Related
So I have made 2 versions of Fibonacci calculator, one is in normal way and one is in recursion.
public static int Fib(int n)
{
if (n == 0) { return 0; }
if(n == 1) { return 1; }
int a = 0;
int b = 1;
int res = 0;
for (int i = 0; i <= n-1; i++)
{
res = a + b;
a = b;
b = res;
}
return res;
}
public static int FibRec(int n)
{
if((n == 0) || (n == 1)) { return n; }
else
{
return FibRec(n-1) + FibRec(n-2);
}
}
When i run both the same time, the recursive version is incorrect.
static void Main()
{
Console.WriteLine(Fib(7)); //return 21
Console.WriteLine(FibRec(7)); //return 13
}
I tried to check for a correct version ones on the internet but strangely all the answer is quite the same as mine. I'm very weak at recursive and I absolutely have no idea what wrong with, so I'd be very grateful if any expert can point out the problem.
Your loop in Fib is incorrect in terms of the number of iterations - and this isn't helped by using non-descript names. FibRec is correct, contrary to your assertion in the question. Note that one way of determining that is to print out (say) the first 10 values of the sequence, which I'd expect to be 0, 1, 1, 2, 3, 5, 8, 13, 21, 34. That helps to find where the problem starts.
I would advice using current (as in, the current result) and next (the next number), and looping n times. That way you don't even need the base conditions, because it all just drops out. You also don't need to keep three variables outside the loop - you can just introduce a third variable inside the loop instead:
static int Fib(int n)
{
int current = 0;
int next = 1;
for (int i = 0; i < n; i++)
{
int tmp = current + next;
current = next;
next = tmp;
}
return current;
}
Note that deconstruction assignment in modern C# allows you to write it even more clearly, reassigning both current and next in a single statement, based on the previous values:
static int Fib(int n)
{
int current = 0;
int next = 1;
for (int i = 0; i < n; i++)
{
(current, next) = (next, current + next);
}
return current;
}
Your case base should return 1, not n
public static int FibRec(int n)
{
if((n == 0) || (n == 1)) { return 1; }
else
{
return FibRec(n-1) + FibRec(n-2);
}
}
When i run both the same time, the recursive version is incorrect.
Fibonacci(7) is 13, so your recursive FibRec method is correct.
I absolutely have no idea what wrong with
There is nothing wrong with it.
I'd be very grateful if any expert can point out the problem.
There is no problem.
There are only some possible stylistic improvements: you could protect against a caller passing a negative number as an argument and you could use a guard clause or a conditional expression.
If you change the condition to n <= 1 or n < 2, then FibRec will not run into an infinite loop if a caller passes a negative number. While that is an elegant way of solving the problem, FibRec will return a wrong result in that case. A better way would be to throw an appropriate Exception, in particular an ArgumentOutOfRangeException in that case:
if (n < 0)
throw new ArgumentOutOfRangeException(
nameof(n),
n,
"Argument must be non-negative."
);
Since throw terminates the method, there is no need for an else. The rest of the method will only be executed if the argument value is valid.
The same applies to return: if you return from the method, the rest of the method will not be executed, therefore, it is not necessary to wrap the rest of the method into an else branch:
public static int FibRec(int n)
{
if (n < 0)
throw new ArgumentOutOfRangeException(
nameof(n),
n,
"Argument must be non-negative."
);
if (n < 2) return n;
return FibRec(n-1) + FibRec(n-2);
}
This is sometimes called a guard clause.
Alternatively, you could use a conditional expression:
public static int FibRec(int n)
{
if (n < 0)
throw new ArgumentOutOfRangeException(
nameof(n),
n,
"Argument must be non-negative."
);
return n < 2 ? n : FibRec(n-1) + FibRec(n-2);
}
In your original version, where you had no error checking, you could then also use an expression-bodied method:
public static int FibRec(int n) =>
n < 2 ? n : FibRec(n-1) + FibRec(n-2);
But in this case, you lose the error checking.
See this .NET Fiddle with all the different variants.
I need to find the sum of primes below two million, but limit of graph reached at 110002. What is limit of graph and how to avoid it.
public class Prime
{
private List<long> primes=new List<long>() {2,3,5,7,11};
public void prime()
{
long limit=150000;
long sum=0;
for(long i=2;i<=limit;i++)
{
if(isPrime(i))
{
primes.Add(i);
}
}
sum = primes.Sum();
Console.WriteLine(sum);
}
private static bool isPrime(long num)
{
Prime pa=new Prime();
Console.WriteLine("Checking for: "+num);
foreach(long j in pa.primes)
{
if(num%j==0)
{
return false;
}
}
return true;
}
}
You shouldn't be creating a new Prime() every time you call isPrime(). By doing that, you are breaking the isPrime() calculation because it's then not using the list of primes that's being added to.
Instead, it always checks for primes against 2, 3, 5, 7, 11 (only) with the result that any number that's a product of two primes greater than 11 is considered to be a prime.
Instead, do it like this:
public class Prime
{
private List<long> primes = new List<long>(){
2,3,5,7,11
};
public void prime()
{
long limit = 150000;
long sum = 0;
for (long i = 2; i <= limit; i++)
{
if (isPrime(i))
{
primes.Add(i);
}
}
sum = primes.Sum();
Console.WriteLine(sum);
}
private bool isPrime(long num)
{
Console.WriteLine("Checking for: " + num);
foreach (long j in primes)
{
if (num % j == 0)
{
return false;
}
}
return true;
}
}
All I did was make isPrime() non-static and removed the use of pa, instead using the primes list field directly.
