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Recursive backtracking in Java for solving a crossword

I need to solve a crossword given the initial grid and the words (words can be used more than once or not at all).
The initial grid looks like that:
++_+++
+____+
___+__
+_++_+
+____+
++_+++
Here is an example word list:
pain
nice
pal
id
The task is to fill the placeholders (horizontal or vertical having length > 1) like that:
++p+++
+pain+
pal+id
+i++c+
+nice+
++d+++
Any correct solution is acceptable, and it's guaranteed that there's a solution.
In order to start to solve the problem, I store the grid in 2-dim. char array and I store the words by their length in the list of sets: List<Set<String>> words, so that e.g. the words of length 4 could be accessed by words.get(4)
Then I extract the location of all placeholders from the grid and add them to the list (stack) of placeholders:
class Placeholder {
int x, y; //coordinates
int l; // the length
boolean h; //horizontal or not
public Placeholder(int x, int y, int l, boolean h) {
this.x = x;
this.y = y;
this.l = l;
this.h = h;
}
}
The main part of the algorithm is the solve() method:
char[][] solve (char[][] c, Stack<Placeholder> placeholders) {
if (placeholders.isEmpty())
return c;
Placeholder pl = placeholders.pop();
for (String word : words.get(pl.l)) {
char[][] possibleC = fill(c, word, pl); // description below
if (possibleC != null) {
char[][] ret = solve(possibleC, placeholders);
if (ret != null)
return ret;
}
}
return null;
}
Function fill(c, word, pl) just returns a new crossword with the current word written on the current placeholder pl. If word is incompatible with pl, then function returns null.
char[][] fill (char[][] c, String word, Placeholder pl) {
if (pl.h) {
for (int i = pl.x; i < pl.x + pl.l; i++)
if (c[pl.y][i] != '_' && c[pl.y][i] != word.charAt(i - pl.x))
return null;
for (int i = pl.x; i < pl.x + pl.l; i++)
c[pl.y][i] = word.charAt(i - pl.x);
return c;
} else {
for (int i = pl.y; i < pl.y + pl.l; i++)
if (c[i][pl.x] != '_' && c[i][pl.x] != word.charAt(i - pl.y))
return null;
for (int i = pl.y; i < pl.y + pl.l; i++)
c[i][pl.x] = word.charAt(i - pl.y);
return c;
}
}
Here is the full code on Rextester.
The problem is that my backtracking algorithm doesn't work well. Let's say this is my initial grid:
++++++
+____+
++++_+
++++_+
++++_+
++++++
And this is the list of words:
pain
nice
My algorithm will put the word pain vertically, but then when realizing that it was a wrong choice it will backtrack, but by that time the initial grid will be already changed and the number of placeholders will be reduced. How do you think the algorithm can be fixed?

This can be solved in 2 ways:
Create a deep copy of the matrix at the start of fill, modify and return that (leaving the original intact).
Given that you already pass around the matrix, this wouldn't require any other changes.
This is simple but fairly inefficient as it requires copying the matrix every time you try to fill in a word.
Create an unfill method, which reverts the changes made in fill, to be called at the end of each for loop iteration.
for (String word : words.get(pl.l)) {
if (fill(c, word, pl)) {
...
unfill(c, word, pl);
}
}
Note: I changed fill a bit as per my note below.
Of course just trying to erase all letter may erase letters of other placed words. To fix this, we can keep a count of how many words each letter is a part of.
More specifically, have a int[][] counts (which will also need to be passed around or be otherwise accessible) and whenever you update c[x][y], also increment counts[x][y]. To revert a placement, decrease the count of each letter in that placement by 1 and only remove letters with a count of 0.
This is somewhat more complex, but much more efficient than the above approach.
In terms of code, you might put something like this in fill:
(in the first part, the second is similar)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]++;
And unfill would look something like this: (again for just the first part)
for (int i = pl.x; i < pl.x + pl.l; i++)
counts[pl.y][i]--;
for (int i = pl.x; i < pl.x + pl.l; i++)
if (counts[pl.y][i] == 0)
c[pl.y][i] = '_';
// can also just use a single loop with "if (--counts[pl.y][i] == 0)"
Note that, if going for the second approach above, it might make more sense to simply have fill return a boolean (true if successful) and just pass c down to the recursive call of solve. unfill can return void, since it can't fail, unless you have a bug.
There is only a single array that you're passing around in your code, all you're doing is changing its name.
See also Is Java "pass-by-reference" or "pass-by-value"?

