So I need to make a diamond_look of numbers using 2D array in Java. I got my results but with null before the diamond. For drawNumDiamond(9) I have to get a diamond look that goes until 5 and back. I know I can make it without using array, but I want to learn more about 2D arrays :this is how it should look like and what are my results
public class Example1{
private static void drawNumDiamond(int h) {
if(h%2 != 0) {
int size = h/2 +1;
int count = 1;
int loop = 1;
String[][] dijamant = new String[h][];
for(int row = 0; row < dijamant.length; row++) {
dijamant[row] = new String[row+1];
for(int kolona=0; kolona<=row; kolona++) {
dijamant[0][0] = "1";
for(int i=0; i< loop;i++) {
dijamant[row][kolona]+= count;
}
}
count++;
loop+=2;
}
for (int k = 0; k < size; k++) {
System.out.printf("%" + h + "s", dijamant[k]);
h++;
System.out.println();
}
h--;
for (int q = size - 2; q>=0; q--) {
h--;
System.out.printf("%" + h + "s", dijamant[q]);
System.out.println();
}
}
}
public static void main(String[] args) {
drawNumDiamond(9);
}
}
The issue is in this line :
dijamant[row][kolona] += count;
if dijamant[row][kolona] is null and count is 2, the result of the string concatenation will be "null2". Try adding the following if statement before to initialize with an empty string :
if (dijamant[row][kolona] == null) {
dijamant[row][kolona] = "";
}
This will get your code working, but there are still things to think about. E.g. you keep setting dijamant[0][0] = "1"; in the loop.
Related
I have two 2d boolean arrays, the smaller array (shape) is going over the larger array (world).
I am having trouble to find a method to find out when the smaller array can "fit" into the larger one.
When I run the code it either just goes through the larger array, never stopping, or stops after one step (incorrectly).
public void solve() {
ArrayList<Boolean> worldList=new ArrayList<>();
ArrayList<Boolean> shapeList=new ArrayList<>();
for (int i = 0; i < world.length; i++) {
for (int k = 0; k < world[i].length; k++) {
worldList.add(world[i][k]);
display(i, k, Orientation.ROTATE_NONE);
for (int j = 0; j < shape.length; j++) {
for (int l = 0; l < shape[j].length; l++) {
shapeList.add(shape[j][l]);
if(shapeList.equals(worldList)) {
return;
}
}
}
}
}
}
A good place to start with a problem like this is brute force for the simplest case. So, for each index in the world list, just check to see if every following index of world and shapes match.
Notice we only iterate to world.size()-shapes.size(), because naturally if shapes is longer than the portion of world we haven't checked, it won't fit.
import java.util.ArrayList;
public class Test {
ArrayList<Boolean> world = new ArrayList<>();
ArrayList<Boolean> shapes = new ArrayList<>();
public static void main(String[] args) {
new Work();
}
public Test() {
world.add(true);
world.add(false);
world.add(false);
world.add(true);
shapes.add(false);
shapes.add(true);
// Arraylists initialized to these values:
// world: T F F T
// shapes: F T
System.out.println(getFitIndex());
}
/**
* Get the index of the fit, -1 if it won't fit.
* #return
*/
public int getFitIndex() {
for (int w = 0; w <= world.size()-shapes.size(); w++) {
boolean fits = true;
for (int s = 0; s < shapes.size(); s++) {
System.out.println("Compare shapes[" + s + "] and world["+ (w+s) + "]: " +
shapes.get(s).equals(world.get(w+s)));
if (!shapes.get(s).equals(world.get(w+s))) fits = false;
}
System.out.println();
if (fits) return w;
}
return -1;
}
}
When we run this code, we get a value of 2 printed to the console, since shapes does indeed fit inside world, starting at world[2].
You can find the row and column of fitting like this
public void fit() {
int h = world.length - shape.length;
int w = world[0].length - shape[0].length;
for (int i = 0; i <= h; i++) {
for (int k = 0; k <= w; k++) {
boolean found = true;
for (int j = 0; j < shape.length && found; j++) {
for (int l = 0; l < shape[j].length && found; l++) {
if (shape[j][l] != world[i + j][k + l])
found = false;
}
}
if (found) {
//Your shape list fit the world list at starting index (i, k)
//You can for example save the i, k variable in instance variable
//Or return then as an object for further use
return;
}
}
}
I am having trouble to create a (recursive) function that prints all possible indexes for a multi dimensional table.
I got the information about the multi-dimensionality as an array.
