We Keep Coding

Calculate the sequence : n!/ (1+1/2+1/3+...+1/n)

Calculate the following sum
1!/1 + 2!/(1+1/2) + 3!/(1+1/2+1/3) + ... + n!/ (1+1/2+1/3+...+1/n), where n > 0.
public static double GetSumSix(int n)
{
double i, result = 0.0;
static double factorial(double n)
{
double res = 1;
for (double i = 2; i <= n; i++)
res *= i;
return res;
}
for (i = 1.0; i <= n; i++)
{
result += factorial(i) / (1.0 / i);
}
return result;
}
Help me please , I don't understand why is my solution not working?

Your denominator logic is incorrect. You could create another function to work out what '1/1+1/2+...+1/n' is and use that in the denominator? right now your code will work out 1+2!*2+3!*3+...
You could actually use something similar to your factorial method
static double GetDenominator(double n)
{
double res = 1;
for (double i = 2; i <= n; i++)
//insert code here
return res;
}

The Lemon's answer is correct, you're not accumulating the denominator of the sequence so what you were calculating was:
f(n) = 1!/1 + 2!/(1/2) + 3!/(1/3) + ... n!/(1/n)
Since both the numerator and denominator of each term are algorithmically linked to the values in the prior term you can simply update them each pass through the loop. This is (slightly) faster and fairly easy to read.
public static double GetSumSix(int n)
{
double factorial = 1;
double denominator = 1;
double accum = 1;
for (int i = 2; i <= n; i++)
{
factorial *= i;
denominator += 1.0d/i;
accum += factorial / denominator;
}
return accum;
}

Your logic is not correct as per your question , also your code won't execute as you have a function inside your GetSumSix function. I have put some helping points in below code so you will understand how the logic works.
using System;
public class Program
{
public static void Main()
{
var Calculate = GetSumSix(3);
Console.WriteLine("The Answer is " + Calculate);
}
public static double GetSumSix(int n)
{
int i;
double result = 0.0;
int factorial = 1;
string calculatedFormula = String.Empty;
string FinalFormat = String.Empty;
//Find n!
for(int x=n;x>=1;x--)
{
factorial *= x;
}
// Find Denominator (1+1/2+1/3+…+1/n)
for (i = 1.0; i <= n; i++)
{
result += GetDenominator(i, ref calculatedFormula);
FinalFormat += calculatedFormula;
}
result = factorial/result;
Console.WriteLine("Calculated Formula is:"+ factorial +"/(" + FinalFormat +")When N is :" + n);
return result;
}
public static double GetDenominator(double n, ref string cal)
{
if (n == 1)
{
cal += n + "+ ";
return 1;
}
else
{
cal = "1/" + n + "+ ";
return 1 / n;
}
}
}
Thanks.

Incrementing t2 variables

I have 2 variables x and y ,and they have to increment by 1, the tow variables start in 1, first x has to increment by 1,, until it reaches 25 and y stay at 1. Once it reaches 25, x has to go back to 1 and y increment to y=2, and the repeat (once x has reached 25 again y will increment by 1). here my implementation but isn't working...
using System;
namespace ConsoleApp3
{
class Program
{
static void Main(string[] args)
{
int x;
int y = 1;
for (x = 1; x < 26; x++)
{
if (x == 25)
{
x = 1;
for (y = 1; y < 30; y++)
{
Console.WriteLine("X = " + x + ", Y = " + y);
}
}
}
}
}
}

It's not standard practice to manipulate x in a for loop. I assume you only want both variables going to 25 (edit: y now goes to 30 like in OP). So the trick is to put y on the outside and x on the inside.
In this case it's easiest to still use for loops instead of while loops.
Try something like this:
using System;
namespace ConsoleApp1
{
class Program
{
static void Main(string[] args)
{
for(int y = 0; y < 30; y++) //y goes up 1 every time x goes up 25
{
for(int x = 0; x < 25; x++) //counts x to 25
{
Console.WriteLine("X=" + x + " Y=" + y);
}
}
}
}
}