You identified it yourself:
it will backtrack, but by that time the initial grid will be already
changed
That grid should be a local matrix, not a global one. That way, when you back up with a return of null, the grid from the parent call is still intact, ready to try the next word in the for loop.
Your termination logic is correct: when you find a solution, immediately pass that grid back up the stack.

can someone help me fix this code(string)-java

I try to write a old maid.
After dealing cards,and sorting, i have two parts of card,one is playerDeck, one is computerDeck. now the pairs need to be removed.but i was stuck at this stage.
for example(just an example )
playerDeck:
'A♡', 'A♢', '8♡', '8♢', '8♠', 'Q♠', '2♠', '4♣', '7♢', '7♣', 'K♣', 'A♡', 'J♡', '9♣', '3♢'
computerDeck:
'3♡','3♣', '10♡','10♠','10♣', '6♡', 'K♡','K♢', 'A♣', 'A♠', '4♢', '7♡','7♠'
String q;
String p;
ArrayStringsTools AA=new ArrayStringsTools();//this is a class that i will use for removing item
for(int i=0;i<playerDeck.length-1;i++){
q=playerDeck[i];
q=q.substring(0,1);//i try to find the first character
p=playerDeck[i+1];//after finding first character, i can compare them,and if they are same, then i can remove them
p=p.substring(0,1);
if(q==p){
AA.removeItemByIndex(playerDeck,26,i);//this is the method that i used for removing same item,i will put this code below
AA.removeItemByIndex(playerDeck,26,i+1);//there are 51 cards in total,player has 26, computer has 25
}
}
public static int removeItemByIndex(String[] arrayOfStrings, int currentSize, int itemToRemove){//this is the method i used for removing item(first is the array of Deck, second is the size of Deck,third is the index of item to remove)
if( arrayOfStrings == null || currentSize > arrayOfStrings.length) {
System.out.println("ArrayStringsTools.removeItemByIndex: wrong call");
return currentSize;
}
if( itemToRemove < 0 || itemToRemove >= currentSize ) {
System.out.println("ArrayStringsTools.removeItem: item "
+ itemToRemove + " out of bounds. Array Unchanged.");
return currentSize;
}
int i;
for( i = itemToRemove; i < currentSize-1; i++){
arrayOfStrings[i] = arrayOfStrings[i+1];
}
arrayOfStrings[i]= null;
return currentSize-1;
i think i wrote correctly, but it doesnt show any difference compared with the origin.
the result should be:
playerDeck: '8♠', 'Q♠', '2♠', '4♣', 'K♣', 'A♡', 'J♡', '9♣', '3♢'
computerDeck:'10♣', '6♡', '4♢'
or is there another way to do this,because when a pair removed,there are two empty spaces, so... I've been struggling for a long time......

To compare the 1st character, change this line
if (q == p) {
to
if (q.charAt(0) == p.charAt(0)) {
Notice that q == p checks to see if the q and p refer to the same string, and do not look at the contents at all. If you want to compare full strings (or any other object that is not a char, an int, or so on) by content, you should use equals: q.equals(p) returns true only if both have the same content.

