Science & Math

Dueling Pennies

A certain strange casino offers only one game. The casino posts a positive integer n on the wall, and the customer flips a fair coin repeatedly until it falls tails. If he has tossed n – 1 times, he pays the house 8n – 1 dollars; if he’s tossed n + 1 times, the house pays him 8n dollars; and in all other cases the payoff is zero.

The probability of tossing the coin exactly n times is 1/2n, so the customer’s expected winnings are 8n/2n + 1 – 8n – 1/2n – 1 = 4n – 1 for n > 1, and 2 for n = 1. So his expected gain is positive.

But suppose it turns out that the casino arrived at the number n by tossing the same fair coin and counting the tosses, up to and including the first tails. This presents a puzzle: “You and the house are behaving in a completely symmetric manner,” writes David Gale in Tracking the Automatic ANT (1998). “Each of you tosses the coin, and if the number of tosses happens to be the consecutive integers n and n + 1, then the n-tosser pays the (n + 1)-tosser 8n dollars. But we have just seen that the game is to your advantage as measured by expectation no matter what number the house announces. How can there be this asymmetry in a completely symmetric game?”

Visual Calculus

https://commons.wikimedia.org/wiki/File:Cycloid_(PSF).png

As a circle rolls along a line, a point on its circumference traces an arch called a cycloid. The arch encloses an area three times that of the circle, a result commonly proven using calculus. Now Armenian mathematician Mamikon Mnatsakanian has devised a “sweeping-tangent theorem” that accomplishes the same proof using intuition:

Imagine a tangent to the rolling circle. As the circle rolls, the tangent sweeps out a series of vectors (approximated here using colors). If these vectors are then gathered to a common point while preserving their length and orientation, they form a sort of bouquet whose size and shape turn out to match exactly those of the original circle. Because the enclosing rectangle has four times the area of the rolling circle (2πr × 2r = 4πr2), this shows that the area under the arch has three times the circle’s area.

All this is proven rigorously in Mnatsakanian’s 2012 book New Horizons in Geometry, written with his Caltech colleague Tom Apostol. The two have now collaborated on some 30 papers showing that many surprising and useful results that heretofore had required integration can now be obtained using intuitive methods that can appeal even to a young student.

That’s a welcome outcome for Mnatsakanian, who found himself stranded in the United States when the Armenian government collapsed in 1990. Apostol writes, “When young Mamikon showed his method to Soviet mathematicians they dismissed it out of hand and said ‘It can’t be right. You can’t solve calculus problems that easily.'”

Hesiod’s Anvil

https://commons.wikimedia.org/wiki/File:Acme_anvil.gif

How far off is heaven? In the Theogony Hesiod gives us a clue:

For a brazen anvil falling down from heaven nine nights and days would reach the earth upon the tenth; and again, a brazen anvil falling from earth nine nights and days would reach Tartarus upon the tenth.

How far can an anvil fall in nine days? Galileo, who taught that “the distances measured by the falling body increase according to the squares of the time,” would have determined that the anvil starts 2.96 × 109 km from earth, a distance greater than that between the sun and Uranus.

But Galileo’s calculation assumes that gravitational force is independent of the object’s distance from the earth. If we assume instead that it varies inversely with the square of the distance between mass centers (and if we ignore all masses except those of the earth and the anvil, and assume that the anvil falls in a straight line), King College mathematician Andrew Simoson calculates that Galileo’s anvil wouldn’t reach us for

hesiod's anvil calculation

Instead, under this new assumption, to reach us in nine days an anvil would start 5.81 × 105 km away — about one and a half times the distance between the earth and the moon.

(Andrew J. Simoson, Hesiod’s Anvil, 2007.)

Rolling Average

http://www.freeimages.com/photo/549165

In a standard 10-frame game of bowling, the lowest possible score is 0 (all gutterballs) and the highest is 300 (all strikes). An average player falls somewhere between these extremes. In 1985, Central Missouri State University mathematicians Curtis Cooper and Robert Kennedy wondered what the game’s theoretical average score is — if you compiled the score sheets for every legally possible game of bowling, what would be the arithmetic mean of the scores?

It turns out it’s pretty low. There are (669)(241) possible games, which is about 5.7 × 1018. If we divide that into the total number of points scored in these games, we get

mean bowling score

which is about 80 (79.7439 …).

