How to print the triangle below:
2 3 5 8 3 8
4 6 9 4 9
7 1 5 1
2 6 2
7 3
4
First you need to start with number 2 and add one to the next one vertically
My code:
int d = 2, n = 6;
for (int line=1; line <= n; line++ ) {
for (int j = 2; j <= line; j++) {
System.out.print(" ");
}
for (int k = line; k <= n; k++) {
System.out.print(d + " ");
d = d + k;
if (d > 9) {
d = d - 9;
}
}
System.out.println();
}
Result:
2 3 5 8 3 8
5 7 1 5 1
7 1 5 1
7 2 7
4 9
6
The pattern is that the value of d has to be calculated initially on every new line based on the value of d in the first instance of the previous line. That is the part that's missed here. You can do that by having a temp variable store the initial value of d on every line and print based on that. I have used a variable tempD here, which can help print the pattern that you require.
int d = 2, n = 6;
int tempD = d - 1;
for (int line = 1; line <= n; line++) {
tempD = tempD + line;
if (tempD > 9) {
tempD = tempD - 9;
}
d = tempD;
for (int j = 2; j <= line; j++) {
System.out.print(" ");
}
for (int k = line; k <= n; k++) {
System.out.print(d + " ");
d = d + k;
if (d > 9) {
d = d - 9;
}
}
System.out.println();
}
I need series of for loops to return a box like below depending on the values of m & n.
Should output:
1 1 1 1 1 1 1
1 2 2 2 2 2 1
1 2 3 3 3 2 1
1 2 2 2 2 2 1
1 1 1 1 1 1 1
Below is my code so far which uses a series of loops to split the box in half and either ascends or descends the value for both the column and row. Where I am stuck is trying to find a way to make these perimeters for the values. Another note that this should be able to work without using any if statements.
int m = 5; //column value
int n = 7; //row value
int column;
for (int row = 0; row <= (m / 2); row++) {
//Ascending
for (column = 1; column < (n / 2); column++) {
int outputNumber = row + 1;
System.out.print(outputNumber + " ");
}
//Fixed
do {
int outputNumber = row + 1;
System.out.print(outputNumber + " ");
}
while (column < 0);
//Descending
for (column = n / 2; column >= 0; column--) {
int outputNumber = row + 1;
System.out.print(outputNumber + " ");
}
System.out.println();
}
for (int row = m / 2; row > 0; row--) {
for (column = 1; column <= n; column++) {
System.out.print(row + " ");
}
System.out.println();
}
Current output for the code above:
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
2 2 2 2 2 2 2
1 1 1 1 1 1 1
Here is my result. I know that it is a bit complicated and quick written, however it solves your problem. I also tested it with other numbers:
int m = 5; //column value
int n = 7; //row value
for (int i = 0; i < m / 2 + 1; i++) {
for (int j = 0; j < i + 1; j++) {
System.out.print(j + 1 + " ");
}
for (int j = 0; j < n - ((i + 1) * 2); j++) {
System.out.print(i + 1 + " ");
}
for (int j = 0; j < i + 1; j++) {
System.out.print(i + 1 - j + " ");
}
System.out.println("");
}
for (int i = m / 2 + 1; i < m; i++) {
for (int j = 0; j < m - i; j++) {
System.out.print(j + 1 + " ");
}
for(int j = 0; j < n - (m - i) * 2; j++) {
System.out.print(m-i + " ");
}
for(int j = 0; j < m - i; j++) {
System.out.print(m - i - j + " ");
}
System.out.println("");
}
Assuming my understanding is correct that the perimeter tiles should be 1, and subsequently each tile should display the minimum number of 'steps' it would take to get there from outside the matrix.
I personally have a hard time picturing/following your solution, so I'm not sure if my suggestion will mesh with your understanding.
First, lets separate the construction of this matrix from the printing of this matrix. I think this keeps things tidy, but that's my style.
Lets say you have c columns and r rows in your matrix. So we'll iterate over every cell with a nested for loop.
for (int r = 0; r < rows; r++) {
for (int c = 0; c < columns; c++) {
int distanceToEdgeOfRow = Math.abs(rows - (r - rows)); //this finds the number of steps to the nearest row end
int distanceToEdgeOfColumn = Math.abs(columns - (c - columns)); //this find the number of steps to the nearest column end
int shortestPath = Math.min(distanceToEdgeOfColumn, distanceToEdgeOfRow); //is it shorter to take the closest row exit or column exit?
//the shortestPath is still off by one, so we need to add 1 to shortestPath to see what should be printed on this tile
matrix[r][c] = shortestPath + 1;
}
}
Here is my mess of a code. I have to write a program that inputs a positive integer greater than 3. Validate that the integer is in fact greater than 3. Then print all possible pairs of positive integers great than whose product is less than or equal to the number entered.
ex. If 24 is the input.
