Image: Wikimedia Commons

In 1965, as they were writing the first draft of 2001: A Space Odyssey, Stanley Kubrick showed Arthur C. Clarke a set of 12 plastic tiles. Each tile consisted of five squares joined along their edges. These are known as pentominoes, and a set of 12 includes every possible such configuration, if rotations and reflections aren’t considered distinct. The challenge, Kubrick explained, is to fit the 12 tiles together into a tidy rectangle. Because 12 five-square tiles cover 60 squares altogether, there are four possible rectangular solutions: 6 × 10, 5 × 12, 4 × 15, and 3 × 20. (A 2 × 30 rectangle would be too narrow to accommodate all the shapes.)

Clarke, who rarely played intellectual games, found that this challenge “can rather rapidly escalate — if you have that sort of mind — into a way of life.” He stole a set of tiles from his niece, spent hundreds of hours playing with it, and even worked the shapes into the design of a rug for his office. “That a jigsaw puzzle consisting of only 12 pieces cannot be quickly solved seems incredible, and no one will believe it until he has tried,” he wrote in the Sunday Telegraph Magazine. It took him a full month to arrange the 12 shapes into a 6 × 10 rectangle — a task that he was later abashed to learn can be done in 2339 different ways. There are 1010 solutions to the 5 × 12 rectangle and 368 solutions to the 4 × 15.
Image: Wikimedia Commons

But “The most interesting case, however, is that of the long, thin rectangle only 3 units wide and 20 long.” Clarke became fascinated with this challenge when Martin Gardner revealed that only two solutions exist. He offered 10 rupees to anyone who could find the solutions, and was delighted when a friend produced them, as he’d calculated that solving the problem by blind permutation would take more than 20 billion years.

Clarke even worked the 3 × 20 problem into his 1975 novel Imperial Earth. Challenged by his grandmother, the character Duncan struggles with the task and declares it impossible. “I’m glad you made the effort,” she says. “Generalizing — exploring every possibility — is what mathematics is all about. But you’re wrong. It can be done. There are just two solutions; and if you find one, you’ll also have the other.”

Can you find them?

Click for Answer

Art and Science

Alexander Fleming, the discoverer of pencillin, grew “germ paintings” of living bacteria on blotting paper. He made this 4-inch portrait, titled “Guardsman,” in 1933.

“If a paper disc is placed on the surface of an agar plate, the nutrient material diffuses through the paper sufficiently to maintain the growth of many microorganisms implanted on the surface of the paper,” he wrote. “At any stage, growth can be stopped by the introduction of formalin. Finally the paper disc, with the culture on its surface, can be removed, dried, and suitably mounted.”

Here’s a gallery. “Even in Fleming’s time this technique failed to receive much attention or approval. Apparently he prepared a small exhibit of bacterial art for a royal visit to St Mary’s by Queen Mary. The Queen was ‘not amused and hurried past it’ even though it included a patriotic rendition of the Union Jack in bacteria.”

Area Magic Squares

On December 30 William Walkington sent this greeting to a circle of magic-square enthusiasts — it’s a traditional magic square (each row, column, and diagonal sums to 15), but the geometric area of each cell corresponds to its number.

He added, “The areas are approximate, and I don’t know if it is possible to obtain the correct areas with 2 vertically slanted straight lines through the square. Perhaps someone will be able to work this out in 2017?”

It’s only January 19, and the answer is already yes — Walter Trump has produced a “third-order linear area magic square” using the numbers 5-13:

There are many further developments, which have opened new questions and challenges, as these discoveries tend to do — see William’s blog post for more information.

(Thanks, William.)

Watching the Detectives

Police exist, and sometimes they scrutinize other members of the constabulary. We might say Police police police. If the observed officers are already being observed by a third set of officers, then we could say Police police police police police, that is, “Police observe police [whom] police police.”

The trouble is that if you say this sentence, “Police police police police police,” to an innocent friend, she might take you to mean “Police [whom] police police … police police.” Police police police police police has one verb, police, and two noun phrases, Police and police police police, and without some guidance there’s no way to tell which noun phrase is intended to begin and which to end the sentence.

It gets worse. Suppose we add two more polices: Police police police police police police police. Now do we mean “Police [whom] police observe observe police [whom] police observe”? Or “Police observe police [whom] police whom police observe observe”? Or something else again?

In general, McGill University mathematician Joachim Lambek finds that if police is repeated 2n + 1 times (n ≥ 1), then the numbers of ways in which the sentence can be parsed is  \frac{1}{\left ( n + 1 \right )}\binom{2n}{n} , the (n + 1)st Catalan number.

Buffalo have their own troubles.

(J. Lambek, “Counting Ambiguous Meanings,” Mathematical Intelligencer 30:2 [March 2008], 4.)

The Mengenlehreuhr

Further to Saturday’s triangular clock post, reader Folkard Wohlgemuth points out that a “set theory clock” has been operating publicly in Berlin for more than 40 years. Since 1995 it has stood in Budapester Straße in front of Europa-Center.