You can use a little bit of recursivity for that too (just adding this for more infos ).
public static void run25(ref int x, ref int y)
{
if(x <= 25)
{
Console.WriteLine($"x = {x}");
x++;
run25(ref x, ref y);
}
if (y <= 25)
{
Console.WriteLine($" >> y = {y}");
y++;
x = 1;
run25(ref x, ref y);
}
else return ;
}
you can use that as follow :
public static void Main (string[] args)
{
int _x = 1;
int _y = 1;
run25(ref _x, ref _y); // once it's done , _x and _y values are 26 .
}

using System;
public class Program
{
public static void Main()
{
for(var y = 1; y <=30 ;y++)
{
for(var x = 1; x<=25;x++)
{
Console.WriteLine("x = {0}, y = {1}", x, y);
}
}
}
}

Laplace Transform And Getting The Frequent Value For Gyro

I'm getting x,y,z values from gyro-sensor. Each variable is being sent 10 values per second. In 3 seconds I have;
x=[30values]
y=[30values]
z=[30values]
Some of the values are too different from the others cause of noise. With laplace transform I need to get the most frequent value from my array.
I need to filter the array with Laplace Transform equation. I need to build the equation in C#. How can I implement the array with the equation?

Since this kind of filter (Laplace) is very specialized to certain area of Engineering and needs a person who has good understanding on both the programming language (in this case is C#) and the filter itself, I would recommend you to use such source, rather than code the filter by yourself.
Here is the snippet of the source code:
class Laplace
{
const int DefaultStehfest = 14;
public delegate double FunctionDelegate(double t);
static double[] V; // Stehfest coefficients
static double ln2; // log of 2
public static void InitStehfest(int N)
{
ln2 = Math.Log(2.0);
int N2 = N / 2;
int NV = 2 * N2;
V = new double[NV];
int sign = 1;
if ((N2 % 2) != 0)
sign = -1;
for (int i = 0; i < NV; i++)
{
int kmin = (i + 2) / 2;
int kmax = i + 1;
if (kmax > N2)
kmax = N2;
V[i] = 0;
sign = -sign;
for (int k = kmin; k <= kmax; k++)
{
V[i] = V[i] + (Math.Pow(k, N2) / Factorial(k)) * (Factorial(2 * k)
/ Factorial(2 * k - i - 1)) / Factorial(N2 - k)
/ Factorial(k - 1) / Factorial(i + 1 - k);
}
V[i] = sign * V[i];
}
}
public static double InverseTransform(FunctionDelegate f, double t)
{
double ln2t = ln2 / t;
double x = 0;
double y = 0;
for (int i = 0; i < V.Length; i++)
{
x += ln2t;
y += V[i] * f(x);
}
return ln2t * y;
}
public static double Factorial(int N)
{
double x = 1;
if (N > 1)
{
for (int i = 2; i <= N; i++)
x = i * x;
}
return x;
}
}
coded by Mr. Walt Fair Jr. in CodeProject.

Reorder digits in integer using C#

I want to ask how I can reorder the digits in an Int32 so they result in the biggest possible number.
Here is an example which visualizes what I am trying to do:
2927466 -> 9766422
12492771 -> 97742211
I want to perform the ordering of the digits without using the System.Linq namespace and without converting the integer into a string value.
This is what I got so far:
public static int ReorderInt32Digits(int v)
{
int n = Math.Abs(v);
int l = ((int)Math.Log10(n > 0 ? n : 1)) + 1;
int[] d = new int[l];
for (int i = 0; i < l; i++)
{
d[(l - i) - 1] = n % 10;
n /= 10;
}
if (v < 0)
d[0] *= -1;
Array.Sort(d);
Array.Reverse(d);
int h = 0;
for (int i = 0; i < d.Length; i++)
{
int index = d.Length - i - 1;
h += ((int)Math.Pow(10, index)) * d[i];
}
return h;
}
This algorithm works flawlessly but I think it is not very efficient.
I would like to know if there is a way to do the same thing more efficiently and how I could improve my algorithm.

