The Game of Mu Torere

The New Zealand game of mū tōrere is illustrated above with a beautiful handmade wooden board. The game seems to have been developed by the Māori people in response to the European game of draughts (checkers). Play is quite different from draughts, however. The game starts as shown above, with Black to move first. Legal moves involve moving a piece to an adjacent empty space:

  • along the periphery (kewai), or
  • from the centre (pūtahi) to the periphery, or
  • from the periphery to the centre, provided the moved piece is adjacent to an opponent’s piece.

Game play continues forever until a draw is called (by mutual consent) or a player loses by being unable to move. Neither player can force a win, in general, so a loss is always the result of a mistake. For each player there is one “big trap” and four “small traps.” This is the “big trap” (Black wins in 5 moves):

  
The board on the left is the “big trap” for White – Black can force a win by moving as shown, which leaves only one move for White.

  
Again, Black moves as shown, which leaves only one move for White.

  
Now, when Black moves as shown, White cannot move, which means that White loses.

Here is one of the four “small traps” for White. The obvious move by Black results in White losing (but avoiding this does not require looking quite so far ahead as with the “big trap”):

Here (click to zoom) is the complete network of 86 game states for mū tōrere (40 board positions which can occur in both a “Black to move” and a “White to move” form, plus 6 other “lost” board positions). Light-coloured circles indicate White to move, and dark-coloured circles Black to move, with the start position in blue at the top right. Red and pink circles are a guaranteed win for Black, while green circles are a guaranteed win for White. Arrows indicate moves, with coloured arrows being forced moves. The diagram (produced in R) does not fully indicate the symmetry of the network. Many of the cycles are clearly visible, however:


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Fibonacci and his birds (solution)

In the previous post, we described Fibonacci’s “problem of the birds” (“the problem of the man who buys thirty birds of three kinds for 30 denari”). In English:

“A man buys 30 birds of three kinds (partridges, doves, and sparrows) for 30 denari. He buys a partridge for 3 denari, a dove for 2 denari, and 2 sparrows for 1 denaro, that is, 1 sparrow for ½ denaro. How many birds of each kind does he buy?”

The man must buy at least one of each kind of bird, or he wouldn’t be buying “birds of three kinds.” Also, he must buy less than 10 partridges, because 10 partridges (at 3 denari each) would use up all his money. Similarly, he must buy less than 15 doves. We can thus make up a table of possible solutions:

Of those 126 possible solutions, only one works out correctly in terms of cost, and that’s the answer. But that’s an unbelievably tedious way of getting the answer, and you’d be rather foolish to try to solve the problem that way. The obvious approach is to use algebra. Write p for the number of partridges bought, d for the number of doves, and s for the number of sparrows. Because the man buys 30 birds, we have the equation:

p + d + s = 30

And because the costs add up to 30 denari, we have:

3 p + 2 d + ½ s = 30

Doubling that second equation gets rid of the annoying fraction:

6 p + 4 d + s = 60

If you’ve done any high school algebra, no doubt you want to subtract the first equation from this, which will eliminate the variable s:

5 p + 3 d = 30

But now what? That gives a relationship between the variables p and d, but there doesn’t seem to be enough information to get specific values for those variables.

Fibonacci solves the problem a different way. His solution is based on a key insight – the man buys 30 birds for 30 denari, so that the birds cost, on average, 1 denaro each. Fibonacci then makes up “packages” of birds averaging 1 denaro each. There are only two ways of doing this. Package A has 1 partridge and 4 sparrows (5 birds for 5 denari), and package B has 1 dove and 2 sparrows (3 birds for 3 denari). The solution will be a combination of those two packages.

Now the man can take 1, 2, 3, 4, or 5 copies of package A, leaving 25, 20, 15, 10, or 5 birds to be made up of package B. But the birds making up package B must be multiple of 3, so that the only possible answer is 3 copies of package A and 5 copies of package B. This means that the man buys 3 partridges, 5 doves, and 3×4 + 5×2 = 22 sparrows. That’s 30 birds and 3×3 + 2×5 + ½×22 = 30 denari.

Now it turns out that, had we kept on going with the algebraic approach, we would have gotten the same answer. We had:

5 p + 3 d = 30

Given that the numbers of partridges and doves (p and d) had to be positive whole numbers, that meant that p had to be a multiple of 3, and d a multiple of 5. That could only be achieved with p = 3 and d = 5.

