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Why multiply Random.Next() by a Constant?

I recently read an article explaining how to generate a weighted random number, and there's a piece of the code that I don't understand:
int r = ((int)(rand.Next() * (323567)) % prefix[n - 1]) + 1;
Why is rand.Next being multiplied by a constant 323567? Would the code work without this constant?
Below is the full code for reference, and you can find the full article here: https://www.geeksforgeeks.org/random-number-generator-in-arbitrary-probability-distribution-fashion/
Any help is appreciated, thank you!!
// C# program to generate random numbers
// according to given frequency distribution
using System;
class GFG{
// Utility function to find ceiling
// of r in arr[l..h]
static int findCeil(int[] arr, int r,
int l, int h)
{
int mid;
while (l < h)
{
// Same as mid = (l+h)/2
mid = l + ((h - l) >> 1);
if (r > arr[mid])
l = mid + 1;
else
h = mid;
}
return (arr[l] >= r) ? l : -1;
}
// The main function that returns a random number
// from arr[] according to distribution array
// defined by freq[]. n is size of arrays.
static int myRand(int[] arr, int[] freq, int n)
{
// Create and fill prefix array
int[] prefix = new int[n];
int i;
prefix[0] = freq[0];
for(i = 1; i < n; ++i)
prefix[i] = prefix[i - 1] + freq[i];
// prefix[n-1] is sum of all frequencies.
// Generate a random number with
// value from 1 to this sum
Random rand = new Random();
int r = ((int)(rand.Next() * (323567)) % prefix[n - 1]) + 1; // <--- RNG * Constant
// Find index of ceiling of r in prefix array
int indexc = findCeil(prefix, r, 0, n - 1);
return arr[indexc];
}
// Driver Code
static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int[] freq = { 10, 5, 20, 100 };
int i, n = arr.Length;
// Let us generate 10 random numbers
// according to given distribution
for(i = 0; i < 5; i++)
Console.WriteLine(myRand(arr, freq, n));
}
}
UPDATE:
I ran this code to check it:
int[] intArray = new int[] { 1, 2, 3, 4, 5 };
int[] weights = new int[] { 5, 20, 20, 40, 15 };
List<int> results = new List<int>();
for (int i = 0; i < 100000; i++)
{
results.Add(WeightedRNG.GetRand(intArray, weights, intArray.Length));
}
for (int i = 0; i < intArray.Length; i++)
{
int itemsFound = results.Where(x => x == intArray[i]).Count();
Console.WriteLine($"{intArray[i]} was returned {itemsFound} times.");
}
And here are the results with the constant:
1 was returned 5096 times.
2 was returned 19902 times.
3 was returned 20086 times.
4 was returned 40012 times.
5 was returned 14904 times.
And without the constant...
1 was returned 100000 times.
2 was returned 0 times.
3 was returned 0 times.
4 was returned 0 times.
5 was returned 0 times.
It completely breaks without it.

The constant does serve a purpose in some environments, but I don't believe this code is correct for C#.
To explain, let's look at the arguments to the function. The first sign something is off is passing n as an argument instead of inferring it from the arrays. The second sign is it's poor practice in C# to deal with paired arrays rather than something like a 2D array or sequence of single objects (such as a Tuple). But those are just indicators something is odd, and not evidence of any bugs.
So let's put that on hold for a moment and explain why a constant might matter by looking a small example.
Say you have three numbers (1, 2, and 3) with weights 3, 2, and 2. This function first builds up a prefix array, where each item includes the chances of finding the number for that index and all previous numbers.
We end up with a result like this: (3, 5, 7). Now we can use the last value and take a random number from 1 to 7. Values 1-3 result in 1, values 4 and 5 result in 2, and values 6 and 7 result in 3.
To find this random number the code now calls rand.Next(), and this is where I think the error comes in. In many platforms, the Next() function returns a double between 0 and 1. That's too small to use to lookup your weighted value, so you then multiply by a prime constant related the platform's epsilon value to ensure you have a reasonably large result that will cover the entire possible range of desired weights (in this case, 1-7) and then some. Now you take the remainder (%) vs your max weight (7), and map it via the prefix array to get the final result.
So the first error is, in C#, .Next() does not return a double. It is documented to return a non-negative random integer between 0 and int.MaxValue. Multiply that by 323567 and you're asking for integer overflow exceptions. Another sign of this mistake is the cast to int: the result of this function is already an int. And let's not even talk the meaningless extra parentheses around (323567).
There is also another, more subtle, error.
Let's the say the result of the (int)(rand.Next() * 323567) expression is 10. Now we take this value and get the remainder when dividing by our max value (%7). The problem here is we have two chances to roll a 1, 2, or 3 (the extra chance is if the original was 8, 9, or 10), and only once chance for the remaining weights (4-7). So we have introduced unintended bias into the system, completely throwing off the expected weights. (This is a simplification. The actual number space is not 1-10; it's 0 * 323567 - 0.999999 * 323567. But the potential for bias still exists as long that max value is not evenly divisible by the max weight value).
It's possible the constant was chosen because it has properties to minimize this effect, but again: it was based on a completely different kind of .Next() function.
Therefore the rand.Next() line should probably be changed to look like this:
int r = rand.Next(prefix[n - 1]) +1;
or this:
int r = ((int)(rand.NextDouble() * (323567 * prefix[n - 1])) % prefix[n - 1]) + 1;
But, given the other errors, I'd be wary of using this code at all.
For fun, here's an example running several different options:
https://dotnetfiddle.net/Y5qhRm
The original random method (using NextDouble() and a bare constant) doesn't fare as badly as I'd expect.

