Simple 9x9 Sudoku brute force solver with intrinsic parallel candidate set processing using bits to represent digits in the [1, 9] range, and bitwise operations to test a candidate against the candidate set, all at once.

It can be upgraded for 16x16 or 25x25 grids.

The algorithm was implemented in Java, in C, as well as in Zig. The description below concerns the Java implementation, even though, the C implementation is quite similar, but without classes. Zig implementation is similar to C's but with an OOP style stack.

The Windows 64 executable supplied can be used to solve arbitrary grids as decribed in the documentation.

Updates done here and corresponding code are reported on Twitter below this tweet. Please follow me on Twitter for updates.

This is the class containing the grid to be solved.

The grid can be initialized using a 9x9 matrix of type char[][] or through a linear string containing all the elements, representating empty elements as 0 (or ' . ' in the C or Zig version), both given line by line. The char[][] is the unique input, however, and it must exist before being able to use any other input format. Even though the 9x9 matrix contains characters (it's a char[][]), the digits are not represented as ASCII or Unicode characters but rather as integers. In other words, the character '0' is actually represented by 0, and so forth.

In the string input format the string is just copied over the existing input char[][] matrix using the static function set. This string uses ASCII representation for the digits which are converted to integers by the function set.

An additional representation is possible, as illustrated in Main.java, by representing the charcater '0' with the character '.' in the string. In this case one adds .replace('.','0') at the end of the string as shown.

Both string input formats are common representations of Sudoku grids on the web.

The main data structure in Grid is matrix which is a 9x9 matrix in identical format as the input matrix for the grid. This is the matrix where the input matrix is copied to.

The main auxiliary data structures are the most interesting part of this class, besides the solver algorithm itself:

• lines - an array with 9 positions, each one, corresponding to a line in the grid, and functioning as a set where each bit represents a digit already present in that line.
• cols - an array with 9 positions, each one, corresponding to a column in the grid, and functioning as a set where each bit represents a digit already present in that column.
• cells - a 3x3 matrix, corresponding to a 3x3 cell that the grid is subdivided, with 9 positions, each one functioning as a set where each bit represents a digit already present in that cell.

• stk - the stack to implement the backtracking algorithm. It uses an array of 81 positions. It uses the push and pop operators as shown in the algorithm below. The push operator not only stores the digit, its binary representation, the line and column (i and j) of the element inserted in a stack node (StkNode), "pushing" the node in the stack, but also inserts the digit in the internal matrix (matrix[i][j]) as well as its binary representation into the auxiliary data structures, thus, updating the candidate set of the new element inserted. The pop operation only removes the node from the stack, but the node is not garbage collected. It remains in the stack as an unused element. Nodes are lazily allocated, as null elements are found while pushing.
• cel - an array with 9 positions, each one is the inverse mapping of the indices in the lines and columns transformed into indices in the 3x3 matrix cells.

All main auxiliary data structures use a common notation to represent a set of digits present in the line, column, or cell, accordingly. A bit is set to one at the position corresponding to a digit present in the set, or set to zero if it's position corresponds to a digit that is absent. By reversing the bits one gets the "candidate set" of digits that are still missing in the corresponding line, column or cell. For a better understanding of this candidate set scheme, please refer to the subsection explaining how digits are represented in binary.

Let's suppose a particular line, column or cell having the digits, 1, 3, 4 and 9. This set is then represented by the following binary number:

100001101 = 0x10D

• the first rightmost bit corresponds to the digit 1, and in this case it's present in the set already.
• the second bit on its left corresponds to the digit 2, and its clearly not present yet since its value is zero.
• bits three and four, corresponding to the digits 3 and 4, respectively, are clearly present, because they are both set to one.
• bits five, six, seven, and eight are all zeros, and thus, digits 5, 6, 7 and 8 are clearly absent in the set.
• bit 9 is 1. Therefore, the digit 9 is also present in the set.

In order to obtain a candidate set for a given matrix[i][j] element of the grid one calculates:

lines[i] | cols[j] | cells[ cel[i] ][ cel[j] ] (1)

The expression in (1) gives a set where all bits containing zeros correspond to the available digits that are possible to be in matrix[i][j]. The candidate set is detected by the absent elements in the set, that is, all bits which are zero.

The interest in this notation is that the concatenation of all three sets is obtained by just using two bitwise or operations.

