SudokuSolver 1.1：用C++实现的数独解题程序 【二】
作者：互联网
本篇是 SudokuSolver 1.1：用C++实现的数独解题程序 【一】 的续篇。
filterRowCandidatesEx 接口
1 u8 CQuizDealer::filterRowCandidatesEx(u8 row)
2 {
3 u8 vals[10] = {0}; // last item denotes sum of zeros
4 u8 base = row * 9;
5 for (u8 col = 0; col < 9; ++col) {
6 vals[col] = m_seqCell[base + col].val;
7 if (vals[col] == 0) {
8 vals[9] += 1;
9 }
10 }
11 if (vals[9] == 0)
12 return RET_PENDING;
13 u8 blkBase = (row / 3) * 3;
14 u8 blkTakens[21] = {0};
15 setBlkRowTakens(blkBase, row % 3, blkTakens);
16 if (blkTakens[0] == 0 && blkTakens[7] == 0 && blkTakens[14] == 0)
17 return RET_PENDING;
18 u8 ret = RET_PENDING;
19 for (u8 idx = 0; idx < 3; ++idx) {
20 if (ret = filterRowByPolicy1(row, idx, vals, blkTakens))
21 return ret;
22 }
23 for (u8 idx = 0; idx < 3; ++idx) {
24 if (ret = filterRowByPolicy2(row, idx, vals, blkTakens))
25 return ret;
26 }
27 return RET_PENDING;
28 }
filterRowCandidatesEx 接口实现对由参数 row 指定的行实施第二类候选值集合收缩算法。第 3 行到第 12 行的代码段是把该行当前各个 cell 的值填到数组 vals 中，并用 vals[9] 记录 0-cell 数量，若 vals[9] 为 0，则说明该行没有 0-cell，即不具备收缩价值，直接返回 RET_PENGING，好让调用者进一步去考察其它行。
第 13 行到第 17 行的代码段是调用 setBlkRowTakens 接口填制 blkTakens 数组。blkTakens 数组实际记录了三个集合。举例说明一下：
000 000 000
023 010 780
100 406 009
400 050 001
900 000 006
060 000 090
...
第二行从左至右每 3 个一节，被划分为三节，这三节分别落在第 1 宫、第 2 宫、第 3 宫。事实上，每一行都可以依次方法划分为三节，每一节对应左、中、右各一个宫。blkTakens 数组的单元，每 7 个一组分成三组，每组记录一个宫中除去指定的行之外已填数字的集合。在上例中，填入 blkTakens 数组的三个集合分别是 {1}、{4, 6}、{9}，对应到 blkTakens 数组的单元取值，则有：
blkTakens[0] = 1，blkTakens[1] = 1；
blkTakens[7] = 2，blkTakens[8] = 4，blkTakens[9] = 6；
blkTakens[14] = 1，blkTakens[15] = 9。
第 2 行有 4 个空位（0-cell），具体分布是第一节 1 个，第二节 2 个，第三节 1 个。而 blkTakens 数组的中宫集合，即 {4, 6} ，里的元素不能填到第 2 行的第二节（不然中宫里会出现两个 4 或两个 6），于是 4 和 6 只能填到第 2 行的第一节和第三节的空位上，而 [2,9] 空位的候选值里不能有 6（因为第 9 列里已经有 6），于是可以推定 [2,9] = 4，进而又有 [2,1] = 6。
仍以上面的例子为例，考察第 4 行，则填入 blkTakens 数组的三个集合分别是 {6,9}、{}、{6,9}，对应到 blkTakens 数组的单元取值，则有：
blkTakens[0] = 2，blkTakens[1] = 9，blkTakens[2] = 6；
blkTakens[7] = 0；
