Char in random position

Question

I`m facing with the problem of putting char in random position.

I have a table full of dots and I have to replace 30% of these dots with *

Size: 10x5

I used function Random.

Random rnd = new Random();

if (rnd.Next() % 10 > 3)
    Console.Write(". ");
else
    Console.Write("* ");

Everything is in 2 loops which hold Length and Height of table (10x5).

But it only makes PROBABILITY of 30% to make * instead of .

It takes good position but every time I start a program there is different amount of *.

It should just have 16 of * (17 - if rounded) every time I start the program

How should I suppose to make 30% always instead of probability?

Answer 1

You have 50 dots. calculate 50*30/100, it becomes 15.

Then generate 15 unique random numbers within range of 0 to 50. those numbers are indexes you have to replace . with *

var indexes = Enumerable.Range(0,50).OrderBy(x => rng.Next()).Take(50*30/100).ToList();

If you are working with 2d index, its fairly easy to convert 1d index into 2d index.

var i = index % 5;
var j = index / 5;

According to what @KonradRudolph said if you don't want to use OrderBy you can check out other ways to shuffle array (or create randomized set) posted here Best way to randomize an array with .NET

Here is more efficient way using Fisher-Yates algorithm that I suggest you to use instead of using OrderBy

var indexes = Enumerable.Range(0, 50).ToArray();
RandomExtensions.Shuffle(rng, indexes);

Answer 2

Write code that does the following:

1. Declare an array with x * y elements
2. Populate the entire array with .
3. Declare a loop with 0.30 * x * y iterations
4. For each iteration, change a randomly selected element from . to * (you must keep looking until you find one that isn't already a *)
5. Output the array, x elements per line

Answer 3

Imagine your array to be just 50 elements and forget about the rectangular shape for now. How would you approach that? Declare X dots and 50-X stars, then randomly order them. Like you would randomly order a 1->50 list of numbers.

Now how to randomly order a list of 1->50 numbers? One simple and intuitive way is imagining shuffling cards. Go through the loop and for each position obtain a random number in 1->50. Swap elements chosen (say for i=1 we got random number 7 => swap elements 1 and 7).

Here, you simply need to map this rectangle to those 50 points, which is trivial enough for 2D.

Answer 4

I would randomly select 30% of the possible positions

// create char array
int arrayRows = 5;
int arrayCols = 10;
char[,] arr= new char[arrayRows ,arrayCols];

// populate array with dots
for (int i = 0; i < arrayRows; i++)
{
 for (int j = 0; j < arrayCols; j++)
  {
  arr[i,j] = '.';
  }
}

Random rnd = new Random();

int numberOfPossiblePositions = arrayRows * arrayCols;

int k = 0;
while (k < numberOfPossiblePositions * 0.3) {
 int position = rnd.Next(numberOfPossiblePositions);

 int colIndex = position / 10;
 int rowIndex = position % 10;

 // if the cell already has * try again
 if (arr[rowIndex,colIndex] == '*') {
   continue;
 }

 arr[rowIndex,colIndex] = '*';
 k++;
}

Answer 5

Create an array that holds x*y/3 starts and the rest are dots. order by random and iterate through it.

This is the array:

Enumerable.Range(0, count).Select(i => new {Text = "*", Order = rnd.Next() })
     .Concat(Enumerable.Range(0, x*y - count)
     .Select(i=>new { Text = ".", Order = rnd.Next() }))
.OrderBy(i => i.Order).Select(i=>i.Text).ToList();

And this is the code for iteration:

Random rnd = new Random();
int x = 10;
int y = 5;
int count = x*y/3;
var allPlaces =
            Enumerable.Range(0, count).Select(i => new {Text = "*", Order = rnd.Next() })
                .Concat(Enumerable.Range(0, x*y - count)
                .Select(i=>new { Text = ".", Order = rnd.Next() }))
            .OrderBy(i => i.Order).Select(i=>i.Text).ToList();

for (var i = 0; i < x; x++)
{
    for (var j = 0; j < y; j++) { Console.Write(allPlaces[i*j + j]); } 
    Console.WriteLine();
}