One can observe how cel inverse mapping works to access the corresponding cell in cells. First, i and j are used as indices in cel. cel[i] and cel[j] give the corresponding line and column in cells. Therefore, cells[cel[i]][cel[j]] corresponds to the cell where matrix[i][j] is contained.

    public void solve() {
        StkNode node;
        int digit = 1, code = 1, inserted;
        int i, j;
        char[] line = matrix[0];
        char c;
        i = j = 0;
        do {
            c = line[j];
            if (c == 0) {
                inserted = lines[i]|cols[j]|cells[cel[i]][cel[j]];
                for ( ; digit != 10 ; digit++, code <<= 1 ) {
                    if (( code & inserted ) == 0 ) {
                        push(i, j, code, digit);
                        digit = code = 1;
                        break;
                    }
                }
                if ( digit == 10 ) {            // no insertion -> backtrack to previous element
                    node = pop();               // pop previous inserted i, j, and digit
                    i = node.i;
                    j = node.j;
                    digit = node.digit;
                    code = node.code;
                    remove(node);               // remove digit from data structures
                    digit++; code <<= 1;        // let's try next digit;
                    line = matrix[i];           // maybe line has changed
                    continue;                   // short-circuit line by line logic
                }
            }
            if ( j == 8 ) {                     // line by line logic
                j = -1; i++;                    // last line element, advance to next line
                if (i < 9) line = matrix[i];    // update line from grid matrix
            }
            j++;                                // advance to next element in the line
        } while (i < 9);
    }

In the binary representation, a digit is always a power of two, since it's a number with only one bit set to 1 at the position corresponding to the digit. The table below shows the correspondence between digits and their binary representation:

The binary representation as exposed in the table above is often called here as the "code" of the digit.

As we can see the variable inserted contains the "candidate set" for a given matrix[i][j]. This algorithm is quite simple but it contains a major drawback. Since the digit is represented with a 1 bit in its corresponding position in variable code, and it accesses the candidate set in a sequential way, it loops until an empty bit is found (( code & inserted ) == 0 )) or if it finds no available candidate (digit == 10).

This means that even if there are no available candidates, the algorithm has to loop over all the remaining bits sequentially. Even if the binary representation allows to deal with the candidate set with all elements in parallel, that is, all elements at once, we still have to access it one by one sequentially even when there are no useful results. This problem is adressed with some partial solutions as shown here and here, but this later employs far too many operations, despite the fact it's a branchless solution. It's only interesting when associated with other optimizations as it has been done in the C version.

Digits are tried in ascending order from 1 to 9 for each element in the grid that is not yet occupied. That's why digit and code variables are both initialized with 1. Every time a new digit is tried against the candidate set, and a successful candidate is found (that is, when ( code & inserted ) == 0 )), the digit is pushed on the stack.

The push function also updates matrix[i][j], lines[i], cols[j] and cells[cel[i]][cel[j]] with the new digit. Please check the code and the description of stk for details.

When no suitable candidate is found (that is, when ( code & inserted ) == 0 ) fails for every candidate tried), then the for loop ends, and digit == 10. In this case, we need to backtrack, that is, remove the current candidate, and advance the previous inserted digit to be the next candidate. This is taken care by the instructions found under the if ( digit == 10 ) statement, where the previous candidate is popped from the stack, removed from matrix and the auxiliary data structures (function remove), and advanced to be the next candidate (digit is incremented and code is shifted left). Notice that this command sequence terminates with a continue statement in order to skip the line by line logic. Since the line and column (i and j) of the element to be dealt next are already known (they were popped from the stack), modifying i or j is not required. Also of note, if all the possible candidates were tried, digit will become 10, the for loop is summarily skipped, and the flow goes back into this code sequence to backtrack once again, dealing with the cases of "cascaded" backtracking sequences.

This completes the backtracking mechanism, allowing, as can be easily infered, to obtain the solution of the input grid in the internal matrix. As shown in Main.java, the solution is printed using the function print.

The logic to check if there are no candidates with no loops is much more involved than what's done in the algorithm above, but its not rocket science. It only requires more effort to use our bit representation in a smarter way.

Every power of two subtracted by one is always equal to a sequence of ones on the right of the position it was previously one (except in the case of 0², since there are no more binary digits on the right of 1). For example, the digit 8 in binary is 128 in our representation. When subtracted by one, that's 127, that is, 8 bits set to 1 on the right of bit 8:

128 - 1 = 010000000 - 1 = 001111111

By reversing every bit of this result one obtains a mask that's unique when all these bits are 1, that is, when there are no candidates from the bit in the current position until the last bit:

~001111111 = 110000000

That is, by executing a bitwise and operation (&) between this mask and a candidate set, and if the result is identical to this mask, we can say there is no available candidates left in the candidate set, starting with the digit we are trying, 8 in this case.