If you want to compare two strings,you can use 'equals',like
if(q.equals(p)){//q==p if true,they are save in the same location-this may not be what you want,and in this code it will be false forever.
}

Converting a recursive method into a non-recursive method using loop in java

So I'm currently making a game where the instructions are to move left or right within an array using the integer stored at a marked index (circle in this case) until we can get the circle to the last index of the array. The last integer of the array is always 0.
For example,
[4] 1 2 3 1 0, here we start at the circle 0 (index)
We move 4 to the right, 4 1 2 3 [1] 0
Then 1 time to the right, 4 1 2 3 1 [0]. Here the game stops and we win.
My code is as follows for a recursive method:
public static boolean rightWing (int circle, int[] game, List<Integer> checkerList){
int last = game.length-1;
if (circle == last){ // base case for recursion
return true;
}
if (circle < 0){ // if we go out of bounds on the left
return false;
}
if (circle > last){ // if we go out of bounds on the right
return false;
}
if (checkerList.contains(circle)){ // check for the impossible case
return false;
}
checkerList.add(circle); // adds the circle value for the last check to checkerList so we can check for the impossible case
int moveRight = circle + game[circle]; // these two integers help the game move according to the value of the int at circle
int moveLeft = circle - game[circle];
return rightWing( moveRight, game, checkerList) || rightWing(moveLeft, game,checkerList);
}
This works great, but the only problem is it's recursive and slow. I'm trying to redesign it using loops and stacks/queues to make it more efficient, but I'm stuck after writing this (in pseudo):
Boolean rightWing (int circle, List<int> game, List<int> checkerList)
Int lastPlace = game.size() - 1
For int i <- 0 to game.size() - 1 do
If i equals lastPlace then // returns true when i is at the last position of the game
Return true
Any input on how to go forward would be appreciated!

The most important bit: when debugging app for the slowness, you should collect some performance data first to identify where your app is spending the most of its time. Otherwise fixing performance is inefficient. You can use jvisualvm it's bundled with jdk.
Data structures rule the world of performance
One thing why it can be slow is because of this:
if (checkerList.contains(circle)){ // check for the impossible case
return false;
}
The more items you have in the list, the slower it becomes. List has linear complexity for the contains method. You can make it constant complexity if you'll use HashSet. E.g. if you have list with 100 elements, this part will be around slower 100 times with List than with HashSet.
Another thing which might be taking some time is boxing/unboxing: each time you put element to the list, int is being wrapped into new Integer object - this is called boxing. You might want to use IntSet to avoid boxing/unboxing and save on the GC time.
Converting to the iterative form
I won't expect this to affect your application speed, but just for the sake of completeness of the answer.
Converting recursive app to iterative form is pretty simple: each of the method parameters under the cover is stored on a hidden stack on each call of your (or others function). During conversion you just create your own stack and manage it manually
public static boolean rightWingRecursive(int circle, int[] game) {
Set<Integer> checkerList = new HashSet<Integer>();
Deque<Integer> statesToExplore = new LinkedList<>();
int last = game.length - 1;
statesToExplore.push(circle);
while (!statesToExplore.isEmpty()) {
int circleState = statesToExplore.pop();
if (circleState == last) { // base case for recursion
return true;
}
if (circleState < 0) { // if we go out of bounds on the left
continue;
}
if (circleState > last) { // if we go out of bounds on the right
continue;
}
if (checkerList.contains(circle)) { // check for the impossible case
continue;
}
checkerList.add(circle); // adds the circle value for the last check to
// checkerList so we can check for the
// impossible case
int moveRight = circle + game[circle]; // these two integers help the
// game move according to the
// value of the int at circle
int moveLeft = circle - game[circle];
statesToExplore.push(moveRight);
statesToExplore.push(moveLeft);
}
return false;
}

