CS student here. I've just received an introduction to loops and I'm not sure I understand them very well. I'm trying to print a triangle of numbers n, such that if n = 4 you'd get something like this:
4
3 7
2 6 9
1 5 8 10
Instead I'm winding up with something like:
4
3 5
Suffice it to say I'm lost. Here's my code:
void drawT3 (int n)
{
int k = 1;
int t = 1;
for (int i=1;i<=n;i++)
{
k = n;
int j;
for (j=1;j<=n-i;j++)
System.out.print(" ");
for (j=1;j<=t;j++)
{
System.out.printf("%3d",k);
k += (n - j);
}
n--;
t++;
System.out.println();
}
}
void printTriangle(int n)
{
// build an auxiliary 2D array
final int t[][] = new int[n][n];
int i = 1;
for (int s = n - 1; s <= 2 * (n - 1); s++)
{
for (int x = s - n + 1; x < n; x++)
{
t[x][s - x] = i++;
}
}
// print the array
for (int y = 0; y < n; y++)
{
for (int x = 0; x < n; x++)
{
if (t[x][y] > 0)
{
System.out.printf("%3d", t[x][y]);
}
else
{
System.out.printf(" ");
}
}
System.out.println(); // start new line
}
}
Build an auxiliary 2D array of size n.
Put numbers into array as human will do, from 1 to n, following the diagonals. s in the code represents x + y sum. That sum is constant for every diagonal. In the first diagonal (the longest one) sum is equal to n - 1. In the second diagonal sum is 1 more, n. In the last "diagonal" (bottom right corner) the sum is 2 * (n - 1). That's exactly our loop: for (int s = n - 1; s <= 2 * (n - 1); s++). Having the sum and x we can obtain y with simple subtraction, y = s - x.
Print the array. Each cell of array is initialized with 0 (int's default value). So, if a cell has zero, we just print 3 spaces, to preserve the shape of triangle.
PS. My code was written for "educational purposes" :) To show how it can be done, in easy way. It's not optimized for speed nor memory.
public static void main(String[] args) {
// TODO code application logic here
triangle(4);
}
static public void triangle(int n){
int x = 0;
for (int i = n;i>0;i--){
System.out.print(i + " ");
x = i+n;
for (int j=0;j<n-i;j++){
System.out.print(x - j + " ");
x = x + n -j;
}
System.out.println("");
}
}
Output for 4:
4
3 7
2 6 9
1 5 8 10
Output for 6:
6
5 11
4 10 15
3 9 14 18
2 8 13 17 20
1 7 12 16 19 21
Observe that there are many ways to print out a triangle of numbers as described above, For example, here are two,
// for n=5,
// 1 2 3 4 5
// 6 7 8 9
// 10 11 12
// 13 14
// 15
And
// 5
// 4 9
// 3 8 12
// 2 7 11 14
// 1 6 10 13 15
And since recursion is Fun!
class triangle
{
//Use recursion,
static int rowUR( int count, int start, int depth )
{
int ndx;
if(count<=0) return start;
//-depth?
for (ndx=0;ndx<depth;ndx++)
{
System.out.print(" ");
}
//how many? 5-depth, 5,4,3,2,1
for( ndx=0; ndx<count; ++ndx )
{
System.out.printf("%3d",start+ndx);
}
System.out.printf("\n");
if( count>0 )
{
rowUR( count-1, ndx+start, depth+1 );
}
return ndx;
}
//Use recursion,
static int rowLR( int count, int start, int depth )
{
int ndx, accum;
if( start < count )
rowLR( count, start+1, depth+1 );
for( ndx=0; ndx<depth; ++ndx )
{
System.out.print(" ");
}
accum=start;
//how many? 5-depth, 1,2,3,4,5
for( ndx=0; ndx<(count-depth); ++ndx )
{
System.out.printf("%3d",accum);
accum+=count-ndx;
}
System.out.printf("\n");
return ndx;
}
public static void main(String[] args)
{
int count=4, depth=0, start=1;
System.out.printf("rowUR\n");
rowUR( count=5, start=1, depth=0 );
System.out.printf("rowLL\n");
rowLL( count=5, start=1, depth=0 );
}
};
int n=4,i,j,k,t;
for (i=n;i>=1;i--)
{
t=i;
k=n;
for(j=1;j<i;j++)
System.out.printf(" "); // for leading spaces
System.out.printf("%3d",i); // for first digit(or number) in each row (in your example these are 4,3,2,1)
for(j=i;j<n;j++)
{
t+=k;
System.out.printf("%3d",t);
k--;
}
System.out.print("\n");
}
OUTPUT:
for n=8
8
7 15
6 14 21
5 13 20 26
4 12 19 25 30
3 11 18 24 29 33
2 10 17 23 28 32 35
1 9 16 22 27 31 34 36
http://ideone.com/C1O1GS
make space around numbers according to your need.