This “might make you feel better about your average,” Cooper and Kennedy conclude. “The mean bowling score is indeed awful even if you are just an occasional bowler. Even though this information is interesting, there are more difficult questions about the game of bowling that could be asked. For example, you might wish to determine the standard deviation of the set of bowling scores and hence know more about the distribution of the set of all bowling scores. But the exact determination of the distibution of the set of scores is, in our opinion, a difficult problem. For example, given an integer k between 0 and 300, how many different bowling games have the score k? This, we leave as an open problem.”

(Curtis N. Cooper and Robert E. Kennedy, “Is the Mean Bowling Score Awful?”, Journal of Recreational Mathematics 18:3, 1985-86.)

Golomb Rulers

This conference room is 11 units long and has folding partitions at positions 2, 7, and 8. This gives it a curious property: For a meeting of any given size, the room can be configured in exactly one way. A meeting of size 6 must use partitions 2 and 8 — no other setup will work exactly.

This is an example of a Golomb ruler, named for USC mathematician Solomon Golomb. It’s called a ruler because the simplest example is a measuring stick: If we’re given a 6-centimeter ruler, we find that we can add 4 marks (at integer positions) so that no two of them are the same distance apart: 0 1 4 6. No shorter ruler can accommodate 4 marks without duplication, so the 0-1-4-6 ruler is said to be “optimal.” It’s also “perfect” because it can measure any distance from 1 to 6.

The conference room is optimal because no shorter room can accommodate 5 walls without equal-sized partitions becoming available, but it’s not perfect, because it can’t accommodate an assembly of size 10. (It turns out that no perfect ruler with five marks is possible.)

Finding optimal Golomb rulers is hard — simply extending an existing ruler tends to produce a new ruler that’s either not Golomb or not optimal. The only way forward, it seems, is to compare every possible ruler with n marks and note the shortest one, an immensely laborious process. Distributed computing projects have found the longest optimal rulers to date — the most recent, with 27 marks, was found in February, five years after the previous record.

Far From Home

https://commons.wikimedia.org/wiki/File:Peacock_served_in_full_plumage_(detail_of_BRUEGHEL_Taste,_Hearing_and_Touch).jpg

This is a detail from the allegorical painting Taste, Hearing and Touch, completed in 1620 by the Flemish artist Jan Brueghel the Elder. If the bird on the right looks out of place, that’s because it’s a sulphur-crested cockatoo, which is native to Australia. The same bird appears in Hearing, painted three years earlier by Brueghel and Peter Paul Rubens.

How did an Australian bird find its way into a Flemish painting in 1617? Apparently it was captured during one of the first Dutch visits to pre-European Australia, perhaps by Willem Janszoon in 1606, who would have carried it to the Dutch East Indies (Indonesia) and then to Holland in 1611. That’s significant — previously it had been thought that the first European images of Australian fauna had been made during the voyages of William Dampier and William de Vlamingh, which occurred decades after Brueghel’s death in 1625.

Warwick Hirst, a former manuscript curator at the State Library of New South Wales, writes, “While we don’t know exactly how Brueghel’s cockatoo arrived in the Netherlands, it appears that Taste, Hearing and Touch, and its precursor Hearing, may well contain the earliest existing European images of a bird or animal native to Australia, predating the images from Dampier’s and de Vlamingh’s voyages by some 80 years.”

(Warwick Hirst, “Brueghel’s Cockatoo,” SL Magazine, Summer 2013.) (Thanks, Ross.)

Warnsdorff’s Rule

The knight’s tour is a familiar task in chess: On a bare board, find a path by which a knight visits each of the 64 squares exactly once. There are many solutions, but finding them by hand can be tricky — the knight tends to get stuck in a backwater, surrounding by squares that it’s already visited. In 1823 H.C. von Warnsdorff suggested a simple rule: Always move the knight to a square from which it will have the fewest available subsequent moves.

This turns out to be remarkably effective: It produces a successful tour more than 85% of the time on boards smaller than 50×50, and more than 50% of the time on boards smaller than 100×100. (Strangely, on a 7×7 board its success rate drops to 75%; see this paper.) The video above shows a successful tour on a standard chessboard; here’s another on a 14×14 board:

While we’re at it: British puzzle expert Henry Dudeney once set himself the task of devising a complete knight’s tour of a cube each of whose sides is a chessboard. He came up with this:

http://www.gutenberg.org/files/16713/16713-h/16713-h.htm#X_340_THE_CUBIC_KNIGHTS_TOURa

If you cut out the figure, fold it into a cube and fasten it using the tabs provided, you’ll have a map of the knight’s path. It can start anywhere and make its way around the whole cube, visiting each of the 364 squares once and returning to its starting point.