Optimal render draw-order function with specified z-index values

I found recently the default renderable sort function in LibGDX wasn't quite up to my needs. (see; Draw order changes strangely as camera moves? )
Essentially a few objects rendered in front when they should render behind.
Fortunately, the renderables in question always have a guarantied relationship. The objects are attached to eachother so when one moves the other moves. One object can be seen as being literally "pinned" to the other, so always in front.
This gave me the idea that if I specified a "z-index" (int) and "groupname" (String) for each object, I could manually take over the draw order, and for things with the same groupname, ensure they are positioned next to eachother in the list, in the order specified by the z-index. (low to high)
//For example an array of renderables like
0."testgroup2",11
1."testgroup",20
2."testgroup2",10
3.(no zindex attribute)
4."testgroup",50
//Should sort to become
0."testgroup",20
1."testgroup",50
2.(no zindex attribute)
3."testgroup2",10
4."testgroup2",11
// assuming the object2 in testgroup2 are closer to the camera, the one without a index second closest, and the rest furthest<br>
//(It is assumed that things within the same group wont be drastically different distances)
I implemented a sort system in libgdx to do this as followed;
/**
* The goal of this sorter is to sort the renderables the same way LibGDX would do normally (in DefaultRenderableSorter)<br>
* except if they have a ZIndex Attribute.<br>
* A Zindex attribute provides a groupname string and a number.<br>
* Renderables with the attribute are placed next to others of the same group, with the order within the group determined by the number<br>
*
* For example an array of renderables like;<br><br>
* 0."testgroup",20<br>
* 1."testgroup2",10<br>
* 2.(no zindex attribute)<br>
* 3."testgroup",50<br>
* <br>Should become;<br><br>
* 0."testgroup",20<br>
* 1."testgroup",50<br>
* 2.(no zindex attribute)<br>
* 3."testgroup2",10<br>
* <br>
* assuming the object in testgroup2 is closer to the camera, the one without a index second closest, and the rest furthest<br>
* (It is assumed that things within the same group wont be drastically different distances)<br>
*
* #param camera - the camera in use to determine normal sort order when we cant place in a existing group
* #param resultList - an array of renderables to change the order of
*/
private void customSorter(Camera camera, Array<Renderable> resultList) {
//make a copy of the list to sort. (This is probably a bad start)
Array <Renderable> renderables = new Array <Renderable> (resultList);
//we work by clearing and rebuilding the Renderables array (probably not a good method)
resultList.clear();
//loop over the copy we made
for (Renderable o1 : renderables) {
//depending of if the Renderable as a ZIndexAttribute or not, we sort it differently
//if it has one we do the following....
if (o1.material.has(ZIndexAttribute.ID)){
//get the index and index group name of it.
int o1Index = ((ZIndexAttribute)o1.material.get(ZIndexAttribute.ID)).zIndex;
String o1GroupName = ((ZIndexAttribute)o1.material.get(ZIndexAttribute.ID)).group;
//setup some variables
boolean placementFound = false; //Determines if a placement was found for this renderable (this happens if it comes across another with the same groupname)
int defaultPosition = -1; //if it doesn't find another renderable with the same groupname, this will be its position in the list. Consider this the "natural" position based on distance from camera
//start looping over all objects so far in the results (urg, told you this was probably not a good method)
for (int i = 0; i < resultList.size; i++) {
//first get the renderable and its ZIndexAttribute (null if none found)
Renderable o2 = resultList.get(i);
ZIndexAttribute o2szindex = ((ZIndexAttribute)o2.material.get(ZIndexAttribute.ID));
if (o2szindex!=null){
//if the renderable we are comparing too has a zindex, then we get its information
int o2index = o2szindex.zIndex;
String o2groupname = o2szindex.group;
//if its in the same group as o1, then we start the processing of placing them nexto eachother
if (o2groupname.equals(o1GroupName)){
//we either place it in front or behind based on zindex
if (o1Index<o2index){
//if lower z-index then behind it
resultList.insert(i, o1);
placementFound = true;
break;
}
if (o1Index>o2index){
//if higher z-index then it should go in front UNLESS there is another of this group already there too
//in which case we just continue (which will cause this to fire again on the next renderable in the inner loop)
if (resultList.size>(i+1)){
Renderable o3 = resultList.get(i+1);
ZIndexAttribute o3szindex = ((ZIndexAttribute)o3.material.get(ZIndexAttribute.ID));
if (o3szindex!=null){
String o3groupname = o3szindex.group;
if (o3groupname!=null && o3groupname.equals(o1GroupName)){
//the next element is also a renderable with the same groupname, so we loop and test that one instead
continue;
}
}
}
// Gdx.app.log("zindex", "__..placeing at:"+(i+1));
//else we place after the current one
resultList.insert(i+1, o1);
placementFound = true;
break;
}
}
}
//if no matching groupname found we need to work out a default placement.
int placement = normalcompare(o1, o2); //normal compare is the compare function in DefaultRenderableSorter.
if (placement>0){
//after then we skip
//(we are waiting till we are either under something or at the end
} else {
//if placement is before, then we remember this position as the default (but keep looking as there still might be matching groupname, which should take priority)
defaultPosition = i;
//break; //break out the loop
}
}
//if we have checked all the renderables positioned in the results list, and none were found with matching groupname
//then we use the defaultposition to insert it
if (!placementFound){
//Gdx.app.log("zindex", "__no placement found using default which is:"+defaultPosition);
if (defaultPosition>-1){
resultList.insert(defaultPosition, o1);
} else {
resultList.add(o1);
}
}
continue;
}
//...(breath out)...
//ok NOW we do placement for things that have no got a ZIndexSpecified
boolean placementFound = false;
//again, loop over all the elements in results
for (int i = 0; i < resultList.size; i++) {
Renderable o2 = resultList.get(i);
//if not we compare by default to place before/after
int placement = normalcompare(o1, o2);
if (placement>0){
//after then we skip
//(we are waiting till we are either under something or at the end)
continue;
} else {
//before
resultList.insert(i, o1);
placementFound = true;
break; //break out the loop
}
}
//if no placement found we go at the end by default
if (!placementFound){
resultList.add(o1);
};
} //go back to check the next element in the incomeing list of renderables (that is, the copy we made at the start)
//done
}
//Copy of the default sorters compare function
//;
private Camera camera;
private final Vector3 tmpV1 = new Vector3();
private final Vector3 tmpV2 = new Vector3();
public int normalcompare (final Renderable o1, final Renderable o2) {
final boolean b1 = o1.material.has(BlendingAttribute.Type) && ((BlendingAttribute)o1.material.get(BlendingAttribute.Type)).blended;
final boolean b2 = o2.material.has(BlendingAttribute.Type) && ((BlendingAttribute)o2.material.get(BlendingAttribute.Type)).blended;
if (b1 != b2) return b1 ? 1 : -1;
// FIXME implement better sorting algorithm
// final boolean same = o1.shader == o2.shader && o1.mesh == o2.mesh && (o1.lights == null) == (o2.lights == null) &&
// o1.material.equals(o2.material);
o1.worldTransform.getTranslation(tmpV1);
o2.worldTransform.getTranslation(tmpV2);
final float dst = (int)(1000f * camera.position.dst2(tmpV1)) - (int)(1000f * camera.position.dst2(tmpV2));
final int result = dst < 0 ? -1 : (dst > 0 ? 1 : 0);
return b1 ? -result : result;
}
As far as I can tell my customSorter function produces the order I want - the renderables now look like they are drawn in the right order.
However, this also seems like a hackjob, and I am sure my sorting algorithm is horrendously inefficient.
I would like advice on how to either;
a) Improve my own algorithm, especially in regards to any quirks to bare in mind when doing cross-platform LibGDX development (ie, array types, memory management in regards to android/web etc)
b) Alternative more efficient solutions having a similar "z index override" of the normal draw-order sorting.
Notes;
. The grouping is necessary. This is because while things are firmly stuck relatively to eachother within a group, groups themselves can also move about in front/behind eachother. (but not between). This makes it tricky to do a "global" override of the draw order, rather then a local one per group.
. If it helps, I can add/change the zindexattribute object in any way.
. I am thinking somehow "pre-storeing" each group of objects in a array could help things, but not 100% sure how.