It would print:
4 = 2 x 2
6 = 2 x 3
8 = 2 x 4
10 = 2 x 5
12 = 2 x 6
14 = 2 x 7
16 = 2 x 8....
9 = 3 x 3
12 = 3 x 4..
24 = 3 x 8...
all the way to
24 = 4 x 6
import java.util.Scanner;
public class Factors {
public static void main(String[] args) {
// Define Variables
Scanner input = new Scanner(System.in);
int i = 0;
int j = 0;
int k = 2;
int product = 0;
// Ask for input/loop
while (i < 3) {
System.out.println("Please enter an integer greater than 3");
i = input.nextInt();
}
while (product < i) {
if (product == i) { j++; k = 2;
for (j = 2; product < i; k++) {
product = j * k;
System.out.println(product + " = " + j + " x " + k);
if (product == i) { j++; k = 2;
}
}
}
}
}
}
public class Factors {
public static void main(String[] args) {
// Define Variables
Scanner input = new Scanner(System.in);
int i = 0;
int product = 0;
// Ask for input/loop
while (i < 3) {
System.out.println("Please enter an integer greater than 3");
i = input.nextInt();
}
for (int j = 2; j < i / 2; j++) {
for (int k = 2; k < i / 2; k++) {
if (j <= k && j * k <= i)
System.out.println(j * k + " = " + j + "*" + k);
}
}
// while (product < i) {
// if (product == i) {
// j++;
// k = 2;
// for (j = 2; product < i; k++) {
// product = j * k;
// System.out.println(product + " = " + j + " x " + k);
// if (product == i) {
// j++;
// k = 2;
// }
// }
// }
// }
}
}
I made a Java program that prints out a pascal triangle, however I can't figure out how to correctly position it.
Program 1
public class Triangle {
public static void main() {
System.out.println("\nTriangle: ");
int row = 11;
long[][] triangle = new long[row][row];
triangle[1][1] = 1;
System.out.print(triangle[1][1] + "\n");
for (int i = 2; i < row; i++) {
for (int n = 1; n < row; n++) {
triangle[i][n] = triangle[i-1][n-1] + triangle[i-1][n];
if (triangle[i][n] > 0) {
System.out.print(triangle[i][n] + " ");
}
}
System.out.println();
}
}
}
Output:
1
1 1
1 2 1
1 3 3 1
Program 2
public class Triangle {
public static void main() {
System.out.println("\nTriangle: ");
int row = 11;
long[][] triangle = new long[row][row];
int x = 1;
while (x < row - 1) {
System.out.print(" ");
x++;
}
triangle[1][1] = 1;
System.out.print(triangle[1][1] + "\n");
for (int i = 2; i < row; i++) {
x = i;
while (x < row - 1) {
System.out.print(" ");
x++;
}
for (int n = 1; n < row; n++) {
triangle[i][n] = triangle[i-1][n-1] + triangle[i-1][n];
if (triangle[i][n] > 0) {
System.out.print(triangle[i][n] + " ");
}
}
System.out.println();
}
}
}
Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1 //(Notice this line is incorrectly positioned)
When the triangle approaches multiple digit numbers, it starts to break down and makes it ugly. Can someone explain how I can display a normal triangle instead of this ugly one?
Dynamic Pascal Triangle generator is here:
import java.io.IOException;
import java.util.Scanner;
public class Main {
static double fact(int n) {
double result = 1;
for (double i = 1; i <= n; i++)
result *= i;
return result;
}
static double combine(int n, int r) {
return ((fact(n)) / (fact(n - r) * fact(r)));
}
static void pascalTriangle(int n) {
int n2 = n;
for (int i = 0; i < n; i++) {
for (int space = 8 * (n2 - 1); space >= 0; space--) {
System.out.printf(" ");
}
for (int j = 0; j <= i; j++) {
System.out.printf("%14.0f", combine(i, j));
System.out.printf(" ");
}
System.out.println();
n2--;
}
}
public static void main(String[] args)
throws IOException, InterruptedException {
#SuppressWarnings("resource")
Scanner sc = new Scanner(System.in);
System.out.print("Enter Number of Lines(n): ");
int n = sc.nextInt();
pascalTriangle(n);
System.out.println("Press any key to exit! ");
sc.nextByte();
}
}
Try this ...
