Related
I'm quite new to programming in C#, and thought that attempting the Euler Problems would be a good idea as a starting foundation. However, I have gotten to a point where I can't seem to get the correct answer for Problem 2.
"Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms."
My code is this:
int i = 1;
int j = 2;
int sum = 0;
while (i < 4000000)
{
if (i < j)
{
i += j;
if (i % 2 == 0)
{
sum += i;
}
}
else
{
j += i;
if (j % 2 == 0)
{
sum += j;
}
}
}
MessageBox.Show("The answer is " + sum);
Basically, I think that I am only getting the last two even numbers of the sequence and adding them - but I don't know how to get all of the even numbers of the sequence and add them. Could someone please help me, whilst trying to progress from my starting point?
P.S. - If there are any really bad layout choices, do say as eliminating these now will help me to become a better programmer in the future :)
Thanks a lot in advance.
I just logged in into my Project Euler account, to see the correct answer. As others say, you forgot to add the initial term 2, but otherwise your code is OK (the correct answer is what your code outputs + 2), so well done!
It is pretty confusing though, I think it would look way clearer if you'd use 3 variables, something like:
int first = 1;
int second = 1;
int newTerm = 0;
int sum = 0;
while (newTerm <= 4000000)
{
newTerm = first + second;
if (newTerm % 2 == 0)
{
sum += newTerm;
}
first = second;
second = newTerm;
}
MessageBox.Show("The answer is " + sum);
You need to set the initial value of sum to 2, as you are not including that in your sum with the current code.
Also, although it may be less efficient memory-usage-wise, I would probably write the code something like this because IMO it's much more readable:
var fibonacci = new List<int>();
fibonacci.Add(1);
fibonacci.Add(2);
var curIndex = 1;
while(fibonacci[curIndex] + fibonacci[curIndex - 1] <= 4000000) {
fibonacci.Add(fibonacci[curIndex] + fibonacci[curIndex - 1]);
curIndex++;
}
var sum = fibonacci.Where(x => x % 2 == 0).Sum();
Use an array fib to store the sequence. Its easier to code and debug. At every iteration, you just need to check if the value is even.
fib[i] = fib[i - 1] + fib[i - 2];
if (fib[i] > 4000000) break;
if (fib[i] % 2 == 0) sum += fib[i];
I've solved this a few years ago, so I don't remember how I did it exactly, but I do have access to the forums talking about it. A few hints and an outright solution. The numbers repeat in a pattern. two odds followed by an even. So you could skip numbers without necessarily doing the modulus operation.
A proposed C# solution is
long sum = 0, i0, i1 = 1, i2 = 2;
do
{
sum += i2;
for (int i = 0; i < 3; i++)
{
i0 = i1;
i1 = i2;
i2 = i1 + i0;
}
} while (i2 < 4000000);
Your code is a bit ugly, since you're alternating between i and j. There is a much easier way to calculate fibonacci numbers by using three variables and keeping their meaning the same all the time.
One related bug is that you only check the end condition on every second iteration and in the wrong place. But you are lucky that the cutoff fit your bug(the number that went over the limit was odd), so this is not your problem.
Another bug is that you check with < instead of <=, but since there is no fibonacci number equal to the cutoff this doesn't cause your problem.
It's not an int overflow either.
What remains is that you forgot to look at the first two elements of the sequence. Only one of which is even, so you need to add 2 to your result.
int sum = 0; => int sum = 2;
Personally I'd write one function that returns the infinite fibonacci sequence and then filter and sum with Linq. Fibonacci().TakeWhile(i=> i<=4000000).Where(i=>i%2==0).Sum()
I used a class to represent a FibonacciNumber, which i think makes the code more readable.
public class FibonacciNumber
{
private readonly int first;
private readonly int second;
public FibonacciNumber()
{
this.first = 0;
this.second = 1;
}
private FibonacciNumber(int first, int second)
{
this.first = first;
this.second = second;
}
public int Number
{
get { return first + second; }
}
public FibonacciNumber Next
{
get
{
return new FibonacciNumber(this.second, this.Number);
}
}
public bool IsMultipleOf2
{
get { return (this.Number % 2 == 0); }
}
}
Perhaps it's a step too far but the end result is a function which reads quite nicely IMHO:
var current = new FibonacciNumber();
var result = 0;
while (current.Number <= max)
{
if (current.IsMultipleOf2)
result += current.Number;
current = current.Next;
}
return result;
However it's not going to be as efficient as the other solutions which aren't newing up classes in a while loop. Depends on your requirements I guess, for me I just wanted to solve the problem and move on to the next one.
Hi i have solved this question, check it if its right.
