Philosopher Mark Bedau points out that, even on a dead planet, the microscopic crystallites that make up clay and mud seem to have the flexibility to adapt and evolve by natural selection:

• Crystals reproduce in the sense that when they become large enough they cleave and pieces break off, becoming seeds for new crystals. A population of reproducing crystals can become the setting for crystal “evolution.”
• All crystals have flaws, which are reproduced randomly and can become a source of novel information that gets expressed in “phenotypic” traits such as shape, growth rate, and the conditions that cause cleaving.
• Each new layer of crystals copies the geometrical arrangement of atoms in the layer below, and defects can be copied in the same way. Grain boundaries and dislocations in one crystal tend to be copied to its “offspring” and subsequent generations. Over time these “mutations” can produce “species” among which nature selects.
• A crystal’s shape, growth rate, and cleaving conditions all affect the rate at which it proliferates, and crystals with different properties will disperse and diffuse differently, which affects the rate at which they reproduce.

So a population of crystals can exhibit reproduction, variation, heredity, and adaptivity. “What is rather surprising is that, in the process, the planet remains entirely devoid of life. Thus, natural selection can take place in an entirely inorganic setting.”

(Mark Bedau, “Can Biological Teleology Be Naturalized?” Journal of Philosophy 88:11 [November 1991], 647-655. I think A.G. Cairns-Smith originated this idea in Seven Clues to the Origin of Life in 1985, and Richard Dawkins took it up in The Blind Watchmaker the following year.)

One April Fool’s Day, when logician Raymond Smullyan was 10 years old, his brother told him, “Today I am going to trick you like you have never been tricked before.”

“Little Raymond waited, and waited, and waited, and nothing happened,” writes Ron Aharoni in Circularity. “To this very day, he is not sure whether his brother tricked him or not.”

Imprisoned during the French Revolution, zoologist Pierre André Latreille found a beetle on the floor of his dungeon. He pointed out to the prison doctor that it was the rare Necrobia ruficollis, described by Johan Christian Fabricius in 1775. Impressed, the doctor sent it to a local naturalist, who knew Latreille’s work and managed to secure the release of Latreille and a cellmate.

All the other inmates were executed within a month.

In 1884, as engineers sawed brownstone out of a quarry near Manchester, Conn., amateur paleontologist Charles Owen realized that one block contained the hind part of a skeleton. He alerted professor Othniel Charles Marsh, who managed to acquire the block, but the corresponding stone containing the skeleton’s fore part had already been carted off for use in a new highway bridge.

In the ensuing years the identity of that bridge was forgotten, but in 1967, as a new highway was being constructed, Yale paleontologist John Ostrom saw an opportunity. He surveyed 60 bridges in the Manchester area and identified one 40-foot span over Hop Creek as the likeliest candidate. Then, in 1969, as that bridge was replaced, he and his colleagues examined more than 300 likely blocks at the site.

They found two 500-pound blocks that showed distinct fossil markings, and in New Haven Ostrom determined that one of the visible bones matched a thigh bone that Marsh had recovered 85 years earlier. The complete skeleton had belonged to a member of Ammosaurus, a genus that roamed the northeastern United States 200 million years ago.

(Thanks, Glenn.)

How can you arrange seven points in a plane so that if any three are chosen, at least two of them will be a unit distance apart?

The solution is shown here — it’s known as a Moser spindle.

In 1998, the BBC reported that four card players at a whist club in Bucklesham, Suffolk, had each been dealt one suit from a shuffled deck. Hilda Golding found herself holding 13 clubs, Hazel Ruffles held 13 diamonds, and Alison Chivers held 13 hearts (and won, as this was trumps). The dummy hand, face down on the table, held 13 spades.

Though there were 55 people in the village hall at the time, some of whom claimed to have witnessed the event, it’s vastly more likely that this was some misunderstanding or a false report. In 1939 Horace Norton of University College London calculated the odds of such a deal arising naturally to be 1 in 2,235,197,406,895,366,368,301,599,999.

In The Mathematics of Games (2013), John D. Beasley writes, “A typical evening’s bridge comprises perhaps twenty deals, so a once-a-week player must play for over one hundred million years to have an even chance of receiving a thirteen-card suit. If ten million players are active once a week, a hand containing a thirteen-card suit may be expected about once every fifteen years, but it is still extremely unlikely that a genuine deal will produce four such hands.”

(Martin Gardner points out somewhere that anecdotal reports of four perfect hands are strangely more frequent than reports of two perfect hands, which is more likely — though, I guess, less newsworthy.)

(Via Martin Cohen, 101 Philosophy Problems, 2002.)

Male side-blotched lizards compete for mates using a three-sided strategy that resembles a game of rock-paper-scissors. Orange-throated males, the strongest, don’t form strong pair bonds but establish large territories and fight blue-throated males outright for females. The blue-throated males, middle-sized, are less aggressive and tend to pair strongly with individual females. Yellow-throated males, the smallest, have a coloration that resembles that of females; this allows them to approach females in the territories of orange-throated males — though this won’t work with females that have formed strong pair bonds with blue-throated males.

So, broadly speaking, orange beats blue, blue beats yellow, and yellow beats orange, an equilibrium of sorts in which each variety has an advantage over another but not over the third.

A pleasing little theorem by Dutch mathematician Frans van Schooten:

Inscribe equilateral triangle ABC in a circle. Now, from a point P on that circle, the length of the longest of segments PA, PB, PC equals the sum of the lengths of the other two segments (in this example, the length of segment PA equals the sum of the lengths of PB and PC).