The clues are things like 1A = 3D * 6A, 2A = Prime, 3A = 6 Gross, so there are at least 2 values you can work out initially. ]]>

I’m trying to find the best solution of the Sudoku problem. I used Mozart in solving it. Now my problem is, I want to implement a backtracking search on the tree using Mozart and the notion of computational spaces.

Can anyone help me…i need the Mozart code for backtracking…

email me at tantanix@yahoo.com ]]>

Assuming a 3×3 grid

1) Put any random set of numbers in the center box…obviously, one each of the digits 1-9

2) Fill in the box above and below the center box with any random number of choice that doesn’t break the Soduko rule.

3) Keep track of all of the options for each cell and give priority to assigning that random number to the CELL THAT HAS THE LEAST OPTIONS AVAILABLE TO IT. In other words….Each cell has 9 optional digits that the cell can contain. As you enter a value in a cell, the cells in the same box, row and column no longer have that digit as an option. Take that option away from all other cells in the box, row and column. They now only have 8 possible entries. As you continue with this process, you will see that some cells may have five options left while other cells have only 3 options. Provide a random number (that is within the available options for that cell) to the cells with less options first.

4) Providing a random number in the top box and then the bottom box and then repeating seems to keep the random generation working smoothly.

5) Once all cells are completed in the center column of boxes, move to either the left or right and complete all cells in column of boxes.

6) To complete the cells in the left or right columns, simply apply one random number (again, within the remaining possible values) for each cell. Give priority to cells that have the least number of options. Try scanning the whole column of boxes and looking for cells with only one option left. Be sure to do those first.

7) Complete the last column of boxes and assign values to the cells in the same way.

WORKS EVERY TIME IF YOU APPLY THE LOGIC ]]>

I develop my sudoku with java prog language

I use a backtracking method to solve the sudoku

trying all the possibility one into the cell.

firstly my program try to find a single probability and if there is no single probability in the cell, the program will start search by row(only able to search by row) and try to find the empty cell(if 0 means empty) then find the probability of the empty cell.

let’s say i have probability of 1, 4, 7, 8 in 1st cell.

it test with number 1 if there is a cell with null probability found then it will backtracking.

now test number 4(i put the condition if there is empty cell and no single probability) then try the second empty cell and so on. when it reach to the 6th cell, it found that all of the possibility is not valid. my program only able to backtracking to the previous cell.

after i check with sudoku generator the valid answer for 1st cell is 7 instead of 4.

my question is what is the condition to fine the one that works(a valid answer) from all of the probability?

can anyone help me with this?

this is my email cyb3rguys@yahoo.com ]]>

New capabilities :

GENERATOR, HELPER, SOLVER

SURPRISES for Sudoku solved without help

HISTORIC records all the plays. Partial replay.

CLINIC to analyse the sudoku. ]]>

Its originality : it does not play for you, it just show you the places where to play.

Beginners will learn rapidly how to practice.

Experts will go directly to the place where they are beginning to think.

Tired people will press “solution” before going to bed.

You need Excel installed on your computer.

Enjoy ]]>

Its originality : it does not play for you, it just show you the places where to play.

Beginners will learn rapidly how to practice.

Experts will go directly to the place where they are beginning to think.

Tired people will press “solution” before going to bed.

You need Excel installed on your computer.

Enjoy ]]>

probably the easiest web interface. 6 free and printable Sudokus a day! ]]>

The generator is harder and I haven’t solved that problem yet. Working on it though. ]]>