This question already has answers here:
2d Array in C# with random numbers
(2 answers)
Closed 4 years ago.
How do you print and save random values in array using Random.
So I need to have 10x10 table of rows containing random values from 0-9 but I seem to can't find a way how to get them printed!
int[,] mas = new int[9,9];
Random rand = new Random();
for (int i = 0; i <mas.Length; i++)
{
mas[i] = rand.Next(0,9);
Console.WriteLine(mas[i]);
for (int k = 0; k < mas.Length; k++)
{
mas[k] = rand.Next(0,9);
Console.WriteLine(mas[k]);
}
}
You have a small issue in your code: The array you create has two dimensions (int[9,9]). Therefore, whenever you want to read or write a cell, you need to provide two coordinates. You only set mas[i] or mas[k].
You should use max[i,k] instead in the inner loop. That way, every combination of coordinates will be tried out.
Unrelated: You mention you want a 10x10 grid of cells, but declare int[9,9]. While array indexing starts at 0, the size starts at 1. For example, if you create an array a = int[2], then it only contains entries at int[0] and int[1].
Similarly, the maximum parameter of the Random.Next(...) function is exclusive, so to get the value 9, you need to pass the maximum value 10.
If you want 2D array (i.e. you have two coordinates: lines - i and columns - j), nested loops is usual choice:
Random rand = new Random();
int[,] mas = new int[9, 9];
for (int i = 0; i < mas.GetLength(0); ++i)
for (int j = 0; j < mas.GetLength(1); ++j)
mas[i, j] = rand.Next(0, 10); // 10 will not be included, range [0..9]
Let's print the array:
for (int i = 0; i < mas.GetLength(0); ++i) {
for (int j = 0; j < mas.GetLength(1); ++j) {
Console.Write(mas[i, j]);
Console.Write(' '); // <- delimiter between columns
}
Console.WriteLine(); // <- delimiter between lines
}
Also, note that mas.length returns the overall length of the array, which will give you an IndexOutOfRangeException using your Code. Use mas.GetLength instead of mas.length
myArray.GetLength(0) -> Gets first dimension size
myArray.GetLength(1) -> Gets second dimension size
So your Code would look like this:
int[,] mas = new int[9, 9];
Random rand = new Random();
for (int i = 0; i < mas.GetLength(0); i++)
{
for (int k = 0; k < mas.GetLength(1); k++)
{
mas[i, k] = rand.Next(0, 9);
Console.Write(mas[i, k] + " ");
}
Console.WriteLine();
}
Probably like this
int[,] mas = new int[10, 10];
Random rand = new Random();
for (int i = 0; i < mas.GetLength(0); i++)
{
for (int k = 0; k < mas.GetLength(1); k++)
{
mas[i,k] = rand.Next(0, 9);
Console.Write(mas[i,k]);
}
Console.WriteLine("\n");
}
There are many things wrong with your code,
You dont know how indexing works - you want array of 10 times 10 but you create one with 9 times 9.
You need to specify both dimenstion when adding number to array.
Also,
Consolo.Writeline(string);
Will always write new line, something you dont want since you want to print 10 numbers, then new line with another then etc.
Related
Im new to programming and struggling with this task:
In array X [20] random numbers from 1 to 30 are entered, in array Y enter only odd numbers from array X.
Print down Y.
int[] x = new int[20];
Random rnd = new Random();
int counter = 0;
int[] y;
for (int i = 0; i < x.Length; i++)
{
x[i] = rnd.Next(1, 30);
if (x[i] % 2 !=0 )
{
y = new int[counter];
counter++;
y[counter] = x[i];
}
}
foreach (int number in y)
{
Console.WriteLine(number);
}
Im having problems to fill Y array with odd numbers without defining length of Y, I tried with adding counter but getting some errors all the time,
If someone can help me with some suggestions that would be helpful, thank you!
This looks like homework, so I guess using a more appropriate collection such as a List<int> is out of the question, just as using Linq.
At y = new int[counter]; you're reinitializing the array. This happens each iteration, so your final array only holds the latest added value, and all values before that will be set to their default: 0.
