and Willie Ruff, “Kepler’s Harmony of the World: A Realization for the Ear,” American Scientist 67:3 [May-June 1979], 286-292.)

A curious puzzle by Dartmouth mathematician Peter Winkler: You’ve just joined the Coin Flippers of America, and fittingly the amount of your dues will be decided by chance. You’ll name a head-tail sequence of length 5, and then a coin will be flipped until that sequence appears in five consecutive flips. Your dues will be the total number of flips in U.S. dollars; for instance, if you choose HHHHH and it takes 36 flips to produce a run of five heads, then your annual dues will be $36. What sequence should you pick?

At first it seems that it shouldn’t matter — any fixed sequence should have the probability (1/2)⁵, or 1/32. But “Not so fast,” Winkler writes. “Overlapping causes problems.” It is true that in an infinite sequence of random flips, the average distance between one occurrence and the next of any fixed sequence is 1/32. But if you choose HHHHH (for example), one occurrence of this outcome gives a huge head start to the next — if the next flip is a tail, then you’re starting over cleanly, but if it’s a head then you’ve already produced the next occurrence.

“If X is the average time needed to get HHHHH starting fresh, the average of 1 + X and 1 is 32,” Winkler writes. “Solving for X yields a startlingly high 62 flips.” To get your expected dues down to $32, you need to pick a sequence where this “head start” effect doesn’t obtain. There are 10 such sequences; one is HHHTT.

(Peter Winkler, “Coin Flipping,” Communications of the ACM 56:11 [November 2013], 120.)

An observation by Oxford University mathematician Nick Trefethen:

A student leaves university in America with a transcript full of information. Even with grade inflation, there are thirty marks of A or A- or B+ or B to look at, each one attached to a different course like Advanced Calculus or 20th Century Philosophy or Introduction to Economics. Grade-point averages are constructed from these transcripts and reported to three digits of accuracy.

An Oxford graduate finishes with no transcript, just a degree result which may be a First, a II.1, a II.2, a Third, a Pass, or a Fail. Failures are more or less nonexistent, and the numbers last year [2000] for the other degrees were 691, 1925, 374, 39, and 3, respectively. The corresponding probabilities are 23%, 63%, 12%, 1%, and 0.1%.

If you add up these probabilities times their base 2 logarithms, all times minus one, you find out how much information there is in an Oxford degree. The result is: 1.37 bits of information.

(From Trefethen’s Index Cards, 2011.)