Dudeney also came up with this puzzle. The square below contains 36 letters. Exchange each letter once with a letter that’s connected with it by a knight’s move so that you produce a word square — a square whose first row and first column comprise the same six-letter word, as do the second row and second column, and so on.

dudeney knight's tour word square puzzle

So, for example, starting with the top row you might exchange T with E, O with R, A with M, and so on. “A little thought will greatly simplify the task,” Dudeney writes. “Thus, as there is only one O, one L, and one N, these must clearly be transferred to the diagonal from the top left-hand corner to the bottom right-hand corner. Then, as the letters in the first row must be the same as in the first file, in the second row as in the second file, and so on, you are generally limited in your choice of making a pair. The puzzle can therefore be solved in a very few minutes.”

Click for Answer

Different Strokes

http://www.freeimages.com/photo/1063759

G.H. Hardy had a famous distaste for applied mathematics, but he made an exception in 1945 with an observation about golf. Conventional wisdom holds that consistency produces better results in stroke play (where strokes are counted for a full round of 18 holes) than in match play (where each hole is a separate contest). So if two players complete a full round with the same total number of strokes, then the more erratic player should do better if they compete hole by hole.

Hardy argues that the opposite is true. Imagine a course on which every hole is par 4. Player A is so deadly reliable that he shoots par on every hole. Player B has some chance x of hitting a “supershot,” which saves a stroke, and the same chance of hitting a “subshot,” losing a stroke. Otherwise he shoots par. Both players will average par and will be equal over a series of full rounds of golf, but the conventional wisdom says that B’s erratic play should give him an advantage if they play each hole as a separate contest.

Hardy’s insight is that the presence of the hole limits a run of good luck, while there’s no such limit on a run of bad luck. “To do a three, B must produce a supershot at one of his first three strokes, while he will take a five if he makes a subshot at one of his first four. He will thus have a net expectation 4x – 3x of loss on the hole, and should lose the match, contrary to common expectation.”

In general he finds that B’s chance of winning a hole is 3x – 9x2 + 10x3, and his chance of losing is 4x – 18x2 + 40x3 – 35x4, so that there’s a balance f(x) = x – 9x2 + 30x3 – 35x4 against him. If x < 0.37 -- that is, in all realistic cases -- the erratic player should lose.

"If experience points the other way -- and I cannot deny it, since I am no golfer -- what is the explanation? I asked Mr. Bernard Darwin, who should be as good a judge as one could find, and he put his finger at once on a likely flaw in the model. To play a 'subshot' is to give yourself an opportunity of a 'supershot' which a more mechanical player would miss: if you get into a bunker you have an opportunity of recovering without loss, and one which you are naturally keyed up to take. Thus the less mechanical player's chance of a supershot is to some extent automatically increased. How far this may resolve the paradox, if it is one, I cannot say, and changes in the model make it unpleasantly complex."

(G.H. Hardy, "A Mathematical Theorem About Golf," Mathematical Gazette, December 1945.)

Five of a Kind

kordemsky multigrade

All Right Then

W.H. Auden won first prize for mathematics at St. Edmund’s School in Hindhead, Surrey, when he was 13. He recalled being asked to learn the following mnemonic around 1919:

Minus times Minus equals Plus;
The reason for this we need not discuss.

Math Notes

1 + 2 = 3
1×2 + 2×3 + 3×4 = 4×5
1×2×3 + 2×3×4 + 3×4×5 + 4×5×6 = 5×6×7
1×2×3×4 + 2×3×4×5 + 3×4×5×6 + 4×5×6×7 + 5×6×7×8 = 6×7×8×9

In general, the sum of the first (n+1) products of consecutive n-tuples of consecutive integers is equal to the product of the next n-tuple.

(Thanks, Peter.)

Montmort’s Theorem

Suppose n students are sitting at n desks in a classroom. They’re asked to stand, mill around at random, and then sit again. What is the probability that at least one student will find herself in her original seat?

Intuition says that the probability ought to drop as the number of students increases, but in fact it remains about the same:

montmort table

In fact, Pierre Rémond de Montmort showed in 1708 that it’s

montmort theorem

… which approaches 1 – 1/e, or about 0.63212. Whether there are 10 students or 10,000, the chance that at least one student returns to her own seat is about 2/3.