You could've seen this by debugging your code by setting breakpoints, stepping through the code and inspecting your variables. You could then also have provided a more proper problem description than "getting some errors".
If you know the input is never larger than 20, you can initialize the output array to the same size and keep a counter of how many values you copied (the latter of which you already do).
Then when printing, only print the elements up till that count with a for loop instead of foreach.
So something like this:
int[] x = new int[20];
int[] y = new int[x.Length];
Random rnd = new Random();
int counter = 0;
for (int i = 0; i < x.Length; i++)
{
x[i] = rnd.Next(1, 30);
if (x[i] % 2 != 0)
{
y[counter] = x[i];
counter++;
}
}
for (int i = 0; i < counter; i++)
{
Console.WriteLine(y[i]);
}
Your problem is that you create a new y array for each odd number you find. You need to create the array only once and then fill it.
Since you don't know how many odd numbers there will be, I suggest to use a List<int> instead:
int[] x = new int[20];
Random rnd = new Random();
List<int> y = new List<int>(); // create the list before the loop
for (int i = 0; i < x.Length; i++)
{
x[i] = rnd.Next(1, 30);
if (x[i] % 2 !=0 )
y.Add(x[i]); // add odd number to list
}
foreach (int number in y)
{
Console.WriteLine(number);
}
See, In your case you are not aware about the number of odd numbers in that random array. so Array will not be a right choice here if you are following the current implementation. If you want the output as array, then Why not a simple LINQ with Where like this example:
First you collect all random numbers to your array as you are doing currently:
int[] randomIntegers = new int[20];
Random rnd = new Random();
for (int i = 0; i < randomIntegers.Length; i++)
{
randomIntegers[i] = rnd.Next(1, 30);
}
Now you have the all random numbers in x now perform the following operation:
int[] oddValues = randomIntegers.Where(a=> a % 2 !=0).ToArray();
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?
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);
Write code that does the following:
Declare an array with x * y elements
Populate the entire array with .
Declare a loop with 0.30 * x * y iterations
For each iteration, change a randomly selected element from . to * (you must keep looking until you find one that isn't already a *)
Output the array, x elements per line
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.
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++;
}
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();
}
I feel like this task should not be done this way...
My sequenceY length is not equal to number of steam numbers because you can't assign int[] length with int that have 0 as a starting value.
Therefore my sequenceY have a lot of 0 inside and I can't print the whole sequence. I even tried adding this after for loop:
sequenceY = new int[steamnumbercounter];
But it didn't work ... Why ?
My other question is how do programmers deal with sequences that have unknown length?
I managed to print only steam numbers but the task says print sequenceY not only part of it.
// 4. sequenceX[20] is made of random numbers from 1 to 30 ,
// sequenceY is made of steam numbers from sequenceX. Print sequneceY.
int[] nizx = new int[20];
int[] nizy = new int[20];
int n = 0;
int steamnumbercounter = 0;
Random rnd = new Random();
for (int i = 0; i < nizx.Length; i++)
{
nizx[i] = rnd.Next(1, 30);
if (nizx[i]%2==0)
{
nizy[n] = nizx[i];
n++;
steamnumbercounter++;
}
Console.Write("{0} , ", nizx[i]);
}
for (int i = 0; i < steamnumbercounter; i++)
{
Console.WriteLine("{0} , ",nizy[i]);
}
Partial code review along with an answer.
But it didn't work ... Why ?
That code didn't work because you're reassigning sequenceY to a completely new value.
My other question is how do programmers deal with sequences that have unknown length?
So, with that known we can do a few things here: create an array and use Array.Resize, use a List<T>, fill the initial array then swap it for one of the right size which is filled.
I'm going to assume a "steam" number is an even one.
Your naming is not good: nizx and nizy don't convey the meaning or line up with the problem.
I'm going to demonstrate the last option (since you stated that you don't know how to use many of the moderately complex parts of .NET in this class yet, which is fine): fill the initial array and swap it for a new one. This will run in O(n^2) time (sorta).
So, let's start with our source array.
int[] sequenceX = new int[20];
Next we'll define our destination array to be the same size as our source array. (This is the maximum number of values that could be stored in it, we'll shrink it later.)
int[] sequenceY = new int[sequenceX.Length];
Then we need a variable to hold how many numbers we found that meet our criteria:
int steamNumbers = 0;
And lastly, our Random.
