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Mathematical Sciences at Oxford

Mathematical Sciences at Oxford. Stephen Drape Access/Schools Liaison Officer Computer Science. Mathematical Science Subjects. Mathematics Mathematics and Statistics Computer Science Mathematics and Computer Science All courses can be 3 or 4 years. Admissions data for 2007 entry.

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Mathematical Sciences at Oxford

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  1. Mathematical Sciences at Oxford Stephen Drape Access/Schools Liaison Officer Computer Science

  2. Mathematical Science Subjects • Mathematics • Mathematics and Statistics • Computer Science • Mathematics and Computer Science All courses can be 3 or 4 years

  3. Admissions data for 2007 entry

  4. Admissions Process • Fill in UCAS and Oxford form • Choose a college or submit an “Open” Application • Interview Test • Based on common core A-Level • Taken before the interview • Interviews • Take place over a few days • Often have many interviews

  5. Entrance Requirements • Essential: A-Level Mathematics • Recommended: Further Maths or a Science • Note it is not a requirement to have Further Maths for entry to Oxford • For Computer Science, Further Maths is perhaps more suitable than Computing or IT • Usual offer is AAA

  6. Course Structure • The first year consists of compulsory courses which act as a foundation to build on • The second year starts off with more compulsory courses • The reminder of the course consists of a variety of specialised options • In the fourth year, students have to study 6 courses from a choice of 40 • There are opportunities for projects

  7. First Year Maths Course • Algebra (Group Theory) • Linear Algebra (Vectors, Matrices) • Calculus • Analysis (Behaviour of functions) • Applied Maths (Dynamics, Probability) • Geometry

  8. Computer Science • Computer Science • Computer Science firmly based on Mathematics • Mathematics and Computer Science • Closer to a half/half split between CS and Maths • Computer Science is part of the Mathematical Science faculty because it has a strong emphasis on theory

  9. Computer Science Mathematics Computing Project work Year 1 Year 2 Year 3 Year 4

  10. Mathematics & Computer Science Mathematics Computing Project work Year 1 Year 2 Year 3 Year 4

  11. Some of the first year courses Functional Programming Design and Analysis of Algorithms Imperative Programming Digital Hardware Calculus Linear Algebra Logic and Proof Discrete Maths

  12. Subsequent Years • The second year is a combination of compulsory courses and options • Many courses have a practical component • Later years have a greater choice of courses • Third and Fourth year students have to complete a project

  13. Compilers Programming Languages Computer Graphics Computer Architecture Intelligent Systems Machine Learning Lambda Calculus Computer Security Category Theory Computer Animation Linguistics Domain Theory Program Analysis Information Retrieval Bioinformatics Formal Verification Some Computer Science Options

  14. Useful Sources of Information • Admissions: • http://www.admissions.ox.ac.uk/ • Mathematical Institute • http://www.maths.ox.ac.uk/ • Computing Laboratory: • http://www.comlab.ox.ac.uk/ • Colleges

  15. What is Computer Science? It’s not about learning new programming languages. It is about understanding why programs work, and how to design them. If you know how programs work then you can use a variety of languages. It is the study of the Mathematics behind lots of different computing concepts.

  16. Power We can use the power notation to write very large in a compact way. 35 = 3 £ 3 £ 3 £ 3 £ 3 = 243 How could we write a program to compute this? 35 exponent base

  17. Writing a program for this Here’s a very simple program: This uses a technique called recursion

  18. Number of multiplications So we’ve just seen that to work out 35 we need 5 multiplications. What about 325? Using the previous program, we would need 25 multiplications. But it is possible to compute this using only 7 multiplications!

  19. Splitting Up the Exponent Split the exponent (power) into powers of 2. 25 = 1+8+16 = 20 + 23 + 24 and so 325 = 31£ 38£ 316 Now working out the powers: 1 £ 3 = 31 (keep) 31£ 31 = 32 32£ 32 = 34 34£ 34 = 38 (keep) 38£ 38 = 316 (keep) So we have 7 multiplications in total

  20. Setting up an algorithm We have three variables: • y (which keeps the powers of 2 that we need) • z (which works out the powers of 2) • k (a counter to count down) Initially we set y = 1, z = base, k = exponent

  21. The algorithm This algorithm is not written in any programming language – it is written in “pseudo-code” so that we can explain what is happening.