Coming to Terms

Antoni Zygmund once asked if the World Series shouldn’t be called the World Sequence? And shouldn’t a combination lock be called a permutation lock? John Von Neumann once had a dog called Inverse. It would sit when told to stand and go when it was told to come. Von Neumann pronounced the term infinite series as infinite serious.

— Michael Stueben, Twenty Years Before the Blackboard, 1998

Illumination

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

Starting in the 1970s, neurobiologist Otto-Joachim Grüsser spent 10 years collating the light sources in 2,124 paintings selected at random from Western art originating between the 14th and 20th centuries. He found that in most paintings considered Western works of art, especially those painted around the time of the Scientific Revolution, the light falls from the left.

“At the beginning of modern Western art during the early Gothic period, a preference for diffuse illumination or light sources distributed around the painted scene was found,” Grüsser noted. “In a minority of paintings from the fourteenth century that show a clear light direction, a bias to the left side is present. This left-sided preference increased at the expense of diffuse or middle light sources up to the sixteenth and seventeenth centuries and declined thereafter. In the twentieth century, the diffuse or middle type of light distribution again became dominant.”

It’s not clear what to make of this. It seems reasonable that a right-handed artist might favor light falling from the left, but why should this vary with time? Grüsser found that the left-handed Leonardo da Vinci applied light sources from varying angles, and Hans Holbein the Younger, also a dominant left-hander, favored light falling from the right.

“From such observations in the works of these two left-handed painters who painted, drew, and wrote with the left hand, one gains the impression that the distribution of left, middle, and right light direction in left-handed painters deviates significantly from the average distribution of light found in the paintings of other contemporary painters. It would be interesting to study the drawings and paintings of other confirmed left-handed artists, who worked exclusively with the left hand.”

(Otto-Joachim Grüsser, Thomas Selke, and Barbara Zynda, “Cerebral Lateralization and Some Implications for Art, Aesthetic Perception, and Artistic Creativity,” in Ingo Rentschler, Barbara Herzberger, and David Epstein, Beauty and the Brain, 1988.)

Hot and Cold

Suppose you have three identical Dewar flasks labeled A, B, and C. (A Thermos is a Dewar flask.) You also have an empty container labeled D, which has thermally perfect conducting walls and which fits inside a Dewar flask.

Pour 1 liter of 80°C water into flask A and 1 liter of 20°C water into flask B. Now, using all four containers, is it possible to use the hot water to heat the cold water so that the final temperature of the cold water is higher than the final temperature of the hot water? How? (You can’t actually mix the hot water with the cold.)

Click for Answer

A Dice Puzzle

Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colors on the second die?

Click for Answer

Presto

Discovered by R.V. Heath in 1950:

Think of two positive integers.

Add them to get a third number.

Add the second number and the third number to get a fourth number.

Continue in this way until you have 10 numbers.

The sum of the 10 numbers is 11 times the seventh number.

Return to Sender

Mathematician Yutaka Nishiyama of the Osaka University of Economics has designed a nifty paper boomerang that you can use indoors. A free PDF template (with instructions in 70 languages!) is here.

Hold it vertically, like a paper airplane, and throw it straight ahead at eye level, snapping your wrist as you release it. The greater the spin, the better the performance. It should travel 3-4 meters in a circle and return in 1-2 seconds. Catch it between your palms.

The Modern Prometheus

jacobson railroad

By 1958 many of the attributes of living things could be found in our technology: locomotion (cars), metabolism (steam engines), energy storage (batteries), perception of stimuli (iconoscopes), and nervous or cerebral activity (computers). The missing element was reproduction: We hadn’t yet created a nonliving artifact that could make copies of itself.

So Brooklyn College chemistry professor Homer Jacobson built one. Using an HO gauge model railroad, he designed an “organism” made of boxcars that could use sensors to select other cars on the track and assemble them on a siding into models of itself. “Head” cars have “brains,” and “tail” cars have “muscles” and “eyes”; together, a head and a tail make an organism in which the head directs the tail to watch for available cars elsewhere on the track and shunt them appropriately onto a siding to create a new organism.

“Any new ‘organisms’ formed continue the propagation in a linear fashion,” Jacobson wrote, “until the environment runs out of parts, or there are no more sidings available, or a mistake is made somewhere in the operation of a cycle, i.e., a ‘mutation.’ Such an effect, like that with living beings, is usually fatal.”