My code is,
#include<iostream.h>
#include<conio.h>
class euler2
{
unsigned long long int a;
public:
void evensum();
};
void euler2::evensum()
{
a=4000000;
unsigned long long int i;
unsigned long long int u;
unsigned long long int initial=0;
unsigned long long int initial1=1;
unsigned long long int sum=0;
for(i=1;i<=a;i++)
{
u=initial+initial1;
initial=initial1;
initial1=u;
if(u%2==0)
{
sum=sum+u;
}
}
cout<<"sum of even fibonacci numbers upto 400000 is"<<sum;
}
void main()
{
euler2 a;
clrscr();
a.evensum();
getch();
}
Here's my implementation:
public static int evenFibonachi(int n)
{
int EvenSum = 2, firstElem = 1, SecondElem = 2, SumElem=0;
while (SumElem <= n)
{
swich(ref firstElem, ref SecondElem, ref SumElem);
if (SumElem % 2 == 0)
EvenSum += SumElem;
}
return EvenSum;
}
private static void swich(ref int firstElem, ref int secondElem, ref int SumElem)
{
int temp = firstElem;
firstElem = secondElem;
secondElem += temp;
SumElem = firstElem + secondElem;
}
I have been hearing a lot about Project Euler so I thought I solve one of the problems in C#. The problem as stated on the website is as follows:
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.
I wrote my code as follows:
class EulerProblem1
{
public static void Main()
{
var totalNum = 1000;
var counter = 1;
var sum = 0;
while (counter < totalNum)
{
if (DivisibleByThreeOrFive(counter))
sum += counter;
counter++;
}
Console.WriteLine("Total Sum: {0}", sum);
Console.ReadKey();
}
private static bool DivisibleByThreeOrFive(int counter)
{
return ((counter % 3 == 0) || (counter % 5 == 0));
}
}
It will be great to get some ideas on alternate implementations with less verbosity/cleaner syntax and better optimizations. The ideas may vary from quick and dirty to bringing out the cannon to annihilate the mosquito. The purpose is to explore the depths of computer science while trying to improve this particularly trivial code snippet.
Thanks
Updated to not double count numbers that are multiples of both 3 and 5:
int EulerProblem(int totalNum)
{
int a = (totalNum-1)/3;
int b = (totalNum-1)/5;
int c = (totalNum-1)/15;
int d = a*(a+1)/2;
int e = b*(b+1)/2;
int f = c*(c+1)/2;
return 3*d + 5*e - 15*f;
}
With LINQ (updated as suggested in comments)
static void Main(string[] args)
{
var total = Enumerable.Range(0,1000)
.Where(counter => (counter%3 == 0) || (counter%5 == 0))
.Sum();
Console.WriteLine(total);
Console.ReadKey();
}
Here's a transliteration of my original F# solution into C#. Edited: It's basically mbeckish's solution as a loop rather than a function (and I remove the double count). I like mbeckish's better.
static int Euler1 ()
{
int sum = 0;
for (int i=3; i<1000; i+=3) sum+=i;
for (int i=5; i<1000; i+=5) sum+=i;
for (int i=15; i<1000; i+=15) sum-=i;
return sum;
}
Here's the original:
let euler1 d0 d1 n =
(seq {d0..d0..n} |> Seq.sum) +
(seq {d1..d1..n} |> Seq.sum) -
(seq {d0*d1..d0*d1..n} |> Seq.sum)
let result = euler1 3 5 (1000-1)
I haven't written any Java in a while, but this should solve it in constant time with little overhead:
public class EulerProblem1
{
private static final int EULER1 = 233168;
// Equal to the sum of all natural numbers less than 1000
// which are multiples of 3 or 5, inclusive.
public static void main(String[] args)
{
System.out.println(EULER1);
}
}
EDIT: Here's a C implementation, if every instruction counts:
#define STDOUT 1
#define OUT_LENGTH 8
int main (int argc, char **argv)
{
const char out[OUT_LENGTH] = "233168\n";
write(STDOUT, out, OUT_LENGTH);
}
Notes:
There's no error handling on the call to write. If true robustness is needed, a more sophisticated error handling strategy must be employed. Whether the added complexity is worth greater reliability depends on the needs of the user.
If you have memory constraints, you may be able to save a byte by using a straight char array rather than a string terminated by a superfluous null character. In practice, however, out would almost certainly be padded to 8 bytes anyway.
Although the declaration of the out variable could be avoided by placing the string inline in the write call, any real compiler willoptimize away the declaration.
The write syscall is used in preference to puts or similar to avoid the additional overhead. Theoretically, you could invoke the system call directly, perhaps saving a few cycles, but this would raise significant portability issues. Your mileage may vary regarding whether this is an acceptable tradeoff.
