A number of times in feedbacks I've discussed the difference, or utility, of a **brute force** approach to solving Sudoku versus **logical strategies**. By **brute force** I mean any back-tracking method that can fill a puzzle board and derive all possible solutions. The *Solution Count* on the solver is an example. The rest of this site is devoted to logical strategies.

I made a decision early on when exploring Sudoku that I was interested in logical strategies since they told me__how__ to solve a puzzle using steps that showed why each cell could be solved in sequence just from the logical consequences of having clues. This felt much more satisfying than guessing, trial and error and simply placing numbers by intuition. However, I agree that intuition is one way a human can beat a computer: it is not to be dismissed.

Brute force is the domain of computers and people quickly developed optimal - or near optimal - ways of demolishing a puzzle by back-tracking. This is a proof by exhaustion since numbers are inserted in sequence, 1 to 9, top-left to bottom right until all ways of satisfying the rules are shown. The advantage is that you can also**count** the number of solutions, so partial or faulty Sudoku puzzles do not phase a brute-force algorithm. Which is why I employ one to check puzzles in the solver.

I made a decision early on when exploring Sudoku that I was interested in logical strategies since they told me

Brute force is the domain of computers and people quickly developed optimal - or near optimal - ways of demolishing a puzzle by back-tracking. This is a proof by exhaustion since numbers are inserted in sequence, 1 to 9, top-left to bottom right until all ways of satisfying the rules are shown. The advantage is that you can also

I don't pretend to know all Sudoku strategies and those I get stuck on are always solved by clever people in the Weekly Unsolvables providing me with new clues. This is why Sudoku is so fascinating. However, there is another aspect to the two solving methods that pertains to programmers, especially optimisers.

The diagram on the right is a crude impression of how brute-force stacks up against logic when the number of clues is taken into account. Simply put, brute force is much quicker when there are many clues - not surprising - whereas it is known that clue density does not - in general - affect the grade or difficulty of a puzzle. You can have very easy low clue puzzles and very hard high clue ones - although the bias to 'hard' is usually with fewer clues.

The diagram on the right is a crude impression of how brute-force stacks up against logic when the number of clues is taken into account. Simply put, brute force is much quicker when there are many clues - not surprising - whereas it is known that clue density does not - in general - affect the grade or difficulty of a puzzle. You can have very easy low clue puzzles and very hard high clue ones - although the bias to 'hard' is usually with fewer clues.

In a low-clue puzzle there is more to crunch through for a logical solver, but it can collapse quickly. Some strategies require an awful lot of searching to identify patterns, or are not yet optimised, so it depends on if they are needed. But generally the time to solve is flat against clue density.

However, brute force quickly escalates in time cost when clues are low. There are two factors. Density, just discussed, but also the orientation of the puzzle. My Solution Count gives a number which is the number of recursive calls made. However, since the brute force always starts in the top left, if the clues are stacked towards the top left, it will find the solutions more quickly. You can test this by rotating the puzzle and trying Solution Count.

For very extreme puzzles like this Jigsaw, I have never run the Solution Count to completion. It could take hours, or years? I don't know.

However, brute force quickly escalates in time cost when clues are low. There are two factors. Density, just discussed, but also the orientation of the puzzle. My Solution Count gives a number which is the number of recursive calls made. However, since the brute force always starts in the top left, if the clues are stacked towards the top left, it will find the solutions more quickly. You can test this by rotating the puzzle and trying Solution Count.

For very extreme puzzles like this Jigsaw, I have never run the Solution Count to completion. It could take hours, or years? I don't know.

If your aim is to crunch through a set of puzzles in record time, then a mixture of both approaches will be optimal. Using just the basic strategies to start a puzzle and then back-tracking if/when this fails does work well. But I prefer to keep aside failures to inspire new logical strategies since they generate new knowledge.

The key is to express the rows, columns and boxes as bit arrays, which can be done in C or C++. At this point I'd like to credit **G.M. Boynton** who posted an algorithm way back in June 2005 (although I can't remember the language and if I ported it to bit arrays). We don't need to scan along a row or column to test if a number is present, we just need to know IF a number is present anywhere in the row, column or box.

unsigned short r[9]; // bit array to signal presence of 1 to 9 on a row unsigned short c[9]; // bit array to signal presence of 1 to 9 on a columns unsigned short b[9]; // bit array to signal presence of 1 to 9 in a box unsigned short vals[9][9];// clues and solutions of each cellAs well as being called in the main function to insert the next number, this function can be used to set up the initial values from the clues. We shift 1 by 'n' bits for whatever row, column and box it is a member of.

void put_number_back( sudoku *s, int y, int x, int n ) { s->r[y] += (1 << n); s->c[x] += (1 << n); s->b[s->boxnum[y][x]-1] += (1 << n); s->vals[y][x] = n+1; }

Imagine you are starting in A1 which is empty and you want to insert 3. If there is a 3 clue already down in J1 then the flag s->c[0] (first column) will contain a 1 at the third bit (1 shifted 2 bits). You can skip placing a 3 there and go onto 4.

bool bRecSlv( Sudoku s, int depth, unsigned int *solutions ) { register unsigned short x,y, n; calls++; for(y=0;y<9;y++) for(x=0;x<9;x++) // For each cell if( s.vals[y][x] == 0 ) // that is still empty { for(n=0;n<9;n++) // For each possible number: { // if any bit mask is set, skip n if( s.r[y] & (1 << n) ) continue; if( s.c[x] & (1 << n) ) continue; if( s.b[s.boxnum[y][x]-1] & (1 << n) ) continue; // Valid to place n in this cell put_number_back( &s, y, x, n ); // and call this function recursively! if( bRecSlv(s,depth+1,solutions) ) { return true; // If solution found AND no more solutions required, // then exit recursive function } } return false; } // No empty cells, must be a solution. Return true if no more required (*solutions)++; return false; }

Note: Although this is a back-tracking algorithm, we don't need another function called take_number_off() since this is handled by the stack. We pass the whole "sudoku" object into the function, not a pointer to it. When we recurse a copy is made. When the branching is finished and we back out, the previous states are already in memory and in order. Enough clues.

## Comments

Comments Talk