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How can I make this a recursive statement in C#?

The program needs to solve powers of 2 in a table showing n and 2n for n = 1 to 10. Here is my program:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Power
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine($"Number Power of 2");
Console.WriteLine("------------------------");
for (long counter = 1; counter <= 10; ++counter)
{
Console.WriteLine($"{counter, 2} {Power(counter), 20}");
}
Console.ReadLine();
}
static long Power(long number, long exponentValue = 2)
{
long result = 1;
for (int i = 1; i <= number; i++)
{
result *= exponentValue;
}
return result;
}
}
}
But it needs to be recursive. How can I make this function recursive?

This code will look like this:
static long Power(long number, long exponentValue = 2)
{
if (number < 1)
{
return 0; // or throw exception because this shouldn't happen
}
if (number == 1)
{
return exponentValue;
}
return exponentValue * Power(number - 1, exponentValue);
}

[RANT]
This one of those artificial problems meant to teach you about recursion, but doesn't do a good job because there are better non-recursive solutions (such as your current one). In fact, you usually want to convert a recursive solution TO a loop, to avoid possible stack-overflow errors!
[/RANT]
In very general terms: To convert a loop to a recursive solution, you need to find a way of computing part of the solution, and then calling the method again to compute the remainder of the solution, until it is completely solved. There must also be a way to detect when no further recursion is needed.
In the case of computing a power, the answer will be:
Pow(X,N) => N * Pow(X,N-1)
And:
Pow(X,1) => X (this will be the state that terminates the recursion).
So think about how this works for X = 2 and N = 3:
Pow(2,3) => 2 * Pow(2,2)
Pow(2,2) => 2 * Pow(2,1)
Pow(2,1) => 2 (the terminating condition)
Here you wind up with 2*2*2 = 8, which is indeed 2^3.
Note how the N value goes down by 1 each time until it reaches 1, at which point no more recursion is required.
Writing this in C# terms:
static long Power(long number, long exponentValue)
{
if (exponentValue == 1)
return number;
else
return number * Power(number, exponentValue - 1);
}
Note: I did not write any error handling in order to focus on the recursive part. Real code would check that exponentValue was >= 1 and so on.
Further note: The correct terminology for the power is "Exponent" so I have used that terminology. You have it the wrong way around, so I would urge you to correct that! An exponent is the number of times the number is multiplied by itself.

static long Power(long number, long exponentValue = 2)
{
if (number == 1)
{
return 1;
}
else
{
return Power(number - 1, exponentValue) * exponentValue;
}
}

Checking whether a number contains numbers 1 to n as factors

I am trying to solve Project Euler Challenge 5, What is the smallest possible number that is evenly divisible by all the numbers from 1 to 20.
My problem is that my isMultiple() method doesn't return true when it should.
* My Code *
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Challenge_5
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine(isMultiple(2520, 10)); //should return True
Console.WriteLine(smallestMultiple(20)); //should return 232792560
Console.ReadLine();
}
static int factorial(int n)
{
int product = 1;
for (int i = 1; i <= n; i++)
{
product *= i;
}
return product; //returns the factorial of n, (n * n-1 * n-2... * 1)
}
static bool isMultiple(int number, int currentFactor)
{
bool returnBool = false;
if (currentFactor == 1)
{
returnBool = true; // if all factors below largestFactor can divide into the number, returns true
}
else
{
if (number % currentFactor == 0)
{
currentFactor--;
isMultiple(number, currentFactor);
}
}
return returnBool;
}
static int smallestMultiple(int largestFactor)
{
for (int i = largestFactor; i < factorial(largestFactor); i+= largestFactor) //goes through all values from the kargestFactor to largestFactor factorial
{
if (isMultiple(i, largestFactor))
{
return i; // if current number can be evenly divided by all factors, it gets returned
}
}
return factorial(largestFactor); // if no numbers get returned, the factorial is the smallest multiple
}
}
}
I know there are much easier ways to solve this, but I want the program to be used to check the lowest multiple of the numbers from 1 to any number, not just 20.
Help would be much appreciated.
EDIT
Thanks to help, i have fixed my code by changing line 42 from
isMultiple(number, currentFactor);
to
returnBool = isMultiple(number, currentFactor);
I also fixed the problem with not getting an accurate return value for smallestMultiple(20);
by changing some of the variables to long instead of int

Your problem is you forgot to use the output of isMultiple in your recursive part
if (number % currentFactor == 0)
{
currentFactor--;
returnBool = isMultiple(number, currentFactor); //you need a to save the value here.
}
Without assigning returnBool there is no way of knowing if the inner isMultiple returned true or not.