Refactoring #mbeckish's very clever solution:
public int eulerProblem(int max) {
int t1 = f(max, 3);
int t2 = f(max, 5);
int t3 = f(max, 3 * 5);
return t1 + t2 - t3;
}
private int f(int max, int n) {
int a = (max - 1) / n;
return n * a * (a + 1) / 2;
}
That's basically the same way I did that problem. I know there were other solutions (probably more efficient ones too) on the forums for project-euler.
Once you input your answer going back to the question gives you the option to go to the forum for that problem. You may want to look there!
The code in DivisibleByThreeOrFive would be slightly faster if you would state it as follows:
return ((counter % 3 == 0) || (counter % 5 == 0));
And if you do not want to rely on the compiler to inline the function call, you could do this yourself by putting this code into the Main routine.
You can come up with a closed form solution for this. The trick is to look for patterns. Try listing out the terms in the sum up to say ten, or twenty and then using algebra to group them. By making appropriate substitutions you can generalize that to numbers other than ten. Just be careful about edge cases.
Try this, in C. It's constant time, and there's only one division (two if the compiler doesn't optimize the div/mod, which it should). I'm sure it's possible to make it a bit more obvious, but this should work.
It basically divides the sum into two parts. The greater part (for N >= 15) is a simple quadratic function that divides N into exact blocks of 15. The lesser part is the last bit that doesn't fit into a block. The latter bit is messier, but there are only a few possibilities, so a LUT will solve it in no time.
const unsigned long N = 1000 - 1;
const unsigned long q = N / 15;
const unsigned long r = N % 15;
const unsigned long rc = N - r;
unsigned long sum = ((q * 105 + 15) * q) >> 1;
switch (r) {
case 3 : sum += 3 + 1*rc ; break;
case 4 : sum += 3 + 1*rc ; break;
case 5 : sum += 8 + 2*rc ; break;
case 6 : sum += 14 + 3*rc ; break;
case 7 : sum += 14 + 3*rc ; break;
case 8 : sum += 14 + 3*rc ; break;
case 9 : sum += 23 + 4*rc ; break;
case 10 : sum += 33 + 5*rc ; break;
case 11 : sum += 33 + 5*rc ; break;
case 12 : sum += 45 + 6*rc ; break;
case 13 : sum += 45 + 6*rc ; break;
case 14 : sum += 45 + 6*rc ; break;
}
You can do something like this:
Func<int,int> Euler = total=>
new List<int>() {3,5}
.Select(m => ((int) (total-1) / m) * m * (((int) (total-1) / m) + 1) / 2)
.Aggregate( (T, m) => T+=m);
You still have the double counting problem. I'll think about this a little more.
Edit:
Here is a working (if slightly inelegant) solution in LINQ:
var li = new List<int>() { 3, 5 };
Func<int, int, int> Summation = (total, m) =>
((int) (total-1) / m) * m * (((int) (total-1) / m) + 1) / 2;
Func<int,int> Euler = total=>
li
.Select(m => Summation(total, m))
.Aggregate((T, m) => T+=m)
- Summation(total, li.Aggregate((T, m) => T*=m));
Can any of you guys improve on this?
Explanation:
Remember the summation formula for a linear progression is n(n+1)/2. In the first case where you have multiples of 3,5 < 10, you want Sum(3+6+9,5). Setting total=10, you make a sequence of the integers 1 .. (int) (total-1)/3, and then sum the sequence and multiply by 3. You can easily see that we're just setting n=(int) (total-1)/3, then using the summation formula and multiplying by 3. A little algebra gives us the formula for the Summation functor.
I like technielogys idea, here's my idea of a modification
static int Euler1 ()
{
int sum = 0;
for (int i=3; i<1000; i+=3)
{
if (i % 5 == 0) continue;
sum+=i;
}
for (int i=5; i<1000; i+=5) sum+=i;
return sum;
}
Though also comes to mind is maybe a minor heuristic, does this make any improvement?
static int Euler1 ()
{
int sum = 0;
for (int i=3; i<1000; i+=3)
{
if (i % 5 == 0) continue;
sum+=i;
}
for (int i=5; i<250; i+=5)
{
sum+=i;
}
for (int i=250; i<500; i+=5)
{
sum+=i;
sum+=i*2;
sum+=(i*2)+5;
}
return sum;
}
Your approach is brute force apprach, The time complexity of the following approach is O(1), Here we
are dividing the given (number-1) by 3, 5 and 15, and store in countNumOf3,countNumOf5, countNumOf15.
