Area Magic Squares

https://carresmagiques.blogspot.fr/2017/01/area-magic-squares-and-tori-of-order-3.html

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:

https://carresmagiques.blogspot.fr/2017/01/area-magic-squares-and-tori-of-order-3.html

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

https://commons.wikimedia.org/wiki/File:Mengenlehreuhr.jpg

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

http://rmm.ludus-opuscula.org/PDF_Files/Pretz_BinaryClock_5_7(5_2016)_low.pdf

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:

http://rmm.ludus-opuscula.org/PDF_Files/Pretz_BinaryClock_5_7(5_2016)_low.pdf

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)

and

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.)

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).

The Humble-Nishiyama Randomness Game

https://pixabay.com

Mathematicians Steve Humble and Yutaka Nishiyama invented this game to highlight a surprising result in probability, based on a principle discovered by Walter Penney.

Two players play the game using an ordinary deck of cards. The cards will be dealt out in a row, one after another. Before the dealing begins, each player claims an ordered sequence of colors that might turn up, for example “red black red” (RBR) or “black red red” (BRR). As the cards are dealt, if three successive cards turn up in one of these sequences, the player who claimed it gets to collect those three cards as a trick, and the dealing continues. When all 52 cards have been dealt, the player who has collected the most tricks wins. (In a typical game, 7 to 9 tricks are won.)

This sounds like a perfectly even game, but in fact the second player has a strategy that will given him a significant advantage. When the first player has chosen his sequence (say, RRB), the second player changes the middle color, adds it to the start of the sequence, discards the last color, and claims the resulting sequence (in this case, BRR). This gives a decided advantage to the second player no matter which sequence his opponent has chosen. In a computer simulation of 1,000 games, Humble and Nishiyama got these results:

BBB vs RBB – RBB wins 995 times, 4 draws, BBB wins once
BBR vs RBB – RBB wins 930 times, 40 draws, BBR wins 30 times
BRB vs BBR – BBR wins 805 times, 79 draws, RBR wins 116 times
RBB vs RRB – RRB wins 890 times, 68 draws, RBB wins 42 times
BRR vs BBR – BBR wins 872 times, 65 draws, BRR wins 63 times
RBR vs RRB – RRB wins 792 times, 85 draws, RBR wins 123 times
RRB vs BRR – BRR wins 922 times, 51 draws, RRB wins 27 times
RRR vs BRR – BRR wins 988 times, 6 draws, RRR wins 6 times

Penney’s original game uses coin flips; cards are preferable because no record-keeping is required and because the finite number of cards in a deck increases the second player’s chances.

(Steve Humble and Yutaka Nishiyama, “Humble-Nishiyama Randomness Game — A New Variation on Penney’s Coin Game,” Mathematics Today, August 2010.)

An Evening Stroll

http://www.math.uwaterloo.ca/tsp/pubs/index.html

Maybe this was inevitable: A team of mathematicians have worked out the most efficient pub crawl in the United Kingdom, connecting 24,727 pubs in the shortest possible closed loop, 45,495,239 meters, or about 28,269 miles. Because it’s a loop, a determined crawler can start at any point and eventually find himself back home.

Despite the pickled application, this represents a serious achievement in computational mathematics, an advance in the so-called traveling salesman problem (TSP), which asks for the shortest route that passes through each of a set of points once and once only. The pub crawl includes more than 100 times the previous record number of stops in a road-distance TSP.

“We, of course, did not have in mind to bring everything mathematics has to bear in order to improve the lot of a wandering pub aficionado,” wrote lead researcher William Cook of the University of Waterloo. “The world has limited resources and the aim of the applied mathematics fields of mathematical optimisation and operations research is to create tools to help us to use these resources as efficiently as possible.”

(Thanks, Danesh.)

Floor, Please?

https://commons.wikimedia.org/wiki/File:Tri%C3%A1ngulo_de_Pascal_sin_r%C3%B3tulo.svg
Image: Wikimedia Commons

Reader Alex Freuman passed this along — a simple method of establishing any row in Pascal’s triangle, attributed to Edric Cane. To establish, for example, the seventh row (after the initial solitary 1), create a row of fractions in which the numerators are 7, 6, 5, 4, 3, 2, 1 and the denominators are 1, 2, 3, 4, 5, 6, 7:

\displaystyle \frac{7}{1} \times \frac{6}{2} \times \frac{5}{3} \times \frac{4}{4} \times \frac{3}{5} \times \frac{2}{6} \times \frac{1}{7}

Now multiply these in sequence, cumulatively, to get the numbers for the seventh row of the triangle:

\displaystyle 1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1

These are the coefficients for

\displaystyle \left ( a+b \right )^{7}=a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} + b^{7}.

Cane writes, “It couldn’t be easier to remember or to implement.” Another example — row 10:

\displaystyle \frac{10}{1} \times \frac{9}{2} \times \frac{8}{3} \times \frac{7}{4} \times \frac{6}{5} \times \frac{5}{6} \times \frac{4}{7} \times \frac{3}{8} \times \frac{2}{9} \times \frac{1}{10}

\displaystyle 1 \quad 10 \quad 45 \quad 120 \quad 210 \quad 252 \quad 210 \quad 120 \quad 45 \quad 10 \quad 1