Let's check the same logic with digit = 5:

~(000010000 - 1) = ~000001111 = 111110000

Then by testing if

111110000 & inserted == 111110000

What this is actually saying is that there are no candidates neither for 5, neither any digit above it. In other words, this is exactly the condition we were looking for.

One could call this as reachable, that is, more formally speaking what we've got is:

reacheable = (~(code-1)) & 0x1ff;

Notice that we have to filter out all bits above bit 9. Then the condition searched would be written like

if ( (inserted & reacheable ) == reacheable ) (2)

In this case if statement (2) can substitute the following if statement in the algorithm:

if ( digit == 10 )

And we should place if statement (2) above the for loop statement instead of the order presented in the algorithm. In this case the for can be written with no final condition, since it would never be reached:

for ( ; ; digit++, code <<= 1 ) (3)

The reason for that is that if there are no candidates, as calculated here, then the condition of the if statement (2) must be true and, therefore, the continue statement relative to the do-while statement is executed before the for statement (3) is ever reached. This obviously short-circuits the for statement (3), since it is now below the if statement (2). If the for statement (3) is reached, the condition in the if statement (2) must have been false. In this situation there will always be a valid candidate and the break command relative to the for statement (3) will be executed, always ending this loop with no need to test the end condition.

Another way to see this optimization is by observing that instead of calculating the mask as explained above, which implies using an intermediate variable reacheable, one can infere an equivalent conclusion by simply discarding this variable and using the following test instead of if statement (2):

if ( (inserted + code ) > 511 ) (2a)

Which we call here an alternative to (2), or (2a) for short.

If there are only ones in inserted starting at the position of the 1 in code, adding code to inserted will result in some value that is obviously beyond 511 (or 0x1ff). Therefore, we can detect the same situation with only the test (2a), not only eliminating the need of calculating the mask, but also the need of the variable reacheable.

The benchmarks to measure algorithm performance were performed on an i7 2.2 Ghz machine in Java and in C. The executable file compiled in C has been done with optimization option -O3 using the gcc compiler on Windows provided in w64devkit, which is a Mingw-w64 gcc compiler that is portable (can be installed by just copying the directory structure in disk, SD card, or thumb drive).

The benchmarks were executed with several different grids, but particularly with this one, which is known to be time consuming in automatic methods, and used to compare speed of different methods on the web:

	1	2	3	4	5	6	7	8	9
1
2
3
4
6
6
7
8
9

The minimal time measured for the optimized algorithm to solve the above grid after several attempts was 10 miliseconds, and the double for the unoptimized algorithm. Nevertheless, the times verified were quite variable as usual in Java while measuring fast algorithms like this. This is the reason it would worth trying to implement it with an entirely compiled language (Java is only compiled when the JIT compiler is triggered) to verify if execution times are less variable. It looks like that for this kind of problem, an enterily compiled language would be more appropriate, since one expects similar times for the same grid running at different times. Unfortunately this is not the case for this Java implementation.

Astonishingly, execution times running the executable compiled in C were only slightly more constant than in Java. The times varied from 1.5 miliseconds to 5.26 miliseconds. However, these variations were considerably much less significant than in Java. Also, C offered roughly about an order of magnitude to about twice less time than the Java implementation of the same optimized algorithm. Several optimizations were devised besides the ones mentioned below. After all these optimizations were applied, one obtained a significant improvement in performance and the Windows 64 executable supplied was generated with the resulting source code.

The parallel test for no candidates allows to discard unnecessary for loop iterations, while also discarding the unecessary end condition of the for loop (since the order of the if statement (2) and the for statement was reversed). Nevertheless, for detecting the first candidate one still has to loop and test the digits one by one sequentially against the inserted set.

But there is a way to calculate the next candidate without any loop. The technique can be illustrated through and example. Supposing the set included = 101011110 (that is {9,7,5,4,3,2}, the set of digits already inserted) and digit = 000000010 (2), one starts by adding both:

   101011110    // included digits set: {9,7,5,4,3,2}
 + 000000010    // digit = 2

Which is equal to 101100000. One now does an exclusive or with included:

   101100000
 ^ 101011110

Which is equal to 000111110. One now adds digit again:

   000111110
 + 000000010

Which is equal to 001000000. The bit representation of the next candidate, is obtained by shifting one position to right:

   001000000 >> 1

Which is equal to 000100000. This corresponds to the digit 6, which is exactly the first zero bit found by applying the for loop (3).