We can also return to the diagrammatic approach. The equation:

5 p + 3 d = 30

describes the diagonal red line in the diagram below. That line only crosses one of the possible solutions, namely the dot corresponding to 3 partridges and 5 doves.

In mathematics, there’s more than one way to skin a cat. Or, in this case, a bird.


Fibonacci and his birds

The mathematician Leonardo of Pisa (better known as Fibonacci) is famous for his rabbits, but I was recently reminded of his “problem of the birds” or “the problem of the man who buys thirty birds of three kinds for 30 denari.” This problem appears in his influential book, the Liber Abaci.

The “problem of the birds” is expressed in terms of Italian currency of the time – 12 denari (singular: denaro) made up a soldo, and 20 soldi made up a lira. In the original Latin, the problem reads:

“Quidam emit aves 30 pro denariis 30. In quibus fuerunt perdices, columbe, et passeres: perdices vero emit denariis 3, columba denariis 2, et passeres 2 pro denario 1, scilicet passer 1 pro denariis ½. Queritur quot aves emit de unoquoque genere.”

In English, that translates to:

“A man buys 30 birds of three kinds (partridges, doves, and sparrows) for 30 denari. He buys a partridge for 3 denari, a dove for 2 denari, and 2 sparrows for 1 denaro, that is, 1 sparrow for ½ denaro. How many birds of each kind does he buy?”

How many birds of each kind does the man buy? It may help to cut out and play with the bird tokens below (click image to zoom). In a similar vein, what if the man buys birds as follows (still purchasing birds of all three kinds, and at the same price)?

  • 4 birds for 6 denari
  • between 6 and 10 birds for twice as many denari as birds
  • 8, 11, 13–14, 16–22, 24–25, or 27 birds for the same number of denari as birds
  • 8 birds for 12 denari
  • 12 birds for 18 denari
  • 16 birds for 12 denari
  • 28 birds for 21 denari
  • 6, 8–9, or 14 birds for 11 denari
  • 7–10, 12, 15, or 18 birds for 13 denari

Solution to the main problem here.


Measuring the Earth this (Southern) Christmas

In around 240 BC, Eratosthenes calculated the circumference of the Earth. The diagram above (from NOAA) shows how he did it. This Christmas, people in the Southern Hemisphere can repeat his work!

Eratosthenes knew that, at the summer solstice, the sun would be directly overhead at Syene (on the Tropic of Cancer) and would shine vertically down a well there. He also knew the distance to Syene.

On 21 December, the sun will be directly overhead on the Tropic of Capricorn at local noon. This table show the time of local noon on 21 December 2017, and the distance to the Tropic of Capricorn, for some Southern Hemisphere cities:

City Local Noon Distance to Tropic (km)
Adelaide 13:14 1270
Auckland 13:19 1490
Brisbane 11:46 450
Buenos Aires 12:52 1240
Darwin 12:45 1220
Hobart 13:09 2160
Johannesburg 12:06 310
Melbourne 13:18 1590
Perth 12:15 940
Santiago 13:41 1110
Sydney 12:53 1160

At exactly local noon, Eratosthenes measured the length (s) of the shadow of a tall column in his home town of Alexandria. He knew the height (h) of the column. He could then calculate the angle between the column and the sun’s rays using (in modern terms) the formula θ = arctan(s / h).

You can repeat Eratosthenes’ calculation by measuring the length of the shadow of a vertical stick (or anything else you know the height of), and using the arctan button on a calculator. Alternatively, the table below show the angles for various shadow lengths of a 1-metre stick. You could also attach a protractor to the top of the stick, run a thread from the to of the stick to the end of the shadow, and measure the angle directly.

The angle (θ) between the stick and the sun’s rays will also be the angle at the centre of the Earth (see the diagram at top). You can then calculate the circumference of the Earth using the distance to the Tropic of Capricorn and the fact that a full circle is 360° (the circumference of the Earth will be d × 360 / θ, where d is the distance to the Tropic of Capricorn).