This produces an answer of 986017447
Also, if you are using any non-standard development applications such as Linqpad, you should state that in the question - since (from other comments) it appears that Linqpad does not support such a large number of output lines. If that is indeed the case, comment out the Console.WriteLine("Checking for: " + num); line to see if that helps.
Limit of graph means that LINQPad output has reached a certain limit. You are simply writing to much to the output window. Run the code outside LINQPad to get rid of it, but please also implement Matthew Watson's answer.
When working with millions it's a very time to think on efficiency:
You don't want to create Prime instances at all, let alone millions
Hardcoded long limit=150000; looks very suspicious
Both loops are inefficient
Let's get rid of slow UI Console.WriteLine until the very end
Let's enumetare primes with a help of a sieve:
private static IEnumerable<long> AllPrimes() {
// Special case: there's only one even prime number
yield return 2;
List<long> knownPrimes = new List<long>(1_000_000);
for (long p = 3; ; p += 2) {
// if p is not prime,
// the smallest non-trivial divisor is less or eqaul to Sqrt(p)
long upTo = (long) (Math.Sqrt(p) + 0.5);
bool isPrime = true;
foreach (long divisor in knownPrimes)
if (p % divisor == 0) {
isPrime = false;
break;
}
else if (divisor > upTo) // we don't want to scan all known primes
break;
if (isPrime) {
knownPrimes.Add(p);
yield return p;
}
}
}
Time to query the AllPrimes() method with a help of Linq:
long result = AllPrimes()
.TakeWhile(p => p < 2_000_000)
.Sum();
Console.Write(result);
Outcome:
142913828922
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;
}
Suppose you have a set of values (1,1,1,12,12,16) how would you generate all possible combinations (without repetition) whose sum is within a predefined range [min,max]. For example, here are all the combinations (of all depths) that have a range between 13 and 17:
1 12
1 1 12
1 1 1 12
16
1 16
This assumes that each item of the same value is indistinguishable, so you don't have three results of 1 12 in the final output. Brute force is possible, but in situations where the number of items is large, the number of combinations at all depths is astronomical. In the example above, there are (3 + 1) * (2 + 1) * (1 + 1) = 24 combinations at all depths. Thus, the total combinations is the product of the number of items of any given value + 1. Of course we can logically throw out huge number of combinations whose partial sum is greater than the max value (e.g. the set 16 12 is already bigger than the max value of 17, so skip any combinations that have a 16 and 12 in them).
I originally thought I could convert the input array into two arrays and increment them kind of like an odometer. But I am getting completely stuck on this recursive algorithm that breaks early. Any suggestions?
{
int uniqueValues = 3;
int[] maxCounts = new int[uniqueValues];
int[] values = new int[uniqueValues];
// easy code to bin the data, just hardcoding for example
maxCounts[0] = 3;
values[0] = 1;
maxCounts[1] = 2;
values[1] = 12;
maxCounts[2] = 1;
values[2] = 16;
GenerateCombinationsHelper(new List<int[]>(), 13, 17, 0, 0, maxCounts, new int[3], values);
}
private void GenerateCombinationsHelper(List<int[]> results, int min, int max, int currentValue, int index, int[] maxValues, int[] currentCombo, int[] values)
{
if (index >= maxValues.Length)
{
return;
}
while (currentCombo[index] < maxValues[index])
{
currentValue += values[index];
if (currentValue> max)
{
return;
}
currentCombo[index]++;
if (currentValue< min)
{
GenerateCombinationsHelper(results, min, max, currentValue, index + 1, maxValues, currentCombo, values);
}
else
{
results.Add((int[])currentCombo.Clone());
}
}
}
Edit
The integer values are just for demonstration. It can be any object that has a some sort of numerical value (int, double, float, etc...)
Typically there will only be a handful of unique values (~10 or so) but there can be several thousands total items.
Switch the main call to:
GenerateCombinationsHelper2(new List<int[]>(), 13, 17, 0, maxCounts, new int[3], values);
and then add this code:
private void GenerateCombinationsHelper2(List<int[]> results, int min, int max, int index, int[] maxValues, int[] currentCombo, int[] values)
{
int max_count = Math.Min((int)Math.Ceiling((double)max / values[index]), maxValues[index]);
for(int count = 0; count <= max_count; count++)
{
currentCombo[index] = count;
if(index < currentCombo.Length - 1)
{
GenerateCombinationsHelper2(results, min, max, index + 1, maxValues, currentCombo, values);
}
else
{
int sum = Sum(currentCombo, values);
if(sum >= min && sum <= max)
{
int[] copy = new int[currentCombo.Length];
Array.Copy(currentCombo, copy, copy.Length);
results.Add(copy);
}
}
}
}
private static int Sum(int[] combo, int[] values)
{
int sum = 0;
for(int i = 0; i < combo.Length; i++)
{
sum += combo[i] * values[i];
}
return sum;
}
It returns the 5 valid answers.
The general tendency with this kind of problem is that there are relatively few values that will show up, but each value shows up many, many times. Therefore you first want to create a data structure that efficiently describes the combinations that will add up to the desired values, and only then figure out all of the combinations that do so. (If you know the term "dynamic programming", that's exactly the approach I'm describing.)