Scott's answer is perfectly valid. Forgive me if I'm wrong, but it sounds like you're a student, so in the interest of education, I thought I'd give you some pointers for cleaning up your code.
When writing recursive functions, it's (in my opinion) usually cleaner to return the recursive call directly if possible, as well as returning base cases directly, rather than storing a value that you return at the end. (It's not always possible, in complicated cases where you have to make a recursive call, modify the return value, and then make another recursive call, but these are uncommon.)
This practice:
makes base cases very obvious increasing readability and forces you to consider all of your base cases
prevents you from forgetting to assign the return value, as you did in your original code
prevents potential bugs where you might accidentally do some erroneous additional processing that alters the result before you return it
reduces the number of nested if statements, increasing readability by reducing preceding whitespace
makes code-flow much more obvious increasing readability and ability to debug
usually results in tail-recursion which is the best for performance, nearly as performant as iterative code
caveat: nowadays most compilers' optimizers will rejigger production code to produce tail-recursive functions, but getting into the habit of writing your code with tail-recursion in the first place is good practice, especially as interpreted scripting languages (e.g. JavaScript) are taking over the world, where code optimization is less possible by nature
The other change I'd make is to remove currentFactor--; and move the subtraction into the recursive call itself. It increases readability, reduces the chance of side effects, and, in the case where you don't use tail-recursion, prevents you from altering a value that you later expect to be unaltered. In general, if you can avoid altering values passed into a function (as opposed to a procedure/void), you should.
Also, in this particular case, making this change removes up to 3 assembly instructions and possibly an additional value on the stack depending on how the optimizer handles it. In long running loops with large depths, this can make a difference*.
static bool isMultiple(int number, int currentFactor)
{
if (currentFactor == 1)
{
// if all factors below largestFactor can divide into the number, return true
return true;
}
if (number % currentFactor != 0)
{
return false;
}
return isMultiple(number, currentFactor - 1);
}
* A personal anecdote regarding deep recursive calls and performance...
A while back, I was writing a program in C++ to enumerate the best moves for all possible Connect-4 games. The maximum depth of the recursive search function was 42 and each depth had up to 7 recursive calls. Initial versions of the code had an estimated running time of 2 million years, and that was using parallelism. Those 3 additional instructions can make a HUGE difference, both for sheer number of additional instructions and the amount L1 and L2 cache misses.

This algorithm came up to my mind right now, so please anyone correct me if i'm wrong.
As composite numbers are made by multiplication of prime numbers (and prime numbers can not be generated from multiplication of any other numbers) so the number must be multiplied to all prime numbers, some numbers like 6 when they are reached our smallest number is dividable (as its already multiplied to 2 and 3), but for those that are not, multiplying by a prime number (that is obviously less than the number itself so should be in our prime list) would make our smallest dividable to that number too. so for example when we get to 4, multiplying by2` (a prime number less than 4) would be enough, 8 and 9 the same way, ...
bool IsPrime(int x)
{
if(x == 1) return false;
for(int i=2;i<=Math.Sqrt(x);i+=2)
if(x%i==0) return false;
return true;
}
int smallest = 1;
List<int> primes = new List<int>();
for(int i=1;i<=20;i++)
if(IsPrime(i))
{
smallest *= i;
primes.Add(i);
}
else if(smallest % i != 0)
for(int j=0;j<primes.Count;j++)
if((primes[j]*smallest)%i == 0)
{
smallest *= primes[j];
break;
}
Edit:
As we have list of prime numbers so the best way to find out if a number is prime or not would be:
bool IsPrime(int x)
{
if(x == 1) return false;
for(int i = 0; i< primes.Count; i++)
if(x%primes[i] == 0) return false;
return true;
}

Fast and low-memory-consumption way to read in pair of numbers from file and process them?