C# - How to place a given number of random mines in an array

I'm new to c#. I have a task to make a type of minesweeper, but which immediately opens a solution.
static void Main(string[] args)
{
Console.Write("Enter the width of the field: ");
int q = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the length of the field: ");
int w = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the number of bombs: ");
int c = Convert.ToInt32(Console.ReadLine());
Random rand = new Random();
var charArray = new char[q, w];
var intArray = new int[q, w];
for (int i = 0; i < q; i++)
{
for (int j = 0; j < w; j++)
{
intArray[i, j] = rand.Next(2);
charArray[i, j] = intArray[i, j] == 0 ? '_' : '*';
Console.Write(charArray[i, j]);
}
Console.WriteLine();
}
}
}
}
Two arrays should be output. Everything should be closed on the first one, that is, there should be only the characters: _ and *
0 - these are places without mines, I replaced them with a symbol _
1 - these are places with mines, I replaced them with an asterisk symbol, but they do not accept the number of mines entered by the user. And it is necessary that there are as many "*" characters as there are mines.
And in the second array there should be an already open solution of the game. That is, the cells next to which there are mines should take a number meaning the number of mines next to this cell.
Please help me..
Compiling the current code

Random random = new Random();
while(c > 0)
{
var rq = random.Next(q);
var rw = random.Next(w);
if(intArray[rq,rw] == 0)
{
intArray[rq, rw] = 1;
c--;
}
}

I would suggest dividing the problem in smaller manageable chunks. For instance, you can place the bombs in a initial step, and on a second step build the solution. You can build the solution at the same time you place the bombs, although for clarity you can do it after.
Naming of variables is also important. If you prefer using single letter variable names, I believe that's fine for the problem limits, however I would use meaningful letters easier to remember. eg: W and H for the width and height of the board, and B for the total number of bombs.
The first part of the problem then can be described as placing B bombs in a WxH board. So instead of having nested for statements that enumerate WxH times, it's better to have a while loop that repeats the bomb placing logic as long as you have remaining bombs.
Once you generate a new random location on the board, you have to check you haven't placed a bomb there already. You can have an auxiliary function HasBomb that checks that:
bool HasBomb(char[,] charArray, int x, int y)
{
return charArray[x,y] == '*';
}
I'll leave error checking out, this function being private can rely on the caller sending valid coordinates.
Then the bomb placing procedure can be something like:
int remainingBombs = B;
while (remainingBombs > 0)
{
int x = rand.Next(W);
int y = rand.Next(H);
if (!HasBomb(charArray, x, y)
{
charArray[x,y] = '*';
remainingBombs--;
}
}
At this point you may figure out another concern. If the number B of bombs to place is larger than the available positions on the board WxH, then you wont be able to place the bombs on the board. You'll have to check for that restriction when requesting the values for W, H and B.
Then in order to create the array with the number of bombs next to each position, you'll need some way to check for all the neighbouring positions to a given one. If the position is in the middle of the board it has 8 neighbour positions, if it's on an edge it has 5, and if it's on a corner it has 3. Having a helper function return all the valid neighbour positions can be handy.
IEnumerable<(int X, int Y)> NeighbourPositions(int x, int y, int W, int H)
{
bool leftEdge = x == 0;
bool topEdge = y == 0;
bool rightEdge = x == W - 1;
bool bottomEdge = y == H - 1;
if (!leftEdge && !topEdge)
yield return (x-1, y-1);
if (!topEdge)
yield return (x, y-1);
if (!rightEdge && !topEdge)
yield return (x+1, y-1);
if (!leftEdge)
yield return (x-1, y);
if (!rightEdge)
yield return (x+1, y);
if (!leftEdge && !bottomEdge)
yield return (x-1, y+1);
if (!bottomEdge)
yield return (x, y+1);
if (!rightEdge && !bottomEdge)
yield return (x+1, y+1)
}
This function uses Iterators and touples. If you feel those concepts are too complex as you said are new to C#, you can make the function return a list with coordinates instead.
Now the only thing left is to iterate over the whole intArray and increment the value on each position for each neighbour bomb you find.
for (int x = 0; x < W; x++)
{
for (int y = 0; y < H; y++)
{
foreach (var n in NeighbourPositions(x, y, W, H))
{
if (HasBomb(charArray, n.X, n.Y))
intArray[x,y]++;
}
}
}

