In my code i have static int method that should return an array of random integers. The problem is when i print an array, only the last number is random int, and other ints are all zeros.
static void Main(string[] args)
{
int a = int.Parse(Console.ReadLine());
int[] array = mtd(a);
for (int i = 0; i < array.Length; i++)
{
Console.WriteLine(array[i]);
}
}
public static int[] mtd(int b)
{
int[] arr = new int[b];
Array.Resize<int>(ref arr, b);
Random rand = new Random();
for (int i = 0; i < arr.Length; i++)
{
arr.SetValue(rand.Next(1, 5), arr.Length - 1);
}
return arr;
}
In the for loop in your method mtd you are always setting the last value of your array arr by using arr.Length - 1. Use i instead:
for (int i = 0; i < arr.Length; i++)
{
arr[i] = rand.Next(1, 5);
}
Or with arr.SetValue:
for (int i = 0; i < arr.Length; i++)
{
arr.SetValue(rand.Next(1, 5), i);
}
Working one :
public static int[] mtd(int b)
{
int[] arr = new int[b];
Random rand = new Random();
for (int i = 0; i < arr.Length; i++)
{
arr.SetValue(rand.Next(1, 5), i);
}
return arr;
}
EDIT : Btw the Array.resize is useless here, you already define the length of your array. ( new int[b] set the length of the array to b )
Your problem was that you never used i, so you just set the value of the last value of your array.
I modified and simplified your code:
public static int[] mtd(int b)
{
int[] arr = new int[b];
Random rand = new Random();
for (int i = 0; i < arr.Length; i++)
arr[i] = rand.Next(1, 5);
return arr;
}
Related
An array of arrays is given. It is necessary to find the minimum element in each column and write the data to a new array. I have done the following implementation, in accordance with the given conditions. But I am having a problem:expected minimum element in each column of the array are incorrect. Where am I making a mistake?
class Program
{
static int[][] Input()
{
Console.Write("n = ");
int n = int.Parse(Console.ReadLine());
int[][] a = new int[n][];
//int[] minA = new int[n];
for (int i = 0; i < n; ++i)
{
a[i] = new int[n];
for (int j = 0; j < n; ++j)
{
Console.Write("a[{0},{1}]= ", i, j);
a[i][j] = int.Parse(Console.ReadLine());
}
}
return a;
}
static void Print(int[] a)
{
foreach (double elem in a)
{
Console.Write("{0} ", elem);
}
}
static void Print2(int[][] a)
{
for (int i = 0; i < a.Length; ++i, Console.WriteLine())
for (int j = 0; j < a[i].Length; ++j)
Console.Write("{0,5} ", a[i][j]);
}
static int[] F(int[][] a)
{
int[] b = new int[a[1].Length];
for (int j = 0; j < a[1].Length; j++)
{
int tempmin = a[0][j];
for (int i = 0; i < a[0].Length; i++)
{
if (a[j][i] <= tempmin)
{
tempmin = a[j][i];
b[j] += tempmin;
}
}
}
return b;
}
static void Main()
{
int[][] myArray = Input();
Print2(myArray);
int[] b = new int[myArray.Length];
b = F(myArray);
Print(b);
}
}
I suggest looping over all lines, while tracking all min columns values:
using System.Linq; // for the final `ToArray()`
...
private static int[] MinColumns(int[][] data) {
if (null == data)
throw new ArgumentNullException(nameof(data));
// List of columns' mins; initially the list is empty
List<int> list = new List<int>();
// for each line (not column!) within jagged array...
foreach (int[] line in data) {
// let's just skip null lines (alternative is to throw exception)
if (null == line)
continue;
// each new line can update columns' max values.
// now we update each column
for (int c = 0; c < line.Length; ++c)
// if index c is too big, i.e.
// the line is too long and some columns appear first time...
if (c >= list.Count)
// ...we just add values of such columns as columns' min
for (int i = list.Count; i <= c; ++i)
list.Add(line[i]);
else
// otherwise we update min values: we compare known min and current value
list[c] = Math.Min(list[c], line[c]);
}
// finally, we convert list into array with ahelp of Linq
return list.ToArray();
}
Note, that here we ignore all holes, e.g. for
int[][] demo = new int[][] {
new int[] {1, 2, 3, 4},
new int[] {5, 6}, // <- hole: we don't have 3d and 4th columns here
new int[] {7, 0, 8},
};
the answer will be {Min(1, 5, 7), Min(2, 6, 0), Min(3, 8), Min (4)} = {1, 0, 3, 4}
Edit: Usage is quite direct; something like this (fiddle yourself)
static void Main()
{
// Get jagged array
int[][] myArray = Input();
// Print it
Print2(myArray);
// Get max for each column
int[] b = MinColumns(myArray);
// Print these maxes
Print(b);
}
I want to randomly generate two matrix arrays so I can later add them up and store them into a third matrix, how would I go about doing this? Nearly totally lost, here's what I have so far.
using System;
namespace question2_addingrandommatrice
{
class Program
{
static void Main(string[] args)
{
Random random = new Random();
int[,] newarray = new int[3, 3];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
int ran2 = random.Next(-10, 10);
int ran1 = random.Next(-10, 10);
newarray[i, j] = ran1, ran2;
}
}
Console.ReadKey();
}
}
}
You are nearly there, you just needed one random.Next
Here is a method that does it for you
private static int[,] GenerateRandomMatrix(int x, int y)
{
var array = new int[x, y];
for (int i = 0; i < array.GetLength(0); i++)
for (int j = 0; j < array.GetLength(1); j++)
array[i, j] = random.Next(-10, 10);
return array;
}
Add pepper and salt to taste
Usage
// 3*3 random matrix
var matrix = GenerateRandomMatrix(3,3);
Additional Resources
Multidimensional Arrays (C# Programming Guide)
You can simply do this.