Okay, so this is my challenge taken from CodeEval. I have to read numbers from a file that is formatted in a standard way, it has a pair of numbers separated by a comma on each line (x, n). I have to read in the pair values and process them, then print out the smallest multiple of n which is greater than or equal to x, where n is a power of 2.
EXACT REQUIREMENT: Given numbers x and n, where n is a power of 2, print out the smallest multiple of n which is greater than or equal to x. Do not use division or modulo operator.
I have come up with a number of solutions, but none of them satisfy the computer's conditions to let me pass the challenge. I only get a partial completion with scores that vary from 30 to 80 (from 100).
I'm assuming that my solutions do not pass the speed but more likely the memory-usage requirements.
I would greatly appreciate it if anyone can enlighten me and offer some better, more efficient solutions.
Here are two of my solutions:
var filePath = #"C:\Users\myfile.txt";
int x;
int n;
using (var reader = new StreamReader(filePath))
{
string numsFile = string.Empty;
while ((numsFile = reader.ReadLine()) != null)
{
var nums = numsFile.Split(',').ToArray();
x = int.Parse(nums[0]);
n = int.Parse(nums[1]);
Console.WriteLine(DangleNumbers(x, n));
}
}
<<<>>>
var fileNums = File.ReadAllLines(filePath);
foreach (var line in fileNums)
{
var nums = line.Split(',').ToArray();
x = int.Parse(nums[0]);
n = int.Parse(nums[1]);
Console.WriteLine(DangleNumbers(x, n));
}
Method to check numbers
public static int DangleNumbers(int x, int n)
{
int m = 2;
while ((n * m) < x)
{
m += 2;
}
return m * n;
}
I'm fairly new to C# and programming but these two ways I found to get the best score from several others I have tried. I'm thinking that it's not too optimal for a new string to be created on each iteration, nor do I know how to use a StringBuilder and get the values into an Int from it.
Any pointers in the right direction would be appreciated as I would really like to get this challenge passed.

The smallest multiple of n that is larger or equal to x is likely this:
if(x <= n)
{
return n;
}
else
{
return x % n == 0 ? x : (x/n + 1) * n;
}
As x and n are integers, the result of x/n will be truncated (or effectively rounded down). So the next integer larger than x that is a multiple of n is (x/n + 1) * n
Since you missed the requirements, the modulo version was the most obvious choice. Though you still got your method wrong. m = 2 would not result in the smallest being returned but it could actually be the double of the smallest if n is already larger than x.
x = 7, n = 8 would get you 16 instead of 8.
Also adding 2 to m would result in a similar problem.
x = 5, n = 2 would get you 8 instead of 6.
use the following method instead:
public static int DangleNumbers(int x, int n)
{
int result = n;
while(result < x)
result += n;
return result;
}
Still capable of begin optimized but at least right according to the (now) stated constraints.

I have tried to improve the solution with some suggestions from you guys and take the variables outside the loop and drop the ToArray() call which was redundant.
static void Main(string[] args)
{
var filePath = #"C:\Users\sorin\Desktop\sorvas.txt";
int x;
int n;
string[] nums;
using (var reader = new StreamReader(filePath))
{
string numsFile = string.Empty;
while ((numsFile = reader.ReadLine()) != null)
{
nums = numsFile.Split(',');
x = int.Parse(nums[0]);
n = int.Parse(nums[1]);
Console.WriteLine(DangleNumbers(x, n));
}
}
}
public static int DangleNumbers(int x, int n)
{
int m = 2;
while ((n * m) < x)
{
m += 2;
}
return m * n;
}
So it looks like this. The thing is that even if now the numbers have slightly improved, I got a lower score.
May it be their system to blame ?