Now we can say that 3 will make AP, within the range of given (number-1) with difference of 3.
suppose you are given number is 16, then
3=> 3, 6, 9, 12, 15= sum1=>45
5=> 5, 10, 15 sum2=> 30
15=> 15 => sum3=15
Add sum= sum1 and sum2
Here 15 is multiple of 3 and 5 so remove sum3 form sum, this will be your answer. **sum=sum-
sum3** please check link of my solution on http://ideone.com/beXsam]
import java.util.*;
class Multiplesof3And5 {
public static void main(String [] args){
Scanner scan=new Scanner(System.in);
int num=scan.nextInt();
System.out.println(getSum(num));
}
public static long getSum(int n){
int countNumOf3=(n-1)/3;//
int countNumOf5=(n-1)/5;
int countNumOf15=(n-1)/15;
long sum=0;
sum=sumOfAP(3,countNumOf3,3)+sumOfAP(5,countNumOf5,5)-sumOfAP(15,countNumOf15,15);
return sum;
}
public static int sumOfAP(int a, int n, int d){
return (n*(2*a +(n -1)*d))/2;
}
}
new List<int>{3,5}.SelectMany(n =>Enumerable.Range(1,999/n).Select(i=>i*n))
.Distinct()
.Sum()
[Update] (In response to the comment asking to explain this algorothm)
This builds a flattened list of multiples for each base value (3 and 5 in this case), then removes duplicates (e.g where a multiple is divisible, in this case, by 3*5 =15) and then sums the remaining values. (Also this is easily generalisable for having more than two base values IMHO compared to any of the other solutions I have seen here.)
I'm having some trouble with this problem in Project Euler.
Here's what the question asks:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
Find the sum of all the even-valued terms in the sequence which do not exceed four million.
My code so far: EDITED WITH NEW CODE THAT STILL DOESN'T WORK.
static void Main(string[] args)
{
int a = 1;
int b = 2;
int Container = 0;
int Sum = 0;
while (b < 4000000)
{
if (a % 2 == 0)
{
Container += a;
}
Sum = a + b;
a = b;
b = Sum;
}
Container += b;
Console.WriteLine(Container.ToString());
Console.ReadLine();
}
One of the fun feature in C# is the "yield" keyword, which is very useful for this kind of thing:
IEnumerable<int> Fibonacci()
{
int n1 = 0;
int n2 = 1;
yield return 1;
while (true)
{
int n = n1 + n2;
n1 = n2;
n2 = n;
yield return n;
}
}
long result=0;
foreach (int i in Fibonacci().TakeWhile(i => i<4000000).Where(i => i % 2 == 0))
{
result+=i;
}
Console.WriteLine(result);
The "traditional" recursive Fibonacci implementation is problematic here because it throws away all the work done along the way to the last requested term. You would have to call such a function over and over in a loop, which would duplicate a lot of work, or you could start with that implementation and add an argument to the recursive function to build up the desired sum result as the final fibonacci term is calculated. I like this much better, because it's still a general purpose fibonacci sequence at the core, rather than one you had to re-write or specialize.
Another approach is to use events (delegates) in a traditional implementation to call a separate method as each term is completed, but as I still like the iterator method better I'll leave the delegate option as an exercise for the reader.
Your problem is not that your code contains a bug; your problem is that your code contains a bug and you don't know how to find it. Solve the second problem first, and then you won't need to ask us when you have a bug, you'll be able to find it yourself.
Learning how to find bugs is hard and takes a lot of practice. Here's how I would approach this problem.
I'd start by simplifying the problem down to something I could do myself. Instead of "what's the sum of the even fib numbers that do not exceed four million?" I'd ask "what's the sum of the even fib numbers that do not exceed 40?" That's easy to work out by hand -- 2 + 8 + 34 = 44.
Now run your program in a debugger, stepping through each line, and see where things go wrong. Does your program actually add up 2, 8 and 34? And if so, does it get the correct result?
int sum = 2;
for(int f1 = 1, f2 = 2, f3 = 0; !((f3 = (f1 + f2)) > 4000000); f1 = f2, f2 = f3)
sum += f3 * (~f3 & 1);
I think the question is written to say that you would add all the even numbers together while the numbers in the sequence don't exceed four million, meaning you would add 3,999,992.
You're checking both a and b on every iteration. So that means you're double counting almost everything.
Edit:
Ok, I see your update. This is pretty basic debugging, and you should really learn to try it yourself. Think about what the values of a and b are when your loop condition stops being true.