blkTakens[14] = 2，blkTakens[15] = 6，blkTakens[16] = 9。
第 4 行有 6 个空位，具体分布是每一节各 2 个。而 blkTakens 数组的左宫集合与右宫集合相等，更一般而言，它们的交集为 {6, 9} ，易知这个交集里的元素只能填到第 4 行的第二节的空位上，而 [4,6] 空位的候选值里不能有 6（因为第 6 列里已经有 6），于是可以推定 [4,6] = 9，进而又有 [4,4] = 6。
上面例子里的第 2 行和第 4 行收缩示例实际就是 filterRowByPolicy2 和 filterRowByPolicy1 里分别实现的方法。
setBlkRowTakens 接口
1 void CQuizDealer::setBlkRowTakens(u8 blkBase, u8 innRow, u8* pTakens)
2 {
3 for (u8 idx = 0; idx < 3; ++idx) {
4 u8 blk = blkBase + idx;
5 u8* pVals = pTakens + idx * 7;
6 u8 colBase = idx * 3;
7 u8 row = block2row(blk, ((innRow + 1) % 3) * 3);
8 setRowTakensInBlk(row * 9, colBase, pVals);
9 row = block2row(blk, ((innRow + 2) % 3) * 3);
10 setRowTakensInBlk(row * 9, colBase, pVals);
11 }
12 }
其中调用的 setRowTakensInBlk 接口为：
1 void CQuizDealer::setRowTakensInBlk(u8 rowBase, u8 colBase, u8* pVals)
2 {
3 for (u8 col = colBase; col < colBase + 3; ++col) {
4 u8 val = m_seqCell[rowBase + col].val;
5 if (val != 0) {
6 u8 sum = pVals[0] + 1;
7 pVals[sum] = val;
8 pVals[0] = sum;
9 }
10 }
11 }
filterRowByPolicy1 接口
1 u8 CQuizDealer::filterRowByPolicy1(u8 row, u8 idx, u8* pVals, u8* pBlkTakens)
2 {
3 u8 colBase = idx * 3;
4 u8 v1 = pVals[colBase], v2 = pVals[colBase + 1], v3 = pVals[colBase + 2];
5 u8 zeroSum = (u8)(v1 == 0) + (u8)(v2 == 0) + (u8)(v3 == 0);
6 if (zeroSum == 0)
7 return RET_PENDING;
8 u8 isVals[7] = {0};
9 if (idx == 0)
10 intersection(pBlkTakens + 7, pBlkTakens + 14, isVals);
11 else if (idx == 1)
12 intersection(pBlkTakens, pBlkTakens + 14, isVals);
13 else
14 intersection(pBlkTakens, pBlkTakens + 7, isVals);
15 if (isVals[0] == 0)
16 return RET_PENDING;
17 return shrinkRowCandidatesP1(isVals, pVals, zeroSum, row, colBase);
18 }
filterRowCandidatesEx 接口里调用 filterRowByPolicy1 的代码段如下：
19 for (u8 idx = 0; idx < 3; ++idx) {
20 if (ret = filterRowByPolicy1(row, idx, vals, blkTakens))
21 return ret;
22 }
沿用上面的例子说明一下 filterRowByPolicy1 接口的实现代码：
000 000 000
023 010 780
100 406 009
400 050 001
900 000 006
060 000 090
...