Height (h) Shadow (s) Angle (θ)
1 0.02
1 0.03
1 0.05
1 0.07
1 0.09
1 0.11
1 0.12
1 0.14
1 0.16
1 0.18 10°
1 0.19 11°
1 0.21 12°
1 0.23 13°
1 0.25 14°
1 0.27 15°
1 0.29 16°
1 0.31 17°
1 0.32 18°
1 0.34 19°
1 0.36 20°
1 0.38 21°
1 0.4 22°
1 0.42 23°
1 0.45 24°
1 0.47 25°
1 0.49 26°
1 0.51 27°
1 0.53 28°
1 0.55 29°
1 0.58 30°
1 0.6 31°
1 0.62 32°
1 0.65 33°
1 0.67 34°
1 0.7 35°
1 0.73 36°
1 0.75 37°
1 0.78 38°
1 0.81 39°
1 0.84 40°
1 0.87 41°
1 0.9 42°
1 0.93 43°
1 0.97 44°
1 1 45°

Mathematics of the Harp

After my second harp post, I thought I’d keep going with some mathematics. In particular I want to answer the question: why does a harp have that shape?


A modern electric lever harp (photo: Athy)

The physics of vibrating strings gives us Mersenne’s laws, which tell us that the frequency of a string of length L is (1 / 2L) √ T / μ , where T is the tension force on the string (in newtons), and μ is the density per unit length (in kg per metre).

For a string of diameter d and density ρ, we can calculate μ = A ρ, where A = π (d / 2)2 is the cross-sectional area. Nylon has a density ρ of 1150 kg/m3. For a nylon string of 1 mm diameter, we get μ = 0.0009 kg/m. Putting 0.448 m (17.6 inches) of that string under a tension of 140 newtons (31.5 pounds force), we get a frequency of (1 / 0.896) √ 140 / 0.0009  = 440 Hz. That is, the string plays the note A.

The important thing here is that the frequency is inversely proportional to the length. Over an octave the frequency doubles, which means that the string length halves. A 36-string harp covers 5 octaves, therefore if all the strings were made of the same material under the same tension, the longest string would be 32 times the length of the shortest. Doing some calculations in R for strings of diameter 0.8 mm under a tension of 140 newtons (31.5 pounds), we would get the following harp, which has strings ranging from 7 to 224 cm in length (note that the strings run from A to A, and the C strings are red):


Image produced in R. Click to zoom.

You can see quite clearly that, starting at the treble end, there is an exponential growth in string length. That makes for a terribly unwieldy instrument, and creates all sorts of problems in playing. In practice, we make bass strings thicker (and usually of heavier material) and we vary the tension as well – although, to make life easier for the harpist, we want the properties of the strings to vary reasonably smoothly. If we re-do our calculations with string diameters varying linearly from 3 mm to 0.6 mm, and tension varying linearly from 210 newtons at the bass end to 60 newtons at the treble end, we get a much more realistic-looking harp:


Image produced in R. Click to zoom.

We can flatten out the curve at the top a little by changing the way we vary the strings (after all, a guitar manages to span 2 octaves with all the strings being the same length). However, we cannot eliminate that curve completely – it is the inevitable result of spanning so many octaves, combined with the mathematics of exponential growth.

      
Left: an exponential curve (red) and a similar polynomial (dashed). Right: the quotient of the two functions (green) compared to a straight line (grey). Click images to zoom.

Mathematically speaking, the modifications to the strings have the effect of dividing an exponential function by some kind of polynomial (as shown above). Over a short range of x values, we can find a polynomial that fits the exponential well, and gives us strings of the same length. Over a wide range of x values, however, the exponential wins out. Furthermore, exponential growth is initially slow (sub-linear), so that (starting at the right of the harp), growth in string length is slower than the linear shift provided by the sloping base, which means that the top of the harp curves down. After a few octaves, growth in string length speeds up, and so the top of the harp curves up again.

A similar situation arises with the strings of a piano, although these are usually hidden from view:


Inside of a piano (photo: Alexandre Eggert)

And to finish, here is one of my favourite classical harpists in action:


Ecological networks and the Australian dingo

I’m excited at the publication of a joint paper on network ecology, with a focus on the Australian dingo: “Trophic cascades in 3D: Network analysis reveals how apex predators structure ecosystems” (by Arian D. Wallach, Anthony H. Dekker, Miguel Lurgi, Jose M. Montoya, Damien A. Fordham & Euan G. Ritchie, and appearing in Methods in Ecology and Evolution).

Associated with this publication is an animation I put together for the paper showing how the ecological network changes if the role of the dingo as apex predator is weakened. I’m grateful to my ecologist co-authors at the opportunity to contribute my mathematical skills to such an interesting project.