How do you generate a Sudoku board with a unique solution? What I thought was to initialize a random board and then remove some numbers. But my question is how do I maintain the uniqueness of a solution?

Write an algorithm that solves a sudoku no matter how many clues it has and by that i mean even if it has 0 clues. That algorithm will help you in many tasks you will need afterwards. The most basic thing it will do is giving you a variaty of solved sudokus that you will be able to use to create the unsolvables with the help of a different function which will remove clues and another which will find the number of solutions each time you remove a clue.– ArgyrisSofroniouMar 25 at 14:43
Here is the way my own SuDoKu program does it:
Start with a complete, valid board (filled with 81 numbers).
Make a list of all 81 cell positions and shuffle it randomly.
As long as the list is not empty, take the next position from the list and remove the number from the related cell.
Test uniqueness using a fast backtracking solver. My solver is  in theory  able to count all solutions, but for testing uniqueness, it will stop immediately when it finds more than one solution.
If the current board has still just one solution, goto step 3) and repeat.
If the current board has more than one solution, undo the last removal (step 3), and continue step 3 with the next position from the list
Stop when you have tested all 81 positions.
This gives you not only unique boards, but boards where you cannot remove any more numbers without destroying the uniqueness of the solution.
Of course, this is only the second half of the algorithm. The first half is to find a complete valid board first (randomly filled!) It works very similar, but "in the other direction":
Start with an empty board.
Add a random number at one of the free cells (the cell is chosen randomly, and the number is chosen randomly from the list of numbers valid for this cell according to the SuDoKu rules).
Use the backtracking solver to check if the current board has at least one valid solution. If not, undo step 2 and repeat with another number and cell. Note that this step might produce full valid boards on its own, but those are in no way random.
Repeat until the board is completely filled with numbers.

I found your algorithm to be particularly simple and effective. Thanks.– krawyotiMay 6 '13 at 15:09

4I'm a bit perplexed by
(3) Use the solver to check if the current board has at least one valid solution.
If you've only added one character to an empty board (in step 2) and then test your solver on in (in step 3), you're essentially solving an empty board. I don't think my solver is that good, and more importantly if it could solve an empty board then the problem of getting a valid solution would already be solved and I could skip to step 4!– The111Aug 9 '13 at 2:53 
@The111: solving an empty board is easy, you can do this even without a computer. But I am looking for a randomly filled board, that's why I don't just stop after step 3. Jan 17 '14 at 18:43

What's the purpose of 3rd point in second algorithm? Is it possible to generate a valid board which does not have any solutions?– LukeMar 23 '16 at 19:00

@Luke: take an arbitrary Sudoku with exactly one solution. Lets assume the upper left corner is free, and if you just apply the rules (in short: each row, column and 3x3 square shall contain the numbers 19), you find out directly that it is allowed to place 1,3,5 and 7 into the upper left corner. But only "1" is allowed in the final solution, so if you place 3,5 or 7, the backtracking solver will show that these three numbers will not lead to the valid final solution. Mar 23 '16 at 22:03
Easy:
 Find all solutions with an efficient backtracking algorithm.
 If there is just one solution, you are done. Otherwise if you have more than one solution, find a position at which most of the solutions differ. Add the number at this position.
 Go to 1.
I doubt you can find a solution that would be much faster than this.

1I think you are right, but how to grade level for the borad generated in this way, it seems no parameter to control the difficut. Aug 9 '11 at 8:46

Well, that's a different question, much more difficult. What is sure is that he more numbers you add, the easier.– TomasAug 9 '11 at 10:06

3No need to find all solutions, it's enough to search for a second one. Mar 20 '17 at 3:38
You can cheat. Start with an existing Sudoku board that can be solved then fiddle with it.
You can swap any row of three 3x3 blocks with any other row. You can swap any column of three 3x3 blocks with another column. Within each block row or block column you can swap single rows and single columns. Finally you can permute the numbers so there are different numbers in the filled positions as long as the permutation is consistent across the whole board.
None of these changes will make a solvable board unsolvable.

but how about uniqueness? how do u choose the blank cells to keep the solution unique? Aug 3 '11 at 12:21

1@kvphxga: You start with a partial board with a unique solution. None of the allowed swaps affect the uniqueness of the solution.– rossumAug 3 '11 at 12:47

Isn't this a horrible solution? If you use one single complete Sudoku board and swap rows and columns, will the solver notice similarities (feel samey) between puzzles? You end up only using an incredibly small number of unique solutions and I fear at some point it wont feel random to the solver. It may be worth the effort to do better than this.– brycSep 10 '20 at 21:40

You swap individual lines within rows/columns and also reassign numbers to positions. If you want, you can have, say, ten different starting grids and pick one at random.– rossumSep 11 '20 at 7:45
Unless P = NP, there is no polynomialtime algorithm for generating general Sudoku problems with exactly one solution.
In his master's thesis, Takayuki Yato defined The Another Solution Problem (ASP), where the goal is, given a problem and some solution, to find a different solution to that problem or to show that none exists. Yato then defined ASPcompleteness, problems for which it is difficult to find another solution, and showed that Sudoku is ASPcomplete. Since he also proves that ASPcompleteness implies NPhardness, this means that if you allow for arbitrarysized Sudoku boards, there is no polynomialtime algorithm to check if the puzzle you've generated has a unique solution (unless P = NP).
Sorry to spoil your hopes for a fast algorithm!