The circular light at the top blinks on or off once per second. Each cell in the top row represents five hours; each in the second row represents one hour; each in the third row represents five minutes (for ease of reading, the cells denoting 15, 30, and 45 minutes past the hour are red); and each cell in the bottom row represents one minute. So the photo above was taken at (5 × 2) + (0 × 1) hours and (6 × 5) + (1 × 1) minutes past midnight, or 10:31 a.m.

Online simulators display the current time in the clock’s format in Flash and Javascript.

If that’s not interesting enough, apparently the clock is a key to the solution of Kryptos, the enigmatic sculpture that stands on the grounds of the CIA in Langley, Va. In 2010 and 2014 sculptor Jim Sanborn revealed to the New York Times that two adjacent words in the unsolved fourth section of the cipher there read BERLIN CLOCK.

When asked whether this was a reference to the Mengenlehreuhr, he said, “You’d better delve into that particular clock.”

Triangular Time

Aachen University physicist Jörg Pretz has devised a binary clock in the shape of a triangular array of 15 lamps. Here’s how to read it:

  • When lit, the top lamp denotes 6 hours.
  • Each lamp on on the second row denotes 2 hours.
  • Each lamp on on the third row denotes 30 minutes.
  • Each lamp on on the fourth row denotes 6 minutes.
  • Each lamp on on the fifth row denotes 1 minute.

So the clock above shows 6 hours + (2 × 2 hours) + (2 × 30 minutes) + (3 × 1 minute) = 11:03. The lamps’ color, red, shows that it’s after noon, or 11:03 p.m. The same array displayed in green would mean 11:03 a.m. A few more examples:

The time value assigned to each lamp is the total time value of the row below if that row contained one additional lamp.

On each row the lamps light up from left to right, so a row with n lamps can display n + 1 states (all lamps off to all lamps on). So for a triangular array with n lamps on the bottom row, the total number of states is

(n + 1) × ((n – 1) + 1) × ((n – 2) + 1) × · · · × (1 + 1) = (n + 1)!

That is, it’s a factorial of a natural number. And by a happy coincidence, the total number of minutes in 12 hours is such a factorial (720 = 6!).

“Thus the whole concept works because our system of time divisions is based on a sexagesimal system, dating back to the Babylonians, rather than a decimal system, as proposed during the French Revolution.”

There’s more info in Pretz’s article, and you can play with the clock using this applet.

(Jörg Pretz, “The Triangular Binary Clock,” Recreational Mathematics Magazine, March 2016.)

The Holdout

Reader Joe Antognini sent this in: Brazilian mathematician Inder Taneja has found a way to render every number from 1 to 11,111 by starting with either of these strings:

1 2 3 4 5 6 7 8 9

9 8 7 6 5 4 3 2 1

and applying any of the operations addition, subtraction, multiplication, division, and exponentiation. Brackets are permitted. For example:

6439 = 1 + 2 × (34 × 5 × 6 + 789)


6439 = 9 × (8 + 7 + 6) + 54 × (32 + 1)

Intriguingly, there’s one hole: There doesn’t seem to be a way to render 10958 from the increasing sequence.

Taneja’s paper is here. (Thanks, Joe.)

01/29/2017 UPDATE: Taneja tells me that, while it can’t be calculated using only basic operations, 10958 can be reached using factorials or square roots. Here are two factorials:

10958 = 1 + 2 + 3!! + (-4 + 5! + 6 – 7) × 89

10958 = 1 × 2 × (3!! – 4! × (5 + 6) + 7! – 8 – 9)

The Real World

I had a growing feeling in the later years of my work at the subject that a good mathematical theorem dealing with economic hypotheses was very unlikely to be good economics: and I went more and more on the rules — (1) Use mathematics as a shorthand language, rather than an engine of inquiry. (2) Keep to them until you have done. (3) Translate into English. (4) Then illustrate by examples that are important in real life. (5) Burn the mathematics. (6) If you can’t succeed in 4, burn 3. This last I did often.

— Alfred Marshall, in a letter to A.L. Bowley, Jan. 27, 1906

Traffic Planning

traffic planning - cars

Towns A and B are connected by two roads. Suppose that two cars connected by a rope of length 2r can travel from A to B without breaking the rope. How can we prove that two circular wagons of radius r, moving along these roads in opposite directions, will necessarily collide?

traffic planning - wagons

This can be solved neatly by creating a configuration space. Map each road onto a unit segment, and set these up as two sides of a square. The northern car’s progress is reflected by a point moving up the left side of the square, and the southern car’s by a point moving from left to right along the bottom. Now the motion of the two cars from A to B is represented by a continuous curve connecting (0,0) and (1,1).

The wagons start from opposite towns, so their motion is represented by a curve from (0,1) to (1,0), and it’s immediately clear that the two curves must intersect. The intersection point corresponds to the collision of the wagons.

This example, by N. Konstantinov, is reportedly common in Russian mathematical folklore; I found it in Serge Tabachnikov’s 2005 book Geometry and Billiards (of all places).