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

https://commons.wikimedia.org/wiki/File:Golomb_ruler_conference_room.svg
Image: Wikimedia Commons

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.)

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