3To be fair, you can generate a few hundred unique puzzles a second using the technique in the selected answer.– GrandpaDec 18 '12 at 12:58

1Well, in this case i would like to see that. Because If you try to generate diabolical sudoku, it is sometimes really long to test all possible possibilities. For easy sudoku with a lot of initial filled digits, I agree.– dyesdyesJan 3 '13 at 18:11

My hopes for fast Zebra puzzle generator almost vanished until I read the beginning of this paper (thank you!) carefully. In solver you need to find a solution (hence the name solver), while in generator you need to generate puzzle  you don't need to actually solve it (the fact that most approaches uses solver as part of generator is another story). I am not saying your first statement is false, I am saying it is not proven in that paper. Jan 14 '15 at 13:53
The solution is divide in to 2 parts:
A. Generating the number pattern 600 billion
B. Generating the masking pattern ~ 7e23 combinations
A ) For Number pattern the fastest way which can generate unique combinations with NO time spent on backtracing or testing
Step 1. Choose an already exiting matrix, I chose the below one as it can be made easily by human without any help from a computing device or solver:
First row is numbers in ascending order
Second row is also in ascending order but start from 4 & roll around
Third row is also in ascending order but start from 7 & roll around
Row 4,5,6: Replace the three cell column with the top right column  2 5 8 and roll within the 3x3 cell for last column
Row 7,8,9: Replace the three cell column with the top right column  3 6 9 and roll within the 3x3 cell for last column
1 2 3 4 5 6 7 8 9
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
2 3 1 5 6 4 8 9 7
5 6 4 8 9 7 2 3 1
8 9 7 2 3 1 5 6 4
3 1 2 6 4 5 9 7 8
6 4 5 9 7 8 3 1 2
9 7 8 3 1 2 6 4 5
Step 2. Shuffle the the digits and replace in all other cells
Step 3. Randomly rearrange columns 1,2 and 3 within themselves
Step 4. Randomly rearrange columns 4,5 and 6 within themselves
Step 5. Randomly rearrange columns 7,8 and 9 within themselves
Step 6. Randomly rearrange rows 1,2 and 3 within themselves
Step 7. Randomly rearrange rows 4,5 and 6 within themselves
Step 8. Randomly rearrange rows 7,8 and 9 within themselves
Step 9. Randomly rearrange in 3 column groups of size 9x3
Step 10. Randomly rearrange in 3 row groups of size 3x9
voila...
5 8 3 1 6 4 9 7 2
7 2 9 3 5 8 1 4 6
1 4 6 2 7 9 3 8 5
8 5 2 6 9 1 4 3 7
3 1 7 4 2 5 8 6 9
6 9 4 8 3 7 2 5 1
4 6 5 9 1 3 7 2 8
2 3 1 7 8 6 5 9 4
9 7 8 5 4 2 6 1 3
B ) For Masking Pattern we need to have a solver algorithm. As we already have a quite unique number grid (which is also solved!) this gives us faster performance for using solver
Step 1: Start with selecting 15 random locations out of the 81.
Step 2: Check with solver whether it has unique solution
Step 3: If solution not unique select additional location. iterate Steps 2 and 3 until unique solution found
This should give you the very unique and fast Sudoku board.

It took some thought but I think that I have it now. Step 2 means e.g. change all the 1's for 5's and 2's for 1's. Steps 38 mean that you can rearrange rows and columns so long as they stay in the same squares. Steps 9 & 10 mean rearrange rows and columns of squares. Did I get it right? Aug 12 '19 at 8:44

