Im writing a recursive function in Java (graph theory) to get all paths in a 4x4 table, beginning at a random starting point. Possible directions are horizontal, vertical & diagonal, but I have a requirement that the same location cannot be visited twice.
The script works fine so far, I get a lot of combinations. The problem is that in the for loop of the function, when there is more than one possible way, then I get wrong results in the second and following loops because the boolean[] tempvisited is not getting back to his old values.
I hope there is someone, that may understand my English and my problem too. Here is my code so far:
// here I define a constant input of values:
String letters = "1548987425461854"
// This matrix shows all possible directions from every startpoint in the matrix:
// from the second value, you may get to the following locations: 1,3,5,6 and 7
private int[][] matrix = {
{1,4,5},
{0,2,4,5,6},
{1,3,5,6,7},
{2,6,7},
{0,1,5,8,9},
{0,1,2,4,6,8,9,10},
{1,2,3,5,7,9,10,11},
{2,3,6,10,11},
{4,5,9,12,13},
{4,5,6,8,10,12,13,14},
{5,6,7,9,11,13,14,15},
{6,7,10,14,15},
{8,9,13},
{8,9,10,12,14},
{9,10,11,13,15},
{10,11,14}
};
// Here begins the recursive function
public List<Combination> depthFirst(int vertex, boolean[] visited, Combination zeichen, List<Combination> combis){
// A temporary list of booleans to mark every value position visited or not
boolean[] tempvisited = new boolean[16];
// combis is the whole list of ways, zeichen is just the actual combination
zeichen.name = zeichen.name + this.letters.charAt(vertex);
combis.add(zeichen.name);
//marks actual value as visited
visited[vertex] = true;
for(int i = 0; i < 16; i++){
tempvisited[i] = visited[i];
}//end for
// going to next possible locations
for (int i = 0; i < this.matrix[vertex].length; i++) {
if (!visited[this.matrix[vertex][i]]) {
combis = depthFirst(this.matrix[vertex][i], tempvisited, zeichen, combis);
}//end if
}//end for
return combis;
}
You have the right idea with tempvisited, making a copy. But you're doing so in the wrong place.
You're setting visited[vertex] = true, which means that the visited you passed in is changing. What you want is for visited to never change. Make a copy of it, and make your changes to that copy.
Also, I notice that you use the same zeichen every time. So if you have a path 3 steps long, your combis list with be 3 copies of the same zeichen. That seems incorrect.
You set visited[vertex] to true before the first for loop; you could reset it to false just before you return. If every call undoes the change it did (directly), then every call will return with visited back to its state when that call was made. No tempvisited needed.
Take a look to this other recursive solution (pseudocode) for the Depth First Search (DFS).
void search(Node root) {
if (root == null) return;
visit(root);
root.visited = true;
foreach (Node n in root.adjacent) {
if (n.visited == false)
search(n);
}
}
Actually you don't need a copy of the visited array. Mark the node as visited right before the reccurrent call of depthFirst and then "unmark" it right after the call. Something like:
for (int i = 0; i < this.matrix[vertex].length; i++) {
if (!visited[this.matrix[vertex][i]]) {
visited[this.matrix[vertex][i]] = true;
combis = depthFirst(this.matrix[vertex][i], tempvisited, zeichen, combis);
visited[this.matrix[vertex][i]] = false;
}//end if
}//end for
Related
I have a 2d array that generates terrain using perlin noise, and then places a (3D) block at a specific height - Before you click away, all I need help with is the 2D array that generates the "height-map". I am trying to figure out whether or not the block next to it is at the same elevation (if it is "neighboring" or not) by checking the values directly up, down, left, and right in the 2D array. If they are equal, then they are at the same elevation, and therefore "neighbors". if the problem that I am running into is that the check is always returning true for all the neighbors, even if the block has no neighbors.
A small example of the perlin noise height map
151514141312121111
151414131313121211
141414131312121211
141313131312121211
131313121212121111
131312121212111111
121212121111111111
111111111110101111
111111111010101111
111111111010101010
111111111010101010
101011101010101010
101010101099109999
991010109999988889
999109999888888999
and here is the checking code, you will have to see the entire file, linked below for context
if (terrain[x][leftColumn] == terrain[x][z]) {
neighbors[2] = true; // left side
}
if (terrain[x][rightColumn] == terrain[x][z]) {
neighbors[3] = true; //right side
}
if (terrain[belowRow][z] == terrain[x][z]) {
neighbors[4] = true; // front side (below)
}
if (terrain[aboveRow][z] == terrain[x][z]) {
neighbors[5] = true; // back side (above)
}
Pastebin: https://www.pastiebin.com/5d5c5416391ec
any help is appreciated, Asher
Move this static variable initialization
boolean[] neighbors = new boolean[]{false, false, false, false, false, false};
inside the inner loop, where you check each block's neighbors, to instantiate a new neighbors array for each individual block. Right now neighbors is a static variable. You never reset the values on the neighbors array so it remains true after each iteration.
edit:
Also
if (belowRow > 1) {
belowRowExists = false;
belowRow = 0;
}
if (rightColumn > - 1) {
rightColumnExists = false;
rightColumn = 0;
}
is wrong, you want to check if the column or row is out of bounds right? Then you want to see if they are >= chunkSize.
