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

The Jeep Problem

https://en.wikipedia.org/wiki/File:Jeep_problem_1.png
Image: Wikipedia

An adventurer wants to explore a desert. He has a jeep that can carry up to 1 unit of fuel at any time and that will travel 1 unit of distance on 1 unit of fuel. As he travels he can leave any amount of the fuel that he’s carrying at any point, as a fuel dump to be picked up later. He starts from a fixed base at the edge of the desert, where there’s an unlimited supply of fuel. How far into the desert can he go if he wants to return safely to the base at the end of each trip?

Surprisingly, with some intelligent planning he can go as far as he likes. The diagram above shows how far he can get with 3 trips:

  1. On the first trip he departs the base with 1 unit of fuel. He drives 1/6 unit into the desert, leaves 2/3 units of fuel at a fuel dump, and returns to the base using the remaining 1/6 unit of fuel.
  2. On the second trip he leaves the base with 1 unit of fuel, drives 1/6 unit to the fuel dump, and draws 1/6 unit from the dump. Now he’s carrying 1 unit of fuel. Then he drives 1/4 unit farther into the desert and leaves 1/2 unit at a new dump. Now he has 1/4 unit of fuel remaining, which is just enough to reach the first fuel dump, where he collects another 1/6 unit of fuel and returns to base.
  3. On the third trip he drives 1/6 unit to reach the first fuel dump, where he tops up with 1/6 unit of fuel (leaving 1/6 unit remaining there). Then he drives 1/4 to the second dump, where he collects 1/4 unit of fuel (again topping up to 1 full unit). (1/4 unit now remains in the second fuel dump.) Now he can drive 1/2 unit distance into the desert before he has to return to the second fuel dump, where he collects the remaining 1/4 unit fuel, which enables him to reach the first fuel dump, where he collects the last 1/6 unit of fuel, which is just enough to get back to base.

So if 3 trips are planned the explorer can travel a round-trip distance of 1 + 1/2 + 1/3 = 11/6 units. You can see the pattern: If the explorer had planned 4 trips, he would set up fuel dumps at distances of 1/8, 1/6, and 1/4 from the base, initially storing 3/4, 2/3, and 1/2 units at each and then drawing 1/8, 1/6, and 1/4 units of fuel from each on each visit. As before, on the final trip he could depart the last fuel dump with a full tank, drive 1/2 unit into the desert, and then return to the base, exhausting each fuel dump on the way. In that case he’d have traveled a round-trip distance of 1 + 1/2 + 1/3 + 1/4 = 25/12 units.

This is just the harmonic series, 1 + 1/2 + 1/3 + 1/4 + 1/5 + …, which is divergent — in principle, at least, the explorer can travel as far as he likes into the desert, provided he plans a large enough series of trips. In practice it would be very difficult, though — both the number of fuel dumps and the total amount of fuel necessary increase exponentially with the distance to be traveled.

Solitons

In 1834, engineer John Scott Russell was experimenting with boats in Scotland’s Union Canal when he made a strange discovery:

I was observing the motion of a boat which was rapidly drawn along a narrow channel by a pair of horses, when the boat suddenly stopped — not so the mass of water in the channel which it had put in motion; it accumulated round the prow of the vessel in a state of violent agitation, then suddenly leaving it behind, rolled forward with great velocity, assuming the form of a large solitary elevation, a rounded, smooth and well-defined heap of water, which continued its course along the channel apparently without change of form or diminution of speed. I followed it on horseback, and overtook it still rolling on at a rate of some eight or nine miles an hour, preserving its original figure some thirty feet long and a foot to a foot and a half in height. Its height gradually diminished, and after a chase of one or two miles I lost it in the windings of the channel. Such, in the month of August 1834, was my first chance interview with that singular and beautiful phenomenon which I have called the Wave of Translation.

They’re known today as solitons. He found that such waves can travel over very large distances, at a speed that depends on their size and width and the depth of the water. Remarkably, as shown above, they emerge from a collision unchanged, simply “passing through” one another.

(Thanks, Steve.)

Kapitza’s Pendulum

Here’s a surprise: A pendulum can be made stable in its inverted position if its support is oscillated rapidly up and down.

Even more improbably, in 1993 David Acheson and Tom Mullin showed that a double and even a triple pendulum can be made to stand up vertically like this if the pivot vibrates at the right frequency.

“The ‘trick’ really did work, and it worked, in fact, far better than we could ever have imagined,” Acheson wrote in 1089 and All That. “We were quite taken aback by just how stable the inverted state could be, and provided the pendulums were kept roughly aligned with one another, we could push them over by as much as 40 degrees or so and they would still gradually wobble back to the upward vertical.”

They published their results in Nature and later appeared on the BBC program Tomorrow’s World, but their demonstration didn’t impress everyone — afterward, an outraged caller upbraided the producers for “lowering their usually high standards” and “falling prey to two tricksters from Oxford.”

Here’s the double pendulum (thanks, Eccles):