PS: I would never suggest to write any pattern code using array unless it is very complicated. array will use extra memory space.
Related
So I've been working on this lab for a while now for my programming class and so far I think I'm on the right track.
However, I'm not quite sure how to mirror the numbers. So pretty much, my code is only printing the top half of the triangle. Anyway here is the actual assignment that was given to us:
Write a program using a Scanner that asks the user for a number n between 1 and 9 (inclusive). The program prints a triangle with n rows. The first row contains only the square of 1, and it is right-justified. The second row contains the square of 2 followed by the square of 1, and is right justified. Subsequent rows include the squares of 3, 2, and 1, and then 4, 3, 2 and 1, and so forth until n rows are printed.
Assuming the user enters 4, the program prints the following triangle to the console:
1
4 1
9 4 1
16 9 4 1
9 4 1
4 1
1
For full credit, each column should be 3 characters wide and the values should be right justified.
Now here is what I have written for my code so far:
import java.util.Scanner;
public class lab6 {
public static void main(String[] args) {
Scanner kybd = new Scanner(System.in);
System.out.println(
"Enter a number that is between 1 and 9 (inclusive): ");
// this is the value that the user will enter for # of rows
int rows = kybd.nextInt();
for (int i = rows; i > 0; i--) {
for (int j = rows; j > 0; j--)
System.out.print((rows - j + 1) < i ?
" " : String.format("%3d", j * j));
System.out.println();
}
}
}
And this is what that code PRINTS when I enter 4:
Enter a number that is between 1 and 9 (inclusive):
4
1
4 1
9 4 1
16 9 4 1
As you can see, I can only get the TOP half of the triangle to print out. I've been playing around trying to figure out how to mirror it but I can't seem to figure it out. I've looked on this website for help, and all over the Internet but I can't seem to do it.
Answer is:
public static void main(String... args) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
for (int i = -rows + 1; i < rows; i++) {
for (int j = -rows; j < 0; j++)
System.out.print(abs(i) > j + rows ? " " : String.format("%3d", j * j));
System.out.println();
}
}
Try think of this as how to find points(carthesians) that are betwean three linear functions(area of triangle that lied betwean):
y = 0 // in loops i is y and j is x
y = x + 4
y = -x -4
And here is example result for 4:
And 9:
In the outer loop or stream you have to iterate from 1-n to n-1 (inclusive) and take absolute values for negative numbers. The rest is the same.
If n=6, then the triangle looks like this:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
Try it online!
int n = 6;
IntStream.rangeClosed(1 - n, n - 1)
.map(Math::abs)
.peek(i -> IntStream.iterate(n, j -> j > 0, j -> j - 1)
// prepare an element
.mapToObj(j -> i > n - j ? " " : String.format("%3d", j * j))
// print out an element
.forEach(System.out::print))
// start new line
.forEach(i -> System.out.println());
See also: Output an ASCII diamond shape using loops
Another alternative :
public static void main(String args[]) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
int row = rows, increment = -1;
while (row <= rows){
for (int j = rows; j > 0; j--) {
System.out.print(rows - j + 1 < row ? " " : String.format("%3d", j * j));
}
System.out.println();
if(row == 1) {
increment = - increment;
}
row += increment;
}
}
The outer loop from 1-n to n-1 inclusive, and the inner decrementing loop from n to 0. The if condition is the absolute value of i should not be greater than n - j.
Try it online!
int n = 6;
for (int i = 1 - n; i <= n - 1; i++) {
for (int j = n; j > 0; j--)
if (Math.abs(i) > n - j)
System.out.print(" ");
else
System.out.printf("%3d", j * j);
System.out.println();
}
Output:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
See also: Invert incrementing triangle pattern
I'm solving an algorithmic task with matrices involved.
I need to count each inner square (smaller matrix) inside a larger one.
Is there any dependencies between original length and each next?
As an example I have this matrix:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
As you may see Rows = 6, Cols = 6, Length = Rows * Cols;
Problem statement: How to calculate the length of inner matrices:
8 9 10 11
14 15 16 17
20 21 22 23
26 27 28 29
And the last one
15 16
21 22
What I could do:
for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
//mirrored row and col e.g. 0 1 2 3 2 1 0
int rowMir = i >= (int)Math.round(M/2.0d) ? (M - 1 - i) : i;
int colMir = j >= (int)Math.round(N/2.0d) ? (N - 1 - j) : j;
int depth = row > col ? col : row; //depth of inner square for each element in iteration. square's border in other words.
