Sudoku is that Japanese puzzle that requires you to fill in a grid of numbers. Here, I describe a general algorithm to solve these puzzles. Also provided is the source code in C++.

You can see a Sudoku board below. The goal is to fill in every empty cell with a number between *1* and *9* so that no number repeats itself on any line, column of zone (three-by-three marked square).

If you’ve never solved a Sudoku before, now would be the time to do it. Wikipedia describes some useful strategies.

How do you approach a problem like this? The simplest way is from input to output. What’s your input? It’s a *9×9* matrix of numbers like this one:

`5 3 0 0 7 0 0 0 0`

6 0 0 1 9 5 0 0 0

0 9 8 0 0 0 0 6 0

8 0 0 0 6 0 0 0 3

4 0 0 8 0 3 0 0 1

7 0 0 0 2 0 0 0 6

0 6 0 0 0 0 2 8 0

0 0 0 4 1 9 0 0 5

0 0 0 0 8 0 0 7 9

How do you store it in the programme? If you said as a matrix, you were close. A matrix is obvious, it’s easy to understand, but it’s a pain to code. Believe me when I say, it’s a lot easier to store it as an array of *9*9* elements. What else do you need? A variable to keep track of how many cells have been filled in (*0* means empty board; *81* means full board). An array of bitsets to keep track of what digits can’t be used in each cell (I’ll explain a little later), and the setter and getter functions. As it happens, it’s also easier if you encapsulate it in a C++ class. Here’s the full code to the programme (sudoku.cpp). I’ll explain it all in a bit.

#include <iostream> #include <fstream> using namespace std; class SudokuBoard; void printB(SudokuBoard sb); typedef unsigned int uint; const uint MAXVAL = 9; const uint L = 9; const uint C = 9; const uint S = L * C; const uint ZONEL = 3; const uint ZONEC = 3; const uint ZONES = ZONEL * ZONEC; const uint lineElements[L][C] = { { 0, 1, 2, 3, 4, 5, 6, 7, 8}, { 9, 10, 11, 12, 13, 14, 15, 16, 17}, {18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32, 33, 34, 35}, {36, 37, 38, 39, 40, 41, 42, 43, 44}, {45, 46, 47, 48, 49, 50, 51, 52, 53}, {54, 55, 56, 57, 58, 59, 60, 61, 62}, {63, 64, 65, 66, 67, 68, 68, 70, 71}, {72, 73, 74, 75, 76, 77, 78, 79, 80} }; const uint columnElements[C][L] = { { 0, 9, 18, 27, 36, 45, 54, 63, 72}, { 1, 10, 19, 28, 37, 46, 55, 64, 73}, { 2, 11, 20, 29, 38, 47, 56, 65, 74}, { 3, 12, 21, 30, 39, 48, 57, 66, 75}, { 4, 13, 22, 31, 40, 49, 58, 67, 76}, { 5, 14, 23, 32, 41, 50, 59, 68, 77}, { 6, 15, 24, 33, 42, 51, 60, 69, 78}, { 7, 16, 25, 34, 43, 52, 61, 70, 79}, { 8, 17, 26, 35, 44, 53, 62, 71, 80} }; const uint zoneElements[S / ZONES][ZONES] = { { 0, 1, 2, 9, 10, 11, 18, 19, 20}, { 3, 4, 5, 12, 13, 14, 21, 22, 23}, { 6, 7, 8, 15, 16, 17, 24, 25, 26}, {27, 28, 29, 36, 37, 38, 45, 46, 47}, {30, 31, 32, 39, 40, 41, 48, 49, 50}, {33, 34, 35, 42, 43, 44, 51, 52, 53}, {54, 55, 56, 63, 64, 65, 72, 73, 74}, {57, 58, 59, 66, 67, 68, 75, 76, 77}, {60, 61, 62, 68, 70, 71, 78, 79, 80} }; class SudokuBoard { public: SudokuBoard() : filledIn(0) { for (uint i(0); i < S; ++i) table[i] = usedDigits[i] = 0; } virtual ~SudokuBoard() { } int const at(uint l, uint c) { // Returns the value at line l and row c if (isValidPos(l, c)) return table[l * L + c]; else return -1; } void set(uint l, uint c, uint val) { // Sets the cell at line l and row c to hold the value val if (isValidPos(l, c) && ((0 < val) && (val <= MAXVAL))) { if (table[l * C + c] == 0) ++filledIn; table[l * C + c] = val; for (uint i = 0; i < C; ++i) // Update lines usedDigits[lineElements[l][i]] |= 1<<val; for (uint i = 0; i < L; ++i) // Update columns usedDigits[columnElements[c][i]] |= 1<<val; int z = findZone(l * C + c); for (uint i = 0; i < ZONES; ++i) // Update columns usedDigits[zoneElements[z][i]] |= 1<<val; } } void solve() { try { // This is just a speed boost scanAndSet(); // Logic approach goBruteForce(); // Brute force approach } catch (int e) { // This is just a speed boost } } void scanAndSet() { int b; bool changed(true); while (changed) { changed = false; for (uint i(0); i < S; ++i) if (0 == table[i]) // Is there a digit already written? if ((b = bitcount(usedDigits[i])) == MAXVAL - 1) { // If there's only one digit I can place in this cell, do int d(1); // Find the digit while ((usedDigits[i] & 1<<d) > 0) ++d; set(i / C, i % C, d); // Fill it in changed = true; // The board has been changed so this step must be rerun } else if (bitcount(usedDigits[i]) == MAXVAL) throw 666; // Speed boost } } void goBruteForce() { int max(-1); // Find the cell with the _minimum_ number of posibilities (i.e. the one with the largest number of /used/ digits) for (uint i(0); i < S; ++i) if (table[i] == 0) // Is there a digit already written? if ((max == -1) || (bitcount(usedDigits[i]) > bitcount(usedDigits[max]))) max = i; if (max != -1) { for (uint i(1); i <= MAXVAL; ++i) // Go through each possible digit if ((usedDigits[max] & 1<<i) == 0) { // If it can be placed in this cell, do SudokuBoard temp(*this); // Create a new board temp.set(max / C, max % C, i); // Complete the attempt temp.solve(); // Solve it if (temp.getFilledIn() == S) { // If the board was completely solved (i.e. the number of filled in cells is S) for (uint j(0); j < S; ++j) // Copy the board into this one set(j / C, j % C, temp.at(j / C, j % C)); return; // Break the recursive cascade } } } } uint getFilledIn() { return filledIn; } private: uint table[S]; uint usedDigits[S]; uint filledIn; bool const inline isValidPos(int l, int c) { return ((0 <= l) && (l < (int)L) && (0 <= c) && (c < (int)C)); } uint const inline findZone(uint off) { return ((off / C / ZONEL) * (C / ZONEC) + (off % C / ZONEC)); } uint const inline bitcount(uint x) { uint count(0); for (; x; ++count, x &= (x - 1)); return count; } }; void printB(SudokuBoard sb) { cout << " | ------------------------------- |" << endl; for (uint i(0); i < S; ++i) { if (i % 3 == 0) cout << " |"; cout << " " << sb.at(i / L, i % L); if (i % C == C - 1) { if (i / C % 3 == 2) cout << " |" << endl << " | -------------------------------"; cout << " |" << endl; } } cout << endl; } int main(int argc, char *argv[]) { SudokuBoard sb; ifstream fin("sudoku4.in"); int aux; for (uint i(0); i < S; ++i) { fin >> aux; sb.set(i / L, i % L, aux); } fin.close(); printB(sb); sb.solve(); printB(sb); return 0; }

