So for this extra credit problem in my calculus class, my other nerdy classmates and I decided that we would build a program to brute force a solution. One of these steps involves permutations. Through this algorithm, I managed to get it to work (I think):
public void genPermutations(int[] list, int k){
System.out.println("List: " + Arrays.toString(list));
System.out.println("----------------------");
if(k > list.length){
System.out.println("Not enough elements!");
return;
}
int[] counts = new int[list.length];
for(int i = 0; i < counts.length; i++){
counts[i] = 1;
}
int[] data = new int[k];
permutationHelper(list, counts, data, 0, k);
}
public void permutationHelper(int[] list, int[] counts, int[] data, int index, int k){
if(index == k){
//System.out.println(Arrays.toString(data));
permutations.add(data);
}else{
for(int i = 0; i < list.length; i++){
if(counts[i] == 0){
continue;
}
data[index] = list[i];
counts[i]--;
permutationHelper(list, counts, data, index + 1, k);
counts[i]++;
}
}
}
I have an ArrayList that stores all of the possible permutations (as integer arrays) that can be made from k elements of the list that I pass into the function. The problem is that if I print all of these permutations outside of the function, say after I call the genPermutations function, every permutation now is the same. But, when I print out the data where the comment is in the permutationHelper function, it correctly lists every possible permutation; I'm just unable to access them within the program later. My question is why are the values changing when I exit the function? Any help would be greatly appreciated.
Here are some pictures:
What is printed where the comment is.
What is printed later in the program.
The code used to print everything outside of the function is:
for(int i = 0; i < permutations.size(); i++){
System.out.println(Arrays.toString(permutations.get(i)));
}
I don't really know if that's necessary to know, but I just thought I'd include it just in case. Thanks in advance.
You're constantly modifying the same array object. Instead of adding different arrays to your list, you're in fact adding a reference to the same array over and over again.
To fix, instead of adding the data array to your list, you would have to add a copy of it, e.g. using Arrays.copyOf():
permutations.add(Arrays.copyOf(data, data.length));
Here the problem is that you are modifying the array after adding it to the list, you are modifying the same object again and again in different iterations. You were getting [3,2,1] in the list is because that was the outcome from last iteration. So as a fix you can use the following code. What it does is it will create a copy of data array and add that to the list.
int[] temp = Arrays.copyOf(data, data.length);
permutations.add(temp);
OR you can use clone() from array as follows.
int[] temp = data.clone();
permutations.add(temp);
Related
I just recently started learning java and today I learned how I can do the so called selection sort. I have been trying for the last 3 hours to do a bucket sort, but there are some parts which I don't know how to code. Important : I am learning java completely by myself with a book. I am not a student and I am doing this as a hobby. I already googled everything I could think of and I didn't find a solution. I don't have a teacher or anybody who I can ask, so yea, any help would be appreciated!
Code:
private int[] bucketSort() {
int[]bucket=new int[maxSize+1];
int[]sortedElements = new int[elementaros.length];
for(int i=0; i<elementaros.length;i++) {
bucket[elementaros[i]]++; //it says that I can't convert from Car to int. How can I add the elements of array elemenators to bucket?
}
int outPos = 0;
for (int j = 0; j < bucket.length; j++){
for (int k = 0; k < bucket[i]; k++){
sortedElements[outPos++] = i;
}
}
return bucket;
}
The idea of the code :
I have an array elements of type Car(Car is another class of my program). It looks like this - Car[] elementaros. int maxSize shows the maximum number of administrable Car objects. What I want to do is the following - I want to sort the elements in the elementaros array alphabetically. I would really really appreciate it if somebody has the time to show me how this would function with an example code or would just give me some tips. As I said - I have nobody who I can ask.
A selection sort is a combination of searching and sorting.
The principle is quite simple but I always prefer a diagram than huge explanations.
Start a pointer at the beginning of your unsorted array. Then, for each value of the array, search for the minimum value (or search for the alphabeticaly ordered car) in your array and switch the position of the founded Car with the pointer (which is a Car too)
Then you can advance the pointer to the next element of the array.
Here is a basic implementation to do this
public static Car[] doSelectionSort(Car[] elementaros) {
for (int i = 0; i < elementaros.length - 1; i++) {
int index = i;
for (int j = i + 1; j < elementaros.length; j++) {
if (elementaros[j].getName().compareTo(elementaros[index].getName()) < 0) {
index = j;
}
}
Car nextOrderedCar = elementaros[index];
elementaros[index] = elementaros[i];
elementaros[i] = nextOrderedCar;
}
return elementaros;
}
Just for example sake, I imagine your object of type Car has a name that we could use for the comparison.
UPDATE 1:
I have read your initial question too quickly and it leds me to answer a total different sorting algorithm. My bad.
