Last time, I began working on the Project Euler problems and I set out guidelines for myself that I would attempt the problem first in C# (that I’m most comfortable with) and then try to solve the problem in some new way. Last time, I used some of the LINQ extension methods that I hadn’t previously used before. Some solutions might be in F#, Ruby, Python, etc, and I might use each “new way” more than once (since 10 lines of code doesn’t make me an expert!).
I present Project Euler Problem Two. Here is the code for the C# solution using LINQ.
1: /// <summary>
2: /// Each new term in the Fibonacci sequence is generated by
3: /// adding the previous two terms.
4: /// By starting with 1 and 2, the first 10 terms will be:
5: /// 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
6: /// Find the sum of all the even-valued terms in the sequence
7: /// which do not exceed four million.
8: ///
9: /// Credit to Bill Wagner for the solution.
10: /// http://tinyurl.com/4846a5
11: /// I knew LINQ had to have a cool answer and I learned a lot
12: /// about what LINQ can do from this little snippet of code.
13: /// </summary>
14: public class ProjectEulerProblemTwo
15: {
16: private Dictionary<int, long> cachedFibonacci =
17: new Dictionary<int, long>();
18:
19: public long SolveProblem()
20: {
21: var evens = (from n in Enumerable.Range(0, 4000000)
22: let value = ComputeFibonacci(n)
23: where (value%2) == 0
24: select ComputeFibonacci(n)).TakeWhile(n => n <= 4000000);
25:
26: return evens.Sum();
27: }
28:
29: public long ComputeFibonacci(int term)
30: {
31: long answer;
32:
33: if (cachedFibonacci.TryGetValue(term, out answer))
34: {
35: return answer;
36: }
37:
38: if (term < 2)
39: {
40: answer = term;
41: }
42: else
43: {
44: answer = ComputeFibonacci(term - 1) + ComputeFibonacci(term - 2);
45: }
46:
47: cachedFibonacci.Add(term, answer);
48:
49: return answer;
50: }
51: }
As I note in the comments, this is basically Bill Wagner’s solution to this problem. His solution was so ingenious and taught me some things about LINQ that I just had to work it in here. The compute Fibonacci method was created because other Project Euler problems in the future are going to use the Fibonacci sequence, so we should have a reproducible way to create the series. I like his use of the dictionary to be able to ask for any point in the series at any time and you will only have to compute as terms that you have not yet calculated. Caching is a good thing.
The thing that was new to me was the let keyword in the LINQ query. What Bill does is store the value of the current n into value and then says that we only want it if it is even. This allows only the even values to be used. The rest of the problem is very simple, but I just wanted to point out those two points of interest.
I also tried this problem in Ruby. You can go to an interactive ruby session in your web browser here. If you type in the following code, you can see that it also works as well.
def computeFibonacci(firstTerm, secondTerm) return firstTerm + secondTerm end firstTerm = 0 secondTerm = 1 answer = 0 while secondTerm < 4000000 newTerm = computeFibonacci(firstTerm, secondTerm) if secondTerm % 2 == 0 answer += secondTerm end firstTerm = secondTerm secondTerm = newTerm end print "Answer is ", answer
There is nothing special about this code. No whiz-bang code maneuvers. Just straight forward Ruby code. I found it very helpful to test it out at the link above and see that it worked. What is funny to me (and will likely get me in trouble) is how similar the basics of Ruby are to VB (or just BASIC) in general. It is easy to pick up because it is familiar to so many people. It was a good introductory exercise into Ruby, however, so I thought that I would share it with you.
If you have any comments or questions, please feel free to leave me a comment here.
(From their website) Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
I decided to take a crack or two at these, even though a trillion people have done these before me (32,022 registered answers as of the time that I’m writing this). My goal is to try to do this in whatever way occurs to me at first, and then to try to solve the problem using some new technique. Over the course of solving these problems, I want to try to dig a little into F#, Ruby, C# 3.0, or whatever seems like it might be cool 
Today, I attack Problem One. I’ve included the answer in the comments, so don’t read on if you don’t want to know.
namespace ProjectEuler
{
///
/// Problem 1
/// 05 October 2001
/// If we list all the natural numbers below 10
/// that are multiples of 3 or 5, we get 3, 5, 6
/// and 9. The sum of these multiples is 23.
///
/// Find the sum of all the multiples of 3 or 5 below 1000.
///
/// The answer is 233168
///
public class Problem1
{
public int SolveProblem()
{
int answer = 0;
for (int i = 1; i < 1000; i++)
{
if (IsMultipleOfThreeOrFive(i))
{
answer += i;
}
}
return answer;
}
public static bool IsMultipleOfThreeOrFive(int number)
{
return (((number % 3) == 0) || ((number % 5) == 0));
}
}
}
Okay, that was a snoozefest. I was reading in the Pro LINQ book (the one that gave me this post) and I found Enumerable.Range. Do the wonders of LINQ never cease? Enumerable.Range will generate a range of numbers for you so that you don't have to do a loop. Then, you could use that range to then perform a lambda on (the divisible by 3 or 5 check) and then just sum the remaining list. It turns out that this kind of thing is *exactly* what LINQ will just knock out for you. Check the resulting code.
using System.Linq;
namespace ProjectEuler
{
///
/// Problem 1
/// 05 October 2001
/// If we list all the natural numbers below 10
/// that are multiples of 3 or 5, we get 3, 5, 6
/// and 9. The sum of these multiples is 23.
///
/// Find the sum of all the multiples of 3 or 5 below 1000.
///
/// The answer is 233168
///
public class Problem1
{
public int SolveProblem()
{
var problemSet = from i in
Enumerable.Range(1, 999)
where (((i % 3).Equals(0)) || ((i % 5).Equals(0)))
select i;
return problemSet.Sum();
}
}
}
Same answer? Check. Learned and used something new in the process? Check. Looks like I win the nerd prize!