Look at the **main** function. It first opens a file and then reads *S* ints from it. *S* is just the number of columns (*C*) multiplied by the number of lines (*L*). It reads the value into an auxiliary variable and then sets it in the SudokuBoard.

How does it set a cell? The relevant code is in **set**. The first line just checks if the parameters are valid (if the value’s not too large, if the specified cell does not exist, etc.). Then it checks if there’s a value already in the cell (there shouldn’t be). If not, it increments the number of filled-in cells.

Now things get intresting. If a certain cell contains the number *n*, it should be obvious that none of the cells on the same line, column or zone as the cell can contain *n*. Look at the board above: because there’s a *5* in cell *1,1*, there can’t be any more fives in any of the cells on the first line, on the first column or in the upper-left zone. This is what the remainder of **set** does. It sets the *n ^{th}* bit in every bitset in whose corresponding cell the number

*n*cannot appear.

Note: For a given cell, it’s trivial to find the line, column and zone in which it happens to be. What’s hard is to find the *other* cells in the same line, column or zone. To keep things simple, use three arrays of arrays that contain the number of the cells on a certain line, column or zone.

The next function of intrest is **solve**. If you’ll look at it, you’ll notice that it contains a **try**…**except** statement. As the comments clearly note, it’s just a speed boost. If you comment it out, the programme will still work (but in some cases *a lot* slower).

**Solve** calls two other functions: **scanAndSet** and **goBruteForce**. These are both algorithms to determine or guess what value should be placed in which cell.

##### scanAndSet

void scanAndSet() { int b; bool changed(true); while (changed) { changed = false; for (uint i(0); i < S; ++i) if (0 == table[i]) // Is there a digit already written? if ((b = bitcount(usedDigits[i])) == MAXVAL - 1) { // If there's only one digit I can place in this cell, do int d(1); // Find the digit while ((usedDigits[i] & 1<<d) > 0) ++d; set(i / C, i % C, d); // Fill it in changed = true; // The board has been changed so this step must be rerun } else if (bitcount(usedDigits[i]) == MAXVAL) throw 666; // Speed boost } }

The basic idea is this: we have a list of cells that need to be completed (those whose value is *0*) and list of digits that **cannot** be placed in each cell. Go through the list of used digits, searching for a cell in which *8* digits cannot be placed (i.e. there’s only one digit that can be placed), and place it.

Now, every time you place a digit, you change the board a bit, restricting the digits that can be placed in other cells. So, you have to do the previous step until you don’t change anything any more.

There’s also a check in there if the number of used digits for any cell is *9* (i.e. no digit can be placed in the cell). If such a cell exists then the board is clearly wrong, so throw an exception (which is caught in the **solve** routine).

##### goBruteForce

void goBruteForce() {

int max(-1); // Find the cell with the _minimum_ number of posibilities (i.e. the one with the largest number of /used/ digits)

for (uint i(0); i < S; ++i)
if (table[i] == 0) // Is there a digit already written?
if ((max == -1) || (bitcount(usedDigits[i]) > bitcount(usedDigits[max])))

max = i;

if (max != -1) {

for (uint i(1); i <= MAXVAL; ++i) // Go through each possible digit
if ((usedDigits[max] & 1<*solve you’ll notice that there are some boards that are not completed by scanAndSet. Why? Because there are some boards that can’t be completed through logic alone (and ours isn’t particularly thorough either).*

To get over this, you have to use a brute-force algorithm. The idea is simple enough: given the list of which digits cannot be placed in each cell, find the cell in which the minimum number of digits **can** be placed. For this cell, for every possible digit, write it down and try to solve the board.

This is where it becomes apparent why a C++ object-oriented approach is a smart move. Instead of writing the try in the current board and then having to keep track of what changes are made, simply clone the current board, fill in a cell and let it solve itself.

That’s it. You might want to try some of the other algorithms suggested on the Net. Good luck. Always open to comments.