1This algorithm only creates very specific kinds of puzzles. As you can see the numbers (5, 8, 3) appear in rows 1, 2 and 3 always as a group. Same goes for all other 3groups. For a general purpose sudoku generator this algorithm is unfortunately not useful. Apr 10 '20 at 17:07
SWIFT 5 version
The simply way, here my code:
First, create the function into [[Int]] array
func getNumberSudoku() > [[Int]] {
// Original number
let originalNum = [1,2,3,4,5,6,7,8,9]
// Create line 1 to 9 and shuffle from original
let line1 = originalNum.shuffled()
let line2 = line1.shift(withDistance: 3)
let line3 = line2.shift(withDistance: 3)
let line4 = line3.shift(withDistance: 1)
let line5 = line4.shift(withDistance: 3)
let line6 = line5.shift(withDistance: 3)
let line7 = line6.shift(withDistance: 1)
let line8 = line7.shift(withDistance: 3)
let line9 = line8.shift(withDistance: 3)
// Final array
let renewRow = [line1,line2,line3,line4,line5,line6,line7,line8,line9]
// Preshuffle for column
let colSh1 = [0,1,2].shuffled()
let colSh2 = [3,4,5].shuffled()
let colSh3 = [6,7,8].shuffled()
let rowSh1 = [0,1,2].shuffled()
let rowSh2 = [3,4,5].shuffled()
let rowSh3 = [6,7,8].shuffled()
// Create the let and var
let colResult = colSh1 + colSh2 + colSh3
let rowResult = rowSh1 + rowSh2 + rowSh3
var preCol: [Int] = []
var finalCol: [[Int]] = []
var prerow: [Int] = []
var finalRow: [[Int]] = []
// Shuffle the columns
for x in 0...8 {
preCol.removeAll()
for i in 0...8 {
preCol.append(renewRow[x][colResult[i]])
}
finalCol.append(preCol)
}
// Shuffle the rows
for x in 0...8 {
prerow.removeAll()
for i in 0...8 {
prerow.append(finalCol[x][rowResult[i]])
}
finalRow.append(prerow)
}
// Final, create the array into the [[Int]].
return finalRow
}
Then usage:
var finalArray = [[Int]]
finalArray = getNumberSudoku()
It's not easy to give a generic solution. You need to know a few things to generate a specific kind of Sudoku... for example, you cannot build a Sudoku with more than nine empty 9number groups (rows, 3x3 blocks or columns). Minimum given numbers (i.e. "clues") in a singlesolution Sudoku is believed to be 17, but number positions for this Sudoku are very specific if I'm not wrong. The average number of clues for a Sudoku is about 26, and I'm not sure but if you quit numbers of a completed grid until having 26 and leave those in a symmetric way, you may have a valid Sudoku. On the other hand, you can just randomly quit numbers from completed grids and testing them with CHECKER or other tools until it comes up with an OK.