Random random = new Random();
Then, we look through all our sequenceX as you did, but we'll update the logic a bit.
for (int i = 0; i < sequenceX.Length; i++)
{
sequenceX[i] = random.Next(1, 30);
if (sequenceX[i] % 2 == 0)
{
sequenceY[steamNumbers] = sequenceX[i];
steamNumbers++;
}
}
So our code looks almost the same as yours, but we have one more thing to do: since you only want sequenceY to contain steamNumbers we have to shrink it or something.
int[] tempSequenceY = sequenceY;
sequenceY = new int[steamNumbers];
for (int i = 0; i < steamNumbers; i++)
{
sequenceY[i] = tempSequenceY[i];
}
Now sequenceY only has your steam numbers in it.
Final code:
int[] sequenceX = new int[20];
int[] sequenceY = new int[sequenceX.Length];
int steamNumbers = 0;
Random random = new Random();
for (int i = 0; i < sequenceX.Length; i++)
{
sequenceX[i] = random.Next(1, 30);
if (sequenceX[i] % 2 == 0)
{
sequenceY[steamNumbers] = sequenceX[i];
steamNumbers++;
}
}
int[] tempSequenceY = sequenceY;
sequenceY = new int[steamNumbers];
for (int i = 0; i < steamNumbers; i++)
{
sequenceY[i] = tempSequenceY[i];
}
// Print your `sequenceY` here.
You could extract this to a method pretty easily as well:
public int[] GetSteamNumbers(int sequenceCount, int randomMinimum, int randomMaximum)
{
int[] sequenceX = new int[sequenceCount];
int[] sequenceY = new int[sequenceX.Length];
int steamNumbers = 0;
Random random = new Random();
for (int i = 0; i < sequenceX.Length; i++)
{
sequenceX[i] = random.Next(randomMinimum, randomMaximum);
if (sequenceX[i] % 2 == 0)
{
sequenceY[steamNumbers] = sequenceX[i];
steamNumbers++;
}
}
int[] tempSequenceY = sequenceY;
sequenceY = new int[steamNumbers];
for (int i = 0; i < steamNumbers; i++)
{
sequenceY[i] = tempSequenceY[i];
}
return sequenceY;
}
And then call it with:
int[] steamNumbers = GetSteamNumbers(20, 1, 30);
Of course, for the more advanced users (this doesn't help you, but it may help others) we can do something as follows using LINQ:
var random = new Random();
var sequenceY = Enumerable.Range(1, 20)
.Select(x => random.Next(1, 30))
.Where(x => x % 2 == 0)
.ToArray();
Which should have the same effect. (Just demonstrating that there are still things in C# to look forward to in the future.)
Disclaimer: I wrote this entire answer outside of the IDE and without actually compiling it, I make no guarantees to the accuracy of the code but the procedure itself should be fairly straight forward.
The thing with arrays in C# is that they are of fixed size.
You'll have to iterate through and re-create it or use a IEnumerable that has dynamic sizes, such as Lists.
Solution here would be to use a List that contains your integers and then you would use nizx.Add(rnd.Next(1, 30));
Elaborating on my comment above: You can create a 'fake' list by concatenating the values you need in a string, separated by commas. The string.Split(',') will give you the resulting array that you need.
Given a string of form "a,b,c,d" string.Split(',') will create the array ["a","b,"c","d"]. The code:
{
int[] nizx = new int[20];
string numberString = string.Empty;
int n = 0;
Random rnd = new Random();
for (int i = 0; i < nizx.Length; i++)
{
nizx[i] = rnd.Next(1, 30);
if (nizx[i] % 2 == 0)
{
numberString += nizx[i] + ",";
n++;
}
}
var numberArray = numberString.Split(',');
for (int i = 0; i < n; i++)
{
Console.WriteLine("{0} , ", numberArray[i]);
}
}
This question already has answers here:
Random number generator only generating one random number
(15 answers)
Closed 7 years ago.