The obvious data structure in C# terms would be a Hashtable whose keys are the totals that the combination adds up to, and whose values are arrays listing the positions of the last elements that can be used in a combination that could add up to that particular total.
How do you build that data structure?
First you start with a Hashtable which contains the total 0 as a key, and an empty array as a value. Then for each element of your array you create a list of the new totals you can reach from the previous totals, and append your element's position to each one of their values (inserting a new one if needed). When you've gone through all of your elements, you have your data structure.
Now you can search that data structure just for the totals that are in the range you want. And for each such total, you can write a recursive program that will go through your data structure to produce the combinations. This step can indeed have a combinatorial explosion, but the nice thing is that EVERY combination produced is actually a combination in your final answer. So if this phase takes a long time, it is because you have a lot of final answers!
Try this algo
int arr[] = {1,1,1,12,12,16}
for(int i = 0;i<2^arr.Length;i++)
{
int[] arrBin = BinaryFormat(i); // binary format i
for(int j = 0;j<arrBin.Length;j++)
if (arrBin[j] == 1)
Console.Write("{0} ", arr[j]);
Console.WriteLine();
}
This is quite similar to the subset sum problem which just happens to be NP-complete.
Wikipedia says the following about NP-complete problems:
Although any given solution to such a problem can be verified quickly,
there is no known efficient way to locate a solution in the first
place; indeed, the most notable characteristic of NP-complete problems
is that no fast solution to them is known. That is, the time required
to solve the problem using any currently known algorithm increases
very quickly as the size of the problem grows. This means that the
time required to solve even moderately sized versions of many of these
problems can easily reach into the billions or trillions of years,
using any amount of computing power available today. As a consequence,
determining whether or not it is possible to solve these problems
quickly, called the P versus NP problem, is one of the principal
unsolved problems in computer science today.
If indeed there is a way to solve this besides brute-forcing through the powerset and finding all subsets which sum up to a value within the given range, then I would be very interested in hearing it.
An idea for another implementation:
Create from the list of numbers a list of stacks, each stack represents a number that appear in the list, and this number is pushed into the stack as many times as he appeared in the numbers list. more so, this list is sorted.
The idea is that you iterate through the stack list, in each stack you pop one number at a time if it doesn't exceed the max value and recall the function, and perform an additional call of skipping the current stack.
This algorithm reduces many redundant computations like trying to add different elements which have the same value when adding this value exceeds the maximal value.
I was able to solve pretty large problems with this algorithm (50 numbers and more), depending on the min and max values, obviously when the interval is very big the number of combinations may be huge.
Here's the code:
static void GenerateLimitedCombinations(List<int> intList, int minValue, int maxValue)
{
intList.Sort();
List<Stack<int>> StackList = new List<Stack<int>>();
Stack<int> NewStack = new Stack<int>();
NewStack.Push(intList[0]);
StackList.Add(NewStack);
for (int i = 1; i < intList.count; i++)
{
if (intList[i - 1] == intList[i])
StackList[StackList.count - 1].Push(intList[i]);
else
{
NewStack = new Stack<int>();
NewStack.Push(intList[i]);
StackList.Add(NewStack);
}
}
GenerateLimitedCombinations(StackList, minValue, maxValue, 0, new List<int>(), 0);
}
static void GenerateLimitedCombinations(List<Stack<int>> stackList, int minValue, int maxValue, int currentStack, List<int> currentCombination, int currentSum)
{
if (currentStack == stackList.count)
{
if (currentSum >= minValue)
{
foreach (int tempInt in CurrentCombination)
{
Console.Write(tempInt + " ");
}
Console.WriteLine(;
}
}
else
{
int TempSum = currentSum;
List<int> NewCombination = new List<int>(currentCombination);
Stack<int> UndoStack = new Stack<int>();
while (stackList[currentStack].Count != 0 && stackList[currentStack].Peek() + TempSum <= maxValue)
{
int AddedValue = stackList[currentStack].Pop();
UndoStack.Push(AddedValue);
NewCombination.Add(AddedValue);
TempSum += AddedValue;
GenerateLimitedCombinations(stackList, minValue, maxValue, currentStack + 1, new List<int>(NewCombination), TempSum);
}
while (UndoStack.Count != 0)
{
stackList[currentStack].Push(UndoStack.Pop());
}
GenerateLimitedCombinations(stackList, minValue, maxValue, currentStack + 1, currentCombination, currentSum);
}
}
Here's a test program:
static void Main(string[] args)
{
Random Rnd = new Random();
List<int> IntList = new List<int>();
int NumberOfInts = 10, MinValue = 19, MaxValue 21;
for (int i = 0; i < NumberOfInts; i++) { IntList.Add(Rnd.Next(1, 10));
for (int i = 0; i < NumberOfInts; i++) { Console.Write(IntList[i] + " "); } Console.WriteLine(); Console.WriteLine();
GenerateLimitedCombinations(IntList, MinValue, MaxValue);
Console.ReadKey();
}
I want to find the prime number between 0 and a long variable but I am not able to get any output.
The program is
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication16
{
class Program
{
void prime_num(long num)
{
bool isPrime = true;
for (int i = 0; i <= num; i++)
{
for (int j = 2; j <= num; j++)
{
if (i != j && i % j == 0)
{
isPrime = false;
break;
}
}
if (isPrime)
{
Console.WriteLine ( "Prime:" + i );
}
isPrime = true;
}
}
static void Main(string[] args)
{
Program p = new Program();
p.prime_num (999999999999999L);
Console.ReadLine();
}
}
}
Can any one help me out and find what is the possible error in the program?