考察第四行，上面代码段里至多会调用 filterRowByPolicy1 接口三次，idx 由 0 开始，逐次加一。
先看 idx = 0 的情形，四个传入参数分别为：row = 3（第四行的行下标为 3）、idx = 0、vals 中记录了第四行各 cell 的取值、blkTakens 中记录了三个集合，即左宫集合 {6, 9}，中宫集合 {}，右宫集合 {6, 9}。这一次调用是求中宫集合与右宫集合的交集，并试图往第四行第一节的空位上填值。
第 3 行到第 7 行的代码段，考察第四行的第一节三个 cell 中有几个空位，若没有空位，则无需填值；此时有两个空位，继续往下走。
第 8 行到第 16 行的代码段，求集合交集。因为 {} ∩ {6, 9} = {}，即交集为空，没有明确要填空的值，直接返回 RET_PENDING（即 0）。回到 filterRowCandidatesEx 接口，idx 加一，再次调用 filterRowByPolicy1。
此时 idx = 1，要试图往第四行第二节填空，第二节有空位，继续考察左宫集合 {6, 9} 与右宫集合 {6, 9} 的交集，交集不为空，进一步调用 shrinkRowCandidatesP1 接口，发现可以对第四行做收缩处理，返回 RET_OK，回到 filterRowCandidatesEx 接口，继续向上级调用返回 RET_OK。
intersection 函数实现求两个集合的交集，具体实现为：
1 void intersection(u8* pA, u8* pB, u8* pOut)
2 {
3 u8 sumA = pA[0], sumB = pB[0];
4 if (sumA == 0 || sumB == 0)
5 return;
6 for (u8 pos = 1; pos <= sumA; ++pos)
7 for (u8 idx = 1; idx <= sumB; ++idx)
8 if (pB[idx] == pA[pos]) {
9 u8 sum = pOut[0] + 1;
10 pOut[sum] = pA[pos];
11 pOut[0] = sum;
12 break;
13 }
14 }
shrinkRowCandidatesP1 接口
1 u8 CQuizDealer::shrinkRowCandidatesP1(u8* pBlk, u8* pRow, u8 zeroSum, u8 row, u8 colBase)
2 {
3 for (u8 col = colBase; col < colBase + 3; ++col) {
4 if (pRow[col] != 0)
5 if (!removeVal(pBlk, pRow[col]))
6 return RET_PENDING;
7 }
8 if (zeroSum < pBlk[0]) {
9 printf("shrinking row %d with ply1[blk:%d] went WRONG: 0Sum [%d] < vSum [%d]\n", (int)row + 1, (int)colBase / 3 + 1, (int)zeroSum, (int)pBlk[0]);
10 return RET_WRONG;
11 }
12 u8 rowBase = row * 9;
13 u8 exVals[4] = {0};
14 if (zeroSum > pBlk[0]) {
15 calcExCols(exVals, pBlk, pRow, rowBase, colBase, true);
16 if (zeroSum - exVals[0] < pBlk[0]) {
17 printf("shrinking row %d with ply1[blk:%d] went WRONG: 0Sum [%d] - xSum [%d] < vSum [%d]\n", (int)row + 1, (int)colBase / 3 + 1, (int)zeroSum, (int)exVals[0], (int)pBlk[0]);
18 return RET_WRONG;
19 }
20 if (zeroSum - exVals[0] > pBlk[0])
21 return RET_PENDING;
22 }
23 u8 ret = RET_PENDING;
24 bool shrunken = false;
25 for (u8 col = colBase; col < colBase + 3; ++col) {
26 if (pRow[col] != 0 || inSet(col, exVals))
27 continue;
28 ret = shrinkCandidates(rowBase + col, pBlk);
29 if (ret == RET_WRONG) {
30 printf("shrinking [%d,%d] with row-ply1 went WRONG\n", (int)row + 1, (int)col + 1);
31 return RET_WRONG;
32 }
33 if (ret == RET_OK)
34 shrunken = true;
35 }
36 if (shrunken) {
37 ret = RET_OK;
38 printf("row %d shrunken ply-1 by blk %d\n", (u8)row + 1, (u8)colBase / 3 + 1);
39 }
40 return ret;
41 }