Random random = new Random();
int[,] newarray = new int[3, 3];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
newarray[i, j] = random.Next(-10, 10); ;
}
I'm trying to fill one dimensional array with random BUT unique numbers (No single number should be same). As I guess I have a logical error in second for loop, but can't get it right.
P.S I'm not looking for a more "complex" solution - all I know at is this time is while,for,if.
P.P.S I know that it's a really beginner's problem and feel sorry for this kind of question.
int[] x = new int[10];
for (int i = 0; i < x.Length; i++)
{
x[i] = r.Next(9);
for (int j = 0; j <i; j++)
{
if (x[i] == x[j]) break;
}
}
for (int i = 0; i < x.Length; i++)
{
Console.WriteLine(x[i);
}
Here is a solution with your code.
int[] x = new int[10];
for (int i = 0; i < x.Length;)
{
bool stop = false;
x[i] = r.Next(9);
for (int j = 0; j <i; j++)
{
if (x[i] == x[j]) {
stop = true;
break;
}
}
if (!stop)
i++;
}
for (int i = 0; i < x.Length; i++)
{
Console.WriteLine(x[i]);
}
A simple trace of the posted code reveals some of the issues. To be specific, on the line…
if (x[i] == x[j]) break;
if the random number is “already” in the array, then simply breaking out of the j loop is going to SKIP the current i value into the x array. This means that whenever a duplicate is found, x[i] is going to be 0 (zero) the default value, then skipped.
The outer i loop is obviously looping through the x int array, this is pretty clear and looks ok. However, the second inner loop can’t really be a for loop… and here’s why… basically you need to find a random int, then loop through the existing ints to see if it already exists. Given this, in theory you could grab the same random number “many” times over before getting a unique one. Therefore, in this scenario… you really have NO idea how many times you will loop around before you find this unique number.
With that said, it may help to “break” your problem down. I am guessing a “method” that returns a “unique” int compared to the existing ints in the x array, may come in handy. Create an endless while loop, inside this loop, we would grab a random number, then loop through the “existing” ints. If the random number is not a duplicate, then we can simply return this value. This is all this method does and it may look something like below.
private static int GetNextInt(Random r, int[] x, int numberOfRandsFound) {
int currentRand;
bool itemAlreadyExist = false;
while (true) {
currentRand = r.Next(RandomNumberSize);
itemAlreadyExist = false;
for (int i = 0; i < numberOfRandsFound; i++) {
if (x[i] == currentRand) {
itemAlreadyExist = true;
break;
}
}
if (!itemAlreadyExist) {
return currentRand;
}
}
}
NOTE: Here would be a good time to describe a possible endless loop in this code…
Currently, the random numbers and the size of the array are the same, however, if the array size is “larger” than the random number spread, then the code above will NEVER exit. Example, if the current x array is set to size 11 and the random numbers is left at 10, then you will never be able to set the x[10] item since ALL possible random numbers are already used. I hope that makes sense.
Once we have the method above… the rest should be fairly straight forward.
static int DataSize;
static int RandomNumberSize;
static void Main(string[] args) {
Random random = new Random();
DataSize = 10;
RandomNumberSize = 10;
int numberOfRandsFound = 0;
int[] ArrayOfInts = new int[DataSize];
int currentRand;
for (int i = 0; i < ArrayOfInts.Length; i++) {
currentRand = GetNextInt(random, ArrayOfInts, numberOfRandsFound);
ArrayOfInts[i] = currentRand;
numberOfRandsFound++;
}
for (int i = 0; i < ArrayOfInts.Length; i++) {
Console.WriteLine(ArrayOfInts[i]);
}
Console.ReadKey();
}
Lastly as other have mentioned, this is much easier with a List<int>…
static int DataSize;
static int RandomNumberSize;
static void Main(string[] args) {
Random random = new Random();
DataSize = 10;
RandomNumberSize = 10;
List<int> listOfInts = new List<int>();
bool stillWorking = true;
int currentRand;
while (stillWorking) {
currentRand = random.Next(RandomNumberSize);
if (!listOfInts.Contains(currentRand)) {
listOfInts.Add(currentRand);
if (listOfInts.Count == DataSize)
stillWorking = false;
}
}
for (int i = 0; i < listOfInts.Count; i++) {
Console.WriteLine(i + " - " + listOfInts[i]);
}
Console.ReadKey();
}
Hope this helps ;-)
The typical solution is to generate the entire potential set in sequence (in this case an array with values from 0 to 9). Then shuffle the sequence.