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.
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;
}
Here is the algorithm (not working) Please let me know where the error is
Thanks
private void checkSouth(Location point, int player) {
//Loop through everything south
boolean isthereAnOppositePlayer=false;
int oppositePlayer=0;
//Set opposite player
if (player==1) {
oppositePlayer=2;
}else{
oppositePlayer=1;
}
for (int i = point.getVertical(); i < 8; i++) {
//Create a location point with the current location being compared
MyLocation locationBeingChecked= new MyLocation();
locationBeingChecked.setHorizontal(point.getHorizontal());
locationBeingChecked.setVertical(i);
int value = board[locationBeingChecked.getVertical()][locationBeingChecked.getHorizontal()];
//If the first checked is the opposite player
if (value==oppositePlayer) {
//Then potential to evaluate more
isthereAnOppositePlayer=true;
}
//If it isn't an opposite player, then break
if(!isthereAnOppositePlayer && value!=0){
break;
}
//If another of the player's piece found or 0, then end
if (isthereAnOppositePlayer && value==player || isthereAnOppositePlayer && value==0) {
break;
//end
}
//Add to number of players to flip
if(isthereAnOppositePlayer && value==oppositePlayer && value!=0){
//add to array
addToPiecesToTurn(locationBeingChecked);
}
}
}
It looks like the locations that got rotated back to the other player are the exact same as those rotated during the first move. I would guess that the array being populated by addToPiecesToTurn is perhaps not being cleared out between each move, so all the previous locations are still in there.
If you are storing the pieces to be turned in an ArrayList, you can use the clear() method to erase the contents of the collection between each turn.
Another possible problem is that you are checking for the opposite player, and then instantly beginning to populate addToPiecesToTurn. However, the pieces in that direction are not necessarily valid to be rotated unless they are "sandwiched" in by a second location containing the current player's piece. I don't think your code is properly checking for that case; when that happens, you'll want to somehow skip flipping those pieces to the other player, such as clearing out the array of piecesToTurn.
Edit: Looking at your current solution where you are implementing every direction separately, you are going to have a lot of duplicated code. If you think about what it means to walk along a certain direction, you can think of it as adjusting the x/y value by a "step" amount. The step amount could be -1 for backwards, 0 for no move, or 1 for forwards. Then you could create a single method that handles all directions without duplicating the logic:
private void checkDirection(Location point, int player, int yStep, int xStep) {
int x = point.getHorizontal() + xStep;
int y = point.getVertical() + yStep;
MyLocation locationBeingChecked = new MyLocation();
locationBeingChecked.setHorizontal(x);
locationBeingChecked.setVertical(y);
while (isValid(locationBeingChecked)) {
// do the logic here
x += xStep;
y += yStep;
locationBeingChecked = new MyLocation();
locationBeingChecked.setHorizontal(x);
locationBeingChecked.setVertical(y);
}
}
You would need to implement isValid to check that the location is valid, i.e., in the board. Then you could call this method for each direction:
// north
checkDirection(curPoint, curPlayer, -1, 0);
// north-east
checkDirection(curPoint, curPlayer, -1, 1);
// east
checkDirection(curPoint, curPlayer, 0, 1);
// etc
This is the sort of problem that is ripe for some unit testing. You could very easily set up a board, play a move, and validate the answer, and the test results would give plenty of insight into where your expectations and reality diverge.
why didn't you use a 2d array ?
each cell would contain an enum : EMPTY, PLAYER_1, PLAYER_2 .
then, in order to go over the cells, you simply use loops for each direction.
for example, upon clicking on a cell , checking to the right would be:
for(int x=pressedLocation.x+1;x<cells[pressedLocation.y].length;++x)
{
Cell cell=cells[pressedLocation.y][x];
if(cell==EMPTY||cell==currentPlayerCell)
break;
cells[pressedLocation.y][x]=currentPlayerCell;
}
checking from top to bottom would be:
for(int y=pressedLocation.y+1;y<cells.length;++y)
{
Cell cell=cells[y][pressedLocation.x];
if(cell==EMPTY||cell==currentPlayerCell)
break;
cells[y][pressedLocation.x]=currentPlayerCell;
}
Hey guys, recently posted up about a problem with my algorithm.
Finding the numbers from a set which give the minimum amount of waste
Ive amended the code slightly, so it now backtracks to an extent, however the output is still flawed. Ive debugged this considerablychecking all the variable values and cant seem to find out the issue.
Again advice as opposed to an outright solution would be of great help. I think there is only a couple of problems with my code, but i cant work out where.
//from previous post:
Basically a set is passed to this method below, and a length of a bar is also passed in. The solution should output the numbers from the set which give the minimum amount of waste if certain numbers from the set were removed from the bar length. So, bar length 10, set includes 6,1,4, so the solution is 6 and 4, and the wastage is 0. Im having some trouble with the conditions to backtrack though the set. Ive also tried to use a wastage "global" variable to help with the backtracking aspect but to no avail.
SetInt is a manually made set implementation, which can add, remove, check if the set is empty and return the minimum value from the set.