You can use this code:
var digit = 2927466;
String.Join("", digit.ToString().ToCharArray().OrderBy(x => x));
Or
var res = String.Join("", digit.ToString().ToCharArray().OrderByDescending(x => x) );

Not that my answer may or may not be more "efficient", but when I read your code you calculated how many digits there are in your number so you can determine how large to make your array, and then you calculated how to turn your array back into a sorted integer.
It would seem to me that you would want to write your own code that did the sorting part without using built in functionality, which is what my sample does. Plus, I've added the ability to sort in ascending or descending order, which is easy to add in your code too.
UPDATED
The original algorithm sorted the digits, now it sorts the digits so that the end result is the largest or smallest depending on the second parameter passed in. However, when dealing with a negative number the second parameter is treated as opposite.
using System;
public class Program
{
public static void Main()
{
int number1 = 2927466;
int number2 = 12492771;
int number3 = -39284925;
Console.WriteLine(OrderDigits(number1, false));
Console.WriteLine(OrderDigits(number2, true));
Console.WriteLine(OrderDigits(number3, false));
}
private static int OrderDigits(int number, bool asc)
{
// Extract each digit into an array
int[] digits = new int[(int)Math.Floor(Math.Log10(Math.Abs(number)) + 1)];
for (int i = 0; i < digits.Length; i++)
{
digits[i] = number % 10;
number /= 10;
}
// Order the digits
for (int i = 0; i < digits.Length; i++)
{
for (int j = i + 1; j < digits.Length; j++)
{
if ((!asc && digits[j] > digits[i]) ||
(asc && digits[j] < digits[i]))
{
int temp = digits[i];
digits[i] = digits[j];
digits[j] = temp;
}
}
}
// Turn the array of digits back into an integer
int result = 0;
for (int i = digits.Length - 1; i >= 0; i--)
{
result += digits[i] * (int)Math.Pow(10, digits.Length - 1 - i);
}
return result;
}
}
Results:
9766422
11224779
-22345899
See working example here... https://dotnetfiddle.net/RWA4XV

public static int ReorderInt32Digits(int v)
{
var nums = Math.Abs(v).ToString().ToCharArray();
Array.Sort(nums);
bool neg = (v < 0);
if(!neg)
{
Array.Reverse(nums);
}
return int.Parse(new string(nums)) * (neg ? -1 : 1);
}

This code fragment below extracts the digits from variable v. You can modify it to store the digits in an array and sort/reverse.
int v = 2345;
while (v > 0) {
int digit = v % 10;
v = v / 10;
Console.WriteLine(digit);
}
You can use similar logic to reconstruct the number from (sorted) digits: Multiply by 10 and add next digit.

I'm posting this second answer because I think I got the most efficient algorithm of all (thanks for the help Atul) :)
void Main()
{
Console.WriteLine (ReorderInt32Digits2(2927466));
Console.WriteLine (ReorderInt32Digits2(12492771));
Console.WriteLine (ReorderInt32Digits2(-1024));
}
public static int ReorderInt32Digits2(int v)
{
bool neg = (v < 0);
int mult = neg ? -1 : 1;
int result = 0;
var counts = GetDigitCounts(v);
for (int i = 0; i < 10; i++)
{
int idx = neg ? 9 - i : i;
for (int j = 0; j < counts[idx]; j++)
{
result += idx * mult;
mult *= 10;
}
}
return result;
}
// From Atul Sikaria's answer
public static int[] GetDigitCounts(int n)
{
int v = Math.Abs(n);
var result = new int[10];
while (v > 0) {
int digit = v % 10;
v = v / 10;
result[digit]++;
}
return result;
}

Brute force Algorithm for creation of Sudoku Board

What I am developing is that initially the entire sudoku board is empty.
One of the random cells(out of 81) is filled with a random value(1-9).
Now I want to fill all the remaining cells using brute force approach.
From what I came to know after googling is that we should start with the first cell and fill it with 1(if it's valid), then fill the second cell with 2(if it's valid, we will begin checking with a number greater than the last filled cell, which in this case is 1, once we reach 9, we reset it with 1).
The thing is that it's not working properly!
Can anyone link me to the exact algorithm.