//In current matrix there are 3 inner squares (0, 1, 2)
}
}
Is it useful at all? I've tried to calculate it basing on values I've got, but no success until now. Googling gave me nothing in this case.
EDIT
The solution I'm looking for is to get size of the matrix for each element dynamically. Example for value at [i][j] I've counted the depth, the inner matrix it belongs to.
You can simply decrease by a step of 2 at each iteration:
int minDim = Math.min(rows, columns);
for(int i = minDim ; i >= 0 ; i -= 2) {
System.out.println("Inner length: " + ((rows - i) * (columns - i)))
}
In case you actually want to print all the inner matrices and get the count then you can do something like this:
public static void main (String[] args)
{
/* Define Matrix */
int matrixSize = 6;
int[][] matrix = new int[][]{{1, 2, 3, 4, 5, 6 },
{7, 8, 9, 10,11,12},
{13,14,15,16,17,18},
{19,20,21,22,23,24},
{25,26,27,28,29,30},
{31,32,33,34,35,36}};
/* Initialize Count Counter */
int count = 0;
/* Count/Print Inner Matrices */
for(int x = 2; x < matrixSize; x++) {
for(int i = 0; i <= matrixSize - x; i++) {
for(int j = 0; j <= matrixSize - x; j++) {
/* Call Print Matrix Function */
printMatrix(matrix, i, j, x);
++count; /* Increment Counter */
}
}
}
/* Print Actual Count */
System.out.println("\nTotal Inner Square Matrices Count: " + count);
}
/**
* Print Matrix
* Arguments: Matrix, Row Index, Column Index, Matrix Size
**/
public static void printMatrix(int[][] matrix, int i, int j, int size) {
System.out.println();
/* Row Iterator */
for(int iIndex = i; iIndex < i + size; iIndex++) {
System.out.println();
/* Column Iterator */
for(int jIndex = j; jIndex < j + size; jIndex++) {
System.out.print(" " + matrix[iIndex][jIndex]);
}
}
}
Output:
1 2
7 8
2 3
8 9
...
Total Inner Matrices: 54
I want my output to be like this e.g. if the user inputs 3:
without using 2d array
1 2 3
1 1 2 3
2 1 4 6
3 3 6 9
My code so far
public void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
int[] cloumnarray = new int[i];
int[] rowarray = new int[i];
for (int z = 0; z <= i - 1; z++) {
cloumnarray[z] = z + 1;
rowarray[z] = z + 1;
}
for (int j = 0; j < i; j++) {
System.out.println(cloumnarray[j] * rowarray[j]);
}
}
I tried different options and can't get this to work properly.
public static void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a <= i; a++) {
for (int b = 0; b <= i; b++) {
// top corner, don't print nothing
if (a == 0 && b == 0) System.out.print("\t");
// top row 0-1, 0-2, 0-3 etc... just 1,2,3...
else if (a == 0) {
System.out.print(b + "\t");
// last line, print extra line break
if (b == i)
System.out.print("\n");
}
// first column 1-0, 2-0, 3-0... just a + space (tabulator)
else if (b == 0) System.out.print(a + "\t");
// any other cases, are candidates to multiply and give result
else System.out.print(a*b + "\t");
}
//look this is out of scope of nested loops, so,
// in each a iteration, print line break :)
System.out.print("\n");
}
}
public static void main(String[] args) throws Exception {
matrixmutilplication();
}
OUTPUT (3)
1 2 3
1 1 2 3
2 2 4 6
3 3 6 9
OUTPUT (5)
1 2 3 4 5
1 1 2 3 4 5
2 2 4 6 8 10
3 3 6 9 12 15
4 4 8 12 16 20
5 5 10 15 20 25
But problem (for me) is the numbers are not padded in the natural order, so, to achieve your goal, exactly as in your demo, will need a bit of padding like this
public static void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a <= i; a++) {
for (int b = 0; b <= i; b++) {
if (a == 0 && b == 0) System.out.print("\t");
else if (a == 0) {
System.out.print(String.format("%3s", b));
if (b == i)
System.out.print("\n");
}
else if (b == 0) System.out.print(a + "\t");
else System.out.print(String.format("%3s", a*b));
}
System.out.print("\n");
}
}
public static void main(String[] args) throws Exception {
matrixmutilplication();
}
OUTPUT (7)
1 2 3 4 5 6 7
1 1 2 3 4 5 6 7
2 2 4 6 8 10 12 14
3 3 6 9 12 15 18 21
4 4 8 12 16 20 24 28
5 5 10 15 20 25 30 35
6 6 12 18 24 30 36 42
7 7 14 21 28 35 42 49
What looks quite good :)
So this should be pretty simple.
public void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a < i; a++) {
for (int b = 0; b < i; b++) {
System.out.print(a*b + "\t");
}
System.out.print("\n");
}
}
Whenever you're working with a matrix involving two arrays (especially if you're trying to a solve a problem that deals with patterns), you want to have a nested for loop like so:
for(int row = 0; row < numSelected; row++) {
for(int col = 0; col < numSelected; col++) {
...