Results:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
import java.util.*;
public class HelloWorld {
static int binCoeff(int n, int k) {
int res = 1;
if (k > n - k)
k = n - k;
for (int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
static void pascalTriangle(int lines) {
for (int i = 0; i < lines; i++) {
for (int j = 0; j <= i; j++)
System.out.print(HelloWorld.binCoeff(i, j) + " ");
System.out.println();
}
}
public static void main(String[] args) {
System.out.println("Results: ");
HelloWorld.pascalTriangle(8);
}
}
/**
* #author Ranjith
*/
public class JavaApplication2 {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
int i;
int x;
int n = 15; //number of rows
String newLine = System.getProperty("line.separator");
for (i = 0; i < n; i++) { //loop to adjust spacing
x = i;
while (x < n - 1) {
System.out.print(" ");
x++;
}
fib(i); //fibonacci function is called
System.out.print(newLine);
}
}
public static void fib(int num) { //fibonacci function
int[] febo = new int[100];
febo[0] = 0;
febo[1] = 1;
for (int i = 2; i < num; i++) {
febo[i] = febo[i - 1] + febo[i - 2];
}
for (int i = 0; i < num; i++) {
System.out.print(febo[i] + " ");
}
}
}
Output:
0
0 1
0 1 1
0 1 1 2
0 1 1 2 3
0 1 1 2 3 5
0 1 1 2 3 5 8
0 1 1 2 3 5 8 13
0 1 1 2 3 5 8 13 21
0 1 1 2 3 5 8 13 21 34
0 1 1 2 3 5 8 13 21 34 55
0 1 1 2 3 5 8 13 21 34 55 89
0 1 1 2 3 5 8 13 21 34 55 89 144
0 1 1 2 3 5 8 13 21 34 55 89 144 233
You can represent such a triangle as a 2d array, where the elements of the first row and column are equal to one, and all other elements are the sum of the previous element in the row and column.
arr[i][j] = arr[i][j-1] + arr[i-1][j];
Then you can position it in the upper left corner as follows:
1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8
1 3 6 10 15 21 28
1 4 10 20 35 56
1 5 15 35 70
1 6 21 56
1 7 28
1 8
1
Try it online!
public static void main(String[] args) {
int n = 9;
// an array of 'n' rows
int[][] arr = new int[n][];
// iterate over the rows of the array
for (int i = 0; i < n; i++) {
// a row of 'n-i' elements
arr[i] = new int[n - i];
// iterate over the elements of the row
for (int j = 0; j < n - i; j++) {
if (i == 0 || j == 0) {
// elements of the first row
// and column are equal to one
arr[i][j] = 1;
} else {
// all other elements are the sum of the
// previous element in the row and column
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
}
}
}
// formatted output
for (int[] row : arr) {
for (int el : row) {
// formatting as a number with a trailing space
System.out.printf("%2d ", el); // two-digit number
// System.out.printf("%3d ", el); // three-digit number
// System.out.printf("%4d ", el); // four-digit number
}
System.out.println();
}
}
See also:
• Pascal's triangle 2d array - formatting printed output
• Print Pascal's Triangle
class pascal {
static void main(int n) {
int a[][] = new int[n][n + 1];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
a[i][j] = 0;
}
}
a[0][1] = 1;
int k = 5;
int p = 0;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n + 1; j++) {
a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
}
}
for (int i = 0; i < a.length; i++) {
for (p = n + -i; p > 0; p--) {
System.out.print(" ");
}
for (int j = 0; j < a[i].length; j++) {
if (a[i][j] != 0) {
System.out.print(a[i][j] + " ");
} else {
System.out.print(" ");
}
}
System.out.println();
}
}
}
I'm trying to print out all possibilities of nCr, which are the combinations when order doesn't matter. So 5C1 there are 5 possibilities: 1 , 2, 3, 4, 5. 5C2 there are 10 possibilities: 1 2, 1 3, 1 4, 1 5, 2 3, 2 4, 2 5, 3 4, 3 5, 4 5.