2

I'd like the add that the problem of the minimum number of prefilled cells necessary to guarantee a unique solution has, since this discussion, been proven to be 17. (Of course, that does not mean that every board can be reducible to 17 cells: it simply means that there is no Sudoku board with 16 prefilled cells that has a unique solution, and there is at least one board with 17 prefilled cells that has a unique solution.) Nov 12 '18 at 14:56
Here is a way to make a classic sudoku puzzle (sudoku puzzle with one and only solution; prefilled squares are symmetrical around the center square R5C5).
1) start with a complete grid (using group filling plus circular shift to get it easily)
2) remove number(s) from two symmetrical squares if the cleared squares can be inferred using the remaining clues.
3) repeat (2) until all the numbers are checked.
Using this method you can create a very easy sudoku puzzle with or without programming. You can also use this method to craft harder Sudoku puzzles. You may want to search "create classic sudoku" on YouTube to have a step by step example.
 Find any filled board of sudoku. (use trivial ones)
int[][] board = new int[][] {
{1,2,3, 4,5,6, 7,8,9},
{4,5,6, 7,8,9, 1,2,3},
{7,8,9, 1,2,3, 4,5,6},
{2,3,1, 5,6,4, 8,9,7},
{5,6,4, 8,9,7, 2,3,1},
{8,9,7, 2,3,1, 5,6,4},
{3,1,2, 6,4,5, 9,7,8},
{6,4,5, 9,7,8, 3,1,2},
{9,7,8, 3,1,2, 6,4,5}
};
 for each number 0 to 9 (say curNum), (i.e 0, 1, 2, 3, 5, 6, 7, 8) take a random number from range (0 to 9), traverse the board, swap curNum with your random number.
void shuffleNumbers() {
for (int i = 0; i < 9; i++) {
int ranNum = random.nextInt(9);
swapNumbers(i, ranNum);
}
}
private void swapNumbers(int n1, int n2) {
for (int y = 0; y<9; y++) {
for (int x = 0; x<9; x++) {
if (board[x][y] == n1)
board[x][y] = 0;
if (board[x][y] == n2)
board[x][y] = n1;
}
}
for (int y = 0; y<9; y++) {
for (int x = 0; x<9; x++) {
if (board[x][y] == 0)
board[x][y] = n2;
}
}
}
 Now shuffle rows. Take the first group of 3 rows , shuffle them , and do it for all rows. (in 9 X 9 sudoku), do it for second group and as well as third.
void shuffleRows() {
int blockNumber;
for (int i = 0; i < 9; i++) {
int ranNum = random.nextInt(3);
blockNumber = i / 3;
swapRows(i, blockNumber * 3 + ranNum);
}
}
void swapRows(int r1, int r2) {
int[] row = board[r1];
board[r1] = board[r2];
board[r2] = row;
}
 Swap columns, again take block of 3 columns , shuffle them, and do it for all 3 blocks
void shuffleCols() {
int blockNumber;
for (int i = 0; i < 9; i++) {
int ranNum = random.nextInt(3);
blockNumber = i / 3;
swapCols(i, blockNumber * 3 + ranNum);
}
}
void swapCols(int c1, int c2) {
int colVal;
for (int i = 0; i < 9; i++){
colVal = board[i][c1];
board[i][c1] = board[i][c2];
board[i][c2] = colVal;
}
}
 swap the row blocks itself (ie 3X9 blocks)
void shuffle3X3Rows() {
for (int i = 0; i < 3; i++) {
int ranNum = random.nextInt(3);
swap3X3Rows(i, ranNum);
}
}
void swap3X3Rows(int r1, int r2) {
for (int i = 0; i < 3; i++) {
swapRows(r1 * 3 + i, r2 * 3 + i);
}
}
 do the same for columns, swap blockwise
void shuffle3X3Cols() {
for (int i = 0; i < 3; i++) {
int ranNum = random.nextInt(3);
swap3X3Cols(i, ranNum);
}
}
private void swap3X3Cols(int c1, int c2) {
for (int i = 0; i < 3; i++) {
swapCols(c1 * 3 + i, c2 * 3 + i);
}
}
Now you are done board should be a valid sudoku board
as for how efficient this algorithm is , Took 3.6 sec to generate a million boards
Yes you can make variable that are used through out process class or global variables so you don't need to make them over and over again eg ranNum
and if you want generate final board with hidden values, read about a bit of backtracking sudoku algorithm and try to remove one element from board until you have a board, that will become unsolvable even if you remove only one element.
if you want to categorised final generated board by difficulty, just count how many numbers are left in board while removing element one by one. The less the number harder it will be to solve
Just a hint least possible hints in sudoku can be 17, but all possible sudoku board not necessarily reducible to 17 hint sudoku
I also think that you will have to explicitly check uniqueness. If you have less than 17 givens, a unique solution is very unlikely, though: None has yet been found, although it is not clear yet whether it might exist.)
But you can also use a SATsolver, as opposed to writing an own backtracking algorithm. That way, you can to some extent regulate how difficult it will be to find a solution: If you restrict the inference rules that the SATsolver uses, you can check whether you can solve the puzzle easily. Just google for "SAT solving sudoku".
One way to generate sudoku faster.
 find an exist sudoku.
 exchange the value with a random group.
 exchange the cell or the column or the rowgrid or the columngrid.
You exchange the value will make the value different, if not exchange the rows or the column, the sudoku isn't changed in the essential.
You can flag the sudoku with 9 grids, the rows and column exchanged must do in the same grid. Like you can exchange row13, row46, row79, don't exchange row14 or row17. You can also exchange the rowgrid(exchange row1~3 with the row4~6 or row7~9).
Solve the sudoku: record the empty with all the possible value, then check the value from 1 to 9. If one value is unique, remove it from the loop.
You can start with any valid (filled) puzzle and modify it to produce a completely different one (again, filled). Instead of permutating groups of numbers, you can swap single cells  there will be no similarity whatsoever between the seed puzzle and the resulting puzzle. I have written a simple program long ago in VB, you can find it here: https://www.charalampakis.com/blog/programmingvbnet/asimplealgorithmforcreatingsudokupuzzlesusingvisualbasic. It can be translated to any language easily.
Then, randomly and gradually remove cells and check if the puzzle is solvable and has a unique solution. You can also rate the puzzle in terms of difficulty depending on the rules needed for the solution. Continue until removing any known cell leads to an unsolvable puzzle.
HTH
You may need code like this:
#pz is a 9x9 numpy array
def PossibleValueAtPosition(pz:[], row:int, col:int):
r=row//3*3
c=col//3*3
return {1,2,3,4,5,6,7,8,9}.difference(set(pz[r:r+3,c:c+3].flat)).difference(set(pz[row,:])).difference(set(pz[:,col]))
def SolvePuzzle(pz:[], n:int, Nof_solution:int):# init Nof_solution = 0
if Nof_solution>1:
return Nof_solution # no need to further check
if n>=81:
Nof_solution+=1
return Nof_solution
(row,col) = divmod(n,9)
if pz[row][col]>0: # location filled, try next location
Nof_solution = SolvePuzzle(pz, n+1, Nof_solution)
else:
l = PossibleValueAtPosition(pz, row,col)
for v in l: # if l = empty set, bypass all
pz[row][col] = v # try to fill a possible value v
Nof_solution = SolvePuzzle(pz, n+1, Nof_solution)
pz[row][col] = 0
return Nof_solution # resume the value, blacktrack