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package recursivebacktracking;
/**
*
* #author User
*/
public class RecBack {
int WASTAGE = 10;
int BESTWASTAGE;
int BARLENGTH = 10;
public void work()
{
int[] nums = {6,1,2,5};
//Order Numbers
SetInt ORDERS = new SetInt(nums.length);
SetInt BESTSET = new SetInt(nums.length);
SetInt SOLUTION = new SetInt(nums.length);
//Set Declarration
for (int item : nums)ORDERS.add(item);
//Populate Set
SetInt result = tryCutting(ORDERS, SOLUTION, BARLENGTH, WASTAGE);
result.printNumbers();
}
public SetInt tryCutting(SetInt possibleOrders, SetInt solution, int lengthleft, int waste)
{
for (int i = 0; i < possibleOrders.numberInSet(); i++) // the repeat
{
int a = possibleOrders.min(); //select next candidate
System.out.println(a);
if (a <= lengthleft) //if accecptable
{
solution.add(a); //record candidate
lengthleft -= a;
WASTAGE = lengthleft;
possibleOrders.remove(a); //remove from original set
if (!possibleOrders.isEmpty()) //solution not complete
{
System.out.println("this time");
tryCutting(possibleOrders, solution, lengthleft, waste);//try recursive call
BESTWASTAGE = WASTAGE;
if ( BESTWASTAGE <= WASTAGE )//if not successfull
{
lengthleft += a;
solution.remove(a);
System.out.println("never happens");
}
} //solution not complete
}
} //for loop
return solution;
}
}
Instead of using backtracking, have you considered using a bitmask algorithm instead? I think it would make your algorithm much simpler.
Here's an outline of how you would do this:
Let N be number of elements in your set. So if the set is {6,1,2,5} then N would be 4. Let max_waste be the maximum waste we can eliminate (10 in your example).
int best = 0; // the best result so far
for (int mask = 1; mask <= (1<<N)-1; ++mask) {
// loop over each bit in the mask to see if it's set and add to the sum
int sm = 0;
for (int j = 0; j < N; ++j) {
if ( ((1<<j)&mask) != 0) {
// the bit is set, add this amount to the total
sm += your_set[j];
// possible optimization: if sm is greater than max waste, then break
// out of loop since there's no need to continue
}
}
// if sm <= max_waste, then see if this result produces a better one
// that our current best, and store accordingly
if (sm <= max_waste) {
best = max(max_waste - sm);
}
}
This algorithm is very similar to backtracking and has similar complexity, it just doesn't use recursion.
The bitmask basically is a binary representation where 1 indicates that we use the item in the set, and 0 means we don't. Since we are looping from 1 to (1<<N)-1, we are considering all possible subsets of the given items.
Note that running time of this algorithm increases very quickly as N gets larger, but with N <= around 20 it should be ok. The same limitation applies with backtracking, by the way. If you need faster performance, you'd need to consider another technique like dynamic programming.
For the backtracking, you just need to keep track of which element in the set you are on, and you either try to use the element or not use it. If you use it, you add it to your total, and if not, you proceeed to the next recursive call without increasing your total. Then, you decrement the total (if you incremented it), which is where the backtracking comes in.
It's very similar to the bitmask approach above, and I provided the bitmask solution to help give you a better understanding of how the backtracking algorithm would work.
EDIT
OK, I didn't realize you were required to use recursion.
Hint1
First, I think you can simplify your code considerably by just using a single recursive function and putting the logic in that function. There's no need to build all the sets ahead of time then process them (I'm not totally sure that's what you're doing but it seems that way from your code). You can just build the sets and then keep track of where you are in the set. When you get to the end of the set, see if your result is better.
Hint2
If you still need more hints, try to think of what your backtracking function should be doing. What are the terminating conditions? When we reach the terminating condition, what do we need to record (e.g. did we get a new best result, etc.)?
Hint3
Spoiler Alert
Below is a C++ implementation to give you some ideas, so stop reading here if you want to work on it some more by yourself.
int bestDiff = 999999999;
int N;
vector< int > cur_items;
int cur_tot = 0;
int items[] = {6,1,2,5};
vector< int > best_items;
int max_waste;
void go(int at) {
if (cur_tot > max_waste)
// we've exceeded max_waste, so no need to continue
return;
if (at == N) {
// we're at the end of the input, see if we got a better result and
// if so, record it
if (max_waste - cur_tot < bestDiff) {
bestDiff = max_waste - cur_tot;
best_items = cur_items;
}
return;
}
// use this item
cur_items.push_back(items[at]);
cur_tot += items[at];
go(at+1);
// here's the backtracking part
cur_tot -= items[at];
cur_items.pop_back();
// don't use this item
go(at+1);
}
int main() {
// 4 items in the set, so N is 4
N=4;
// maximum waste we can eliminiate is 10
max_waste = 10;
// call the backtracking algo
go(0);
// output the results
cout<<"bestDiff = "<<bestDiff<<endl;
cout<<"The items are:"<<endl;
for (int i = 0; i < best_items.size(); ++i) {
cout<<best_items[i]<<" ";
}
return 0;
}