(Homer Jacobson, “On Models of Reproduction,” American Scientist, September 1958.)

The Vacuum Airship

https://commons.wikimedia.org/wiki/File:Flying_boat.png

A conventional balloon rises because its airbag displaces a large volume of air. But the gas that fills the bag has some weight; it, along with the weight of the gondola, reduces the balloon’s total lift.

Realizing this, Italian monk Francesco Lana de Terzi in 1670 proposed a “vacuum airship,” a balloon whose airbag was filled with nothing at all. Since a vacuum weighs nothing, this should maximize the vehicle’s lift — the vacuum could displace a large volume of air without itself adding any weight.

In principle this might work; the problem is that the vacuum would tend to collapse its container, and building a shell sturdy enough to withstand it would leave us with a ship too heavy to lift. It’s not clear whether any material or structure could overcome this problem.

Stamps and Math

https://www.macaupost.gov.mo/Philately/XVersion/ProductList.aspx?admcode=MAC&emicode=201408&lang=en-us

Lee Sallows tells me that the postal system of Macau is releasing a new series of stamps based on magic squares. The full set will touch on everything from the Roman SATOR square to Dürer’s Melencolia. Details are here.

Charmingly, the values of the stamps will be 1, 2, …, 9 Macau patacas, so that the sheet of the nine stamps will itself form a classic Lo Shu magic square. Lee’s contribution, above, is a Nasik 2D geomagic square of order 3 — not only are all the rows and columns magic, but so are all six diagonals, including the four “broken” diagonals.

Somewhat related: In 2000 Finland issued seven stamps in classic tangram shapes, featuring images of science and education. (One of the small triangles, barely visible here, is a Sierpinski gasket.) Only three of the seven shapes are denominated postage, but I should think the temptation is overwhelming to arrange all seven on an envelope in the shape of a little man or a fish or something. I wonder what the post office makes of that.

http://philaquelymoi.blogspot.com/2014/06/stamps-with-interactive-games-update.html

Stretch Goals

stretch goals puzzle

Two circles intersect. A line AC is drawn through one of the intersection points, B. AC can pivot around point B — what position will maximize its length?

Click for Answer

Cancel That

Howard C. Saar of Albion, Mich., pointed out an innovative solution to this problem in Recreational Mathematics Magazine, April 1962:

log(3x + 2) + log(4x – 1) = 2log11

Divide each side of the equation by the word “log”:

(3x + 2) + (4x – 1) = (2)(11)

7x = 21

x = 3

… which is correct.

Catch as Catch Can

Claude Shannon, the father of information theory, took an active interest in juggling. He used to juggle balls while riding a unicycle through the halls of Bell Laboratories, and he built the first juggling robot from an Erector set in the 1970s. (The machine above mimics W.C. Fields, who himself juggled in vaudeville before turning to comedy.)

Noting that juggling seems to appeal to mathematics-minded people, Shannon offered the following theorem:

claude shannon juggling theorem

F is flight time, the time the ball spends in the air
D is “dwell time,” the time it spends in the hand
V is vacancy, the time a hand spends empty
B is the number of balls
H is the number of hands

“Theorem 1 allows one to calculate the range of possible periods (time between hand throws) for a given type of uniform juggle and a given time of flight,” he wrote. “A juggler can change this period, while keeping the height of his throws fixed, by increasing dwell time (to increase the period) or reducing dwell time to reduce the period. The total mathematical range available for a given flight time can be obtained by setting D = 0 for minimum range and V = 0 for maximum range in Theorem 1. The ratio of these two extremes is independent of the flight time and dependent only on the number of balls and hands.”

To measure dwell times, Shannon actually created a “jugglometer” in which a juggler wore copper mesh over his fingers and juggled foil-covered lacrosse balls; catching the ball closed a connection between the fingers and started a clock. “Preliminary results from testing a few jugglers indicate that, with ball juggling, vacant time is normally less than dwell time, V ranging in our measurements from fifty to seventy per cent of D.”

Shannon noted that juggling gets dramatically harder as the number of balls increases. He worked out a foolproof solution, at least in theory. A light ray that starts at one focus of an ellipse will be reflected to the other focus. If the ellipse is rotated around its major axis, it will create an egglike shell with two foci. Now if a juggler stands with a hand at each focus, then a ball thrown from either hand, in any direction, will bounce off the shell and arrive at the other hand!

(“Scientific Aspects of Juggling,” in Claude Elwood Shannon: Collected Papers, 1993.)