Climate change threatens the world's glaciers, which is why scientists simulate them on computers. Based on mathematical models, these computer simulations help predict how glaciers are likely to change in the future, depending on environmental factors.
The Aletsch glacier in Switzerland.
But you can also run these simulations backwards in time, to see how a glacier behaved in the past. This is what a mathematician and a glaciologist have just done, not in order to understand glaciers, but in order to solve a mystery. In March 1926 four young men, three of whom were brothers, set off on a tour from the Aletsch glacier in Switzerland, the longest glacier in the Alps. They never returned. It's likely they got caught up in a blizzard that raged in the mountains for several days. Despite an extensive search their bodies could not be found and nobody knew where on their trip they met their end. Eighty-six years later, in the summer of 2012, two English alpinists found the remains of the three brothers and some of their equipment.
The mathematician Guillaume Jouvet of Freie Universität Berlin and the glaciologist Martin Funk of the ETH in Zürich realised that they could use a computer model of glaciers, which Jouvet had developed during his PhD in 2012, to find out where the men had met their death. The model was the first to represent the three-dimensional flow of a glacier, including the velocity of the flow beneath its surface. Starting from the place the men's bones were found, the scientists simulated the evolution of the glacier backwards in time. They found that the bodies probably moved by around 10.5km in the 86 years since they had been swallowed by the ice, at an average speed of 122 meters per year. In 1980 they were buried some 250m below the surface of the ice. And the place of the men's demise could be narrowed down to an area of 1600m by 300m.
Jouvet and Funk hope that they can solve other glacial mysteries using their method. For example, in 1964 a plane carrying US military crashed on the Gauli glacier in Switzerland. Crew and passengers could be rescued, but the plane disappeared in the ice without a trace. Using their model, Juvet and Funk hope to predict when and where it will be released from the ice. And who knows what other icy stories their model will reveal in the future.
It might be grey and raining outside, but inside the Museum of London and Barnard's Inn Hall, Gresham College have some fascinating lectures planned to brighten the darkest day! First up Caroline Crawford will explore the lives of stars on Wednesday 5 February at 1pm at the Museum of London, how they are born, evolve over billions of years and dramatically burn, revealing clues to our own origins. It seems Hollywood isn't the only place where the stars crash and burn.
Prince Charming? Or just a slimy toad?
And surely the best way to cheer up a dreary day is to fall in love. Tony Mann will explain the mechanics of computer dating and how maths can help us find our heart's desire (or at least a charming dinner companion) in Finding stable matches: the mathematics of computer dating on Monday 17 February at 6pm, at Barnard's Inn Hall.
And if you are using a coin to decide whether to get out of bed or stay under the duvet, you need to hear what Raymond Flood has to say about Probability and its limits on Tuesday 18 February at 1pm at the Museum of London. He'll explain why we know a coin will come up heads roughly half the time over many tosses, but we can't tell you whether yours will land heads or tails tomorrow morning. If it lands heads and you make it to his lecture, you'll find out much more!
There's no need to register for these free public events, just come a bit early to get a seat. You can find out about all the other fascinating talks on the Gresham College website and you can get in the mood by reading about astronomy, dating and probability on Plus.
Vending machines that don't return change are annoying, especially if the prices they demand aren't nice round figures you can make up with a single coin. If that's the case, then there's nothing but to cram through your wallet, fishing out the right coins to make up the amount exactly. What's the best way of doing that? Without noticing many of us probably follow this recipe: find the biggest (as in largest denomination) coin that fits into the amount, then the next-biggest that fits into the remainder, and so on, until you (hopefully) hit the required sum. As an example, if you're being asked for 85p, you probably fish a 50p coin out first, then a 20p coin, then a 10p coin, and finally a 5p coin. And what if you haven't got all of the coins just mentioned in your wallet? In that case you follow the same recipe using what you've got.
This greedy recipe (greedy because you always go for the biggest coin that fits) seems to offer the best solution in that it seems to involve the fewest number of coins to make up the amount you need. For example, supposing you do have a 20p coin, but decide for to go for two 10p coins instead, you increase the number of coins to make up 85p from four to five. So the greedy algorithm seems useful, not just for people struggling with vending machines, but also for cashiers returning change to customers.
But is greed really always the best option? It turns out that this depends on the coins that are available. Imagine, for example, you need to make up 8p. Greed would tell you to go for a 5p coin, then a 2p coin and then a 1p coin. And that's indeed the smallest number of coins to make up 8p with if you are using Pound Sterling, Euros, US Dollars, and most other currencies. But now imagine a currency that in addition to these denominations also has a 4p coin. Then you could make up 8p with two of those, beating the greedy strategy by one. Such a currency system might seem silly but it's not unheard-of: the pre-decimal British coinage system was one for which the greedy recipe failed when it came to minimising the number of coins needed to make up a given amount.
Santa's just putting his hat on and button up his jolly red coat when an exasperated elf runs up to him: "Santa! We've got a problem! There have been so many good children this year we can't put all the presents into your sleigh!"
The problem is that Santa's sleigh has a weight limit, and can only carry 2 tonnes of present. Each present has a different weight, and each present has a different star value too: a star for each good deed of the toy's recipient. Santa obviously wants to reward as many good deeds as possible, maximising the total star value of the presents in the sleigh. But he can't exceed the maximal weight or those presents are going nowhere! So which should he put in his sleigh?