Example:
int[]{6,6} would be a 2-dimensional table with 6x6 = 36 fields, so the result would be [0,0],[0,1],[1,1],[1,2],[2,2]... and so on.
Another example:
int[]{2,2,2} would be a 3-dimensional table with 8 possible indexes: [0,0,0],[0,0,1],[0,1,1]... and so on
I'm trying to do this in Java 7.
Edit: As requested, what I got so far. Code is producing OutOfBound Exception.
loop(new int[2], new int[]{6, 6}, 0);
private void loop(int[] index, int[] dimensionSizes, int dimensionIndex) {
if (index[dimensionIndex] < dimensionSizes[dimensionIndex] && dimensionIndex < dimensionSizes.length) {
System.out.println(Arrays.toString(index));
index[dimensionIndex] = index[dimensionIndex] + 1;
if (index[dimensionIndex] < dimensionSizes[dimensionIndex]) {
loop(index, dimensionSizes, dimensionIndex);
} else {
loop(index, dimensionSizes, dimensionIndex + 1);
}
}
}
I think this code could respond to your question:
public static void printAllIndex(int[] dimensions) {
int[] index = new int[dimensions.length];
int stepNumber = 0;
// Initialization
for (int i : index) { index[i] = 0; } // init index to 0
for (int d : dimensions) { stepNumber += d; } // count number of iteration needed
System.out.println(Arrays.toString(index)); // print first index [0,0,...]
for(int s = 0; s <= stepNumber - 1; s++) {
boolean allEquals = true;
int value = index[index.length - 1];
for (int i = index.length - 1; i >= 0; i--) {
if(index[i] != value) {
index[i]++;
allEquals = false;
break;
}
}
if (allEquals) { index[index.length - 1]++; }
System.out.println(Arrays.toString(index));
}
}
I need to find all the permutations for a given n(user input) without backtracking.
What i tried is:
import java.util.Scanner;
import java.util.Vector;
class Main {
private static int n;
private static Vector<Vector<Integer>> permutations = new Vector<>();
private static void get_n() {
Scanner user = new Scanner(System.in);
System.out.print("n = ");
n = user.nextInt();
}
private static void display(Vector<Vector<Integer>> permutations) {
for (int i = 0; i < factorial(n) - 1; ++i) {
for (int j = 0; j < n; ++j) {
System.out.print(permutations.elementAt(i).elementAt(j) + " ");
}
System.out.println();
}
}
private static int factorial(int n) {
int result = 1;
for (int i = 1; i <= n; ++i) {
result *= i;
}
return result;
}
private static int max(Vector<Integer> permutation) {
int max = permutation.elementAt(0);
for (int i = 1; i < permutation.size(); ++i)
if (permutation.elementAt(i) > max)
max = permutation.elementAt(i);
return max;
}
// CHECKS FOR ELEMENT COUNT AND 0 - (n-1) APPARITION
public static int validate_permutation(Vector<Integer> permutation) {
// GOOD NUMBER OF ELEMENTS
if (max(permutation) != permutation.size() - 1)
return 0;
// PROPER ELEMENTS APPEAR
for (int i = 0; i < permutation.size(); ++i)
if (!permutation.contains(i))
return 0;
return 1;
}
private static Vector<Integer> next_permutation(Vector<Integer> permutation) {
int i;
do {
i = 1;
// INCREMENT LAST ELEMENT
permutation.set(permutation.size() - i, permutation.elementAt(permutation.size() - i) + 1);
// IN A P(n-1) PERMUTATION FOUND n. "OVERFLOW"
while (permutation.elementAt(permutation.size() - i) == permutation.size()) {
// RESET CURRENT POSITION
permutation.set(permutation.size() - i, 0);
// INCREMENT THE NEXT ONE
++i;
permutation.set(permutation.size() - i, permutation.elementAt(permutation.size() - i) + 1);
}
} while (validate_permutation(permutation) == 0);
// OUTPUT
System.out.print("output of next_permutation:\t\t");
for (int j = 0; j < permutation.size(); ++j)
System.out.print(permutation.elementAt(j) + " ");
System.out.println();
return permutation;
}
private static Vector<Vector<Integer>> permutations_of(int n) {
Vector<Vector<Integer>> permutations = new Vector<>();
// INITIALIZE PERMUTATION SET WITH 0
for (int i = 0; i < factorial(n); ++i) {
permutations.addElement(new Vector<>());
for(int j = 0; j < n; ++j)
permutations.elementAt(i).addElement(0);
}
for (int i = 0; i < n; ++i)
permutations.elementAt(0).set(i, i);
for (int i = 1; i < factorial(n); ++i) {
// ADD THE NEXT PERMUTATION TO THE SET
permutations.setElementAt(next_permutation(permutations.elementAt(i - 1)), i);
System.out.print("values set by permutations_of:\t");
for (int j = 0; j < permutations.elementAt(i).size(); ++j)
System.out.print(permutations.elementAt(i).elementAt(j) + " ");
System.out.println("\n");
}
System.out.print("\nFinal output of permutations_of:\n\n");
display(permutations);
return permutations;
}
public static void main(String[] args) {
get_n();
permutations.addAll(permutations_of(n));
}
}
Now, the problem is obvious when running the code. next_permutation outputs the correct permutations when called, the values are set correctly to the corresponding the vector of permutations, but the end result is a mass copy of the last permutation, which leads me to believe that every time a new permutation is outputted by next_permutation and set into the permutations vector, somehow that permutation is also copied over all of the other permutations. And I can't figure out why for the life of me.