Therefore, assuming code as the bit representation of the digit, one calculates the next candidate doing:

    code = (((code + inserted) ^ inserted) + code) >> 1; // branchless code calculation

The problem is that one only obtains the bit representation of the digit, not the digit itself. As, one can see, digit is necessary to be able to use this technique.

To obtain the digit from its code, one "assembles" the bit configuration of the digit from its bit representation (code) as follows:

    digit = ( code >> 8 ) |
            (( code & 0x40 ) >> 6 ) | 
            (( code & 0x140) >> 5 ) |
            (( code & 0xf0 ) >> 4 ) |
            (( code & 0x20 ) >> 3 ) |
            (( code & 0x14 ) >> 2 ) |
            (( code & 0x0c ) >> 1 ) |
            ( code & 3);

This conversion is not only complex to understand, but also requires a high number of operations. Trying out this code and the brachless calculation of the next candidate as shown previously, the minimal time in C passed from 1.5 to 1.4 miliseconds, which apparently wouldn't seem to justify the effort.

However, after multiple further opimizations, including using register variables, the minimal running time was reduced to 1.2 miliseconds. This corresponds to a speedup of roughly 20%, which starts to become quite consequential. It's clear that this is also consequence of the highly "imperative" way of implementing this algorithm which manifestly highly benefits the C implementation, that in itself is more easily optimizable by employing extremely low level gimmicks that are absent in Java.

Another way to do the calculation above is using tables. For example, in C:

    unsigned short c1[] = { 0, 1, 2, 0, 3 };
    unsigned short c2[] = { 0, 4, 5, 0, 6 };
    unsigned short c3[] = { 0, 7, 8, 0, 9 };

One can compose the digit from its bit representation code in the following way:

    digit = c1[code & 7] | c2[(code >> 3) & 7] | c3[code >> 6];

This code is more understandable than the previous one. If the digit is 1, 2 or 3, one simply filters the first 3 bits of codeand index the table c1 with this result. Position 3 is invalid since code has only 1 bit set, and, thus, it can't be 3. Notwithstanding, the resulting operation can be zero, in the case the binary representation doesn't have any bit set in that range. In this case, to satisfy the branchless logic, the table value is 0. If the digit is 4, 5 or 6, one shifts code to the right 3 positions and filters the first 3 bits and index the table c2 with this result. The same logic applies to digits 7, 8 and 9, using table c3. Since one doesn't know which one is correct, one simply apply a binary or operation with the 3 results, after all only one of them contains the good digit. The other two will be zero.

Trying this solution instead of the previous, had a significant impact in the minimal execution time of the compiled C code, that was reduced to practically 1 millisecond, that is, an optimization of more than 30%, since the initial minimal time in our comparisons was 1.5 milliseconds.

The several optimizations proposed are complex to understand and most of them do not result in a significant speed up. The initial algorithm and in the Java and C codes, are more clear and relatively easy to understand after the binary representation is understood.

The idea of parallelizing the code by dealing with the whole candidate set at once just using binary representation is promising. However, it falls short if one was thinking in using its intrisic parallelism in the entire algorithm. As seen above, the approach allows branchless solutions for the sequential search of a candidate from an arbitrary digit value, which only partially exploits this intrisic paralellism. Notwithstanding, it's heavily relying on the integer addition carry propagation mechanism, which is actually a sequential mechanism, but implemented highly efficiently in hardware. This is just additional ingenuity, but not the same approach. The actual problem in this partial solution is that it's highly complex and requires a high number of operations. Thus, it highly diverges from the extreme simplicity of the original algorithm. Fortunately, associated with numerous other low level optimizations in C language, it contributed to a significant speedup (as can be seen here), and a better speedup as well as less complexity (as seen here).

A comparative test between the Java implementation and an identical C inplementation has given a considerable advantage to the C implementation, not only in terms of raw performance, but also in terms of less variability in times measured for solving the same grid, even though, variable execution times were also present in the C implementation. This was expected since Java activates the JIT compiler not quite regularly in codes that are executed in short ammounts of time like this one.

Given the extremely short execution times, the low level nature of the original algorithm, and the considerable amount of low level optmizations that are possible in C language, one may confortably conclude that C is the most appropriate language to use the algorithm, since it will provide faster answers. This means, that the C implementation can be seen as the ideal engine for an interactive program where the grid can be entered through a GUI and that the solution must be supplied in real time when it is requested by the user.