The answers here are mostly about generating random x and random y put in loop and trying to put the mine into empty cells. It is ok solution, but if you think of it, it is not that sufficient. Every time you try to find a new random cell, there is chance that cell is already a mine. This is pretty much alright, if you don't put too many mines into your field, but if you put some greater number of mines, this event can occur quite often. This means that the loop might take longer than usually. Or, theoretically, if you wanted to put 999 mines into 1000 cell field, it would be really hard for the loop to fill all the necessary cells, especially for the last mine. Now, I am not saying that the solutions here are bad, I think, it's really alright solution for many people. But if someone wanted a little bit efficient solution, I have tried to crate it.
Solution
In this solution, you iterate each cell and try to calculate a probability of the mine to be placed there. I have come up with this easy formula, which calculates the probability:
Every time you try to place a mine into one cell, you calculate this formula and compare it to random generated number.
bool isMine = random.NextDouble() < calcProbability();

multiplicative persistence - recursion?

I'm working on this:
Write a function, persistence, that takes in a positive parameter num
and returns its multiplicative persistence, which is the number of
times you must multiply the digits in num until you reach a single
digit.
For example:
persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit
persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2
persistence(4) == 0 // because 4 is already a one-digit number
This is what I tried:
public static int Persistence(long n)
{
List<long> listofints = new List<long>();
while (n > 0)
{
listofints.Add(n % 10);
n /= 10;
}
listofints.Reverse();
// list of a splited number
int[] arr = new int[listofints.Count];
for (int i = 0; i < listofints.Count; i++)
{
arr[i] = (int)listofints[i];
}
//list to array
int pro = 1;
for (int i = 0; i < arr.Length; i++)
{
pro *= arr[i];
}
// multiply each number
return pro;
}
I have a problem with understanding recursion - probably there is a place to use it. Can some1 give me advice not a solution, how to deal with that?

It looks like you've got the complete function to process one iteration. Now all you need to do is add the recursion. At the end of the function call Persistence again with the result of the first iteration as the parameter.
Persistence(pro);
This will recursively call your function passing the result of each iteration as the parameter to the next iteration.
Finally, you need to add some code to determine when you should stop the recursion, so you only want to call Persistence(pro) if your condition is true. This way, when your condition becomes false you'll stop the recursion.
if (some stop condition is true)
{
Persistence(pro);
}

Let me take a stab at explaining when you should consider using a recursive method.
Example of Factorial: Factorial of n is found by multiplying 1*2*3*4*..*n.
Suppose you want to find out what the factorial of a number is. For finding the answer, you can write a foreach loop that keeys multiplying a number with the next number and the next number until it reaches 0. Once you reach 0, you are done, you'll return your result.
Instead of using loops, you can use Recursion because the process at "each" step is the same. Multiply the first number with the result of the next, result of the next is found by multiplying that next number with the result of the next and so on.
5 * (result of rest)
4 * (result of rest )
3 * (result of rest)
...
1 (factorial of 0 is 1).---> Last Statement.
In this case, if we are doing recursion, we have a terminator of the sequence, the last statement where we know for a fact that factorial of 0 = 1. So, we can write this like,
FactorialOf(5) = return 5 * FactorialOf(4) = 120 (5 * 24)
FactorialOf(4) = return 4 * FactorialOf(3) = 24 (4 * 6)
FactorialOf(3) = return 3 * FactorialOf(2) = 6 (3 * 2)
FactorialOf(2) = return 2 * FactorialOf(1) = 2 (2 * 1)
FactorialOf(1) = return 1 * FactorialOf(0) = 1 (1 * 1)
FactorialOf(0) = Known -> 1.
So, it would make sense to use the same method over and over and once we get to our terminator, we stop and start going back up the tree. Each statement that called the FactorialOf would start returning numbers until it reaches all the way to the top. At the top, we will have our answer.
Your case of Persistence
It calls for recursive method as well as you are taking the result and doing the same process on it each time.
Persistence(39) (not single) = return 1 + Persistence(3 * 9 = 27) = 3
Persistence(27) (not single) = return 1 + Persistence(2 * 7 = 14) = 2
Persistence(14) (not single) = return 1 + Persistence(1 * 4 = 4) = 1
Persistence(4) (single digit) = Known -> 0 // Terminator.
At the end of the day, if you have same process performed after each calculation / processing with a termination, you can most likely find a way to use recursion for that process.

