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]]>The post Wrap-up and thanks appeared first on Exploring Maths 2017.

]]>There was some shock when Prof. Paterson pocketed his substantial winnings as the Evil-Eavesdropper at the end of the Alice/Bob cryptography demonstration on Friday. I’m happy to clarify that he later returned all the money to his victims, who therefore got a free (and probably unforgettable) lesson on the importance of key management in cryptography.

The event could not be run without the voluntary help of our maths students. So a big thank you to **Awala Alero, Andy Bassett, Josie Browne, Melodie Gray, Ceejay Hammond, Harjit Hullait, Sebastian Koba, Amanda Lonergan, Sayali Mandke, Gurjeevan Sembi, Ewa Wlodarska** and **Laura Webber** for their tireless efforts over the two days.

Huge thanks are also owed to **Laurence O’Toole**, who did far more work than me in organizing the event, to our administrators **Jenny Lee** and **Valerie Nicol**, and to our technical expert, **Tristan Findley**, who you will have seen as video-camera operator. (Tristan also designed our beautiful poster.)

Yet more thanks are owed to members of the School of Mathematics and Information Security for contributing long or short talks to the day.

Finally here are the promised solutions to the two ‘Brain-Teasers’ in the programme.

In the pond below Founder’s Building live 30 fish. Each fish is either red, green or blue. When two fish

of the same colour meet, they undergo a freakish metamorphosis, and change into fish of each of the other two colours. When two fish of different colours meet, they both change into fish of the third colour.

In the morning there are 15 red fish, 7 blue and 8 green. Is it possible that at some time there are 30 red fish in the pond?

**Solution.** No. Consider the difference between the number of red fish and the number of blue fish. At the start it is 8. When two red fish meet, we lose two red fish and gain one blue fish (and also gain one, irrelevant green fish), so the difference goes down by 3. When two blue fish meet, we lose two blue fish and gain one red fish (and also gain one, irrelevant green fish), so the difference goes up by 3. When two green fish meet the difference is unchanged. So the difference is always the same, working modulo 3. Since mod 3, whereas mod 3, the difference can never be 30.

Consider this chain of claimed equalities:

Where is the mistake?

**Solution.** This is harder than it seems. The problem lies in the implicit assumption in the second equality that . It is impossible to define a square-root function for which this equality always holds. (Of course it holds when and are both positive, but this does not prove it holds in general, and in fact it does not.) Here is some further reading.

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]]>The post Colin Wright’s Talk ‘Juggling Theory and Practice’ appeared first on Exploring Maths 2017.

]]>The connection between juggling and mathematics comes from a remarkable mathematical way to specify juggling patterns, called site swap notation. The idea is this: encode each throw by its height.

A throw of height 3 gives enough time for two more throws before the ball comes back down, and generally, a throw of height *n* gives time for *n*-1 throws before the ball has to be caught. If hands alternate it follows that throws of odd height are from one hand to another, while throws throws of even height are thrown and caught by the same hand.

For example 33333 is five throws of the three ball cascade, the most basic juggling pattern. The Wikipedia article linked to above gives more examples.

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]]>The post Exploring Maths 2015 is now open! appeared first on Exploring Maths 2017.

]]>Highlights include Dr Colin Wright, who will talk about the amazing mathematics behind juggling in *Juggling: Theory and Practice*, and our own Professor Kenny Paterson, who will perform a live demonstration of the mathematics behind public cryptography in *Cryptographyâ€”Everywhere!* There will also be a packed programme of quizzes, short talks and other small group sessions.

We look forward to welcoming you to Royal Holloway in July 2015.

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