You can do this faster using a nearly optimal trial division sieve in one (long) line like this:
Enumerable.Range(0, Math.Floor(2.52*Math.Sqrt(num)/Math.Log(num))).Aggregate(
Enumerable.Range(2, num-1).ToList(),
(result, index) => {
var bp = result[index]; var sqr = bp * bp;
result.RemoveAll(i => i >= sqr && i % bp == 0);
return result;
}
);
The approximation formula for number of primes used here is π(x) < 1.26 x / ln(x). We only need to test by primes not greater than x = sqrt(num).
Note that the sieve of Eratosthenes has much better run time complexity than trial division (should run much faster for bigger num values, when properly implemented).
Try this:
void prime_num(long num)
{
// bool isPrime = true;
for (long i = 0; i <= num; i++)
{
bool isPrime = true; // Move initialization to here
for (long j = 2; j < i; j++) // you actually only need to check up to sqrt(i)
{
if (i % j == 0) // you don't need the first condition
{
isPrime = false;
break;
}
}
if (isPrime)
{
Console.WriteLine ( "Prime:" + i );
}
// isPrime = true;
}
}
You only need to check odd divisors up to the square root of the number. In other words your inner loop needs to start:
for (int j = 3; j <= Math.Sqrt(i); j+=2) { ... }
You can also break out of the function as soon as you find the number is not prime, you don't need to check any more divisors (I see you're already doing that!).
This will only work if num is bigger than two.
No Sqrt
You can avoid the Sqrt altogether by keeping a running sum. For example:
int square_sum=1;
for (int j=3; square_sum<i; square_sum+=4*(j++-1)) {...}
This is because the sum of numbers 1+(3+5)+(7+9) will give you a sequence of odd squares (1,9,25 etc). And hence j represents the square root of square_sum. As long as square_sum is less than i then j is less than the square root.
People have mentioned a couple of the building blocks toward doing this efficiently, but nobody's really put the pieces together. The sieve of Eratosthenes is a good start, but with it you'll run out of memory long before you reach the limit you've set. That doesn't mean it's useless though -- when you're doing your loop, what you really care about are prime divisors. As such, you can start by using the sieve to create a base of prime divisors, then use those in the loop to test numbers for primacy.
When you write the loop, however, you really do NOT want to us sqrt(i) in the loop condition as a couple of answers have suggested. You and I know that the sqrt is a "pure" function that always gives the same answer if given the same input parameter. Unfortunately, the compiler does NOT know that, so if use something like '<=Math.sqrt(x)' in the loop condition, it'll re-compute the sqrt of the number every iteration of the loop.
You can avoid that a couple of different ways. You can either pre-compute the sqrt before the loop, and use the pre-computed value in the loop condition, or you can work in the other direction, and change i<Math.sqrt(x) to i*i<x. Personally, I'd pre-compute the square root though -- I think it's clearer and probably a bit faster--but that depends on the number of iterations of the loop (the i*i means it's still doing a multiplication in the loop). With only a few iterations, i*i will typically be faster. With enough iterations, the loss from i*i every iteration outweighs the time for executing sqrt once outside the loop.
That's probably adequate for the size of numbers you're dealing with -- a 15 digit limit means the square root is 7 or 8 digits, which fits in a pretty reasonable amount of memory. On the other hand, if you want to deal with numbers in this range a lot, you might want to look at some of the more sophisticated prime-checking algorithms, such as Pollard's or Brent's algorithms. These are more complex (to put it mildly) but a lot faster for large numbers.
There are other algorithms for even bigger numbers (quadratic sieve, general number field sieve) but we won't get into them for the moment -- they're a lot more complex, and really only useful for dealing with really big numbers (the GNFS starts to be useful in the 100+ digit range).
First step: write an extension method to find out if an input is prime
public static bool isPrime(this int number ) {
for (int i = 2; i < number; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}
2 step: write the method that will print all prime numbers that are between 0 and the number input
public static void getAllPrimes(int number)
{
for (int i = 0; i < number; i++)
{
if (i.isPrime()) Console.WriteLine(i);
}
}
It may just be my opinion, but there's another serious error in your program (setting aside the given 'prime number' question, which has been thoroughly answered).
Like the rest of the responders, I'm assuming this is homework, which indicates you want to become a developer (presumably).
You need to learn to compartmentalize your code. It's not something you'll always need to do in a project, but it's good to know how to do it.
Your method prime_num(long num) could stand a better, more descriptive name. And if it is supposed to find all prime numbers less than a given number, it should return them as a list. This makes it easier to seperate your display and your functionality.
If it simply returned an IList containing prime numbers you could then display them in your main function (perhaps calling another outside function to pretty print them) or use them in further calculations down the line.
So my best recommendation to you is to do something like this:
public void main(string args[])
{
//Get the number you want to use as input
long x = number;//'number' can be hard coded or retrieved from ReadLine() or from the given arguments
IList<long> primes = FindSmallerPrimes(number);
DisplayPrimes(primes);
}
public IList<long> FindSmallerPrimes(long largestNumber)
{
List<long> returnList = new List<long>();
//Find the primes, using a method as described by another answer, add them to returnList
return returnList;
}
public void DisplayPrimes(IList<long> primes)
{
foreach(long l in primes)
{
Console.WriteLine ( "Prime:" + l.ToString() );
}
}
Even if you end up working somewhere where speration like this isn't needed, it's good to know how to do it.