shrinkRowCandidatesP1 接口实现考虑的情形比前述例子里的情形要复杂一些。
第 3 行到第 11 行的代码段是对传入的交集集合 pBlk 做进一步的削减处理，因为交集中某个数字如果已经在要填入的对应行的对应节中出现，则需要从交集中剔除（removeVal 函数实现在上一篇里有）。做完剔除处理后，集合里的数字都是待填值，此时若空位数小于待填值的数量，则必然会导致无法求解的局面，因此返回 RET_WRONG，好让上层调用尽快做出调整。
第 12 行到第 22 行的代码段的逻辑还要复杂一些，通过调用 calcExCols 接口考察对应行的对应节中的空位的候选值情况，若某个空位的候选值集合和做完剔除处理的待填值集合无交集，则把该空位的列下标记入 exVals 数组。exVals 里记录的空位是不能填值的，因此，此时若空位数减去 exVals 里的空位数小于待填值的数量，还是会导致无法求解的局面，同样需要返回 RET_WRONG。
第 23 行到第 40 行的代码段是遍历对应行的对应节的空位，调用 shrinkCandidates 接口试图对空位的候选值做收缩处理。
上面用到的 inSet 函数实现为：
1 bool inSet(u8 val, u8* pVals)
2 {
3 for (u8 idx = 1; idx <= pVals[0]; ++idx)
4 if (val == pVals[idx])
5 return true;
6 return false;
7 }
calcExCols 接口
1 void CQuizDealer::calcExCols(u8* pEx, u8* pBlk, u8* pRow, u8 rowBase, u8 colBase, bool ply1)
2 {
3 u8 lower = (ply1 ? colBase : 0);
4 u8 upper = (ply1 ? colBase + 3 : 9);
5 for (u8 col = lower; col < upper; ++col) {
6 if (ply1) {
7 if (pRow[col] != 0)
8 continue;
9 }
10 else {
11 if (pRow[col] != 0 || (col >= colBase && col <= colBase + 2))
12 continue;
13 }
14 u8 vals[10] = {0};
15 intersection(m_seqCell[rowBase + col].candidates, pBlk, vals);
16 if (vals[0] == 0) {
17 u8 sum = pEx[0] + 1;
18 pEx[sum] = col;
19 pEx[0] = sum;
20 }
21 }
22 }
calcExCols 接口为 shrinkRowCandidatesP1 和 shrinkRowCandidatesP2 共用。
shrinkCandidates 接口
1 u8 CQuizDealer::shrinkCandidates(u8 cellPos, u8* pVals)
2 {
3 u8 vals[10] = {0};
4 intersection(m_seqCell[cellPos].candidates, pVals, vals);
5 if (vals[0] == 0)
6 return RET_WRONG;
7 if (vals[0] != m_seqCell[cellPos].candidates[0]) {
8 for (u8 idx = 0; idx <= pVals[0]; ++idx)
9 m_seqCell[cellPos].candidates[idx] = vals[idx];
10 if (vals[0] == 1)
11 return RET_OK;
12 }
13 return RET_PENDING;
14 }
shrinkCandidates 接口对下标为 cellPos 的 0-cell 做候选值收缩处理，逻辑不难，不做额外解释。
filterRowByPolicy2 接口
1 u8 CQuizDealer::filterRowByPolicy2(u8 row, u8 idx, u8* pVals, u8* pBlkTakens)
2 {
3 u8 colBase = idx * 3;
4 u8 v1 = pVals[colBase], v2 = pVals[colBase + 1], v3 = pVals[colBase + 2];
5 u8 zeroSum = pVals[9] - (u8)(v1 == 0) - (u8)(v2 == 0) - (u8)(v3 == 0);
6 if (zeroSum == 0)
7 return RET_PENDING;
8 u8* pBlk = pBlkTakens + (idx * 7);
9 if (pBlk[0] == 0)
10 return RET_PENDING;
11 return shrinkRowCandidatesP2(pBlk, pVals, zeroSum, row, colBase);
12 }
filterRowByPolicy2 和 filterRowByPolicy1 类似。
继续沿用上面的例子说明一下 filterRowByPolicy2 接口的实现代码：
000 000 000
023 010 780
100 406 009
400 050 001
900 000 006
060 000 090
...