I tried both set, setElementAt, and an implementation where I don't initialize the permutations vector fist, but add the permutations as they are outputted by next_permutation with add() and I hit the exact same problem. Is there some weird way in which Java handles memory? Or what would be the cause of this?
Thank you in advance!
permutations.setElementAt(next_permutation(permutations.elementAt(i - 1)), i);
This is literally setting the vector at permutations(i) to be the same object as permutations[i-1]. Not the same value - the exact same object. I think this the source of your problems. You instead need to copy the values in the vector.
I have to make a 3 way merge sort of an array. the array length is a in a power of 3, i.e. 3,9,27 etc. So I can use only one split function and not "left","mid","right".
Would like to get an answer how to repair it and why does not it work.
I have written the code, however don't know how to get it to work.
Here it is:
EDITED THE CODE, STILL DOES NOT WORK
public class Ex3 {
public static void main(String[] args) { //main function
Scanner in = new Scanner(System.in); //scanner
int size = in.nextInt();
int[] arr = new int[size];
for (int i = 0; i<arr.length; i++){
arr[i] = in.nextInt();
}
in.close();
arr = merge3sort (arr); //send to the function to merge
for (int i = 0; i<arr.length; i++){ //printer
System.out.print(arr[i]+ " ");
}
}
static int[] split(int[] m, int thirdNum) { //split function that splits to 3 arrays
int third[] = new int[m.length/3];
int third1[]=new int[m.length/3];
int third2[]=new int[m.length/3];
for(int i = 0; i<=m.length/3; i++)
third[i]=m[i];
for(int i=0; i<=m.length/3;i++)
third1[i]=m[i+thirdNum];
for(int i=0; i<=m.length/3;i++)
third2[i]=m[i+2*thirdNum];
return merge(third,third1,third2);
//return null;
}
static int minOf3(int[] a3) { //function that finds out how what is the index of the smallest number
int num0 = a3[0];
int num1 = a3[1];
int num2 = a3[2];
int idx = 0;
if(num0<num1 && num1<num2)
idx=0;
if(num1<num0 && num0<num2)
idx=1;
else
idx=2;
return idx;
}
static int[] merge(int[] th0, int[] th1, int[] th2) { //function that sorts the numbers between 3 arrays
int len0=th0.length;
int len1=th1.length;
int len2=th2.length;
int[] united = new int[len0+len1+len2];
int ind = 0; int i0=0; int i1=0; int i2=0;
while(i0<len0 && i1<len1 && i2<len2){
if(th0[i0]<th1[i1]){
if(th0[i0]<th2[i2]){
united[ind]=th0[i0];
i0=i0+1;
}//end inner if
else{
united[ind]=th2[i2];
i2=i2+1;
}//end inner else
}//end outer if
else{
united[ind]=th1[i1];
i1=i1+1;
}//end outer else
ind=ind+1;
}//end while
for (int i = i0; i < len0; i = i + 1) {
united[ind] = th0[i];
ind = ind + 1;
}
for (int i = i1; i < len1; i = i + 1) {
united[ind] = th1[i];
ind = ind + 1;
}for (int i = i2; i < len2; i = i + 1) {
united[ind] = th2[i];
ind = ind + 1;
}
return united;
}
static int[] merge3sort(int[] m) { //function that glues all together
if (m.length == 1) {
return m;
}
else{
return merge(merge3sort(split(m,m.length/3)),merge3sort(split(m,m.length/3)),merge3sort(split(m,m.length/3))); }
}
I get the following exception:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0
at ololosh1.Ex3.split(Ex3.java:27)
at ololosh1.Ex3.merge3sort(Ex3.java:98)
at ololosh1.Ex3.main(Ex3.java:15)
Look at this part of your code:
for(int i = 0; i<=m.length/3; i++)
third[i]=m[i];
for(int i=0; i<=m.length/3;i++)
third1[i]=m[i+thirdNum];
for(int i=0; i<=m.length/3;i++)
third2[i]=m[i+2*thirdNum];
Arrays are indexed from 0 to length-1. Each third* array has length m.length/3. Therefore their index can only go up to m.length/3 - 1. Yet you are indexing up to and including m.length/3.