The elves start out trying all different combinations of presents to find the best one. This would have been fine if there only were a few items, but they quickly realise it is totally impractical when you have millions of presents to deal with. If they carry on this way Christmas Eve will have long passed and no presents at all will be delivered!
Such a brute force solution is unacceptable, and not only to the elves and Santa. Mathematicians would also find this an unacceptable approach, and ask whether there is an algorithm
– a recipe for finding a solution – which works for any number of items,
and so that the time it takes to complete the algorithm grows with the
number of items in a reasonable, non-explosive fashion.
Mathematicians have a clear definition of "reasonable and non-
explosive" in this context: the time it takes to complete the
algorithm should grow with the number N of items no faster than the
polynomial N to the power of K, for some integer K, grows with N.
That's still pretty rapid growth, especially if K is large, but at
least it's not exponential.
So does such a polynomial time algorithm exist for our problem? The
answer is that nobody knows - not yet - though most mathematicians
believe that there isn't. In fact, if you can prove or disprove that a
polynomial time algorithm exists you will have answered the question behind door #22 too, as they both can be turned into NP complete problems.
Optimisation problems such as this one, which is known as the
knapsack problem, crop up in real life all the time. Thankfully Santa is well read in mathematical literature and knows an algorithm they can use that won't take too long and, while it not give the
very best combination of starred pressies, it will get sufficiently close.
So rest assured, if you've been really, really good, your present is probably now packed on the sleigh, ready for the Santa's big delivery run. Merry Christmas!
You can find out more about NP complete problems on Plus.
When Kepler was decorating his Christmas tree you can be sure his Christmas baubles weren't boring and round. Oh no, we're sure they would have been beautiful and elliptical!
Ah, the humble ellipse. It might look just like a squashed circle but it is
so much more! One of the conic sections discovered by the ancient Greeks,
it generalises the circle using eccentricity, a measure of how squashed it
is. A circle is an ellipse with eccentricity zero, a very squashed nearly
flat ellipse will have eccentricity close to one. An ellipse has two foci
(points that lie within the ellipse), with the handy property that the sum
of the distances from each focus to a point on the ellipse is equal to the
ellipse's width. But not just a pretty shape, it turned out to be Kepler's
answer to life, the universe and everything - he discovered that the
heavenly bodies didn't move in circles, as had been thought for hundreds of
years, but instead had elliptical orbits around the sun. Kepler tried to
convince himself of the perfection of this design of the heavens by
describing the interaction of nature with the heavens introducing a
distortion to pull the motion from a perfect circle into an ellipse.
It's that time of the year again when Santa needs to think about how to deliver all those presents to the people on Earth. Unfortunately his reindeers are a bit weak this year, having suffered some severe bouts of flu, and he knows he can't ask them to travel more than a million kilometres in total. Is there a route around the Earth that visits all those cities, towns and villages he needs to visit, but is less than a million kilometres long?
A bit tired this year.
That, it turns out, is a very tricky question. One way of answering it would be to try out all different routes and work out their lengths. But since there are 7 billion people on Earth and he needs to visit every single dwelling that's totally unfeasible. It would take him until long after the Universe has ended to try all possible routes. Is there a clever algorithm Santa could use instead?
Obviously, the time it will take any algorithm to answer the question will depend on the number of places that need visiting — the more places, the longer it takes. So the question is whether there's an algorithm for which the time taken grows in a manageable, non-explosive way as the number of places grows — it's what complexity theorists call a polynomial time algorithm. If Santa were to ask such a complexity theorist he would find out that the class of all problems that can be solved in polynomial time has a name: it's called P.
Now Santa doesn't know if his problem of finding a route belongs to the class P. What's pretty obvious, though, is that if someone gave him a potential solution — a route they claim visits every place but is less than a million kilometres long —
it would be easy to check whether that's true: he'd simply have to add up all the individual distances between places. A computer could do that quite easily. This means that Santa's problem belongs to a class of problems called NP. It consists of those problems for which potential answers can be checked in polynomial time, even though we might not know how to solve them from scratch in polynomial time.
But now suppose that Santa has a brainwave. Sitting down with a computer he constructs a polynomial time algorithm that finds the kind of route he is after. He would then have shown that his problem is not only a member of the NP class, but also of the P class. Amazingly, it would also hand Santa a polynomial time algorithm for all other problems in the NP class — because it turns out that his problem is a version of the famous travelling salesman problem, and that all the other problems in the NP class can be reduced to it. What is more, his algorithm would win Santa $1 million: the question of whether the P class and the NP class of problems are equal is considered one of the hardest in mathematics, which is why the Clay Mathematics Institute as offered $1 million to anyone who solves it.
Unfortunately Santa doesn't have the time to do any of this, and he's not a computer expert anyway. So he just takes an educated guess at what route to take and gets going, hoping for the best. It's a shame really because if he'd been able to show that P=NP he would have held the key to many of the world's secret communications, including everyone's bank details that are sent encrypted over the internet. This is because NP problems are used as the key for such encryptions, with the secrecy relying on the fact that the problems are very hard to crack. You can find out more in the Plus articles The travelling salesman and Safety in numbers.