考察第二行，filterRowCandidatesEx 接口里至多会调用 filterRowByPolicy2 接口三次，idx 由 0 开始，逐次加一。
先看 idx = 0 的情形，四个传入参数分别为：row = 3（第四行的行下标为 3）、idx = 0、vals 中记录了第四行各 cell 的取值、blkTakens 中记录了三个集合，即左宫集合 {1}，中宫集合 {4, 6}，右宫集合 {9}。这一次调用是考察左宫集合 {1}，并试图往第二行的第二节和第三节的空位上填值。
第 3 行到第 7 行的代码段，考察第二行的第二节和第三节共有几个空位，若没有空位，则无需填值；此时有三个空位，继续往下走。
第 8 行到第 10 行的代码段，求 idx = 0 对应的集合，即左宫集合 {1}。不为空，继续调用 shrinkRowCandidatesP2 接口做进一步处理，后者会因为左宫集合 {1} 里的 1 已经出现在第二行第二节里而返回 RET_PENDING。
回到 filterRowCandidatesEx 接口，idx 加一，再次调用 filterRowByPolicy2。
此时 idx = 1，要试图往第二行的第一节和第三节填空，此两节共有两个空位，而 idx = 1 对应中宫集合 {4, 6}，不为空，进一步调用 shrinkRowCandidatesP2 接口，后者发现可以对第二行做收缩处理，返回 RET_OK，回到 filterRowCandidatesEx 接口，继续向上级调用返回 RET_OK。
shrinkRowCandidatesP2 接口
1 u8 CQuizDealer::shrinkRowCandidatesP2(u8* pBlk, u8* pRow, u8 zeroSum, u8 row, u8 colBase)
2 {
3 u8 rowBase = row * 9;
4 for (u8 col = 0; col < 9; ++col) {
5 if (col >= colBase && col <= colBase + 2)
6 continue;
7 if (pRow[col] != 0)
8 if (!removeVal(pBlk, pRow[col]))
9 return RET_PENDING;
10 }
11 if (zeroSum < pBlk[0]) {
12 printf("shrinking row %d with ply2[blk:%d] went WRONG: 0Sum [%d] < vSum [%d]\n", (int)row + 1, (int)colBase / 3 + 1, (int)zeroSum, (int)pBlk[0]);
13 return RET_WRONG;
14 }
15 u8 exVals[7] = {0};
16 if (zeroSum > pBlk[0]) {
17 calcExCols(exVals, pBlk, pRow, rowBase, colBase, false);
18 if (zeroSum - exVals[0] < pBlk[0]) {
19 printf("shrinking row %d with ply2[blk:%d] went WRONG: 0Sum [%d] - xSum [%d] < vSum [%d]\n", (int)row + 1, (int)colBase / 3 + 1, (int)zeroSum, (int)exVals[0], (int)pBlk[0]);
20 return RET_WRONG;
21 }
22 if (zeroSum - exVals[0] > pBlk[0])
23 return RET_PENDING;
24 }
25 u8 ret = RET_PENDING;
26 bool shrunken = false;
27 for (u8 col = 0; col < 9; ++col) {
28 if (pRow[col] != 0 || (col >= colBase && col <= colBase + 2) || inSet(col, exVals))
29 continue;
30 ret = shrinkCandidates(rowBase + col, pBlk);
31 if (ret == RET_WRONG) {
32 printf("shrinking [%d,%d] with row-ply2 went WRONG\n", (int)row + 1, (int)col + 1);
33 return RET_WRONG;
34 }
35 if (ret == RET_OK)
36 shrunken = true;
37 }
38 if (shrunken) {
39 ret = RET_OK;
40 printf("row %d shrunken ply-2 by blk %d\n", (u8)row + 1, (u8)colBase / 3 + 1);
41 }
42 return ret;
43 }
shrinkRowCandidatesP2 接口实现和 shrinkRowCandidatesP1 类似。
filterColCandidatesEx 接口
1 u8 CQuizDealer::filterColCandidatesEx(u8 col)
2 {
3 u8 vals[10] = {0}; // last item denotes sum of zeros
4 for (u8 row = 0; row < 9; ++row) {
5 u8 celIdx = row * 9 + col;
6 vals[row] = m_seqCell[celIdx].val;
