Very good post from Paper Trail:
So why study computer science? The job prospects at the end are usually pretty good – because, if nothing else, you can become a pretty good software developer fairly quickly – but they are unknown. My argument is that the study of computer science is enough of an incentive enough to make it worthwhile.
A recent article written by Dr. Robert B.K. Dewar and Dr. Edmond Schonberg (both from AdaCore Inc.) is generating some discussion on the state of Computer Science (CS) education in the United States. In “Computer Science Education: Where Are the Software Engineers of Tomorrow?“, Dewar and Schonberg claim that U.S. universities are training unqualified and easily replaceable programmers.
“It is our view that Computer Science (CS) education is neglecting basic skills, in particular in the areas of programming and formal methods. We consider that the general adoption of Java as a first programming language is in part responsible for this decline. We examine briefly the set of programming skills that should be part of every software professional’s repertoire.”
The comment about Java’s adoption annoyed some Java aficionados, but in a recent interview, Robert Dewar adds that the problem goes far beyond the choice of Java as the first programming language. The real problem is that CS programs are being dumbed down, so that they become more accessible and popular. In result, they “are not rigorous enough and don’t promote in-depth thinking and problem solving”.
“A lot of it is, ‘Let’s make this all more fun.’ You know, ‘Math is not fun, let’s reduce math requirements. Algorithms are not fun, let’s get rid of them. Ewww – graphic libraries, they’re fun. Let’s have people mess with libraries. And [forget] all this business about ‘command line’ – we’ll have people use nice visual interfaces where they can point and click and do fancy graphic stuff and have fun.”
Although the paper is concerned with the American reality, I believe we have the same problem in Europe — at least, and as far as I know, in the UK and in Portugal. However, in my opinion, the problem starts before university. The maths’s programs in secondary schools are also being simplified (or dumbed down, if you prefer) and many important concepts, like logic and proofs, are being ignored.
In result, first-year students usually have a poor background on maths and problem solving. In fact, most of them have never seen a proof and don’t even understand the importance of mathematical reasoning. With poor reasoning abilities, they become intellectually less curious, accepting things as they are presented, and they have tremendous difficulties creating new algorithms, or convincing someone that their own algorithms are correct.
Moreover, once they are in the university, one of two things happens:
Last November I have solved Problem 15 of Project Euler (a counting problem involving paths in square grids), and, although the problem admits a simple solution, some of the solutions presented in their forums are very complicated. Thus, I thought it would be a good idea to present my solution, as I consider it very simple.
Starting in the top left corner of a $2{\times}2$ grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a $20{\times}20$ grid?
In order to make the problem more interesting, let us investigate the more general problem of counting the number of routes in an $n{\times}n$ grid. Our argument is based on three observations:
From these three observations, we can transform the problem to the following:
How many different strings of size $2{\times}n$, consisting of $n$ Rs and n Ds, are there?
The solution is now very simple, because the positioning of $n$ Ds (or Rs) determines the positioning of the other $n$ Rs (or Ds). Hence, the number we are interested in is the number of ways in which we can choose $n$ positions from $2{\times}n$ available positions. The answer, using the traditional notation for the binomial coefficient, is:
\[
{2n \choose n} = \frac{(2n)!}{n! \times n!}~~~~.
\]
Instantiating n with 20, we get the answer to the initial problem of the $20{\times}20$ grid.
The generalization to an $m{\times}n$ grid is also simple. The only difference is that the strings have length $m+n$. Using the same reasoning as above, the number of paths through an ${m{\times}n}$ grid is:
\[
{m+n \choose n} = \frac{(m+n)!}{m!\times n!}~~~~.
\]
Final note: If you want to access the forum of the problem, you have to solve it.