I've run into a strange issue. When I run the following code, each of my text boxes get's filled with the same randomly generated number.
public void diceAdd()
{
int[] die = new int[4];
for(int i = 0; i < total.Length; i++)
{
for (int r = 0; r < die.Length; r++)
{
//Simulates rolling a 6 sided die
Random rand = new Random();
randomNumber = rand.Next(1, 7);
die[r] = randomNumber;
}
int smallest = die[0];
for (int c = 1; c < die.Length; ++c)
{
if (die[c] < smallest)
{
smallest = die[c];
}
}
total[i] = die[0] + die[1] + die[2] + die[3] - smallest;
}
strTxt.Text = total[0].ToString();
dexTxt.Text = total[1].ToString();
conTxt.Text = total[2].ToString();
intTxt.Text = total[3].ToString();
wisTxt.Text = total[4].ToString();
chaTxt.Text = total[5].ToString();
The thing is, if I add this messagebox
MessageBox.Show(i.ToString());
after
total[i] = die[0] + die[1] + die[2] + die[3] - smallest;
the numbers each get unique outputs, as intended.
I'm thinking it has something to do with threading, but figured I'd ask here before messing something up.
you are recreating the random number generator every time in the loop, create one before the loop:
Random rand = new Random();
for(int i = 0; i < total.Length; i++)
{
...
}
see also here and here, it explains why the numbers don't change
For some coursework I need to generate a normal magic square through brute force, this is part of the code. fyi; I'm not allowed to use any classes other than the commonly used. (I'm probably pushing my luck with Math.Pow)
I have the following method to generate a 2 dimensional of size NxN:
static int[,] GenerateSquare(int n)
{
int[,] sqarray = new int[n,n];
int[] rndarray = new int[n];
//puts completely random integers from 1 to n^2 in all elements of the square array (sqarray)
for (int i = 0; i < n; i++)
{
rndarray = FullRndArray(n);
for (int j = 0; j < n; j++)
{
sqarray[i, j] = rndarray[j];
}
}
return sqarray;
}
The FullRndArray() method is below:
static int[] FullRndArray(int n)
{
//creates an array of size n and fills with random intigers between 1 and n^2
int[] rndarray = new int[n];
Random rnd = new Random();
int ntothe2 = Convert.ToInt32(Math.Pow(n, 2));
for (int i = 0; i < n; i++)
rndarray[i] = rnd.Next(1, ntothe2 + 1);
return rndarray;
}
The problem is that when i run this code, the contents of each line is random, but each line of the square is the same as the last (ie 1-1, 1-2, 1-3 are the same as 2-1, 2-2, 2-3 and the same as 3-1, 3-2, 3-3 respectively). Yet when i go through debugger line by line, i end up with a perfectly random set of numbers in every space. Can anyone explain this error to me please?
This is the culprit:
Random rnd = new Random();
Random numbers are generated starting from a seed: same seed means same not-so-random sequence of numbers. Random uses the current time as seed, so when you run your code it runs so fast that you are creating two Randoms with the same seed, which then produce two identical rows. On the other hand, when you are debugging you let enough time pass and everything goes as expected.
The solution is to create an instance of Random, static or at the beginning of GenerateSquare, and use that one for the whole process.
I got exactly the same behavior that you describe.
This seems to work :
static int[,] GenerateSquare(int n)
{
int[,] sqarray = new int[n, n];
int[] rndarray = new int[n];
Random rnd = new Random();
//puts completely random integers from 1 to n^2 in all elements of the square array (sqarray)
for (int i = 0; i < n; i++)
{
rndarray = FullRndArray(n, rnd);
for (int j = 0; j < n; j++)
{
sqarray[i, j] = rndarray[j];
}
}
return sqarray;
}
static int[] FullRndArray(int n, Random rnd)
{
//creates an array of size n and fills with random intigers between 1 and n^2
int[] rndarray = new int[n];
int ntothe2 = Convert.ToInt32(Math.Pow(n, 2));
for (int i = 0; i < n; i++)
rndarray[i] = rnd.Next(1, ntothe2 + 1);
return rndarray;
}
I understand we must use the Random.Next() method on the same instance in order to be "really" random (as described in the BlackBear answer).
It probably worked during debug because you induced time laps between steps.