Here's an implementation of the backtracking approach:
import java.util.Random;
public class Sudoku {
public static void main(String[] args) {
Random rand = new Random();
int r = rand.nextInt(9);
int c = rand.nextInt(9);
int value = rand.nextInt(9) + 1;
Board board = new Board();
board.set(r, c, value);
System.out.println(board);
solve(board, 0);
System.out.println(board);
}
private static boolean solve(Board board, int at) {
if (at == 9*9)
return true;
int r = at / 9;
int c = at % 9;
if (board.isSet(r, c))
return solve(board, at + 1);
for (int value = 1; value <= 9; value++) {
if (board.canSet(r, c, value)) {
board.set(r, c, value);
if (solve(board, at + 1))
return true;
board.unSet(r, c);
}
}
return false;
}
static class Board {
private int[][] board = new int[9][9];
private boolean[][] rs = new boolean[9][10];
private boolean[][] cs = new boolean[9][10];
private boolean[][][] bs = new boolean[3][3][10];
public Board() {}
public boolean canSet(int r, int c, int value) {
return !isSet(r, c) && !rs[r][value] && !cs[c][value] && !bs[r/3][c/3][value];
}
public boolean isSet(int r, int c) {
return board[r][c] != 0;
}
public void set(int r, int c, int value) {
if (!canSet(r, c, value))
throw new IllegalArgumentException();
board[r][c] = value;
rs[r][value] = cs[c][value] = bs[r/3][c/3][value] = true;
}
public void unSet(int r, int c) {
if (isSet(r, c)) {
int value = board[r][c];
board[r][c] = 0;
rs[r][value] = cs[c][value] = bs[r/3][c/3][value] = false;
}
}
public String toString() {
StringBuilder ret = new StringBuilder();
for (int r = 0; r < 9; r++) {
for (int c = 0; c < 9; c++)
ret.append(board[r][c]);
ret.append("\n");
}
return ret.toString();
}
}
}