Using the first option of reading line by line (rather than reading all lines) is clearly going to use less memory (except potentially in the case where the file is very small (eg "1,1") in which case the overhead of the reader may cause problems but at that point the memory used is probably irrelevant.
Likewise declaring the variables outside the loop is generally better but in this case since the objects are value types I'm not sure it makes a difference.
Lastly the most efficient way of doing your DangleNumbers method is probably using bitwise logic operators and the fact that n is always a power of 2. Here is my attempt:
public static int DangleNumbers3(int x, int n)
{
return ((x-1) & ~(n-1))+n;
}
Essentially it relies on the fact that in binary a power of n is always a 1 followed by zero or more zeros. Thus a multiple of n will always end in that same number of zeros. So if n has M zeros after the one then you can take the binary form of x and if it already ends in M zeros then you have your answer. Otherwise you zero out the last M digits at which point you have the multiple of n that is just under x and then you add 1.
In the code ~(n-1) is a bitmask that has M zeros at the end and the leading digits are all 1. Thus when you AND it with a number it will zero out the trailing digits. I apply this to (x-1) to avoid having to do the check for if it is already the answer and have special cases.
It is important to note that this only works because of the special form of n as a power of 2. This method avoids the need for any loops and thus should run much faster (it has five operations total and no branching at all compared to other looping methods which will tend to have at the very least an operation and a comparison per loop.

Factorial function - design and test

I'm trying to nail down some interview questions, so I stared with a simple one.
Design the factorial function.
This function is a leaf (no dependencies - easly testable), so I made it static inside the helper class.
public static class MathHelper
{
public static int Factorial(int n)
{
Debug.Assert(n >= 0);
if (n < 0)
{
throw new ArgumentException("n cannot be lower that 0");
}
Debug.Assert(n <= 12);
if (n > 12)
{
throw new OverflowException("Overflow occurs above 12 factorial");
}
int factorialOfN = 1;
for (int i = 1; i <= n; ++i)
{
//checked
//{
factorialOfN *= i;
//}
}
return factorialOfN;
}
}
Testing:
[TestMethod]
[ExpectedException(typeof(OverflowException))]
public void Overflow()
{
int temp = FactorialHelper.MathHelper.Factorial(40);
}
[TestMethod]
public void ZeroTest()
{
int factorialOfZero = FactorialHelper.MathHelper.Factorial(0);
Assert.AreEqual(1, factorialOfZero);
}
[TestMethod]
public void FactorialOf5()
{
int factOf5 = FactorialHelper.MathHelper.Factorial(5);
Assert.AreEqual(5*4*3*2*1,factOf5);
}
[TestMethod]
[ExpectedException(typeof(ArgumentException))]
public void NegativeTest()
{
int factOfMinus5 = FactorialHelper.MathHelper.Factorial(-5);
}
I have a few questions:
Is it correct? (I hope so ;) )
Does it throw right exceptions?
Should I use checked context or this trick ( n > 12 ) is ok?
Is it better to use uint istead of checking for negative values?
Future improving: Overload for long, decimal, BigInteger or maybe generic method?
Thank you

It looks right to me, but it would be inefficient with larger numbers. If you're allowing for big integers, the number will keep growing with each multiply, so you would see a tremendous (asymptotically better) increase in speed if you multiplied them hierarchically. For example:
bigint myFactorial(uint first, uint last)
{
if (first == last) return first;
uint mid = first + (last - first)/2;
return myFactorial(first,mid) * myFactorial(1+mid,last);
}
bigint factorial(uint n)
{
return myFactorial(2,n);
}
If you really want a fast factorial method, you also might consider something like this:
Factor the factorial with a modified Sieve of Eratosthenes
Compute the powers of each prime factor using a fast exponentiation algorithm (and fast multiplication and square algorithms)
Multiply all the powers of primes together hierarchically

Yes, it looks right
The exceptions seem OK to me, and also as an interviewer, I can't see myself being concerned there
Checked. Also, in an interview, you'd never know that 12 just happened to be the right number.
Uint. If you can enforce something with a signature instead of an exception, do it.
You should just make it long (or bigint) and be done with it (int is a silly choice of return types here)
Here are some follow-up questions I'd ask if I were your interviewer:
Why didn't you solve this recursively? Factorial is a naturally recursive problem.
Can you add memoization to this so that it does a faster job computing 12! if it's already done 11!?
Do you need the n==0 case here?
As an interviewer, I'd definitely have some curveballs like that to throw at you. In general, I like the approach of practicing with a whiteboard and a mock interviewer, because so much of it is being nimble and thinking on your feet.

