Brute force Sudoku algorithm

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Sudoku algorithm, brute force

For several days I have tried to write a brute force algorithm for solving sudoku, my problem is that I never realy get the algorithm to work 100 %, can someone please direct me and give some help ?

The Algorithm is located in Square class, recursive function.

public abstract class Square {

private Square next;

private Box box;
private Row row;
private Columne columne;

private int value;

Square(int value, Box box, Row row, Columne columne) {
    this.value = value;
    this.box = box;
    this.row = row;
    this.columne = columne;
}

void setNumberMeAndTheRest(Board board) {
    if(getNext() == null) {
        System.out.println("next == null");
        for(int i = 1; i <= board.getDimension(); i++) {
            if(legalValue(i)) {
                setValue(i);
            }
        }
        board.saveSolution();
        return;
    } else {
        if(this instanceof DefinedSquare) {
            getNext().setNumberMeAndTheRest(board);

        } else {
            for(int i = 1; i <= board.getDimension(); i++) {
                if(legalValue(i)) {
                    setValue(i);
                    getNext().setNumberMeAndTheRest(board);
                }
            }
            return;
        }
    }
}

int getValue() {
    return value;
}

void setValue(int value) {
    this.value = value;
}

void setNext(Square next) {
    this.next = next;
}

public Square getNext() {
    return next;
}

/**
 * Checks if value is legal in box, row and column.
 * @param value to check.
 * @return true if value is legal, else false.
 */
boolean legalValue(int value) {
    if(box.legalValue(value) && row.legalValue(value) && columne.legalValue(value)) {
        return true;
    }
    return false;
}

It helps to know what's wrong - like input and the expected and received output — dfb, Apr 17 '12 at 14:43
Brute force will not solve all Sudoku puzzles. Sometimes, you have to guess a number, and be able to backtrack to the guess if you guessed wrong. — Gilbert Le Blanc, Apr 17 '12 at 14:44
@GilbertLeBlanc you could solve it buy trying every number in every available slot and checking if it worked, so you can solve by brute force, just not nicely — Jacxel, Apr 17 '12 at 14:45
@GilbertLeBlanc - huh? Brute force to me would be all possible combination of numbers in every position, which surely would find the answer. Sounds like you're talking about a greedy algorithm maybe? This looks like what's going on above too but I'm ot sure — dfb, Apr 17 '12 at 14:47
For a brute force algorithm, I would expect to see some recursion, iterating over each row to fill in the answers, and a separate function that tests to see if it's solved. — Marcus Adams, Apr 17 '12 at 14:51
@Jacxel & spinning_plate: You do realize that you're talking about roughly 9 to the 81st power combinations, right? I don't even want to guess how long that would take. Usually when people talk about solving Sudoku by "brute force", they're talking about not coding the solving "rules" that a human would use. — Gilbert Le Blanc, Apr 17 '12 at 14:52
@GilbertLeBlanc - Of course, I'm just answering the question as it was asked. The code looks like it's backtracking, acutally — dfb, Apr 17 '12 at 14:53

dfb · Answer 1 · 2012-04-17T16:03:09.203

2

I think your problem may lie here

    for(int i = 1; i <= board.getDimension(); i++) {
        if(legalValue(i)) {
            setValue(i);
            getNext().setNumberMeAndTheRest(board);
        }
    }

If legalValue(i) returns true independent of the current state of i, then you're back tracking, if not, you're not backtracking

What most backtracking looks like is osmething like htis

    for(int i = 1; i <= board.getDimension(); i++) {
        if(legalValue(i)) {
            setValue(i);
                // boolean indicating whether solution was found
            if(getNext().setNumberMeAndTheRest(board))
               return true;
            else
               unsetValue(i)
        }
    }

We need more code to know if legalValue returns false when square i is already set

Try this to see if I'm on the right track or post all of your code

    System.out.println("STARTING ITERATION")
    for(int i = 1; i <= board.getDimension(); i++) {

        if(legalValue(i)) {
            System.out.println("GOING " + i)
            setValue(i);
            getNext().setNumberMeAndTheRest(board);
        }
    }
    System.out.println("ENDING ITERATION")

If it fills out the grid and then stops without backtracking, your problem is that you calling setValue(i) and then calling legalValue(i+1) and it is return false because the value is alraedy set, not because it's not legal. If this is so, you need an equivalent 'unset' after the reucrsion

edited Apr 17 '12 at 16:03

answered Apr 17 '12 at 15:01

dfb

13,133
2
31
52

And unsetValue(i) is doing what ? – user265767 Apr 17 '12 at 15:06
Nulling out the value. My fear is that legalValue returns false for the next legal value when the square has been filled – dfb Apr 17 '12 at 15:09
Specifically, whats the Box.LegalValue function look like – dfb Apr 17 '12 at 15:10
Box, Row and Columne looks like this public boolean legalValue(int value) { for(Square s : squars) { if(s.getValue() == value) { return false; } } return true; } – user265767 Apr 17 '12 at 15:22
remove 'box.legalValue(value)' in the legalValue function and tell me what happens – dfb Apr 17 '12 at 15:47
I have tried to remove box.legalValue(value) both with the original code and "spinning_plate" suggestion, the latest option gives a total different answer like this: 421653 132546 513264 265431 654320 300102 – user265767 Apr 17 '12 at 15:56
I am not sure that I understand unsetValue(i) totally, do unsetValue(i) the same as setValue(i) ? – user265767 Apr 17 '12 at 15:59
I can't help you with the specifics because I don't have all the code- I need you to answer this question on the original code above - on each level of recursion, how many times is setValue called - see my eidts – dfb Apr 17 '12 at 16:00
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/10183/discussion-between-user265767-and-spinning-plate) – user265767 Apr 17 '12 at 16:09
Here is the output: STARTING ITERATION GOING 4 STARTING ITERATION GOING 2 STARTING ITERATION GOING 5 STARTING ITERATION GOING 6 STARTING ITERATION GOING 1 STARTING ITERATION ENDING ITERATION ENDING ITERATION ENDING ITERATION ENDING ITERATION ENDING ITERATION ENDING ITERATION – user265767 Apr 17 '12 at 18:22
entire code https://github.com/cbeberge/Sudoku/tree/master/src – user265767 Apr 17 '12 at 18:23

vaughandroid · Answer 2 · 2012-04-17T15:04:27.853

From a quick look at your algorithm, it looks as though it only ever tries a single possible value in each square. When it reaches a square where it can't find a legal value, it just gives up. It needs some mechanism of backtracking and trying alternative legal values in squares that it has previously filled.

As an example, here's a mini 4x4 puzzle:

  1 |    
    | 2  
---------
    |   4
3   |

Your algorithm, from what I can tell, will get this far then quit:

2 1 | 3 X 
    | 2  
---------
    |   4
3   |

Instead of quitting, it ought to go back and change either of the 2 values it has inserted.

Brute force Sudoku algorithm

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