I found an implementation here that does the trick:
williamfiset bucket sort
explanation of bucket sort plus different implementation
Hope this helps.
I'm currently working on a homework assignment for a beginner-level class and I need help building a program that tests if a sodoku solution presented as an int[][] is valid. I do this by creating helper methods that check both rows, columns and grids.
To check the column I call a method called getColumn that returns a column[]. When I test it out it works fine. I then pass it out on a method called uniqueEntries that makes sure that there are no duplicates.
Problem is, when I call my getColumn method, it returns an array consisting of only one number (for example 11111111, 22222222, 33333333). I have no idea why it does that. Here is my code:
int[][] sodokuColumns = new int[length][length];
for(int k = 0 ; k < sodokuPuzzle.length ; k++) {
sodokuColumns[k] = getColumn(sodokuPuzzle, k);
}
for (int l = 0; l < sodokuPuzzle.length; l++) {
if(uniqueEntries(sodokuColumns[l]) == false) {
columnStatus = false;
}
}
my helper is as follows
public static int[] getColumn(int[][] intArray, int index) {
int[] column = new int[intArray.length];
for(int i = 0 ; i < intArray.length ; i++) {
column[i] = intArray[i][index];
}
return column;
}
Thanks !
You said:
when I call my getColumn method, it returns an array consisting of only one number (for example 11111111, 22222222, 33333333).
I don't see any issue with your getColumn method other than the fact it's not even needed because getColumn(sodokuPuzzle, k) is the same as sodokuPuzzle[k]. If you're going to conceptualize your 2D array in such a way that your first index is the column then for your purpose of checking uniqueness you only need to write a method to get rows.
The issue you're having would seem to be with another part of your code that you did not share. I suspect there's a bug in the logic that accepts user input and that it's populating the puzzle incorrectly.
Lastly a tip for checking uniqueness (if you're allowed to use it) would be to create a Set of some kind (e.g. HashSet) and add all of your items (in your case integers) to that set. If the set has the same size as your original array of items then the items are all unique, if the size differs there are duplicates.
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
The goal of this is to find an object in the array and then remove it? is it possible??
Changing the length of an array is not possible. Recall that array is a static data structure whose size is determined before hand. Increasing or decreasing is not supported in this data structure. The fact that one has to increase or decrease the size depending on the usecase means that they have picked up the wrong data structure. They should perhaps go with an ArrayList.
Anyway, coming back to your question, you can simulate the 'size decrease' by maintaining a variable which you let track the array index and decrease the size of this variable. This lets you give the impression of shrinking the array.
The code you have provided does the same. Note however, that you should be using this modified index to track the contents of your array.
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
Whenever a particular bag at a given index equals to the item under question i.e., 'a', we swap elements so that the current bag element to be removed moves to the last and also we reduce the size of our new index - numElements by 1 to simulate this.
If you have the full code with you, please consider adding the following snippet at the end of that program to understand this more:
// Simulation of the array shrinking.
for(int i = 0; i < numElements; i++)
{
System.out.println( bag[i] );
}
// Movement of uninteresting elements to the end of the array.
for(int i = 0; i < bag.length; i++)
{
System.out.println( bag[i] );
}
It's not possible to change the length of an array. You can overwrite the element you wish to remove with the last element of the array and then copy the first bag.length - 1 elements of your array to a new array whose length is bag.length - 1.
for(int i = 0; i < bag.length; i++) {
if(bag[i].equals(a)) {
bag[i] = bag[bag.length-1];
bag = Arrays.copyOf (bag, bag.length - 1);
break;
}
}
public static String[] removeElements(String[] input) {
List<String> result = new ArrayList<String>();
String deleteValue = "somevalue";
for(String item : input)
if(!deleteValue .equals(item))
result.add(item);
return result.toArray(input);
}
This is one method you can fit this into your program.
You cannot decrease the size of an array. okay no problem! you can create your own data structure which supports that right?
Now, create a class named say MyArray with functions like increaseLenght(int) and decreseLength(int). Try it if you want to, will be fun for sure..
You cannot reduce the size of an array. Arrays are fixed length. What you can do is have a variable that indicates how many entries of the array you are using. This is what you are doing with numElements. The standard class ArrayList is implemented like this. The data is kept in an array and a private field size is used. With an ArrayList, when you remove an element, all the elements to the right are shifted left. However I also like your idea.
I would suggest 2 changes.
Make the last element null instead. If you are removing the element, why does it still need to be in the array?
Use numElements - 1 rather than bag.length-1 as the array could be bigger.
