SudokuSolver 2.0:用C++实现的数独解题程序 【二】
本篇是 SudokuSolver 2.0: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:
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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:
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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:
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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 2.0 2021/10/2 by readalps\n\n" void showOrderList() { printf(STR_VER); ... }