Here is the calendar for March (**click for hi-res image**). Pi Day is coming up, along with some science festivals. See more calendars here.

# Author Archives: Tony

# 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 two “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.

Here is one of the two “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:

# Solar cars – From the Outback to the Oregon Trail

As I watch the lead-up to the American Solar Challenge next July, above are the entries that previously raced in the World Solar Challenge in Australia last year. Left to right from the top, they are Michigan (came 2^{nd}), Iowa State University / PrISUm (raced in the Cruiser class), Western Sydney (came 6^{th}), Illini (participated in the Adventure class), Principia (raced, but trailered after 2390 km), and Minnesota (raced in the Cruiser class).

Several different approaches are visible here to building a car for both races. It’s easier to do so in the Cruiser class, for a start. Illini chose to build their car for the ASC and race it non-competitively in Australia. Principia built their car for both events (unveiling it at FSGP 2017). Michigan and Western Sydney built their cars specifically for the WSC, and may have to adapt the cars for the American race (last ASC, Michigan had to modify their car, and then incurred a daily 6-minute race penalty because the modifications made the car too wide).

My ASC race information page will be updated from time to time with information on the progress of these and all the other teams.

# 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 3denari, a dove for 2denari, and 2 sparrows for 1denaro, that is, 1 sparrow for ½denaro. 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:

*p*+ 2

*d*+ ½

*s*= 30

Doubling that second equation gets rid of the annoying fraction:

*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*:

*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:

*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:

*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 3denari, a dove for 2denari, and 2 sparrows for 1denaro, that is, 1 sparrow for ½denaro. 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.

# Famous radiation leaks

The chart above shows approximate radiation releases (in becquerels) for some major nuclear disasters. It should be interpreted with caution, since some radioisotopes are more dangerous than others. For example, the releases from Three Mile Island were largely noble gases (mostly xenon), and that incident appears to have had few detectable environmental or health effects. Ticks on the vertical axis of the chart go up logarithmically, in steps of ×1000. For comparison, radium and bananas are also listed.

- Chernobyl (1986)
- Releases from the Mayak plant (1949–1957), culminating in the Kyshtym disaster
- Fukushima (2011)
- Three Mile Island (1979)
*Kursk*submarine (2000)- Windscale fire (1957)

# American Solar Challenge status check

Just a quick status check on the lead-up to the American Solar Challenge next July. As my race information page indicates, we still have 39 solar car teams from 8 countries registered. But when we check the status of pre-race documents due last December, another story emerges:

The histogram is tri-modal. First, there is a group of 21 teams that are on schedule, or close to it. Most of them also show good signs of progress on social media. Then there are 12 teams that are well behind schedule. And finally, there are 6 teams that have submitted almost none of the required documents, raising serious doubts as to whether they will actually turn up.

My race information page will be updated from time to time with information on the progress of these teams. In July, I plan to blog about the race itself.