所以我最近看到了英国GCHQ发布的puzzle：
它涉及解决25x25非图形：
非图形是图像逻辑谜题，其中网格中的单元格必须根据网格侧面的数字着色或留空，以显示隐藏的图片。在这种拼图类型中，数字是离散断层扫描的一种形式，用于测量任何给定行或列中有多少完整的填充正方形线。例如，“4 8 3”的线索意味着按顺序存在四个，八个和三个实心正方形的集合，在连续的组之间至少有一个空白正方形。“
当然，我有兴趣尝试编写一个可以为我解决的程序。我正在考虑从第0行开始的递归回溯算法，并且对于给定来自行线索的信息的该行的每个可能的排列，它放置下一行的可能组合并验证它是否是给定列的有效放置线索。如果是，则继续，如果没有回溯，直到所有行都放置在有效配置中，或者所有可能的行组合都已用完。
我在几个5x5谜题上进行了测试，效果很好。问题是计算25x25 GCHQ难题需要很长时间。我需要一些方法来使这个算法更有效 - 足以让它解决上面链接的难题。有什么想法吗？
这是我为每行生成一组行可能性的代码以及求解器的代码（注意*它使用了一些非标准库，但这不应该减损这一点）：
// The Vector<int> input is a list of the row clues eg. for row 1, input = {7,3,1,1,7}. The
// int currentElemIndex keeps track of what block of the input clue we are dealing with i.e
// it starts at input[0] which is the 7 sized block and for all possible places it can be
// placed, places the next block from the clue recursively.
// The Vector<bool> rowState is the state of the row at the current time. True indicates a
// colored in square, false indicates empty.
// The Set< Vector<bool> >& result is just the set that stores all the possible valid row
// configurations.
// The int startIndex and endIndex are the bounds on the start point and end point between
// which the function will try to place the current block. The endIndex is calculated by
// subtracting the width of the board from the sum of the remaining block sizes + number
// of blocks remaining. Ie. if for row 1 with the input {7,3,1,1,7} we were placing the
// first block, the endIndex would be (3+1+1+7)+4=16 because if the first block was placed
// further than this, it would be impossible for the other blocks to fit.
// BOARD_WIDTH = 25;
// The containsPresets funtion makes sure that the row configuration is only added to the
// result set if it contains the preset values of the puzzle (the given squares
// at the start of the puzzle).
void Nonogram::rowPossibilitiesHelper(int currentElemIndex, Vector<bool>& rowState,
Vector<int>& input, Set< Vector<bool> >& result,
int startIndex, int rowIndex) {
if(currentElemIndex == input.size()) {
if(containsPresets(rowState, rowIndex)) {
result += rowState;
}
} else {
int endIndex = BOARD_WIDTH - rowSum(currentElemIndex+1, input);
int blockSize = input[currentElemIndex];
for(int i=startIndex; i<=endIndex-blockSize; i++) {
for(int j=0; j<blockSize; j++) {
rowState[i+j] = true; // set block
}
rowPossibilitiesHelper(currentElemIndex+1, rowState, input, result, i+blockSize+1, rowIndex); // explore
for(int j=0; j<blockSize; j++) {
rowState[i+j] = false; // unchoose
}
}
}
}
// The function is initally passed in 0 for the rowIndex. It gets a set of all possible
// valid arrangements of the board and for each one of them, sets the board row at rowIndex
// to the current rowConfig. Is then checks if the current configuration so far is valid in
// regards to the column clues. If it is, it solves the next row, if not, it unmarks the
// current configuration from the board row at rowIndex.