7 if (vals[row] == 0) {
8 vals[9] += 1;
9 }
10 }
11 if (vals[9] == 0)
12 return RET_PENDING;
13 u8 blkBase = col / 3;
14 u8 blkTakens[21] = {0};
15 setBlkColTakens(blkBase, col % 3, blkTakens);
16 if (blkTakens[0] == 0 && blkTakens[7] == 0 && blkTakens[14] == 0)
17 return RET_PENDING;
18 u8 ret = RET_PENDING;
19 for (u8 idx = 0; idx < 3; ++idx) {
20 if (ret = filterColByPolicy1(col, idx, vals, blkTakens))
21 return ret;
22 }
23 for (u8 idx = 0; idx < 3; ++idx) {
24 if (ret = filterColByPolicy2(col, idx, vals, blkTakens))
25 return ret;
26 }
27 return RET_PENDING;
28 }
filterColCandidatesEx 接口实现和 filterRowCandidatesEx 接口的实现实质是一样的，相当于对 9×9 方阵旋转 90 度后做一次 filterRowCandidatesEx。
setBlkColTakens 和 setColTakensInBlk 接口
1 void CQuizDealer::setBlkColTakens(u8 blkBase, u8 innCol, u8* pTakens)
2 {
3 for (u8 idx = 0; idx < 3; ++idx) {
4 u8 blk = blkBase + idx * 3;
5 u8* pVals = pTakens + idx * 7;
6 u8 rowBase = idx * 27;
7 u8 col = block2col(blk, (innCol + 1) % 3);
8 setColTakensInBlk(col, rowBase, pVals);
9 col = block2col(blk, (innCol + 2) % 3);
10 setColTakensInBlk(col, rowBase, pVals);
11 }
12 }
1 void CQuizDealer::setColTakensInBlk(u8 col, u8 rowBase, u8* pVals)
2 {
3 for (u8 idx = rowBase; idx < rowBase + 27; idx += 9) {
4 u8 val = m_seqCell[idx + col].val;
5 if (val != 0) {
6 u8 sum = pVals[0] + 1;
7 pVals[sum] = val;
8 pVals[0] = sum;
9 }
10 }
11 }
filterColByPolicy1 和 shrinkColCandidatesP1 接口
1 u8 CQuizDealer::filterColByPolicy1(u8 col, u8 idx, u8* pVals, u8* pBlkTakens)
2 {
3 u8 segRowBase = idx * 3;
4 u8 v1 = pVals[segRowBase], v2 = pVals[segRowBase + 1], v3 = pVals[segRowBase + 2];
5 u8 zeroSum = (u8)(v1 == 0) + (u8)(v2 == 0) + (u8)(v3 == 0);
6 if (zeroSum == 0)
7 return 0;
8 u8 isVals[7] = {0};
9 if (idx == 0)
10 intersection(pBlkTakens + 7, pBlkTakens + 14, isVals);
11 else if (idx == 1)
12 intersection(pBlkTakens, pBlkTakens + 14, isVals);
13 else
14 intersection(pBlkTakens, pBlkTakens + 7, isVals);
15 if (isVals[0] == 0)
16 return RET_PENDING;
17 return shrinkColCandidatesP1(isVals, pVals, zeroSum, col, segRowBase);
18 }
1 u8 CQuizDealer::shrinkColCandidatesP1(u8* pBlk, u8* pCol, u8 zeroSum, u8 col, u8 segRowBase)
2 {
3 for (u8 idx = segRowBase; idx < segRowBase + 3; ++idx) {
4 if (pCol[idx] != 0)
5 if (!removeVal(pBlk, pCol[idx]))
6 return RET_PENDING;
7 }
8 if (zeroSum < pBlk[0]) {
9 printf("shrinking col %d with ply1[vblk:%d] went WRONG: 0Sum [%d] < vSum [%d]\n", (int)col + 1, (int)segRowBase / 3 + 1, (int)zeroSum, (int)pBlk[0]);