Joel, I wrote a very some similiar code; I'm posting it anyways:
static IEnumerable<int> Fibonacci(int maximum)
{
int auxiliar = 0;
int previous = 0;
int current = 1;
while (current < maximum)
{
auxiliar = previous;
previous = current;
current = auxiliar + current;
yield return current;
}
}
Console.WriteLine(Fibonacci(4000000).Where(number => number % 2 == 0).Sum());
The trickier way:
//1: Allow declaring of recursive functions
private delegate Func<T, R> FuncRec<T, R>(FuncRec<T, R> f);
static Func<T, R> RecFunction<T, R>(Func<Func<T, R>, Func<T, R>> f)
{
FuncRec<T, R> funcRec = r => t => f(r(r))(t);
return funcRec(funcRec);
}
//Define the factorial function
public static readonly Func<ulong, ulong> Fibonacci
= RecFunction<UInt64, UInt64>(fib => n =>
(n == 1 || n == 0)
? n
: fib(n - 1) + fib(n - 2));
//Make a "continous" version
static IEnumerable<ulong> ContinousFibonacci()
{
ulong count = 0;
while(true)
{
ulong n = Fibonacci(count);
count++;
yield return n;
}
}
//Linq result
static void Main(string[] args)
{
ulong result = ContinousFibonacci()
.TakeWhile(r => r < 4000000)
.Where(IsEven)
.Aggregate<ulong, ulong>(0,(current, s) => (s + current));
Console.WriteLine(result);
Console.ReadLine();
}
///The Functional-Style method of allowing one to create recursive functions such as above was made by Bart De Smet. See http://bartdesmet.net/blogs/bart/archive/2009/11/08/jumping-the-trampoline-in-c-stack-friendly-recursion.aspx
Here's a nice way to find Fibonnaci numbers.
IEnumerable<BigInteger> Fibs()
{
for(BigInteger a = 0,b = 1;;b = a + (a = b))
yield return b;
}
// count(user input) of Fibonacci numbers
int[] array = new int[20];
array[0] = 0;
array[1] = 1;
Console.WriteLine(array[0] + "\n" + array[1]);
for (int i = 2; i < 20; i++)
{
array[i] = array[i - 1] + array[i - 2];
Console.WriteLine(array[i]);
}
I am trying to write a method to calculate the sum of the odd numbers in all the numbers less than the given number. so eg. CalcOdd(7) would return 5 + 3 + 1 = 9. CalcOdd (10) would return 9 + 7 + 5 + 3 + 1 = 25 etc
The method needs to take in a number, subtract 1, then recursively work backwards adding all odd numbers until it reaches 0. This is what I have so far.
private static int CalcOdd(int n)
{
if (n <= 1)
return 1;
else
if (n % 2 == 0)
n--;
return n + CalcOdd(n - 2);
}
It doesn't work so well, it includes the number passed in in the addition which is not what I want. Can anyone suggest a better way of doing this ? I would also loke to be able to port the answer to work for even numbers and add the option to include the original passed in number in the answer.
Many thanks
Why would you use recursion here? Just loop; or better, figure out the math to do it in a simple equation...
The fact is that C# doesn't make for excellent deep recursion for things like maths; the tail-call isn't really there at the moment.
Loop approach:
private static int CalcOdd(int n)
{
int sum = 0, i = 1;
while (i < n)
{
sum += i;
i += 2;
}
return sum;
}
You could do this with recursion as you say, but if you wish to do it quicker, then I can tell you that the sum of the n first odd numbers is equal to n*n.
private static int CalcOdd(int n) {
if (n<=1)
return 0;
if (n%2 == 1)
n--;
int k = n/2;
return k*k;
}
The reason this works is:
Every even number is of the form 2k, and the odd number before it is 2k-1.
Because 2*1-1 = 1, there are k odd numbers below 2k.
If n is odd, we don't want to include it, so we simply go down to the even number below it and we automatically have what we want.
Edited to fix broken code.
the sum of odd numbers less than a given number is a perfect square.
get the whole part of (n/2) to get the number of odd number less than itself.
square that and voila!
private static int CalcSumOdd(int n)
{
int i;
int.tryParse(n / 2, out i);
return i*i;
}
for even numbers its:
int i = n/2;
return i*(i+1);
correction. The above "even number sum" includes the original number "n". ie fn(12) = 42 = 2 + 4 + 6 + 8 + 10 + 12
if you want to exclude it, you should either unilaterally exclude it, or remove it with logic based on a passed in parameter.
Here is a correction,
int CalcOdd(int n)
{
n--; // <----
if (n <= 1)
return 0; // <----
else
if (n % 2 == 0)
n--;
return n + CalcOdd(n); // <----
}
i'm new here but this seems like a silly recursion exercise, given it can be done with a simple equation:
int sum(n,isEven,notFirst) {
int c=1; //skip the else
if (isEven) c=2;
if (notFirst) n-=2;
return ((n+c)*((n+c)/2))/2; }
classic discrete math sum series..
sum from 1 to 100 (odds and evens) is ((100+1)*(100/2))=5050
edit: in my code here, if you're calculating the sum of odds with n being even, or vice versa, it doesn't work, but i'm not going to put the work into that (and slop the code) right now. i'll assume your code will take care of that by the time it hits the function.. for example 7/2 isn't an int (obviously)
Why use recursion?
private Int32 CalcOdd(Int32 value)
{
Int32 r = 0;
{
while (value >= 1)
{
value--;
if (value % 2 != 0)
{
r += value;
}
}
}
return r;
}
Use a helper function. CalcOdd consists of testing n to see if it is even or odd; if it is even, return helper(n); if it is odd, return helper(n-2).