I made functions that print what I want for r = 2, r = 3, and r = 4, and I sort of see the pattern, but I cant seem to make a working method for variable r:
public void printCombinationsChoose2(int n, int k) //for when k = 2
{
for (int a = 1; a < n; a++)
{
for (int b = a + 1; b <= n; b++)
{
System.out.println("" + a + " " + b);
}
}
}
public void printCombinationsChoose3(int n, int k) //for when k = 3
{
for (int a = 1; a < n - 1; a++)
{
for (int b = a + 1; b < n; b++)
{
for (int c = b + 1; c <= n; c++)
{
System.out.println("" + a + " " + b + " " + c);
}
}
}
}
public void printCombinationsChoose4(int n, int k) //for when k = 4
{
for (int a = 1; a < n - 2; a++)
{
for (int b = a + 1; b < n - 1; b++)
{
for (int c = b + 1; c < n; c++)
{
for (int d = c + 1; d <= n; d++)
{
System.out.println("" + a + " " + b + " " + c + " " + d);
}
}
}
}
}
public void printCombinations(int n, int k) //Doesn't work
{
int[] nums = new int[k];
for (int i = 1; i <= nums.length; i++)
nums[i - 1] = i;
int count = 1;
while (count <= k)
{
for (int a = nums[k - count]; a <= n; a++)
{
nums[k - count] = a;
for (int i = 0; i < nums.length; i++)
System.out.print("" + nums[i] + " ");
System.out.println();
}
count++;
}
}
So I think the layout of my last method is right, but I'm just not doing the right things, because when I call printCominbations(5, 2), it prints
1 2
1 3
1 4
1 5
1 5
2 5
3 5
4 5
5 5
when it should be what I said earlier for 5C2.
Edit
The last example was bad. This is a better one to illustrate what it's doing wrong: printCombinations(5, 3) gives this:
1 2 3
1 2 4
1 2 5
1 2 5
1 3 5
1 4 5
1 5 5
1 5 5
2 5 5
3 5 5
4 5 5
5 5 5
How do I get it to be:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
How about this:
public class Test {
public static void main(final String[] args) {
print_nCr(7, 4);
}
public static final void print_nCr(final int n, final int r) {
int[] res = new int[r];
for (int i = 0; i < res.length; i++) {
res[i] = i + 1;
}
boolean done = false;
while (!done) {
System.out.println(Arrays.toString(res));
done = getNext(res, n, r);
}
}
/////////
public static final boolean getNext(final int[] num, final int n, final int r) {
int target = r - 1;
num[target]++;
if (num[target] > ((n - (r - target)) + 1)) {
// Carry the One
while (num[target] > ((n - (r - target)))) {
target--;
if (target < 0) {
break;
}
}
if (target < 0) {
return true;
}
num[target]++;
for (int i = target + 1; i < num.length; i++) {
num[i] = num[i - 1] + 1;
}
}
return false;
}
}
The key to this solution for me was to look at the problem as a numbering system and you want to increase a number by one and every time you reach an upper bound, you just carry the excess to the left one and ... You just need to implement the increasing algorithm correctly...
The first point where your code deviates from the expectation is here:
...
1 2 5
1 2 5 <-- first bad output
1 3 5
...
So ask yourself three things:
What should have happened in that line of code with the given state of the variables?
Why doesn't do my code exactly that?
What must be changed to achieve that?
The answer for the first part is like this:
It should have incremented the 2 to 3 and it should have set the following numbers to
4, 5, ... similar to the initialisation of nums.
The second and third part is your part again.
BTW: When you come back because you need more help, please explain in detail what you have deduced so far and clean up and shorten the question quite a bit.
OK... What is the solution when we know we need loops, but not the number of them?? RECURSION...
You need to use a recursive implementation. Have this in mind: ANYTIME, you need loops but the number of the nested loops can only be known at runtime, based on the specific parameters of the problem, you should use recursive methods... I'll give you some time to try it yourself, I'll be back to give you the final implementation...
I have done it in c++
#include <iostream>
using namespace std;
#define ARR_LIMIT 100
int arr[ARR_LIMIT];
void _ncr(int N,int R, int n,int r , int start )
{
if(r>0)
{
for(int i = start ; i <= start + (n-r); i++)
{
arr[R-r] = i;
_ncr(N,R,N-i, r-1, i+1 );
}
}
else
{
for(int i=0;i<R;i++)
{
cout << arr[i] << " ";
if(i==R-1)
cout<<"\n";
}
}
}
void ncr(int n,int r)
{
//Error checking of parameters
bool error = false;
if( n < 1)
{
error = true;
cout<< "ERROR : n should be greater 0 \n";
}
if( r < 1)
{
error = true;
cout<< "ERROR : r should be greater 0 \n";
}
if(r > n)
{
error = true;
cout<< "ERROR : n should be greater than or equal to r \n";
}
// end of Error checking of parameters
if(error)
return;
else
_ncr(n,r,n,r,1);
}
int main()
{
int n,r;
cout << "Enter n : ";
cin >> n;
cout << "Enter r : ";
cin >> r;
ncr(n,r);
return 0;
}
The layout of function printCombination() seems wrong. The while loop will iterate two times, for count = 1 and count = 2.
When count = 1, only values in nums[0][here] will change, since in this case k - count = 1.
Hence,
1,2
1,3
1,4 and
1,5.
And when count = 2, only values in nums[here][1] will change, since here k - count = 0.
Hence
1,5
2,5
3,5
4,5 and
5,5