With these changes it becomes:
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
bag[i] = bag[numElements-1];
bag[numElements-1] = null;
numElements--;
break;
}
}
I need to delete elements from the array points. This is how I do this. The problem is that pts.length is always the same, and the removed elements have the value null. Therefore at some moment I receive the error message java.lang.NullPointerException.
for (int i = 0; i < points.length; i++) {
int ind = r.nextInt(pts.length);
TSPPoint pt = points[ind];
pts = removeElements(points,ind);
solPoints[i] = pt;
System.out.println(pts.length);
}
private static TSPPoint[] removeElements(TSPPoint[] input, int ind) {
List<TSPPoint> result = new LinkedList<TSPPoint>();
for(int i=0; i<input.length; i++)
if(i != ind)
result.add(input[i]);
return (TSPPoint[]) result.toArray(input);
}
What your code seems to (be supposed to) do is to remove random elements from the original array of points and append them to the pts array, i.e., create a permutation of points.
If this is the case, I suggest converting your array to a List and using Collections.shuffle.
#nrathaus has found the bug for you. It's simply that you've got your arrays confused (you're passing points into removeElements, but using pts everywhere else).
But if memory churn is being an issue, there's a much more efficient way to implement removeElements, using System.arraycopy rather than a temporary LinkedList.:
private static TSPPoint[] removeElements(TSPPoint[] input, int ind) {
TSPPoint[] rv;
if (ind >= 0 && ind < input.length) {
// New array will be one smaller
rv = new TSPPoint[input.length - 1];
if (rv.length > 0) {
// Copy the bit before the element we delete
if (ind > 0) {
System.arraycopy(input, 0, rv, 0, ind);
}
// Copy the rest
System.arraycopy(input, ind + 1, rv, ind, input.length - ind);
}
}
else {
// No change
rv = input;
}
return rv;
}
Mind you, if you're doing this a lot, creating and releasing all of these arrays may not be ideal. Using a List throughout may be better.
Wow, for deleting every element you recreate the rest of the array in a LinkedList, which you then turn into an array... The performance of this code, both in time and space, is terrible and so is readability, maintainability and testability.
Why not drop using arrays all together?.. convert points into an ArrayList and use remove(index) directly on that list and use an iterator of the llist to remove multiple elements as you iterate through?
I'm trying to write a simple game where an enemy chases the player on a grid. I'm using the simple algorithm for pathfinding from the Wikipedia page on pathfinding. This involves creating two lists with each list item containing 3 integers. Here's test code I'm trying out to build and display such a list.
When I run the following code, it prints out the same numbers for each array in the ArrayList. Why does it do this?
public class ListTest {
public static void main(String[] args) {
ArrayList<Integer[]> list = new ArrayList<Integer[]>();
Integer[] point = new Integer[3];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//Added this line to confirm that Integer[] point is actually
//being filled with 3 random ints.
System.out.println(point[0] + "," + point[1] + "," + point[2]);
}
System.out.println();
//My current understanding is that this section should step through
//ArrayList list and retrieve each Integer[] point added above. It runs, but only
//the values of the last Integer[] point from above are displayed 10 times.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
}
First of all, several of the other answers are misleading and/or incorrect. Note that an array is an object. So you can use them as elements in a list, no matter whether the arrays themselves contain primitive types or object references.
Next, declaring a variable as List<int[]> list is preferred over declaring it as ArrayList<int[]>. This allows you to easily change the List to a LinkedList or some other implementation without breaking the rest of your code because it is guaranteed to use only methods available in the List interface. For more information, you should research "programming to the interface."
Now to answer your real question, which was only added as a comment. Let's look at a few lines of your code:
Integer[] point = new Integer[3];
This line creates an array of Integers, obviously.
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//...
}
Here you assign values to the elements of the array and then add a reference to the array to your List. Each time the loop iterates, you assign new values to the same array and add another reference to the same array to the List. This means that the List has 10 references to the same array which has been repeatedly written over.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
Now this loop prints out the same array 10 times. The values in the array are the last ones set at the end of the previous loop.
To fix the problem, you simply need to be sure to create 10 different arrays.
One last issue: If you declare it as Iterator<Integer[]> it (or Iterator<int[]> it), you do not need to cast the return value of it.next(). In fact this is preferred because it is type-safe.
Finally, I want to ask what the ints in each array represent? You might want to revisit your program design and create a class that holds these three ints, either as an array or as three member variables.
I would highly recommend to enclose the integer array of 3 numbers into a meaningful class, that would hold, display and control an array of 3 integers.
Then in your main, you can have an growing ArrayList of objects of that class.
You have an extra ) here:
element = (int[])it.next()); //with the extra parenthesis the code will not compile
should be:
element = (int[])it.next();
Besides the problem in the other answer, you cal it.next() two times, that cause the iterator move forward two times, obviously that's not what you want. The code like this:
element = (int[])it.next());
String el = (String)element;
But actually, I don't see you used el. Although it's legal, it seems meaningless.