Sulphur-crested cockatoos

And for a break from solar cars, here are some sulphur-crested cockatoos enjoying the Australian winter sun:


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.

A new species of thrush for 2016

The Himalayan forest thrush (Zoothera salimalii – photo by Craig Brelsford above) is a newly described species of bird. It was formerly grouped with Z. mollissima, which breeds above the tree line. In contrast, Z. salimalii breeds in coniferous forests up to the tree line, from Tibet to Vietnam. In addition, Z. salimalii has a different song from its alpine cousin, as well as differing genetically. This is enough to make it a distinct species.

With improvements in technologies for DNA and other analyses, we are starting to see many such new species “carved out” of existing ones.

Creating an Australian Feather Map

Photo: Louise Docker, 2007

Australian scientists are encouraging people to collect feathers found on the ground or in the water in wetlands (with details of where they were collected). After analysis using mass spectrometry and high resolution X-ray fluorescence, a feather map will be constructed. All aspiring citizen scientists, young and old, can get involved and follow the project on Instagram. It looks like a great way to monitor bird populations in wetlands!

World Solar Challenge: Weather in Darwin

One very helpful input to race strategy in a solar car race is weather expertise. How much sunshine can we expect? And when can we expect it? In 2013, Solar Team Twente took along an expert from the Joint Meteorological Group of the Royal Netherlands Air Force to help with that. This year, Punch Powertrain Solar Team (team 8) is taking along an expert from the the Royal Meteorological Institute of Belgium, who will be blogging his insights and experiences.

For those without his specialist expertise, forget everything you thought you knew about Spring and Summer, Autumn and Winter. Darwin has 7 seasons, as the Larrakia People tell us, and the World Solar Challenge begins towards the end of Dalirrgang (the “Build Up” – click image above for multimedia tutorial). Dalirrgang is a kind of overture to the rainy season (the “Wet”). Traditionally, Dalirrgang is the time to hunt the Magpie goose (photo by Djambalawa below).

Long-term weather forecasts suggest that the World Solar Challenge this year might in fact begin on a partly sunny day, with a little rain, but that’s very uncertain, this far ahead.

Visiting Vanuatu

Recently I took my own advice and visited the island nation of Vanuatu. I had a great time! Since the islands are volcanic and surrounded by coral reefs, the beach sand ranges from pure white to basaltic black, with an intermediate grey-brown in some cases, like the beach in my photo above.

Vanuatu has a range of interesting wildlife (though no native land mammals other than bats). Birds of Vanuatu include the Vanuatu kingfisher (Todiramphus farquhari, above), which I did not see. There are 120 other bird species, including visiting seabirds. Butterflies of Vanuatu (of which I saw many) include the Monarch butterfly (Danaus plexippus) and several subspecies of the Canopus Swallowtail (Papilio fuscus, below).

Underwater, Vanuatu provides wonderful opportunities to see marine life while diving or snorkelling. The Flickr photographs below are by Diane Brook (click images to zoom):