I used a method without backtracing, although the while loop might be it. To quote an algorithm book I've read "Nothing in recursion can't be duplicated using iteration".
I've been using my eyes to inspect this, and since I can't wrap my head around the recursive method, even though recursion is relatively understood:
This method, I kinda wrote with some guidance, had a bug in the grid checker, when I found it, it seems to be working now. I'm positing it 'cause it's hard to find complete-and-working code. IOS SDK.
#define WIDTH 9
#define HEIGHT 9
#interface ViewController ()
//- (BOOL) numberConflicts:(int)testNum;
- (BOOL) number:(int)n conflictsWithRow:(int)r;
- (BOOL) number:(int)n conflictsWithColumn:(int)c;
- (BOOL) number:(int)n conflictsWithSquareInPointX:(int)x andPointY:(int)y;
- (BOOL) number:(int)n conflictsAtGridPointX:(int)xPoint andPointY:(int)yPoint;
- (int) incrementSudokuValue:(int)v;
#end
static int sudoku[WIDTH][HEIGHT];
#implementation ViewController
- (void)viewDidLoad
{
[super viewDidLoad];
/// Initialize it
for (int x = 0; x < WIDTH; x++)
{
for (int y = 0; y < HEIGHT; y++)
{
sudoku[x][y] = 0;
}
}
///
int tries = 0;
for (int j = 0; j < HEIGHT; j++)
{
for (int i = 0; i < WIDTH; i++)
{
int num = arc4random()%9 + 1;
while ([self number:num conflictsAtGridPointX:i andPointY:j])
{
num = [self incrementSudokuValue:num];
tries++;
if (tries > 10) { //restart the column
tries = 0;
for(int count = 0; count < WIDTH; count++)
{
sudoku[count][j] = 0;
}
i = 0;
}
}
if(sudoku[i][j] == 0)
sudoku[i][j] = num;
tries = 0;
for (int y = 0; y < HEIGHT; y++)
{
for (int x = 0; x < WIDTH; x++)
{
printf("%i ", sudoku[x][y]);
}
printf("\n");
}
printf("\n");
}
}
for (int x = 0; x < WIDTH; x++)
{
for (int y = 0; y < HEIGHT; y++)
{
printf("%i ", sudoku[y][x]);
}
printf("\n"); //newline
}
// Do any additional setup after loading the view, typically from a nib.
}
- (void)didReceiveMemoryWarning
{
[super didReceiveMemoryWarning];
// Dispose of any resources that can be recreated.
}
- (BOOL) number:(int)n conflictsWithRow:(int)r;
{
for (int y = 0; y < HEIGHT; y++) {
if (sudoku[y][r] == n) {
return YES;
}
}
return NO;
}
- (BOOL) number:(int)n conflictsWithColumn:(int)c;
{
for (int x = 0; x < WIDTH; x++) {
if (sudoku[c][x] == n) {
return YES;
}
}
return NO;
}
- (BOOL) number:(int)n conflictsAtGridPointX:(int)xPoint andPointY:(int)yPoint;
{
if ([self number:n conflictsWithRow:yPoint])
{
return YES;
}
if ([self number:n conflictsWithColumn:xPoint])
{
return YES;
}
if ([self number:n conflictsWithSquareInPointX:xPoint andPointY:yPoint]) {
return YES;
}
return NO;
}
- (BOOL) number:(int)n conflictsWithSquareInPointX:(int)x andPointY:(int)y;
{
int leftX = x - (x % 3); //used to use int division
// leftX *= 3;
int topY = y - (y % 3);
// topY *= 3;
int rightX = leftX + 2;
int bottomY = topY + 2;
for(int subY = topY; subY <= bottomY; subY++) //bug was here, used < instead of less N equal to...
{
for ( int subX = leftX; subX <= rightX; subX++)
{
if (sudoku[subX][subY] == n) {
return YES;
}
}
}
NSLog(#"Testing grid at %i, %i", x/3, y/3);
NSLog(#"LeftX: %i TopY: %i", leftX, topY);
return NO;
}
- (int) incrementSudokuValue:(int)v;
{
if (v < 9) {
v++;
return v;
}
return 1;
}
Note: The header file is empty, paste this into iOS single View application if you desire.
Caution: might loop infinitely( and above does sometimes, but is very fast), may want another more global "tries" variable, and restart the algorithm as a safety, or give it a seed/do both
edit: the below should be safe from infinite loops, if the source grid is solvable (or nonexistant)
#define WIDTH 9
#define HEIGHT 9
#interface ViewController ()
//- (BOOL) numberConflicts:(int)testNum;
- (BOOL) number:(int)n conflictsWithRow:(int)r;
- (BOOL) number:(int)n conflictsWithColumn:(int)c;