In the for cycle you can start with for (int i = 2...). Multiplying by 1 is quite useless.
I would have throw a single ArgumentOutOfRangeException for both < 0 and > 12. The Debug.Assert will mask the exception when you are using your unit test (you would have to test it in Release mode).

Finding perfect numbers (optimization)

I coded up a program in C# to find perfect numbers within a certain range as part of a programming challenge . However, I realized it is very slow when calculating perfect numbers upwards of 10000. Are there any methods of optimization that exist for finding perfect numbers? My code is as follows:
using System;
using System.Collections.Generic;
using System.Linq;
namespace ConsoleTest
{
class Program
{
public static List<int> FindDivisors(int inputNo)
{
List<int> Divisors = new List<int>();
for (int i = 1; i<inputNo; i++)
{
if (inputNo%i==0)
Divisors.Add(i);
}
return Divisors;
}
public static void Main(string[] args)
{
const int limit = 100000;
List<int> PerfectNumbers = new List<int>();
List<int> Divisors=new List<int>();
for (int i=1; i<limit; i++)
{
Divisors = FindDivisors(i);
if (i==Divisors.Sum())
PerfectNumbers.Add(i);
}
Console.Write("Output =");
for (int i=0; i<PerfectNumbers.Count; i++)
{
Console.Write(" {0} ",PerfectNumbers[i]);
}
Console.Write("\n\n\nPress any key to continue . . . ");
Console.ReadKey(true);
}
}
}

Use the formula
testPerfect = 2n-1(2n - 1)
to generate possiblities then check wether the number is in fact perfect.
try this for some bedtime reading

Do perfect numbers change? No. Look here. Surely, they should be calculated once and then stored.
In your case, the only results will be
6
28
496
8128
The next one is 33550336. Outside your range.

Just the obvious one from me: you don't need to check every divisor. No point looking for divisors past inputNo/2. That cuts down half of the calculations, but this is not an order of magnitude faster.

One way to solve things like this involves building a huge array in memory of every number, and then crossing numbers out.

if your still looking for something to calculate perfect numbers.
this goes through the first ten thousand pretty quick, but the 33 million number is a little slower.
public class Perfect {
private static Perfect INSTANCE = new Perfect();
public static Perfect getInstance() {
return INSTANCE;
}
/**
* the method that determines if a number is perfect;
*
* #param n
* #return
*/
public boolean isPerfect(long n) {
long i = 0;
long value = 0;
while(++i<n){
value = (0 == n%i?value+i:value);
}
return n==value;
}
}

For anyone interested in a LINQ based approach, the following method worked quite well and efficiently for me in determining whether or not a caller supplied integer value is a perfect number.
bool IsPerfectNumber(int value)
{
var isPerfect = false;
int maxCheck = Convert.ToInt32(Math.Sqrt(value));
int[] possibleDivisors = Enumerable.Range(1, maxCheck).ToArray();
int[] properDivisors = possibleDivisors.Where(d => (value % d == 0)).Select(d => d).ToArray();
int divisorsSum = properDivisors.Sum();
if (IsPrime(divisorsSum))
{
int lastDivisor = properDivisors.Last();
isPerfect = (value == (lastDivisor * divisorsSum));
}
return isPerfect;
}
For simplicity and clarity, my implementation for IsPrime(), which is used within IsPerfectNumber(), is omitted.

To continue from Charles Gargent's answer there is a very quick way to check if a Mersenne Number a.k.a. 2^n - 1 is prime. It is called the Lucas-Lehmer test
The basic pseudocode though (taken from the Wikipedia page) is:
// Determine if Mp = 2p − 1 is prime for p > 2
Lucas–Lehmer(p)
var s = 4
var M = 2p − 1
repeat p − 2 times:
s = ((s × s) − 2) mod M
if s == 0 return PRIME else return COMPOSITE