First of all do never copy a list if not needed. The list with renderables could be really huge since it also could contain resources. Copying will be very very slow. If you need something local and you need performance try to make it final since it can improve the performance.
So a simple approach would be the default sorting of Java. You need to implement a Comperator for your class for example the Class with z index could look like this:
public class MyRenderable {
private float z_index;
public MyRenderable(float i)
{
z_index = i;
}
public float getZ_index() {
return z_index;
}
public void setZ_index(float z_index) {
this.z_index = z_index;
}
}
If you want a faster sort since your list wont change that much on runtime you could implement a insertion sort since it does a faster job if the list is kind of presorted. If it is not pre sorted it does take longer but in general it should only be the first sort call where it is alot disordered in your case.
private void sortList(ArrayList<MyRenderable> array) {
// double starttime = System.nanoTime();
for (int i = 1; i < array.size(); i++) {
final MyRenderable temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).getZ_index() < temp.getZ_index()) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
// System.out.println("Time taken: " + (System.nanoTime() - starttime));
}
To use this method you simply call it with your Array
sortList(renderbales);
In your case you need to take care of the ones that do not have a Z index. Maybe you could give them a 0 since they'll get sorted at the right position(i guess). Else you can use the given methods in z case and the regular in no z case as you do already.
After the conversation in the comments. I dont think it is a good idea to push everything into one list. It's hard to sort and would be very slow. A better approach would be a list of groups. Since you want to have groups, programm a group. Do not use String names, use IDs or types (way more easy to sort and it doesn't really matter). So a simple group would be this:
public class Group{
//think about privates and getters or methods to add things which also checks some conditions and so on
public int groupType;
public ArrayList<MyRenderable> renderables;
}
And now all your groups into a list. (this contains all your renderbales then)
ArrayList<Group> allRenderables = new ArrayList<>();
Last but not least sort the groups and sort the renderables. Since i dont think that your group ids/names will change on runtime, sort them once or even use a SortedSet instead of a ArrayList. But basically the whole sorting looks like this:
for(Group g: allRenderables)
sortRenderables(g.renderables); //now every group is sorted
//now sort by group names
sortGroup(allRenderables);
With the following insertionsorts as shown above
public static void sortRenderables(ArrayList<MyRenderable> array) {
for (int i = 1; i < array.size(); i++) {
final MyRenderable temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).getZ_index() < temp.getZ_index()) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
}
public static void sortGroup(ArrayList<Group> array) {
for (int i = 1; i < array.size(); i++) {
final Group temp = array.get(i);
int j = i - 1;
while (j >= 0 && array.get(j).groupType < temp.groupType) {
array.set(j + 1, array.get(j));
j--;
}
array.set(j + 1, temp);
}
}

