Computer Science at Oxford
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Transcript Computer Science at Oxford
Computer Science at Oxford
Stephen Drape
Access/Schools Liaison Officer
Four myths about Oxford
There’s little chance of getting in
It’s expensive
College choice is very important
You have to be very bright
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Myth 1: Little chance of getting in
False!
Statistically: you have a 20–40% chance
Admissions data for 2007 entry:
Applications
Acceptances
%
Comp Sci
82
24
29.3%
Maths & CS
52
16
30.8%
Maths
828
173
20.9%
Maths & Stats
143
29
20.3%
Physics
695
170
24.5%
Chemistry
507
190
37.5%
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Myth 2: It’s very expensive
False!
Most colleges provide cheap accommodation for
three years.
College libraries and dining halls also help you
save money.
Increasingly, bursaries help students from poorer
backgrounds.
Most colleges and departments are very close to
the city centre – low transport costs!
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Myth 3: College Choice Matters
False!
If the college you choose is unable to offer you a
place because of space constraints, they will pass
your application on to a second, computerallocated college.
Application loads are intelligently redistributed in
this way.
Lectures are given centrally by the department as
are many classes for courses in later years.
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Myth 3: College Choice Matters
However…
Choose a college that you like
as you have to live and work
there for 3 or 4 years – visit if
possible.
Look at accommodation &
facilities offered.
Choose a college that has a
tutor in your subject.
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Myth 4: You have to be bright
True!
We find it takes special qualities to benefit from
the kind of teaching we provide.
So we are looking for the very best in ability and
motivation.
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Our courses
Computer Science
Computer Science firmly based on Mathematics
Mathematics and Computer Science
Closer to a half/half split between CS and Maths
Courses can be for three or four years
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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
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Admissions Process
Fill in UCAS 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
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Computer Science
Year 1
Year 2
Year 3
Year 4
Mathematics
Computing
Project work
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Mathematics & Computer Science
Year 1
Year 2
Year 3
Year 4
Mathematics
Computing
Project work
12
Some of the different courses
Functional Programming
Design and Analysis of Algorithms
Imperative Programming
Digital Hardware
Calculus
Linear Algebra
Logic and Proof
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Useful Sources of Information
Admissions:
www.admissions.ox.ac.uk/
Computing Laboratory:
www.comlab.ox.ac.uk/admissions/
Mike Spivey’s info pages:
web2.comlab.ox.ac.uk/oucl/prospective/ugrad/csatox/
College admissions tutors
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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.
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Simple Design Methodology
Try a simple version first
Produce some test cases
Prove it correct
Consider efficiency (time taken and space
needed)
Make improvements (called refinements)
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Fibonacci Numbers
The first 10 Fibonacci numbers (from 1) are:
1,1,2,3,5,8,13,21,34,55
The Fibonacci numbers occurs in nature, for
example: plant structures, population numbers.
Named after Leonardo of Pisa
who was nicked named “Fibonacci”
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The rule for Fibonacci
The next number in the sequence is worked out
by adding the previous two terms.
1,1,2,3,5,8,13,21,34,55
The next numbers are therefore
34 + 55 = 89
55 + 89 = 144
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Using algebra
To work out the nth Fibonacci number, which
we’ll call fib(n), we have the rule:
fib(n) = fib(n – 1) + fib(n – 2)
We also need base cases:
fib(0) = 0
fib(1) = 1
This sequence is defined using previous terms
of the sequence – it is an example of a
recursive definition.
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Properties
The sequence has a relationship with the
Golden Ratio
Fibonacci numbers have a variety of properties
such as
fib(5n) is always a multiple of 5
in fact, fib(a£b) is always a multiple of fib(a)
and fib(b)
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Writing a computer program
Using a language called Haskell, we can write
the following function:
> fib(0) = 0
> fib(1) = 1
> fib(n) = fib(n-1) + fib(n-2)
which looks very similar to our algebraic
definition
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Working out an example
Suppose we want to find fib(5)
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Our program would do this…
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What’s happening?
The program blindly follows the definition of fib,
not remembering any of the other values.
So, for
(fib(3) + fib(2)) + fib(3)
the calculation for fib(3) is worked out twice.
The number of steps needed to work out fib(n)
is proportional to n – it takes exponential time.
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Refinements
Why this program is so inefficient is because at
each step we have two occurrences of fib
(termed recursive calls).
When working out the Fibonacci sequence, we
should keep track of previous values of fib and
make sure that we only have one occurrence of
the function at each stage.
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Writing the new definition
We define
> fibtwo(0) = (0,1)
> fibtwo(n) = (b,a+b)
>
where (a,b) =
fibtwo(n-1)
> newfib(n) = fst(fibtwo(n))
The function fst means take the first number
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Explanation
The function fibtwo actually works out:
fibtwo(n) = (fib(n), fib(n+1))
We have used a technique called tupling – this
means that we keep extra results at each stage
of a calculation.
This version is much more efficient that the
previous one (it is linear time).
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An example of the new function
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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 and techniques.
We should make sure that our algorithms are
correct.
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Comments and Questions
If there’s any comments or questions then
please ask.
There should be staff and students available
to talk to
Thank You
30
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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
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Creating an algorithm
For our example, we had three steps:
1) Find the factors
2) Find those factors in common
3) Find the highest factor in common
These steps allow us to construct an algorithm.
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Creating a program
We are going to use a programming language
called Haskell.
Haskell is used throughout the 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.
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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.
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Function for Step 1
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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).
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Function for Step 2
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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).
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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.
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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.
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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?
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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.
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A Faster Algorithm
This algorithm was apparently first given by the
famous mathematician Euclid around 300 BC.
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An example of this algorithm
hcf(308,1001)
= hcf(308,693)
= hcf(308,385)
= hcf(308,77)
= hcf(231,77)
= hcf(154,77)
= hcf(77,77)
= 77
The algorithm works
because any factor
of a and b is also a
factor of a – b
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Writing this algorithm in Haskell
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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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Computer Science
Year 1
Core CS 1
(75%)
Core Maths
(25%)
Year 2
Core CS 2
(50%)
CS Options
(50%)
Year 3
Year 4
(optional)
CS Options
(25%)
Advanced
Options
(50%)
Project
(25%)
Advanced
Options
(66%)
Project
(33%)
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Mathematics & Computer Science
Year 1
Core 1
(100%)
Year 2
Core 2
(58%)
Maths
Options
(17%)
CS Options
(25%)
Year 3
Maths
Options
(25-75%)
CS Options
(25-75%)
Year 4
(optional)
Maths
Options
(25-75%)
CS Options
(25-75%)
Optional
Project
(33%)
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