The helper function must handle three cases:
1) n is less than 1; in this case return 0.
2) n is even, in this case return helper(n-1).
3) n is odd, in this case return n+helper(n-1).
public static int CalcOdd(int n) {
// Find the highest even number. (Either n, or n-1.)
// Divide that by 2, and the answer should be the square of that number.
n = (n & 0x3FFFFFFE) >> 1;
return (int)Math.Pow(n, 2);
}
private static int CalcOdd(int n) {
n -= 1;
if ((n & 1) == 0) n--;
if (n <= 1) return 1;
return n + CalcOdd(n - 1);
}
But I would say doing loops is better and cleaner.
private static int CalcOdd(int n) {
int i, r = 1;
for (i = 3; i < n; i+=2)
r += i;
return r;
}
Since you want the option of including or excluding the first answer (and, keeping your "recursion" constraint in mind):
int calcOdd(int n, bool includeN)
{
if( !includeN )
return calcOdd(n-1, true);
if(n<=1)
return 1;
else
if(n%2 == 0)
n--;
return n+calcOdd(n-1, true);
}
The includeFirst, if passed as true, will include n in the calculations. Otherwise, the next layer down will start "including N".
Granted, as others have said, this is a horribly inefficient use of recursion, but... If you like recursion, try Haskell. It's a language built almost entirely on the concept.
int CalcOdd(int n)
{
n -= 1;
if (n <= 0)
return 0;
if (n % 2 == 0)
n--;
return n + CalcOdd(n);
}
This function is also recursive, and it has parameters which makes you able to decide wether to do even or odd number and wether you want to include the first number or not. If you are confused as to how it works, remember that bools also can be seen as 1 (true) and 0 (false)
int Calc(int n, bool even = false, bool includeFirst = false)
{
n -= !includeFirst;
if (n <= 0)
return 0;
if (n % 2 == even)
n--;
return n + Calc(n - includeFirst, even);
}
Håkon, I have ported your code to c# in VS 2008 as follows
static int Calc(int n, bool bEven, bool bIncludeFirst)
{
int iEven = Bool2Int(bEven);
int iIncludeFirst = Bool2Int(bIncludeFirst);
n -= 1 - iIncludeFirst;
if (n <= 0)
return 0;
if (n % 2 == iEven)
n--;
return n + Calc(n - iIncludeFirst, bEven, bIncludeFirst);
}
private static int Bool2Int(bool b)
{
return b ? 1 : 0;
}
It seems to be working. Now is there anything I can do to optomise ? i.e. I dont want to have to parse those bools to ints every time etc ?
I'd isolate the 'make it odd' part from the 'sum every other descending number' part: (forgive the Python)
def sumEveryTwoRecursive(n):
if n <= 0:
return 0
return n + sumEveryTwoRecursive(n - 2)
def calcOdd(n):
return sumEveryTwoRecursive(n - (2 if n % 2 else 1))
Just because there isn't one here yet, I've decided to use the LINQ hammer on this nail...
(borrowed from Nick D and Jason's pair programmed answer here)
void Main()
{
GetIterator(7, true, false).Sum().Dump();
// Returns 9
GetIterator(10, true, false).Sum().Dump();
// Returns 25
}
public IEnumerable<int> GetIterator(int n, bool isOdd, bool includeOriginal)
{
if (includeOriginal)
n++;
if (isOdd)
return GetIterator(n, 1);
else
return GetIterator(n, 0);
}
public IEnumerable<int> GetIterator(int n, int odd)
{
n--;
if (n < 0)
yield break;
if (n % 2 == odd)
yield return n;
foreach (int i in GetIterator(n, odd))
yield return i;
}
#include <iostream>
using namespace std;
int sumofodd(int num);
int main()
{
int number,res;
cin>>number;
res=sumofodd(number);
cout<<res;
return 0;
}
int sumofodd(int num)
{ if(num<1) return 0;
if (num%2==0) num--;
return num+sumofodd(num-1);
}
All numbers that divide evenly into x.