10 return RET_WRONG;
11 }
12 u8 exVals[4] = {0};
13 if (zeroSum > pBlk[0]) {
14 calcExRows(exVals, pBlk, pCol, col, segRowBase, true);
15 if (zeroSum - exVals[0] < pBlk[0]) {
16 printf("shrinking col %d with ply1[vblk:%d] went WRONG: 0Sum [%d] - xSum [%d] < vSum [%d]\n", (int)col + 1, (int)segRowBase / 3 + 1, (int)zeroSum, (int)exVals[0], (int)pBlk[0]);
17 return RET_WRONG;
18 }
19 if (zeroSum - exVals[0] > pBlk[0])
20 return RET_PENDING;
21 }
22 u8 ret = RET_PENDING;
23 bool shrunken = false;
24 for (u8 row = segRowBase; row < segRowBase + 3; ++row) {
25 if (pCol[row] != 0 || inSet(row, exVals))
26 continue;
27 ret = shrinkCandidates(row * 9 + col, pBlk);
28 if (ret == RET_WRONG) {
29 printf("shrinking [%d,%d] with col-ply1 went WRONG\n", (int)row + 1, (int)col + 1);
30 return RET_WRONG;
31 }
32 if (ret == RET_OK)
33 shrunken = true;
34 }
35 if (shrunken) {
36 ret = RET_OK;
37 printf("col %d shrunken ply-1 by vblk %d\n", (u8)col + 1, (u8)segRowBase / 3 + 1);
38 }
39 return ret;
40 }
filterColByPolicy2 和 shrinkColCandidatesP2 接口
1 u8 CQuizDealer::filterColByPolicy2(u8 col, u8 idx, u8* pVals, u8* pBlkTakens)
2 {
3 u8 segRowBase = idx * 3;
4 u8 v1 = pVals[segRowBase], v2 = pVals[segRowBase + 1], v3 = pVals[segRowBase + 2];
5 u8 zeroSum = pVals[9] - (u8)(v1 == 0) - (u8)(v2 == 0) - (u8)(v3 == 0);
6 if (zeroSum == 0)
7 return 0;
8 u8* pBlk = pBlkTakens + (idx * 7);
9 if (pBlk[0] == 0)
10 return RET_PENDING;
11 return shrinkColCandidatesP2(pBlk, pVals, zeroSum, col, segRowBase);
12 }
1 u8 CQuizDealer::shrinkColCandidatesP2(u8* pBlk, u8* pCol, u8 zeroSum, u8 col, u8 segRowBase)
2 {
3 for (u8 row = 0; row < 9; ++row) {
4 if (row >= segRowBase && row <= segRowBase + 2)
5 continue;
6 if (pCol[row] != 0)
7 if (!removeVal(pBlk, pCol[row]))
8 return RET_PENDING;
9 }
10 if (zeroSum < pBlk[0]) {
11 printf("shrinking col %d with ply2[vblk:%d] went WRONG: 0Sum [%d] < vSum [%d]\n", (int)col + 1, (int)segRowBase / 3 + 1, (int)zeroSum, (int)pBlk[0]);
12 return RET_WRONG;
13 }
14 u8 exVals[7] = {0};
15 if (zeroSum > pBlk[0]) {
16 calcExRows(exVals, pBlk, pCol, col, segRowBase, false);
17 if (zeroSum - exVals[0] < pBlk[0]) {
18 printf("shrinking col %d with ply2[vblk:%d] went WRONG: 0Sum [%d] - xSum [%d] < vSum [%d]\n", (int)col + 1, (int)segRowBase / 3 + 1, (int)zeroSum, (int)exVals[0], (int)pBlk[0]);
19 return RET_WRONG;
20 }
21 if (zeroSum - exVals[0] > pBlk[0])
22 return RET_PENDING;
23 }