EDIT_ADD: If Will Ness is correct that the question's purpose is just to output a continuous stream of primes for as long as the program is run (pressing Pause/Break to pause and any key to start again) with no serious hope of every getting to that upper limit, then the code should be written with no upper limit argument and a range check of "true" for the first 'i' for loop. On the other hand, if the question wanted to actually print the primes up to a limit, then the following code will do the job much more efficiently using Trial Division only for odd numbers, with the advantage that it doesn't use memory at all (it could also be converted to a continuous loop as per the above):
static void primesttt(ulong top_number) {
Console.WriteLine("Prime: 2");
for (var i = 3UL; i <= top_number; i += 2) {
var isPrime = true;
for (uint j = 3u, lim = (uint)Math.Sqrt((double)i); j <= lim; j += 2) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime) Console.WriteLine("Prime: {0} ", i);
}
}
First, the question code produces no output because of that its loop variables are integers and the limit tested is a huge long integer, meaning that it is impossible for the loop to reach the limit producing an inner loop EDITED: whereby the variable 'j' loops back around to negative numbers; when the 'j' variable comes back around to -1, the tested number fails the prime test because all numbers are evenly divisible by -1 END_EDIT. Even if this were corrected, the question code produces very slow output because it gets bound up doing 64-bit divisions of very large quantities of composite numbers (all the even numbers plus the odd composites) by the whole range of numbers up to that top number of ten raised to the sixteenth power for each prime that it can possibly produce. The above code works because it limits the computation to only the odd numbers and only does modulo divisions up to the square root of the current number being tested.
This takes an hour or so to display the primes up to a billion, so one can imagine the amount of time it would take to show all the primes to ten thousand trillion (10 raised to the sixteenth power), especially as the calculation gets slower with increasing range. END_EDIT_ADD
Although the one liner (kind of) answer by #SLaks using Linq works, it isn't really the Sieve of Eratosthenes as it is just an unoptimised version of Trial Division, unoptimised in that it does not eliminate odd primes, doesn't start at the square of the found base prime, and doesn't stop culling for base primes larger than the square root of the top number to sieve. It is also quite slow due to the multiple nested enumeration operations.
It is actually an abuse of the Linq Aggregate method and doesn't effectively use the first of the two Linq Range's generated. It can become an optimized Trial Division with less enumeration overhead as follows:
static IEnumerable<int> primes(uint top_number) {
var cullbf = Enumerable.Range(2, (int)top_number).ToList();
for (int i = 0; i < cullbf.Count; i++) {
var bp = cullbf[i]; var sqr = bp * bp; if (sqr > top_number) break;
cullbf.RemoveAll(c => c >= sqr && c % bp == 0);
} return cullbf; }
which runs many times faster than the SLaks answer. However, it is still slow and memory intensive due to the List generation and the multiple enumerations as well as the multiple divide (implied by the modulo) operations.
The following true Sieve of Eratosthenes implementation runs about 30 times faster and takes much less memory as it only uses a one bit representation per number sieved and limits its enumeration to the final iterator sequence output, as well having the optimisations of only treating odd composites, and only culling from the squares of the base primes for base primes up to the square root of the maximum number, as follows:
static IEnumerable<uint> primes(uint top_number) {
if (top_number < 2u) yield break;
yield return 2u; if (top_number < 3u) yield break;
var BFLMT = (top_number - 3u) / 2u;
var SQRTLMT = ((uint)(Math.Sqrt((double)top_number)) - 3u) / 2u;
var buf = new BitArray((int)BFLMT + 1,true);
for (var i = 0u; i <= BFLMT; ++i) if (buf[(int)i]) {
var p = 3u + i + i; if (i <= SQRTLMT) {
for (var j = (p * p - 3u) / 2u; j <= BFLMT; j += p)
buf[(int)j] = false; } yield return p; } }
The above code calculates all the primes to ten million range in about 77 milliseconds on an Intel i7-2700K (3.5 GHz).
Either of the two static methods can be called and tested with the using statements and with the static Main method as follows:
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
static void Main(string[] args) {
Console.WriteLine("This program generates prime sequences.\r\n");
var n = 10000000u;
var elpsd = -DateTime.Now.Ticks;
var count = 0; var lastp = 0u;
foreach (var p in primes(n)) { if (p > n) break; ++count; lastp = (uint)p; }
elpsd += DateTime.Now.Ticks;
Console.WriteLine(
"{0} primes found <= {1}; the last one is {2} in {3} milliseconds.",
count, n, lastp,elpsd / 10000);
Console.Write("\r\nPress any key to exit:");
Console.ReadKey(true);
Console.WriteLine();
}
which will show the number of primes in the sequence up to the limit, the last prime found, and the time expended in enumerating that far.
EDIT_ADD: However, in order to produce an enumeration of the number of primes less than ten thousand trillion (ten to the sixteenth power) as the question asks, a segmented paged approach using multi-core processing is required but even with C++ and the very highly optimized PrimeSieve, this would require something over 400 hours to just produce the number of primes found, and tens of times that long to enumerate all of them so over a year to do what the question asks. To do it using the un-optimized Trial Division algorithm attempted, it will take super eons and a very very long time even using an optimized Trial Division algorithm as in something like ten to the two millionth power years (that's two million zeros years!!!).