I put in 4 it returns: 4, 2, 1
edit: I know it sounds homeworky. I'm writing a little app to populate some product tables with semi random test data. Two of the properties are ItemMaximum and Item Multiplier. I need to make sure that the multiplier does not create an illogical situation where buying 1 more item would put the order over the maximum allowed. Thus the factors will give a list of valid values for my test data.
edit++:
This is what I went with after all the help from everyone. Thanks again!
edit#: I wrote 3 different versions to see which I liked better and tested them against factoring small numbers and very large numbers. I'll paste the results.
static IEnumerable<int> GetFactors2(int n)
{
return from a in Enumerable.Range(1, n)
where n % a == 0
select a;
}
private IEnumerable<int> GetFactors3(int x)
{
for (int factor = 1; factor * factor <= x; factor++)
{
if (x % factor == 0)
{
yield return factor;
if (factor * factor != x)
yield return x / factor;
}
}
}
private IEnumerable<int> GetFactors1(int x)
{
int max = (int)Math.Ceiling(Math.Sqrt(x));
for (int factor = 1; factor < max; factor++)
{
if(x % factor == 0)
{
yield return factor;
if(factor != max)
yield return x / factor;
}
}
}
In ticks.
When factoring the number 20, 5 times each:
GetFactors1-5,445,881
GetFactors2-4,308,234
GetFactors3-2,913,659
When factoring the number 20000, 5 times each:
GetFactors1-5,644,457
GetFactors2-12,117,938
GetFactors3-3,108,182
pseudocode:
Loop from 1 to the square root of the number, call the index "i".
if number mod i is 0, add i and number / i to the list of factors.
realocode:
public List<int> Factor(int number)
{
var factors = new List<int>();
int max = (int)Math.Sqrt(number); // Round down
for (int factor = 1; factor <= max; ++factor) // Test from 1 to the square root, or the int below it, inclusive.
{
if (number % factor == 0)
{
factors.Add(factor);
if (factor != number/factor) // Don't add the square root twice! Thanks Jon
factors.Add(number/factor);
}
}
return factors;
}
As Jon Skeet mentioned, you could implement this as an IEnumerable<int> as well - use yield instead of adding to a list. The advantage with List<int> is that it could be sorted before return if required. Then again, you could get a sorted enumerator with a hybrid approach, yielding the first factor and storing the second one in each iteration of the loop, then yielding each value that was stored in reverse order.
You will also want to do something to handle the case where a negative number passed into the function.
The % (remainder) operator is the one to use here. If x % y == 0 then x is divisible by y. (Assuming 0 < y <= x)
I'd personally implement this as a method returning an IEnumerable<int> using an iterator block.
Very late but the accepted answer (a while back) didn't not give the correct results.
Thanks to Merlyn, I got now got the reason for the square as a 'max' below the corrected sample. althought the answer from Echostorm seems more complete.
public static IEnumerable<uint> GetFactors(uint x)
{
for (uint i = 1; i * i <= x; i++)
{
if (x % i == 0)
{
yield return i;
if (i != x / i)
yield return x / i;
}
}
}
As extension methods:
public static bool Divides(this int potentialFactor, int i)
{
return i % potentialFactor == 0;
}
public static IEnumerable<int> Factors(this int i)
{
return from potentialFactor in Enumerable.Range(1, i)
where potentialFactor.Divides(i)
select potentialFactor;
}
Here's an example of usage:
foreach (int i in 4.Factors())
{
Console.WriteLine(i);
}
Note that I have optimized for clarity, not for performance. For large values of i this algorithm can take a long time.
Another LINQ style and tying to keep the O(sqrt(n)) complexity
static IEnumerable<int> GetFactors(int n)
{
Debug.Assert(n >= 1);
var pairList = from i in Enumerable.Range(1, (int)(Math.Round(Math.Sqrt(n) + 1)))
where n % i == 0
select new { A = i, B = n / i };
foreach(var pair in pairList)
{
yield return pair.A;
yield return pair.B;
}
}
Here it is again, only counting to the square root, as others mentioned. I suppose that people are attracted to that idea if you're hoping to improve performance. I'd rather write elegant code first, and optimize for performance later, after testing my software.
Still, for reference, here it is:
public static bool Divides(this int potentialFactor, int i)
{
return i % potentialFactor == 0;
}
public static IEnumerable<int> Factors(this int i)
{
foreach (int result in from potentialFactor in Enumerable.Range(1, (int)Math.Sqrt(i))
where potentialFactor.Divides(i)
select potentialFactor)
{
yield return result;
if (i / result != result)
{
yield return i / result;
}
}
}
Not only is the result considerably less readable, but the factors come out of order this way, too.
I did it the lazy way. I don't know much, but I've been told that simplicity can sometimes imply elegance. This is one possible way to do it:
public static IEnumerable<int> GetDivisors(int number)
{
var searched = Enumerable.Range(1, number)
.Where((x) => number % x == 0)
.Select(x => number / x);
foreach (var s in searched)
yield return s;
}
EDIT: As Kraang Prime pointed out, this function cannot exceed the limit of an integer and is (admittedly) not the most efficient way to handle this problem.
