Mirror Therapy

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When a limb is paralyzed and then amputated, the patient may perceive a “phantom limb” in its place that is itself paralyzed — the brain has “learned” that the limb is paralyzed and has not received any feedback to the contrary.

University of California neuroscientist V.S. Ramachandran found a simple solution: The patient holds the intact limb next to a mirror, looks at the reflected image, and makes symmetric movements with both the good and the phantom limb. In the reflected image, the brain is now able to “see” the phantom limb moving. The impression of paralysis lifts, and the patient can now move the phantom limb out of painful positions.

A 2018 review called the technique “a valid, simple, and inexpensive treatment for [phantom-limb pain].”

A Perfect Bore

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If we assume the existence of an omniscient and omnipotent being, one that knows and can do absolutely everything, then to my own very limited self, it would seem that existence for it would be unbearable. Nothing to wonder about? Nothing to ponder over? Nothing to discover? Eternity in such a heaven would surely be indistinguishable from hell.

— Isaac Asimov, “X” Stands for Unknown, 1984

An Even Dozen

https://commons.wikimedia.org/wiki/File:Football_(soccer_ball).svg

The surface of a standard soccer ball is covered with 20 hexagons and 12 pentagons. Interestingly, while we might vary the number of hexagons, the number of pentagons must always be 12.

That’s because the Euler characteristic of a sphere is 2, so VE + F = 2, where V is the number of vertices, or corners, E is the number of edges, and F is the number of faces. If P is the number of pentagons and H is the number of hexagons, then the total number of faces is F = P + H; the total number of vertices is V = (5P + 6H) / 3 (we divide by 3 because three faces meet at each vertex); and the total number of edges is E = (5P + 6H) / 2 (dividing by 2 because two faces meet at each edge). Putting those together gives

\displaystyle V-E+F={\frac {5P+6H}{3}}-{\frac {5P+6H}{2}}+P+H={\frac {P}{6}},

and since the Euler characteristic is 2, this means P must always be 12.

A Late Arrival

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Image: Wikimedia Commons

There’s only one normal magic hexagon — only one hexagonal arrangement of cells like the one shown here that can be filled with the consecutive integers 1 to n such that every row, in all three directions, sums to the same total. (This excludes the trivial example of a single cell on its own, as well as rotations and reflections of the hexagon shown here.)

Amazingly, it appears that this lonely example may not have been discovered until 1888! In a 1988 letter to the Mathematical Gazette, Martin Gardner reported that the magic hexagon was given as a problem in the German magazine Zeitschrift für mathematischen und naturwissenschaftlichen Unterricht (Volume 19, page 429) in 1888. The proposer was identified only as von Haselberg, of Stralsund; his solution was published in the next volume.

Gardner writes, “The structure could easily have been discovered by mathematicians in ancient times, but as of now, this is the earliest known publication.”

(Martin Gardner, “The History of the Magic Hexagon,” Mathematical Gazette 72:460 [June 1988], 133.)

A Discreet Correspondence

In Ulysses, Leopold Bloom’s locked private drawer at 7 Eccles Street contains, among other things:

3 typewritten letters, addressee, Henry Flower, c/o P.O. Westland Row, addresser, Martha Clifford, c/o P.O. Dolphin’s Barn: the transliterated name and address of the addresser of the 3 letters in reversed alphabetic boustrophedontic punctated quadrilinear cryptogram (vowels suppressed) N. IGS./WI. UU. OX/W. OKS. MH/Y. IM …

This actually works: Quadrilinear means that the cipher is to be set in four lines; reversed alphabetic means that the key is a = z, b = y, etc.; and boustrophedontic is a term from paleography meaning that the writing runs right and left in alternate lines. So the cryptogram and its solution look like this:

N . I G S .
m a r t h a

W I . U U . O X
d r o f f i l c

W . O K S . M H
d o l p h i n s

Y . I M
b a r n

Apparently Joyce or Bloom forgot that the last line should run right to left.

(From David Kahn, The Codebreakers, 1967.)

Russell’s Decalogue

In a 1951 article in the New York Times Magazine, Bertrand Russell laid out “the Ten Commandments that, as a teacher, I should wish to promulgate”:

  1. Do not feel absolutely certain of anything.
  2. Do not think it worthwhile to produce belief by concealing evidence, for the evidence is sure to come to light.
  3. Never try to discourage thinking, for you are sure to succeed.
  4. When you meet with opposition, even if it should be from your husband or your children, endeavor to overcome it by argument and not by authority, for a victory dependent upon authority is unreal and illusory.
  5. Have no respect for the authority of others, for there are always contrary authorities to be found.
  6. Do not use power to suppress opinions you think pernicious, for if you do the opinions will suppress you.
  7. Do not fear to be eccentric in opinion, for every opinion now accepted was once eccentric.
  8. Find more pleasure in intelligent dissent than in passive agreement, for, if you value intelligence as you should, the former implies a deeper agreement than the latter.
  9. Be scrupulously truthful, even when truth is inconvenient, for it is more inconvenient when you try to conceal it.
  10. Do not feel envious of the happiness of those who live in a fool’s paradise, for only a fool will think that it is happiness.

“The essence of the liberal outlook in the intellectual sphere is a belief that unbiased discussion is a useful thing and that men should be free to question anything if they can support their questioning by solid arguments,” he wrote. “The opposite view, which is maintained by those who cannot be called liberals, is that the truth is already known, and that to question it is necessarily subversive.”

A Bimagic Queen’s Tour

walkington semi-bimagic queen's tour
Image: William Walkington (CC BY-NC-SA 4.0)

A queen’s tour is the record of a chess queen’s journey around an empty board in which she visits each of the squares once. If the squares are numbered by the order in which she visits them, then the resulting square is magic if the numbers in each rank and file sum to the same total. It’s bimagic if the squares of these numbers also produce a consistent total.

William Walkington has just found the first bimagic queen’s tour, which also appears to be the first bimagic tour of any chess piece. (William Roxby Beverley published the first magic knight’s tour in The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science in 1848.)

Note that here the long diagonals don’t produce the magic sum, as they would in a magic square. This constraint is normally dropped in a magic tour — in fact, of the 140 magic knight tours possible on an 8×8 board, none have two long magic diagonals, and no bimagic queen’s tour with qualifying diagonals is possible on such a board either.

More details, and an interesting description of the search, are on William’s blog. He has been told that a complete list of such bimagic queen’s tours is within reach of a computer search, though the field is daunting — there are more than 1.7 billion essentially different semi-bimagic squares possible on an 8×8 board, and each allows more than 400 million permutations.