private static Random rng = new Random();
public static void Shuffle(int[] items)
{
int n = list.Length;
while (n > 1) {
n--;
int k = rng.Next(n + 1);
int temp = items[k];
items[k] = items[n];
items[n] = temp;
}
}
static void Main(string[] args)
{
int[] x = new int[10];
for(int i = 0; i<x.Length; i++)
{
x[i] = i;
}
Shuffle(x);
for(int i = 0; i < x.Length; i++)
{
Console.WritLine(x[i]);
}
}
//alternate version of Main()
static void Main(string[] args)
{
var x = Enumerable.Range(0,10).ToArray();
Shuffle(x);
Console.WriteLine(String.Join("\n", x));
}
You can simply do this:
private void AddUniqueNumber()
{
Random r = new Random();
List<int> uniqueList = new List<int>();
int num = 0, count = 10;
for (int i = 0; i < count; i++)
{
num = r.Next(count);
if (!uniqueList.Contains(num))
uniqueList.Add(num);
}
}
Or:
int[] x = new int[10];
Random r1 = new Random();
int num = 0;
for (int i = 0; i < x.Length; i++)
{
num = r1.Next(10);
x[num] = num;
}
i'm code:
Random rand = new Random();
int[] arr = new int[4];
for (int i = 0; i < 4; i++)
{
for (int k = 0; k < 4; k++)
{
int rad = rand.Next(1, 5);
if (arr[k] != rad)
arr[i] = rad;
}
}
for (int i = 0; i < 4; i++)
MessageBox.Show(arr[i].ToString());
I wanna production numbers from 1 to 4 am and unequal with each other.
tnx.
Think it the other way round:
instead of:
generating a number and then check it if it is not duplicate
you make it such that:
you already have a set of non-duplicate numbers, then you take it
one-by-one - removing the possibilities of duplicates.
Something like this will do:
List<int> list = Enumerable.Range(1, 4).ToList();
List<int> rndList = new List<int>();
Random rnd = new Random();
int no = 0;
for (int i = 0; i < 4; ++i) {
no = rnd.Next(0, list.Count);
rndList.Add(list[no]);
list.Remove(list[no]);
}
The result is in your rndList.
This way, no duplicate will occur.
Create an array with unique elements and then shuffle it, like in the code below, it will shuffle an array in uniformly random order it uses Fisher-Yates shuffle algorithm:
int N = 20;
var theArray = new int[N];
for (int i = 0; i < N; i++)
theArray[i] = i;
Shuffle(theArray);
public static void Shuffle(int[] a) {
if (a == null) throw new ArgumentNullException("Array is null");
int n = a.Length;
var theRandom = new Random();
for (int i = 0; i < n; i++) {
int r = i + theRandom.Next(n-i); // between i and n-1
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
Explanation and template version of algorithm could be found in this post with nice answer from Jon Skeet.
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
{
T[] elements = source.ToArray();
// Note i > 0 to avoid final pointless iteration
for (int i = elements.Length-1; i > 0; i--)
{
// Swap element "i" with a random earlier element it (or itself)
int swapIndex = rng.Next(i + 1);
T tmp = elements[i];
elements[i] = elements[swapIndex];
elements[swapIndex] = tmp;
}
// Lazily yield (avoiding aliasing issues etc)
foreach (T element in elements)
{
yield return element;
}
}
Create a list containing the numbers you want, then shuffle them:
var rnd = new Random();
List<int> rndList = Enumerable.Range(1, 4).OrderBy(r => rnd.Next()).ToList();
If you want an array instead of a list:
var rnd = new Random();
int[] rndArray = Enumerable.Range(1, 4).OrderBy(r => rnd.Next()).ToArray();
int[] arr = new int[5];
int i = 0;
while (i < 5)
{
Random rand = new Random();
int a = rand.Next(1,6);
bool alreadyexist = false;
for (int j = 0; j < 5; j++)
{
if (a == arr[j])
{
alreadyexist = true;
}
}
if (alreadyexist == false)
{
arr[i] = a;
i++;
}
}
for (int k = 0; k < 5; k++)
{
MessageBox.Show(arr[k].ToString());
}
Why this code is throw an exception?
int n = 10;
int[] arr = null;
for (int i = 0; i < n; i++)
{
arr[i] = i * 2;
}
Any one can help?
You have to init your array first int[] arr = new int[n];
int n = 10;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
{
arr[i] = i * 2;
}
You have to initialize an array. This also has to be a fixed size, you cannot expand the lend of an array like you can with a List
you could do:
int n = 10;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
{
arr[i] = i * 2;
}
This will initialise an array of length 10 then populate numbers.
Or with a list:
int n = 10;
List<int> arr = new List<int>();
for (int i = 0; i < n; i++)
{
arr.Add(i * 2);
}
Notice we do not define the length of the list like we do for the array. We then go on and add elements as we want rather than a maximum of n.
You have to declare a new array with defined values instead of null. Like the code below. :).
int n = 10;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
{
arr[i] = i * 2;
}