Wouldn't it also make sense to start at 2 and head towards an upper limit value that's continuously being recalculated based on the number you've just checked? See N/i (where N is the Number you're trying to find the factor of and i is the current number to check...) Ideally, instead of mod, you would use a divide function that returns N/i as well as any remainder it might have. That way you're performing one divide operation to recreate your upper bound as well as the remainder you'll check for even division.
Math.DivRem
http://msdn.microsoft.com/en-us/library/wwc1t3y1.aspx
If you use doubles, the following works: use a for loop iterating from 1 up to the number you want to factor. In each iteration, divide the number to be factored by i. If (number / i) % 1 == 0, then i is a factor, as is the quotient of number / i. Put one or both of these in a list, and you have all of the factors.
And one more solution. Not sure if it has any advantages other than being readable..:
List<int> GetFactors(int n)
{
var f = new List<int>() { 1 }; // adding trivial factor, optional
int m = n;
int i = 2;
while (m > 1)
{
if (m % i == 0)
{
f.Add(i);
m /= i;
}
else i++;
}
// f.Add(n); // adding trivial factor, optional
return f;
}
I came here just looking for a solution to this problem for myself. After examining the previous replies I figured it would be fair to toss out an answer of my own even if I might be a bit late to the party.
The maximum number of factors of a number will be no more than one half of that number.There is no need to deal with floating point values or transcendent operations like a square root. Additionally finding one factor of a number automatically finds another. Just find one and you can return both by just dividing the original number by the found one.
I doubt I'll need to use checks for my own implementation but I'm including them just for completeness (at least partially).
public static IEnumerable<int>Factors(int Num)
{
int ToFactor = Num;
if(ToFactor == 0)
{ // Zero has only itself and one as factors but this can't be discovered through division
// obviously.
yield return 0;
return 1;
}
if(ToFactor < 0)
{// Negative numbers are simply being treated here as just adding -1 to the list of possible
// factors. In practice it can be argued that the factors of a number can be both positive
// and negative, i.e. 4 factors into the following pairings of factors:
// (-4, -1), (-2, -2), (1, 4), (2, 2) but normally when you factor numbers you are only
// asking for the positive factors. By adding a -1 to the list it allows flagging the
// series as originating with a negative value and the implementer can use that
// information as needed.
ToFactor = -ToFactor;
yield return -1;
}
int FactorLimit = ToFactor / 2; // A good compiler may do this optimization already.
// It's here just in case;
for(int PossibleFactor = 1; PossibleFactor <= FactorLimit; PossibleFactor++)
{
if(ToFactor % PossibleFactor == 0)
{
yield return PossibleFactor;
yield return ToFactor / PossibleFactor;
}
}
}
Program to get prime factors of whole numbers in javascript code.
function getFactors(num1){
var factors = [];
var divider = 2;
while(num1 != 1){
if(num1 % divider == 0){
num1 = num1 / divider;
factors.push(divider);
}
else{
divider++;
}
}
console.log(factors);
return factors;
}
getFactors(20);
In fact we don't have to check for factors not to be square root in each iteration from the accepted answer proposed by chris fixed by Jon, which could slow down the method when the integer is large by adding an unnecessary Boolean check and a division. Just keep the max as double (don't cast it to an int) and change to loop to be exclusive not inclusive.
private static List<int> Factor(int number)
{
var factors = new List<int>();
var max = Math.Sqrt(number); // (store in double not an int) - Round down
if (max % 1 == 0)
factors.Add((int)max);
for (int factor = 1; factor < max; ++factor) // (Exclusice) - Test from 1 to the square root, or the int below it, inclusive.
{
if (number % factor == 0)
{
factors.Add(factor);
//if (factor != number / factor) // (Don't need check anymore) - Don't add the square root twice! Thanks Jon
factors.Add(number / factor);
}
}
return factors;
}
Usage
Factor(16)
// 4 1 16 2 8
Factor(20)
//1 20 2 10 4 5
And this is the extension version of the method for int type:
public static class IntExtensions
{
public static IEnumerable<int> Factors(this int value)
{
// Return 2 obvious factors
yield return 1;
yield return value;
// Return square root if number is prefect square
var max = Math.Sqrt(value);
if (max % 1 == 0)
yield return (int)max;
// Return rest of the factors
for (int i = 2; i < max; i++)
{
if (value % i == 0)
{
yield return i;
yield return value / i;
}
}
}
}
Usage
16.Factors()
// 4 1 16 2 8
20.Factors()
//1 20 2 10 4 5
Linq solution:
IEnumerable<int> GetFactors(int n)
{
Debug.Assert(n >= 1);
return from i in Enumerable.Range(1, n)
where n % i == 0
select i;
}