How to split a 2D array in different sized pieces for a crossword?

Question

I am experimenting with a generator for a crossword puzzle but I am getting stuck when it is time to split it up into different pieces. I have a 2D array where I store the crossword like this:

int SIZE = 10; //This can be higher for bigger crosswords
Character[][] crossword = new Character[SIZE][SIZE];

I then add a few words to this crossword and for example end up with the following array (. = empty square):

..........
..........
..C...H...
..A...O...
..TIGER...
......S...
...DOVE...
..........
..........
..........

How do I split this 2D array so that I end up with pieces that contains at least 2 letters but not a whole word. The letters must also be next to each other horizontally or vertically but not diagonally. For example I could end up with the following pieces:

C    TIG    H    ER    DOV
A           O     S
                  E

The following piece is NOT valid because the letters is not horizontally or vertically next to eachother.

  O
GE
  S

My first atempt to split this up was by doing the following:

int chunksize = 2; //This should vary depending on how big the pieces should be
List<Character[][]> subArrays = new ArrayList<>();
for(int i = 0; i < SIZE; i += chunksize){
    for(int j = 0; j < SIZE; j += chunksize){
        Character[][] sub = new Character[chunksize][chunksize];
        sub[0][0] = crossword[i][j];
        sub[0][1] = crossword[i][j + 1];
        sub[1][0] = crossword[i + 1][j];
        sub[1][1] = crossword[i + 1][j + 1];
        if(sub[0][0] != null || sub[0][1] != null || sub[1][0] != null || sub[1][1] != null){
            subArrays.add(sub);
        }
    }
}

However, this might create pieces that contains only one letter or pieces where the letters is not next to each other. I do not know how I should solve this problem which is why I come here for help.

Answer 1

Domino packing

The following approach creates as many size-2 blocks as possible. Afterwards, any remaining individual letters need to be attached to a neighbouring block somehow -- eg, by picking one of the neighbouring blocks at random.

Create a graph with a vertex for each position occupied by a letter, and an edge between any pair of vertically or horizontally adjacent letter positions. Now compute a maximum matching on this graph: this chooses a maximum-size edge subset such that no vertex is incident on more than one edge. These edges correspond to the size-2 blocks.

If you imagine the grid as a chessboard, you'll notice that every square is either white or black, and no edge connects two white cells or two black cells: this means that the graph is bipartite, which in turn means that you can use the O(|E|*sqrt(|V|))-time Hopcroft-Karp algorithm , which is faster and simpler than Edmonds's algorithm for general graphs.

Answer 2

What I would suggest you do is to convert each row and each column into a String of it's own.

eg: First 3 Rows

..........
..........
..C...H...

First 3 columns

..........
..........
..CAT.....

You could do this by using for loops for example:

for(int x = 0; x < SIZE; x++){
    //Loop through rows and columns.
    //(eg:crossword[x][y] in this loop will extract a row of values)
    //(eg:crossword[y][x] in this loop will extract a column of values)
    for(int y = 0; y < SIZE; y++){
        //Code to build each row/column string
    }
    //Add extracted Strings to an ArrayList?
}

After you have these string you could split them by using: (Assuming . is still your deliminator)

s.split("\\.");

This will leave you with Arrays of the string from each row and columns

Extracts C and H as separate Strings from

..........
..........
..C...H...

Extracts CAT as a single String from

..........
..........
..CAT.....

This should allow you to check their length against chunksize and return the combinations you want.

Hopefully I have understood the problem correctly or it has at least given you some ideas that could be used to develop the solution you're looking for.