- (BOOL) number:(int)n conflictsWithSquareInPointX:(int)x andPointY:(int)y;
- (BOOL) number:(int)n conflictsAtGridPointX:(int)xPoint andPointY:(int)yPoint;
- (int) incrementSudokuValue:(int)v;
#end
static int sudoku[WIDTH][HEIGHT];
#implementation ViewController
- (BOOL) fillGridWithNext:(int)next;
{
for (int y = 0; y < HEIGHT; y++)
{
for (int x = 0; x < WIDTH; x++)
{
if (sudoku[x][y] != 0)
{
if (x == 8 && y == 8) {
return YES;
}
continue;
}
for (int count = 0; count < (HEIGHT-1); count++)
{
if ([self number:next conflictsAtGridPointX:x andPointY:y])
{
next = [self incrementSudokuValue:next];
}
else
{
sudoku[x][y] = next;
if( [self fillGridWithNext:arc4random()%9+1])
{
return YES;
}
}
}
sudoku[x][y] = 0;
return NO;
}
}
return NO;
}
- (void)viewDidLoad
{
[super viewDidLoad];
/// Initialize it
for (int x = 0; x < WIDTH; x++)
{
for (int y = 0; y < HEIGHT; y++)
{
sudoku[x][y] = 0;
}
}
sudoku[0][0]=9;
int next;
next = (arc4random()%9)+1;
if( [self fillGridWithNext:next]) //seeded
{
NSLog(#"Solved");
}
else
{
NSLog(#"No solution");
}
for (int x = 0; x < WIDTH; x++)
{
for (int y = 0; y < HEIGHT; y++)
{
printf("%i ", sudoku[y][x]);
}
printf("\n"); //newline
}
// Do any additional setup after loading the view, typically from a nib.
}
- (void)didReceiveMemoryWarning
{
[super didReceiveMemoryWarning];
// Dispose of any resources that can be recreated.
}
- (BOOL) number:(int)n conflictsWithRow:(int)r;
{
for (int y = 0; y < HEIGHT; y++) {
if (sudoku[y][r] == n) {
return YES;
}
}
return NO;
}
- (BOOL) number:(int)n conflictsWithColumn:(int)c;
{
for (int x = 0; x < WIDTH; x++) {
if (sudoku[c][x] == n) {
return YES;
}
}
return NO;
}
- (BOOL) number:(int)n conflictsAtGridPointX:(int)xPoint andPointY:(int)yPoint;
{
if ([self number:n conflictsWithRow:yPoint])
{
return YES;
}
if ([self number:n conflictsWithColumn:xPoint])
{
return YES;
}
if ([self number:n conflictsWithSquareInPointX:xPoint andPointY:yPoint]) {
return YES;
}
return NO;
}
- (BOOL) number:(int)n conflictsWithSquareInPointX:(int)x andPointY:(int)y;
{
int leftX = x - (x % 3); //used to use int division
// leftX *= 3;
int topY = y - (y % 3);
// topY *= 3;
int rightX = leftX + 2;
int bottomY = topY + 2;
for(int subY = topY; subY <= bottomY; subY++) //bug was here, used < instead of less N equal to...
{
for ( int subX = leftX; subX <= rightX; subX++)
{
if (sudoku[subX][subY] == n) {
return YES;
}
}
}
NSLog(#"Testing grid at %i, %i", x/3, y/3);
NSLog(#"LeftX: %i TopY: %i", leftX, topY);
return NO;
}
- (int) incrementSudokuValue:(int)v;
{
if (v < 9) {
v++;
return v;
}
return 1;
}
#end
Summary: The first version is flawed but (mostly) gets the job done. It generates every row at random, if the row is invalid, it wipes and starts over. This will wipe out source grids, and can go forever, but works most of the time.
The lower code uses recursion. I don't think it backtracks properly, but it has solved empty and semi-seeded grids on my tests. I think I need to save a "state" grid to backtrack with, but I'm not doing this. I'm posting both since they both answer "Brute force"... on my own, I should study recursion, I can't explain why the lower one works, I personally could use help with doing it.
Note: The first one finishes in a blink or so when it does finish... if speed means more than reliability to your application (somewhat counter-intuitive in this case, with the infinite looping, heh).

This simple random walk algorithm should work too (but is inefficient- use at your own risk!!!):
EDIT: - added fix for unresolvable solutions.
For each empty cell in grid
array = Get_Legal_Numbers_for_cell(row,col);
If (array is empty) {
Clear_All_cells()
} else {
number = Random_number_from(array);
Put_Number_in_Cell(number);
}
EDIT 2
If someone are interested here are described methods for solving sudoku with random-based search.