It isn't much wonder that his desktop machine just sat and stalled when he tried it!!!! If he had tried a smaller range such as one million, he still would have found it takes in the range of seconds as implemented.
The solutions I post here won't cut it either as even the last Sieve of Eratosthenes one will require about 640 Terabytes of memory for that range.
That is why only a page segmented approach such as that of PrimeSieve can handle this sort of problem for the range as specified at all, and even that requires a very long time, as in weeks to years unless one has access to a super computer with hundreds of thousands of cores. END_EDIT_ADD
Smells like more homework. My very very old graphing calculator had a is prime program like this. Technnically the inner devision checking loop only needs to run to i^(1/2). Do you need to find "all" prime numbers between 0 and L ? The other major problem is that your loop variables are "int" while your input data is "long", this will be causing an overflow making your loops fail to execute even once. Fix the loop variables.
One line code in C# :-
Console.WriteLine(String.Join(Environment.NewLine,
Enumerable.Range(2, 300)
.Where(n => Enumerable.Range(2, (int)Math.Sqrt(n) - 1)
.All(nn => n % nn != 0)).ToArray()));
The Sieve of Eratosthenes answer above is not quite correct. As written it will find all the primes between 1 and 1000000. To find all the primes between 1 and num use:
private static IEnumerable Primes01(int num)
{
return Enumerable.Range(1, Convert.ToInt32(Math.Floor(Math.Sqrt(num))))
.Aggregate(Enumerable.Range(1, num).ToList(),
(result, index) =>
{
result.RemoveAll(i => i > result[index] && i%result[index] == 0);
return result;
}
);
}
The seed of the Aggregate should be range 1 to num since this list will contain the final list of primes. The Enumerable.Range(1, Convert.ToInt32(Math.Floor(Math.Sqrt(num)))) is the number of times the seed is purged.
ExchangeCore Forums have a good console application listed that looks to write found primes to a file, it looks like you can also use that same file as a starting point so you don't have to restart finding primes from 2 and they provide a download of that file with all found primes up to 100 million so it would be a good start.
The algorithm on the page also takes a couple shortcuts (odd numbers and only checks up to the square root) which makes it extremely efficient and it will allow you to calculate long numbers.
so this is basically just two typos, one, the most unfortunate, for (int j = 2; j <= num; j++) which is the reason for the unproductive testing of 1%2,1%3 ... 1%(10^15-1) which goes on for very long time so the OP didn't get "any output". It should've been j < i; instead. The other, minor one in comparison, is that i should start from 2, not from 0:
for( i=2; i <= num; i++ )
{
for( j=2; j < i; j++ ) // j <= sqrt(i) is really enough
....
Surely it can't be reasonably expected of a console print-out of 28 trillion primes or so to be completed in any reasonable time-frame. So, the original intent of the problem was obviously to print out a steady stream of primes, indefinitely. Hence all the solutions proposing simple use of sieve of Eratosthenes are totally without merit here, because simple sieve of Eratosthenes is bounded - a limit must be set in advance.
What could work here is the optimized trial division which would save the primes as it finds them, and test against the primes, not just all numbers below the candidate.
Second alternative, with much better complexity (i.e. much faster) is to use a segmented sieve of Eratosthenes. Which is incremental and unbounded.
Both these schemes would use double-staged production of primes: one would produce and save the primes, to be used by the other stage in testing (or sieving), much above the limit of the first stage (below its square of course - automatically extending the first stage, as the second stage would go further and further up).
To be quite frank, some of the suggested solutions are really slow, and therefore are bad suggestions. For testing a single number to be prime you need some dividing/modulo operator, but for calculating a range you don't have to.
Basically you just exclude numbers that are multiples of earlier found primes, as the are (by definition) not primes themselves.
I will not give the full implementation, as that would be to easy, this is the approach in pseudo code. (On my machine, the actual implementation calculates all primes in an Sytem.Int32 (2 bilion) within 8 seconds.
public IEnumerable<long> GetPrimes(long max)
{
// we safe the result set in an array of bytes.
var buffer = new byte[long >> 4];
// 1 is not a prime.
buffer[0] = 1;
var iMax = (long)Math.Sqrt(max);
for(long i = 3; i <= iMax; i +=2 )
{
// find the index in the buffer
var index = i >> 4;
// find the bit of the buffer.
var bit = (i >> 1) & 7;
// A not set bit means: prime
if((buffer[index] & (1 << bit)) == 0)
{
var step = i << 2;
while(step < max)
{
// find position in the buffer to write bits that represent number that are not prime.
}
}
// 2 is not in the buffer.
yield return 2;
// loop through buffer and yield return odd primes too.
}
}
The solution requires a good understanding of bitwise operations. But it ways, and ways faster. You also can safe the result of the outcome on disc, if you need them for later use. The result of 17 * 10^9 numbers can be safed with 1 GB, and the calculation of that result set takes about 2 minutes max.
I know this is quiet old question, but after reading here:
Sieve of Eratosthenes Wiki
This is the way i wrote it from understanding the algorithm:
void SieveOfEratosthenes(int n)
{
bool[] primes = new bool[n + 1];
for (int i = 0; i < n; i++)
primes[i] = true;
for (int i = 2; i * i <= n; i++)
if (primes[i])
for (int j = i * 2; j <= n; j += i)
primes[j] = false;
for (int i = 2; i <= n; i++)
if (primes[i]) Console.Write(i + " ");
}
In the first loop we fill the array of booleans with true.