void Nonogram::solveHelper(int rowIndex) {
if(rowIndex == BOARD_HEIGHT) {
printBoard();
} else {
for(Vector<bool> rowConfig : rowPossisbilities(rowIndex)) {
setBoardRow(rowConfig, rowIndex);
if(isValidConfig(rowIndex)) { // set row
solveHelper(rowIndex+1); // explore
}
unsetBoardRow(rowIndex); // unset row
}
}
}
答案 0 :(得分：2)
我已经用Java创建了一个解决方案，对于你的示例谜题（25x25）在大约50ms中解决了它。
完整的代码和输入示例：Github
R, C // number of rows, columns
int[][] rows; // for each row the block size from left to right (ex: rows[2][0] = first blocksize of 3 row)
int[][] cols; // for each column the block size from top to bottom
long[] grid; // bitwise representation of the board with the initially given painted blocks
置换也以按位表示形式存储。如果填充第一列等，第一位设置为true。 这既节省时间又节省空间。对于计算，我们首先计算可以添加的额外空格的数量。
这是number_of_columns - sum_of_blocksize - (number_of_blocks-1)
Dfs覆盖所有可能的放置额外空间的排列。如果它与最初给定的绘制块匹配，请参阅calcPerms并将它们添加到可能的排列列表中。
rowPerms = new long[R][];
for(int r=0;r<R;r++){
LinkedList<Long> res = new LinkedList<Long>();
int spaces = C - (rows[r].length-1);
for(int i=0;i<rows[r].length;i++){
spaces -= rows[r][i];
}
calcPerms(r, 0, spaces, 0, 0,res);
rowPerms[r] = new long[res.size()];
while(!res.isEmpty()){
rowPerms[r][res.size()-1]=res.pollLast();
}
}
...
// row, current block in row, extra spaces left to add, current permutation, current number of bits to shift
static void calcPerms(int r, int cur, int spaces, long perm, int shift, LinkedList<Long> res){
if(cur == rows[r].length){
if((grid[r]&perm)==grid[r]){
res.add(perm);
}
return;
}
while(spaces>=0){
calcPerms(r, cur+1, spaces, perm|(bits(rows[r][cur])<<shift), shift+rows[r][cur]+1,res);
shift++;
spaces--;
}
}
static long bits(int b){
return (1L<<b)-1; // 1 => 1, 2 => 11, 3 => 111, ...
}
[Trivial：] 我们将使用预先计算的排列，因此我们不需要每行进行任何额外的验证。
因此，我为每个行和列保留当前块大小colIx的索引，以及该大小colVal的位置。
这是通过前一行的值和索引计算的：
样品：
static void updateCols(int row){
long ixc = 1L;
for(int c=0;c<C;c++,ixc<<=1){
// copy from previous
colVal[row][c]=row==0 ? 0 : colVal[row-1][c];
colIx[row][c]=row==0 ? 0 : colIx[row-1][c];
if((grid[row]&ixc)==0){
if(row > 0 && colVal[row-1][c] > 0){
// bit not set and col is not empty at previous row => close blocksize
colVal[row][c]=0;
colIx[row][c]++;
}
}else{
colVal[row][c]++; // increase value for set bit
}
}
}
现在我们可以使用这些索引/值来确定下一行中预期哪些位为false / true。
用于验证的已用数据结构：
static long[] mask; // per row bitmask, bit is set to true if the bit has to be validated by the val bitmask
static long[] val; // per row bitmask with bit set to false/true for as expected for the current row
当设置前一行中的位时，我们希望当前行中的位设置为true，当且仅当前当前大小仍然小于当前索引的预期大小时。否则它必须为0，因为你想在当前行切断它。
或者当最后的块大小已经用于列时，我们无法启动新块。因此位必须为0。
static void rowMask(int row){
mask[row]=val[row]=0;
if(row==0){
return;
}
long ixc=1L;
for(int c=0;c<C;c++,ixc<<=1){
if(colVal[row-1][c] > 0){
// when column at previous row is set, we know for sure what has to be the next bit according to the current size and the expected size
mask[row] |= ixc;
if(cols[c][colIx[row-1][c]] > colVal[row-1][c]){
val[row] |= ixc; // must set
}
}else if(colVal[row-1][c] == 0 && colIx[row-1][c]==cols[c].length){
// can not add anymore since out of indices
mask[row] |= ixc;
}
}
}
这使得实际的dfs部分像您自己一样简单。 如果行掩码适合当前配置，我们可以更新列索引/值并遍历到下一行，最终到达R行。
static boolean dfs(int row){
if(row==R){
return true;
}
rowMask(row); // calculate mask to stay valid in the next row
for(int i=0;i<rowPerms[row].length;i++){
if((rowPerms[row][i]&mask[row])!=val[row]){
continue;
}
grid[row] = rowPerms[row][i];
updateCols(row);
if(dfs(row+1)){
return true;
}
}
return false;
}