}
}
That way, each cell in the matrix will be covered. Now using that, you can try multiplying the row index and the col index and storing that to the correct cell.
How do I make this loop properly? it right now So it loops but it does not loop properly. It does this
Here are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 [1]
How many positions do you want to shift?: 2
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3 [3]
How many positions do you want to shift?: 4
the [] are where its suppose to ask me for my input instead of me just putting in a input
its suppose to run like this:
re are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
How many positions do you want to shift?: 1
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3
How many positions do you want to shift?: 4
System.out.println("Here are the numbers:");
for (i=0; i<numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
while (x != input.nextInt()){
System.out.printf("How many positions do you want to shift?: ");
int shiftTimes=input.nextInt();
for( i = 0; i < shiftTimes; ++i)
shift.Shifter(numberArray);
for(j = 0; j < numberArray.length; j++)
System.out.printf(numberArray[j]+" ");
}
}
}
Also How Do I make it exit the program when I enter in a invalid number and how do I get get it to read a negative value and get it to shift left
Edit: heres my shifter code
public static void Shifter(int[] list)
{
int i;
if (list.length < 2) return;
int last = list[list.length - 1];
for(i = list.length - 1; i > 0; i--) {
list[i] = list[i - 1];
}
list[0] = last;
}
This should work for right shift. It should work with inputs larger then array length as well.
for (int i = shiftTimes%numberArray.length; i > 0; i--) {
System.out.print(numberArray[numberArray.length - i] + " ");
}
for (int i = 0; i < numberArray.length - shiftTimes%numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
Reversing this logic should produce a left shift approach.
An invalid input would be the length of the array (because the result will be the same) or 0 because that doesn't do anything:
if (shiftTimes == numberArray.length || shiftTimes == 0) {
// present error to user
}
UPDATE: Putting the logic in your function. Also updated the invalid input check.
public static void Shifter(int[] list, int input)
{
for (int i = input%list.length; i > 0; i--) {
System.out.print(list[list.length - i] + " ");
}
for (int i = 0; i < list.length - input%list.length; i++) {
System.out.print(list[i] + " ");
}
}
The function call would be:
Shifter(numberArray, shiftTimes);
I need to print the following pattern and i almost did with the coding part.
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1
Following is the program I tried
public class MyPattern {
public static void main(String[] args) {
for (int i = 0; i <= 7; i++) {
for (int j = 1; j <= 7 - i; j++) {
System.out.print(" ");
}
for (int j = 0; j <= i; j++) {
int n = (int) Math.pow(2.0D, j);
if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
}
for (int j = i - 1; j >= 0; j--) {
int n = (int) Math.pow(2.0D, j);
if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
}
System.out.print('\n');
}
}
}
When running the program I got the following output
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1
But I need the output aligned to left (as specified first). Please help.
Well it's clearly caused by this part of code:
for (int j = 1; j <= 7 - i; j++) {
System.out.print(" ");
}
Have you tried running it without it?
if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
Could also just be, as it does not matter what n is - it will all just do the same.
System.out.print(" " + n);
Comment the line:
//System.out.print(" ");
In the first for loop.
I hope this code helps you understand a few things.
// Make it ready for the loop, no point calling Math.pow() every loop - expensive
import static java.lang.Math.pow;
public class MyPattern {
public void showTree(int treeDepth) {
// Create local method fields, we try to avoid doing this in loops
int depth = treeDepth;
String result = "", sysOutput = "";
// Look the depth of the tree
for( int rowPosition = 0 ; rowPosition < depth ; rowPosition++ ) {
// Reset the row result each time
result = "";
// Build up to the centre (Handle the unique centre value here)
for( int columnPosition = 0 ; columnPosition <= rowPosition ; columnPosition++ )
result += (int) pow(2, columnPosition) + " ";
// Build up from after the centre (reason we -1 from the rowPosition)
for ( int columnPosition = rowPosition - 1 ; columnPosition >= 0 ; columnPosition-- )
result += (int) pow(2, columnPosition) + " ";
// Add the row result to the main output string
sysOutput += result.trim() + "\n";
}
// Output only once, much more efficient
System.out.print( sysOutput );
}
// Good practice to put the main method at the end of the methods
public static void main(String[] args) {
// Good practice to Create Object of itself
MyPattern test = new MyPattern();
// Call method on object (very clear this way)
test.showTree(5);
}
}