Second for loop will start from 2 since 1 is not a prime number and will check if prime number is still not changed and then assign false to the index of j.
last loop we just printing when it is prime.
Very similar - from an exercise to implement Sieve of Eratosthenes in C#:
public class PrimeFinder
{
readonly List<long> _primes = new List<long>();
public PrimeFinder(long seed)
{
CalcPrimes(seed);
}
public List<long> Primes { get { return _primes; } }
private void CalcPrimes(long maxValue)
{
for (int checkValue = 3; checkValue <= maxValue; checkValue += 2)
{
if (IsPrime(checkValue))
{
_primes.Add(checkValue);
}
}
}
private bool IsPrime(long checkValue)
{
bool isPrime = true;
foreach (long prime in _primes)
{
if ((checkValue % prime) == 0 && prime <= Math.Sqrt(checkValue))
{
isPrime = false;
break;
}
}
return isPrime;
}
}
Prime Helper very fast calculation
public static class PrimeHelper
{
public static IEnumerable<Int32> FindPrimes(Int32 maxNumber)
{
return (new PrimesInt32(maxNumber));
}
public static IEnumerable<Int32> FindPrimes(Int32 minNumber, Int32 maxNumber)
{
return FindPrimes(maxNumber).Where(pn => pn >= minNumber);
}
public static bool IsPrime(this Int64 number)
{
if (number < 2)
return false;
else if (number < 4 )
return true;
var limit = (Int32)System.Math.Sqrt(number) + 1;
var foundPrimes = new PrimesInt32(limit);
return !foundPrimes.IsDivisible(number);
}
public static bool IsPrime(this Int32 number)
{
return IsPrime(Convert.ToInt64(number));
}
public static bool IsPrime(this Int16 number)
{
return IsPrime(Convert.ToInt64(number));
}
public static bool IsPrime(this byte number)
{
return IsPrime(Convert.ToInt64(number));
}
}
public class PrimesInt32 : IEnumerable<Int32>
{
private Int32 limit;
private BitArray numbers;
public PrimesInt32(Int32 limit)
{
if (limit < 2)
throw new Exception("Prime numbers not found.");
startTime = DateTime.Now;
calculateTime = startTime - startTime;
this.limit = limit;
try { findPrimes(); } catch{/*Overflows or Out of Memory*/}
calculateTime = DateTime.Now - startTime;
}
private void findPrimes()
{
/*
The Sieve Algorithm
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
*/
numbers = new BitArray(limit, true);
for (Int32 i = 2; i < limit; i++)
if (numbers[i])
for (Int32 j = i * 2; j < limit; j += i)
numbers[j] = false;
}
public IEnumerator<Int32> GetEnumerator()
{
for (Int32 i = 2; i < 3; i++)
if (numbers[i])
yield return i;
if (limit > 2)
for (Int32 i = 3; i < limit; i += 2)
if (numbers[i])
yield return i;
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
// Extended for Int64
public bool IsDivisible(Int64 number)
{
var sqrt = System.Math.Sqrt(number);
foreach (var prime in this)
{
if (prime > sqrt)
break;
if (number % prime == 0)
{
DivisibleBy = prime;
return true;
}
}
return false;
}
private static DateTime startTime;
private static TimeSpan calculateTime;
public static TimeSpan CalculateTime { get { return calculateTime; } }
public Int32 DivisibleBy { get; set; }
}
public static void Main()
{
Console.WriteLine("enter the number");
int i = int.Parse(Console.ReadLine());
for (int j = 2; j <= i; j++)
{
for (int k = 2; k <= i; k++)
{
if (j == k)
{
Console.WriteLine("{0}is prime", j);
break;
}
else if (j % k == 0)
{
break;
}
}
}
Console.ReadLine();
}
static void Main(string[] args)
{ int i,j;
Console.WriteLine("prime no between 1 to 100");
for (i = 2; i <= 100; i++)
{
int count = 0;
for (j = 1; j <= i; j++)
{
if (i % j == 0)
{ count=count+1; }
}
if ( count <= 2)
{ Console.WriteLine(i); }
}
Console.ReadKey();
}
U can use the normal prime number concept must only two factors (one and itself).
So do like this,easy way
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace PrimeNUmber
{
class Program
{
static void FindPrimeNumber(long num)
{
for (long i = 1; i <= num; i++)
{
int totalFactors = 0;
for (int j = 1; j <= i; j++)
{
if (i % j == 0)
{
totalFactors = totalFactors + 1;
}
}
if (totalFactors == 2)
{
Console.WriteLine(i);
}
}
}
static void Main(string[] args)
{
long num;
Console.WriteLine("Enter any value");
num = Convert.ToInt64(Console.ReadLine());
FindPrimeNumber(num);
Console.ReadLine();
}
}
}
This solution displays all prime numbers between 0 and 100.
int counter = 0;
for (int c = 0; c <= 100; c++)
{
counter = 0;
for (int i = 1; i <= c; i++)
{
if (c % i == 0)
{ counter++; }
}
if (counter == 2)
{ Console.Write(c + " "); }
}
This is the fastest way to calculate prime numbers in C#.