  22. A real program This is written in a programming language called Haskell. It’s used throughout the Computer Science course at Oxford.

  23. Try an example Let’s try an example and see how this algorithm works out 325. At the start, we take y=1, z=3 and k=25

  24. Steps 1

  25. Steps 2

  26. Comparison of number of steps Here’s a table showing the size of the power and the number of multiplications needed using our fast algorithm:

  27. So What? • Many computer algorithms need to be able to work out powers efficiently. • The RSA encryption algorithm needs to do calculations such as: • For security, e and n need to be 2048 bits (which means bigger than 22048) which has over 600 digits.

  28. Algorithm Design When designing algorithms, we have to consider a number of things: Our algorithm should be efficient – that is, where possible, it should not take too long or use too much memory. We should look at ways of improving existing algorithms. We may have to try a number of different approaches. We should make sure that our algorithms are correct.

  29. An Interview Type Problem You have an urn which contains 23 white beans and 34 black beans. You take out two beans from the jar: • if the beans are the same colour then you put a black bean (from a large pile of beans that you have) into the jar • otherwise, you put a white bean into the jar You repeat this process until there is only one bean left. What colour is it?

  30. Finding the Highest Common Factor Example: Find the HCF of 308 and 1001. 1) Find the factors of both numbers: 308 – [1,2,4,7,11,14,22,28,44,77,154,308] 1001 – [1,7,11,13,77,91,143,1001] 2) Find those in common [1,7,11,77] 3) Find the highest Answer = 77

  31. Creating an algorithm For our example, we had three steps: Find the factors Find those factors in common Find the highest factor in common These steps allow us to construct an algorithm.

  32. Creating a program We are going to use a programming language called Haskell. Haskell is used throughout the Computer Science course at Oxford. It is very powerful as it allows you write programs that look very similar to mathematical equations. You can easily prove properties about Haskell programs.

  33. Step 1 We need produce a list of factors for a number n – call this list factor(n). A simple way is to check whether each number d between 1 and n is a factor of n. We do this by checking what the remainder is when we divide n by d. If the remainder is 0 then d is a factor of n. We are done when d=n. We create factor lists for both numbers.

  34. Function for Step 1

  35. Step 2 Now that we have our factor lists, which we will call f1 and f2, we create a list of common factors. We do this by looking at all the numbers in f1 to see if they are in f2. We there are no more numbers in f1 then we are done. Call this function: common(f1,f2).

  36. Function for Step 2

  37. Step 3 Now that we have a list of common factors we now check which number in our list is the biggest. We do this by going through the list remembering which is the biggest number that we have seen so far. Call this function: highest(list).

  38. Function for Step 3 If list is empty then return 0, otherwise we check whether the first member of list is higher than the rest of list.

  39. Putting the three steps together To calculate the hcf for two numbers a and b, we just follow the three steps in order. So, in Haskell, we can define Remember that when composing functions, we do the innermost operation first.

  40. Problems with this method Although this method is fairly easy to explain, it is quite slow for large numbers. It also wastes quite a lot of space calculating the factors of both numbers when we only need one of them. Can we think of any ways to improve this method?

  41. Possible improvements Remember factors occur in pairs so that we actually find two factors at the same time. If we find the factors in pairs then we only need to check up to n. We could combine common and highest to find the hcf more quickly (this kind of technique is called fusion). Could use prime numbers.

  42. A Faster Algorithm This algorithm was apparently first given by the famous mathematician Euclid around 300 BC.

  43. An example of this algorithm The algorithm works because any factor of both a and b is also a factor of a – b hcf(308,1001) = hcf(308,693) = hcf(308,385) = hcf(308,77) = hcf(231,77) = hcf(154,77) = hcf(77,77) = 77

  44. Writing this algorithm in Haskell

  45. An even faster algorithm hcf(1001,308) 1001 = 3 × 308 + 77 = hcf(308,77) 308 = 4 × 77 = hcf(77,0) = 77

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