Once you get your application working correctly, you really should clean it up. There is a lot of redundancy. For example, you are using the expression m.length/3 multiple times in method split() but you are also passing that same value to it as an argument.
Hello I'm new to programming and registered to this forum :)
So I created a little program with nested for loops that prints out all combinations of five numbers which can have a value from 0 to 5. With nested for-loops this works fine. But isn't there a cleaner solution? I tried it with calling the for loop itself, but my brain doesn't get the solution.. :(
//my ugly solution
int store1, store2, store3, store4, store5;
for (int count = 0; count <= 5; count++) {
store1 = count;
for (int count2 = 0; count2 <= 5; count2++) {
store2 = count2;
for (int count3 = 0; count3 <= 5; count3++) {
store3 = count3;
for (int count4 = 0; count4 <= 5; count4++) {
store4 = count4;
System.out
.println(store1 + " " + store2 + " " + store4);
}
//I'm trying around with something like this
void method1() {
for (int count = 0; count <= 5; count++) {
list.get(0).value = count;
count++;
method2();
}
}
void method2() {
for (int count = 0; count <= 5; count++) {
list.get(1).value = count;
count++;
method1();
}
}
Usually when people try to use recursion or functional, using a loop is simpler or faster. However, in this case recursion is the simpler option in combination with a loop.
public static void method(List<Integer> list, int n, int m) {
if (n < 0) {
process(list);
} else {
for(int i = 0; i < m; i++) {
list.set(n, i);
method(list, n-1, m);
}
}
}
I know that you are trying combinations but this might help.
Permutation with repetitions
When you have n things to choose from ... you have n choices each time!
When choosing r of them, the permutations are:
n × n × ... (r times) = n^r
//when n and r are known statically
class Permutation
{
public static void main(String[] args)
{
char[] values = {'a', 'b', 'c', 'd'};
int n = values.length;
int r = 2;
int i = 0, j = 0;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
System.out.println(values[j] + " " + values[i]);
}
}
}
}
//when n and r are known only dynamically
class Permutation
{
public static void main(String[] args)
{
char[] values = {'a', 'b', 'c', 'd'};
int n = values.length;
int r = 2;
int i[] = new int[r];
int rc = 0;
for(int j=0; j<Math.pow(n,r); j++)
{
rc=0;
while(rc<r)
{
System.out.print(values[i[rc]] + " ");
rc++;
}
System.out.println();
rc = 0;
while(rc<r)
{
if(i[rc]<n-1)
{
i[rc]++;
break;
}
else
{
i[rc]=0;
}
rc++;
}
}
}
}
Something like this?
// Print all sequences of len(list)+n numbers that start w/ the sequence in list
void method( list, n ) {
if ( list.length == n )
// print list
else for ( int c=0; c<=5; c++ ) {
// add c to end of list
method( list, n );
// remove c from end of list
}
}
Initial call would be method( list, 5 ) where list is initially empty.
here another interative but less elegant version
while (store1 < 6) {
store5++;
if (store5 == 6) {
store5 = 0;
store4++;
}
if (store4 == 6) {
store4 = 0;
store3++;
}
if (store3 == 6) {
store3 = 0;
store2++;
}
if (store2 == 6) {
store2 = 0;
store1++;
}
System.out.println(store1 + " " + store2 + " " + store3 + " " + store4 + " " + store5 + " ");
}
The simplest code I can think of would tackle the problem with an entirely different approach:
public class TestA {
public static void main(String[] argv) {
for (int i=0; i<(6 * 6 * 6 * 6 * 6); ++i) {
String permutation = Integer.toString(i, 6);
System.out.println("00000".substring(permutation.length()) + permutation);
}
}
}
From your text (not your code) I gather you have 5 places and 6 symbols, which suggests there are 6 to the 5th power combinations. So the code just counts through those numbers and translates the number to the output combination.
Since this can also be viewed as a number system with base 6, it makes use of Integer.toString which already has formatting code (except the leading zeros) for this. Leading zeros are added where missing.