Create Fibonacci series using lambda operator

Question

I am trying to solve a question in Project Euler, which is creating a fibonacci series until 4 million, and add the even numbers that come in the series, this is obviously very easy task and I answer it in 2 mins,

int result=2;
int first=1;
int second=2;
int i=2;

while (i < 4000000)
{
    i = first + second;

    if (i % 2 == 0)
    {
       result += i;
    }

    first = second;
    second = i;
 }

 Console.WriteLine(result);

but I want to do it using a lambda expression

My effort is going like

DelType del = (oldVal, newVal) =>((oldVal==0?1:newVal  + newVal==1?2:oldVal+newVal) % 2 == 0) ? oldVal + newVal : 0;

int a=del(0, 1);

Kindly suggest how to get this done

Answer 1

My first answer was having misread the question completely, but now I've reread it (thanks MagnatLU!) I'd suggest that this isn't a good fit for lambda expressions. However, it's a brilliant fit for a combination of iterator blocks and LINQ:

// Using long just to avoid having to change if we want a higher limit :)
public static IEnumerable<long> Fibonacci()
{
    long current = 0;
    long next = 1;
    while (true)
    {
        yield return current;
        long temp = next;
        next = current + next;
        current = temp;
    }
}

...

long evenSum = Fibonacci().TakeWhile(x => x < 4000000L)
                          .Where(x => x % 2L == 0L)
                          .Sum();

Answer 2

Use this recursive function

Func<int, int> fib = null;
fib = (x) => x > 1 ? fib(x-1) + fib(x-2) : x;

Example Usage:

Console.WriteLine(fib(10));

Answer 3

Here's oneliner as a lambda expression:

Func<int, int, int, int, int> fib = null;
fib = (n, a, b, res) => n == 0 ? res : fib(n - 1, b, a + b, n % 2 == 0 ? res : res + a + b);
// usage: fib(n, 1, 0, 0)

It uses O(n) stack space and O(n) time on x86 and O(1) stack space on x64 (due to tail recursion optimisation on x64 JIT), so it will fail for n=400000 on 32-bit system.

Edit: it is counting even elements of series from the end, not the beginning, but you should get the idea how to do it as λ with tailrec.

Answer 4

Similar to the Func two-liner, now with local functions it can be done like this:

int Fibonacci(int i) => i <= 1 ? i : Fibonacci(i - 1) + Fibonacci(i - 2);

Answer 5

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;

public class Fibonacci : IEnumerable<int>{
    delegate Tuple<int,int> update(Tuple<int,int> x);
    update func = ( x ) => Tuple.Create(x.Item2, x.Item1 + x.Item2);

    public IEnumerator<int> GetEnumerator(){
        var x = Tuple.Create<int,int>(0,1);
        while (true){
            yield return x.Item1;
            x = func(x);
        }
    }
    IEnumerator IEnumerable.GetEnumerator() {
        return GetEnumerator();
    }
}

class Sample {
    static public void Main(){
        int result= (new Fibonacci()).TakeWhile(x => x < 4000000).Where(x => x % 2 == 0).Sum();
        Console.WriteLine(result);//4613732
   }
} 

OTHER

public static class Seq<T>{
    public delegate Tuple<T,T> update(Tuple<T,T> x);

    static public IEnumerable<T> unfold(update func, Tuple<T,T> initValue){
        var value = initValue;
        while (true){
            yield return value.Item1;
            value = func(value);
        }
    }
}

class Sample {
    static public void Main(){
        var fib = Seq<int>.unfold( x => Tuple.Create<int,int>(x.Item2, x.Item1 + x.Item2), Tuple.Create<int,int>(0,1));
        int result= fib.TakeWhile(x => x < 4000000).Where(x => x % 2 == 0).Sum();
        Console.WriteLine(result);
   }
} 

Answer 6

I know this is an old question, but I was working on the same problem today and arrived at this concise functional-style solution with O(n) running time:

static int FibTotal(int limit, Func<int, bool> include, int last = 0, int current = 1)
{
    if (current < limit)
        return FibTotal(limit, include, current, last + current) + 
                               (include(current) ? current : 0);
    else
        return 0;
}

You can also get a nice one-line solution if you first define this convenience class (maybe something like this already exists in the .NET framework, but I was unable to find it):

public static class Sequence
{
    public static IEnumerable<T> Generate<T>(T seed, Func<T, T> next)
    {
        while (true)
        {
            yield return seed;
            seed = next(seed);
        }
    }
}

The solution then becomes:

var result = Sequence.Generate(Tuple.Create(1, 1), 
                               t => Tuple.Create(t.Item2, t.Item1 + t.Item2))
                     .Select(t => t.Item1)
                     .TakeWhile(i => i < 4000000)
                     .Where(i=> i % 2 == 0)
                     .Sum();

Answer 7

public static void fibSeriesEx3()
{
    List<int> lst = new List<int> { 0, 1 };
    for (int i = 0; i <= 10; i++)
    {
        int num = lst.Skip(i).Sum();
        lst.Add(num);

        foreach (int number in lst)
            Console.Write(number + " ");
            Console.WriteLine();
    }
}

Answer 8

One liner that works:

Func<int, int, int, int, int> fib = null;
fib = (a, b, counter, n) => counter < n ? fib(b, a + b, counter+1, n) : a;
        //print the 9th fib num
        Console.WriteLine(fib(0,1,1,9)); 

output: 21

Answer 9

You know you can do:

Func<int,int,int> func = (first, second) => {  
                                          var result=2;
                                          int i=2;
                                          while (i < 4000000)
                                          {
                                              i = first + second;
                                              if (i % 2 == 0)
                                              {
                                                  result += i;
                                              }
                                              first = second;
                                              second = i;
                                          }
                                          return result;
                                        };

Answer 10

Just in case you wanted a pure recursive lambda solution, take a look at this answer for a couple of links to articles showing how it's done.

However, that uber stuff is too mad for me, so I'd better follow another answer that is already here.