You definitely can invoke your multiplication call recursively.
You will need initial sate (0 multiplications) and keep calling your method until you reach your stop condition. Then you return the last iteration you've got up to as your result and pass it through all the way up:
int persistence(int input, int count = 0) {} // this is how I would define the method
// and this is how I see the control flowing
var result = persistence(input: 39, count: 0) {
//write a code that derives 27 out of 39
//then keep calling persistence() again, incrementing the iteration count with each invocation
return persistence(input: 27, count: 1) {
return persistence(input: 14, count: 2) {
return persistence(input: 4, count: 3) {
return 3
}
}
}
}
the above is obviously not a real code, but I'm hoping that illustrates the point well enough for you to explore it further

Designing a simple recursive solution usually involves two steps:
- Identify the trivial base case to which you can calculate the answer easily.
- Figure out how to turn a complex case to a simpler one, in a way that quickly approaches the base case.
In your problem:
- Any single-digit number has a simple solution, which is persistence = 1.
- Multiplying all digits of a number produces a smaller number, and we know that the persistence of the bigger number is greater than the persistence of the smaller number by exactly one.
That should bring you to your solution. All you need to do is understand the above and write that in C#. There are only a few modifications that you need to make in your existing code. I won't give you a ready solution as that kinda defeats the purpose of the exercise, doesn't it. If you encounter technical problems with codifying your solution into C#, you're welcome to ask another question.

public int PerRec(int n)
{
string numS = n.ToString();
if(numS.Length == 1)
return 0;
var number = numS.ToArray().Select(x => int.Parse(x.ToString())).Aggregate((a,b) => a*b);
return PerRec(number) + 1;
}
For every recursion, you should have a stop condition(a single digit in this case).
The idea here is taking your input and convert it to string to calculate that length. If it is 1 then you return 0
Then you need to do your transformation. Take all the digits from the string representation(in this case from the char array, parse all of them, after getting the IEnumerable<int>, multiply each digit to calculate the next parameter for your recursion call.
The final result is the new recursion call + 1 (which represents the previous transformation)
You can do this step in different ways:
var number = numS.ToArray().Select(x => int.Parse(x.ToString())).Aggregate((a,b) => a*b);
convert numS into an array of char calling ToArray()
iterate over the collection and convert each char into its integer representation and save it into an array or a list
iterate over the int list multiplying all the digits to have the next number for your recursion
Hope this helps

public static int Persistence(long n)
{
if (n < 10) // handle the trivial cases - stop condition
{
return 0;
}
long pro = 1; // int may not be big enough, use long instead
while (n > 0) // simplify the problem by one level
{
pro *= n % 10;
n /= 10;
}
return 1 + Persistence(pro); // 1 = one level solved, call the same function for the rest
}
It is the classic recursion usage. You handle the basic cases, simplify the problem by one level and then use the same function again - that is the recursion.
You can rewrite the recursion into loops if you wish, you always can.

Getting top values from a list in c sharp

So I'm having some trouble with a C# program that is meant to sum the 8 highness value from a list.
The program works by declaring a variable currenthigh, which stores a value from the gradelist. It then compares itself to the value abshigh to see if it is greater than the established highest value. If it is, it sets currenthigh as the new highest value.
Once the loop has run through the list and confirmed the highest value, it adds it to the uppertotal variable and uses the ejector variable to remove it from the list. The program then iterates, this time without the previous highest value. It iterates 8 times so that in the end the top 8 values are added to uppertotal.
Trouble is, the highest variable remains in the list, despite the code having instructions to delete it, so it just adds the highest value to itself 8 times.
int currenthigh = 0;
int abshigh = 0;
int ejector = 0;
int uppertotal = 0;
for (int g = 0; g < 8; g++)
{
for (int z = 0; z < gradelist.Count; z++)
{
Console.WriteLine("PASS STARTED");
currenthigh = Convert.ToInt32((gradelist[z]));
Console.WriteLine("currenthigh" + currenthigh);
if (currenthigh > abshigh)
{
abshigh = currenthigh;
ejector = z;
}
}
Console.WriteLine("ejector" + ejector);
uppertotal = uppertotal + currenthigh;
gradelist.RemoveAt(ejector);
Console.WriteLine("PASS COMPLETE");
Console.WriteLine("RESETING");
}
Note - gradelist is a list of integers containing at least 12 items at all time.