24 u8 ret = 0;
25 bool shrunken = false;
26 for (u8 row = 0; row < 9; ++row) {
27 if (pCol[row] != 0 || (row >= segRowBase && row <= segRowBase + 2) || inSet(row, exVals))
28 continue;
29 ret = shrinkCandidates(row * 9 + col, pBlk);
30 if (ret == RET_WRONG) {
31 printf("shrinking [%d,%d] with col-ply2 went wrong\n", (int)row + 1, (int)col + 1);
32 return RET_WRONG;
33 }
34 if (ret == RET_OK)
35 shrunken = true;
36 }
37 if (shrunken) {
38 ret = RET_OK;
39 printf("col %d shrunken ply-2 by vblk %d\n", (u8)col + 1, (u8)segRowBase / 3 + 1);
40 }
41 return ret;
42 }
calcExRows 接口
1 void CQuizDealer::calcExRows(u8* pEx, u8* pBlk, u8* pCol, u8 col, u8 segRowBase, bool ply1)
2 {
3 u8 lower = (ply1 ? segRowBase : 0);
4 u8 upper = (ply1 ? segRowBase + 3 : 9);
5 for (u8 row = lower; row < upper; ++row) {
6 if (ply1) {
7 if (pCol[row] != 0)
8 continue;
9 }
10 else {
11 if (pCol[row] != 0 || (row >= segRowBase && row <= segRowBase + 2))
12 continue;
13 }
14 u8 vals[10] = {0};
15 intersection(m_seqCell[row * 9 + col].candidates, pBlk, vals);
16 if (vals[0] == 0) {
17 u8 sum = pEx[0] + 1;
18 pEx[sum] = row;
19 pEx[0] = sum;
20 }
21 }
22 }
其他细微代码修改
printCandidates 函数
去掉了末尾的一行语句，即 printf("\n");
showQuiz 接口
1 void CQuizDealer::showQuiz()
2 {
3 ...
30 printf("\nSteps:%u\nCandidates:\n", m_steps);
31 u8 count = 1;
32 u8 foremost = 0;
33 for (std::set<u8>::iterator it = m_setBlank.begin(); it != m_setBlank.end(); ++it, ++count) {
34 u8 pos = *it;
35 u8* pVals = m_seqCell[pos].candidates;
36 if (foremost == 0 && pVals[0] == 1)
37 foremost = pos;
38 printCandidates(pos, pVals);
39 if (count % 3 == 0)
40 printf("\n");
41 else
42 printf(" ");
43 }
44 if (count % 3 != 1)
45 printf("\n");
46 if (foremost != 0)
47 printf("The foremost cell with one candidate at [%d,%d]\n", (int)(foremost / 9 + 1), (int)(foremost % 9 + 1));
48 if (m_guessLevel != 0)
49 printf("\nAt guess level %d [%d,%d] %d\n", (int)m_guessLevel, (int)(m_guessPos / 9 + 1), (int)(m_guessPos % 9 + 1), (int)m_guessValPos);
50 }
parse 和 guess 接口
开头增加一条语句：++m_steps;
nextGuess 接口
1 void CQuizDealer::nextGuess()
2 {
3 ...
11 ++m_steps;
12 Snapshot* pTop = m_stkSnap.top();
13 ...
step 接口
去掉语句：nextGuess(); 之后的一条语句：m_steps++;
showOrderList 函数
// Sudoku Solver 1.0 2021/9/20
#define STR_VER "Sudoku Solver 1.1 2021/10/2 by readalps\n\n"
void showOrderList()
{
printf(STR_VER);
...
}
进一步的实例分析放到下一篇。
标签：1.1,idx,pVals,SudokuSolver,RET,C++,row,col,u8 来源： https://www.cnblogs.com/readalps/p/15364222.html