How to check a matrix postion value in java?

Given a matrix filled in with all its values, there's the need of pulling out data randomly in order to create a random matrix (intialized with null values each position). The issue lies while checking if a position (within the randomMatrix) is different from null, as shown below:
public void randomLogic(String[][] givenMatrix){
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
int positioni = this.randInt(0, 1);
int positionj = this.randInt(0, 1);
int x = this.randInt(0, 3);
int y = this.randInt(0, 2);
/*In here lies the reported issue while checking if empty. */
while (!this.randomMatrix[x][y].isEmpty()) {
x = this.randInt(0, 3);
y = this.randInt(0, 2);
this.randomMatrix[x][y] = givenMatrix[positioni][positionj];
}
}
}
}
I've also tried with the following: while(this.randomMatrix[x][y] != null) and the code breaks down right away. All the solving-logic work out as expected (because if I ommit that part it works with flaws but works) the only problem is that validation. Regardless of what position is being evaluated it always stops working.
Is there any other way of checking a matrix position value?

are you trying to pull random values (and consequently, some will be repeated) from a givenMatrix OR do you want to specifically randomize the givenMatrix?
If it's the latter, I would approach it differently.
First of all, gather all values of the givenMatrix in one linear-array or list.
Then, you can randomize this array, and you'll be left with a big array of rows*columns items, with all the values from the original matrix already randomized.
Then, you can systematically fill the new randomized matrix, taking each element of the array in order, till you complete each row and column of the new matrix.
I would choose this approach, because it's easier, and it will finish in a fixed number of steps; rather than filling random positions, because maybe you land 2 times in the same place to fill, and you'll have to randomize until you hit all the spots.
(Also, if you don't care of repeating the values for the new matrix, then instead of taking each element of the randomized array, you can simply pick a random position of this array each time)

Have you declared String[][] randomMatrix as a class level member variable? If you have declared the randomMatrix correctly then the isEmpty() and != null checks should work.