void PrimeNumber(long number)
{
bool IsprimeNumber = true;
long value = Convert.ToInt32(Math.Sqrt(number));
if (number % 2 == 0)
{
IsprimeNumber = false;
}
for (long i = 3; i <= value; i=i+2)
{
if (number % i == 0)
{
// MessageBox.Show("It is divisible by" + i);
IsprimeNumber = false;
break;
}
}
if (IsprimeNumber)
{
MessageBox.Show("Yes Prime Number");
}
else
{
MessageBox.Show("No It is not a Prime NUmber");
}
}
class CheckIfPrime
{
static void Main()
{
while (true)
{
Console.Write("Enter a number: ");
decimal a = decimal.Parse(Console.ReadLine());
decimal[] k = new decimal[int.Parse(a.ToString())];
decimal p = 0;
for (int i = 2; i < a; i++)
{
if (a % i != 0)
{
p += i;
k[i] = i;
}
else
p += i;
}
if (p == k.Sum())
{ Console.WriteLine ("{0} is prime!", a);}
else
{Console.WriteLine("{0} is NOT prime", a);}
}
}
}
There are some very optimal ways to implement the algorithm. But if you don't know much about maths and you simply follow the definition of prime as the requirement:
a number that is only divisible by 1 and by itself (and nothing else), here's a simple to understand code for positive numbers.
public bool IsPrime(int candidateNumber)
{
int fromNumber = 2;
int toNumber = candidateNumber - 1;
while(fromNumber <= toNumber)
{
bool isDivisible = candidateNumber % fromNumber == 0;
if (isDivisible)
{
return false;
}
fromNumber++;
}
return true;
}
Since every number is divisible by 1 and by itself, we start checking from 2 onwards until the number immediately before itself. That's the basic reasoning.
You can do also this:
class Program
{
static void Main(string[] args)
{
long numberToTest = 350124;
bool isPrime = NumberIsPrime(numberToTest);
Console.WriteLine(string.Format("Number {0} is prime? {1}", numberToTest, isPrime));
Console.ReadLine();
}
private static bool NumberIsPrime(long n)
{
bool retVal = true;
if (n <= 3)
{
retVal = n > 1;
} else if (n % 2 == 0 || n % 3 == 0)
{
retVal = false;
}
int i = 5;
while (i * i <= n)
{
if (n % i == 0 || n % (i + 2) == 0)
{
retVal = false;
}
i += 6;
}
return retVal;
}
}
An easier approach , what i did is check if a number have exactly two division factors which is the essence of prime numbers .
List<int> factorList = new List<int>();
int[] numArray = new int[] { 1, 0, 6, 9, 7, 5, 3, 6, 0, 8, 1 };
foreach (int item in numArray)
{
for (int x = 1; x <= item; x++)
{
//check for the remainder after dividing for each number less that number
if (item % x == 0)
{
factorList.Add(x);
}
}
if (factorList.Count == 2) // has only 2 division factors ; prime number
{
Console.WriteLine(item + " is a prime number ");
}
else
{Console.WriteLine(item + " is not a prime number ");}
factorList = new List<int>(); // reinitialize list
}
Here is a solution with unit test:
The solution:
public class PrimeNumbersKata
{
public int CountPrimeNumbers(int n)
{
if (n < 0) throw new ArgumentException("Not valide numbre");
if (n == 0 || n == 1) return 0;
int cpt = 0;
for (int i = 2; i <= n; i++)
{
if (IsPrimaire(i)) cpt++;
}
return cpt;
}
private bool IsPrimaire(int number)
{
for (int i = 2; i <= number / 2; i++)
{
if (number % i == 0) return false;
}
return true;
}
}
The tests:
[TestFixture]
class PrimeNumbersKataTest
{
private PrimeNumbersKata primeNumbersKata;
[SetUp]
public void Init()
{
primeNumbersKata = new PrimeNumbersKata();
}
[TestCase(1,0)]
[TestCase(0,0)]
[TestCase(2,1)]
[TestCase(3,2)]
[TestCase(5,3)]
[TestCase(7,4)]
[TestCase(9,4)]
[TestCase(11,5)]
[TestCase(13,6)]
public void CountPrimeNumbers_N_AsArgument_returnCountPrimes(int n, int expected)
{
//arrange
//act
var actual = primeNumbersKata.CountPrimeNumbers(n);
//assert
Assert.AreEqual(expected,actual);
}
[Test]
public void CountPrimairs_N_IsNegative_RaiseAnException()
{
var ex = Assert.Throws<ArgumentException>(()=> { primeNumbersKata.CountPrimeNumbers(-1); });
//Assert.That(ex.Message == "Not valide numbre");
Assert.That(ex.Message, Is.EqualTo("Not valide numbre"));
}
}
in the university it was necessary to count prime numbers up to 10,000 did so, the teacher was a little surprised, but I passed the test. Lang c#
void Main()
{
int number=1;
for(long i=2;i<10000;i++)
{
if(PrimeTest(i))
{
Console.WriteLine(number+++" " +i);
}
}
}
List<long> KnownPrime = new List<long>();
private bool PrimeTest(long i)
{
if (i == 1) return false;
if (i == 2)
{
KnownPrime.Add(i);
return true;
}
foreach(int k in KnownPrime)
{
if(i%k==0)
return false;
}
KnownPrime.Add(i);
return true;
}
for (int i = 2; i < 100; i++)
{
bool isPrimeNumber = true;
for (int j = 2; j <= i && j <= 100; j++)
{
if (i != j && i % j == 0)
{
isPrimeNumber = false; break;
}
}
if (isPrimeNumber)
{
Console.WriteLine(i);
}
}