This happens because you do not remove the highest value from gradelist. Pay attention, you put Z in ejector, but Z is an index in gradelist and when you try to remove it, you remove nothing because there is no Z in gradelist as a member! Instead of
gradelist.RemoveAt(ejector);
you should do this:
gradelist.RemoveAt(gradelist[ejector]);
But I'd recommend you completely different approach. If you just want to get your uppertotal which is the sum of top 8 members in gradlist, use this simple trick:
uppertotal += gradelist.OrderByDescending(p => p).Take(8).Sum();

Your code is extremely big for the relatively simple task.
You can select the top X of a list as follows:
top_list = gradelist.OrderByDescending(i => i).Take(X)
When working with lists/collections System.Linq is your friend

parallel bidirectional selection sort

I wrote code to implement a bidirectional selection sort in parallel. I used c#, and the the parallel.invoke function. 2 loops were invoked in parallel, one to find the minimum, and one to find the max. Yet, it doesn't sort. I was wondering is the problem because this type of sort can't handle being done in parallel, since each loop relies on data existing in the other loop?...or is there simply something wrong with my code?
Parallel.Invoke(
() =>
{
for (int i=0; i < array.Length / 2; i++)
{
int m;
int min = i;
for (m = i + 1; m < array.Length - 1; m++)
if (array[m] < array[min])
min = m;
//swap
int temp = array[min];
array[min] = array[m];
array[m] = temp;
}
},
() =>
{
for (int m = 0; m < array.Length / 2; m++)
{
int length = array.Length - 1;
int max = length - m;
int k;
for (k = length--; k > 0; k--)
if (array[k] > array[max])
max = k;
//swap
int temp = array[max];
array[max] = array[k];
array[k] = temp;
}
});

I think it's easier if you search the minimum and maximum within the same loop in 1 thread: (java-code, but I assume you'll understand the principle)
int minimum, maximum;
int k = size();
for(int i = 0; i < size(); ++i)
{
--k;
minimum = i;
maximum = k;
if(k - i <= 0)
break;
for(int j = i; j <= k; ++j)
{
if(greaterThan(minimum, j))
minimum = j;
if(lessThan(maximum, j))
maximum = j;
}
if(minimum != i)
{
swap(minimum, i);
if(maximum == i)
maximum = minimum;
}
if(maximum != k)
swap(maximum, k);
}
The problem with your code is this:
Say this is the array:
[5, 4, 3, 2, 1]
Iteration 0: The first thread will find the smallest element to put on index 0
The first thread finds the minimum element at index 4
Iteration 0: The second thread will find the largest element to put on index 4
The second thread finds the maximum element at index 0
You will already see that this will not end well, as both threads will perform a swap between index 0 and 4 resulting in the same situation as it is now.
Another problem is if your first thread is looping from m -> array.length - 1. If at the same time thread 2 moves the minimum element (which it doesn't need, because it's searching the maximum) from index k to "max" via a swap. With index "max" being < "m". That means the first thread will never find the next minimum value because it was moved before its position.
EDIT: After consideration, I don't think it's possible to implement a straightforward parallel version of selection sort. The version I first recommended was indeed not going to work due to the algorithm finding the same minimum every time because it didn't change the input-array.
What is possible is to only perform selection sort with thread 1 on the first half of the array (and only allow it to find the minimum in that half) and the second half of the array is for the second thread. And then in the end you can merge both halfs with a mergesort-algorithm.
This way you can always use more than 2 threads; say "p" amount of threads for example. Each responsible for N/p part of the input array with "N" being the inputsize. And in the end you just merge every part with a mergesort-algorithm. I never implemented it myself, so I can't say if it would be efficient, but I assume there will be better algorithms to parallelize out there (like mergesort itself).
PS: About the code posted above. I assume everything seems rather straightforward except this part:
if(maximum == i)
maximum = minimum;
That's to deal with a situation like this:
. . . i . . . k
[1, 4, 3, 1, 5]
so with i = 1 and k = 3 (the indices).
The algorithm will find:
maximum = index 1
minimum = index 3
After swapping the minimum value with the one on index i, the situation changes like this:
. . . i . . . k
[1, 1, 3, 4, 5]
Meaning the maximum value (integer 4) actually moved from index "i" to index "minimum". If we would perform a swap(maximum, k), it would have a bad result. That's why we need to update the index of the maximum element if it was positioned at index i.