用C++实现的数独解题程序 SudokuSolver 2.2 及实例分析

SudokuSolver 2.2 程序实现

根据 用C++实现的数独解题程序 SudokuSolver 2.1 及实例分析 里分析，对 2.1 版做了一些改进和尝试。

CQuizDealer 类声明部分的修改

class CQuizDealer
{
public:
    ...
    void run(ulong tilsteps = 0);
    void setOnlyGrpMode() {m_onlyGrp = true;}
    ...
private:
    ...
    CQuizDealer() : m_state(STA_UNLOADED), ..., m_onlyGrp(false) {};
    inline void incSteps() {++m_steps;}
    inline bool IsDone(u8 ret) {return (ret == RET_OK || ret == RET_WRONG);}...
    u8 filterOneGroup(u8* pGrp);
    bool completeShrinkByGrp(u8* pGrp, u8* pTimes);
    u8 incompleteShrinkByGrp(u8 valSum, u8* pCelSumExs, u8* pGrp);
    bool sameCandidates(u8 cel1, u8 cel2);
    ...
    ulong m_steps;
    bool m_onlyGrp;
    ...
};

新增的 setOnlyGrpMode 接口用于控制只采用第一类收缩算法求解。

filterCandidates 接口实现修改

u8 CQuizDealer::filterCandidates()
{
    incSteps();
    u8 ret = RET_PENDING;
    for (u8 row = 0; row < 9; ++row)
        if (ret = filterRowGroup(row))
            return ret;
    for (u8 col = 0; col < 9; ++col)
        if (ret = filterColGroup(col))
            return ret;
    for (u8 blk = 0; blk < 9; ++blk)
        if (ret = filterBlkGroup(blk))
            return ret;
    if (!m_onlyGrp) {
        for (u8 row = 0; row < 9; ++row) {
            ret = filterRowCandidatesEx(row);
            if (IsDone(ret))
                return ret;
        }
        for (u8 col = 0; col < 9; ++col) {
            ret = filterColCandidatesEx(col);
            if (IsDone(ret))
                return ret;
        }
    }
    if (ret == RET_SHRUNKEN) {
        printf("incomplete shrink met, filter again\n");
        return filterCandidates();
    }
return ret;
}

增加了 onlyGrp 控制，即上面所说的是否只采用第一类收缩算法求解。末尾增加的条件递归调用，源自上一篇的实例分析。

filterRowGroup 接口实现修改

u8 CQuizDealer::filterRowGroup(u8 row)
{
    u8 celIdxs[10] = {0}; // first item denotes sum of zeros
    u8 base = row * 9;
    for (u8 col = 0; col < 9; ++col) {
        if (m_seqCell[base + col].val == 0) {
            celIdxs[0] += 1;
            celIdxs[celIdxs[0]] = base + col;
        }
    }
    if (celIdxs[0] == 0)
        return RET_PENDING;
    u8 ret = filterOneGroup(celIdxs);
    if (ret == RET_OK)
        printf("%u) row %d complete shrunken by group\n", m_steps, (int)row + 1);
    else if (ret == RET_WRONG)
        printf("%u) row %d shrink by group went WRONG\n", m_steps, (int)row + 1);
    else if (ret == RET_SHRUNKEN)
        printf("%u) row %d incomplete shrunken by group\n", m_steps, (int)row + 1);
    return ret;
}

filterColGroup 和 filterBlkGroup 接口修改类似。

filterOneGroup 接口新实现

 1 u8 CQuizDealer::filterOneGroup(u8* pGrp)
 2 {
 3     u8 times[20] = {0};
 4     if (completeShrinkByGrp(pGrp, times))
 5         return RET_OK;
 6
 7     u8 size = pGrp[0];
 8     u8 celSumExs[100] = {0};
 9     for (u8 idx = 1; idx <= size; ++idx) {
10         u8 valSum = m_seqCell[pGrp[idx]].candidates[0];
11         u8 base = valSum * 10;
12         celSumExs[base] += 1;
13         u8 pos = base + celSumExs[base];
14         celSumExs[pos] = pGrp[idx];
15     }
16
17     for (u8 idx = 2; idx <= 6; ++idx) {
18         u8 ret = incompleteShrinkByGrp(idx, celSumExs, pGrp);
19         if (ret != RET_PENDING)
20             return ret;
21     }
22     return RET_PENDING;
23 }

filterOneGroup 接口的原有实现只支持完全收缩，现在放到了新增 completeShrinkByGrp 接口里：

bool CQuizDealer::completeShrinkByGrp(u8* pGrp, u8* pTimes)
{
    u8 size = pGrp[0];
    for (u8 idx = 1; idx <= size; ++idx) {
        u8 sum = m_seqCell[pGrp[idx]].candidates[0];
        for (u8 inn = 1; inn <= sum; ++inn) {
            u8 val = m_seqCell[pGrp[idx]].candidates[inn];
            pTimes[val] += 1;
            pTimes[val + 9] = pGrp[idx];
        }
    }
    bool ret = false;
    for (u8 val = 1; val <= 9; ++val) {
        if (pTimes[val] == 1) {
            ret = true;
            u8 celIdx = pTimes[val + 9];
            m_seqCell[celIdx].candidates[0] = 1;
            m_seqCell[celIdx].candidates[1] = val;
        }
    }
    return ret;
}

新增 incompleteShrinkByGrp 接口实现

 1 u8 CQuizDealer::incompleteShrinkByGrp(u8 valSum, u8* pCelSumExs, u8* pGrp)
 2 {
 3     u8 base = valSum * 10;
 4     if (pCelSumExs[base] < valSum)
 5         return RET_PENDING;
 6     u8 itemSum = 0;
 7     Item items[9];
 8     for (u8 pos = 1; pos <= pCelSumExs[base]; ++pos) {
 9         u8 idx = 0;
10         for (; idx < itemSum; ++idx)
11             if (sameCandidates(pCelSumExs[base + pos], items[idx].celIdxs[0])) {
12                 items[idx].celIdxs[items[idx].sameSum] = pCelSumExs[base + pos];
13                 items[idx].sameSum++;
14                 break;
15             }
16         if (idx == itemSum) {
17             items[itemSum].sameSum = 1;
18             items[itemSum].celIdxs[0] = pCelSumExs[base + pos];
19             ++itemSum;
20         }
21     }
22     for (u8 idx = 0; idx < itemSum; ++idx) {
23         if (items[idx].sameSum > valSum)
24             return RET_WRONG;
25     }
26     bool shrunken = false;
27     for (u8 idx = 0; idx < itemSum; ++idx) {
28         if (items[idx].sameSum < valSum)
29             continue;
30         for (u8 pos = 1; pos <= pGrp[0]; ++pos) {
31             if (inSet(pGrp[pos], (u8*)&(items[idx])))
32                 continue;
33             u8 isVals[10] = {0};
34             u8 cel1 = items[idx].celIdxs[0];
35             u8 cel2 = pGrp[pos];
36             intersection(m_seqCell[cel1].candidates, m_seqCell[cel2].candidates, isVals);
37             if (isVals[0] == 0)
38                 continue;
39             shrunken = true;
40             for (u8 valIdx = 1; valIdx <= isVals[0]; ++valIdx)
41                 if (!removeVal(m_seqCell[cel2].candidates, isVals[valIdx]))
42                     return RET_WRONG;
43         }
44     }
45     return (shrunken ? RET_SHRUNKEN : RET_PENDING);
46 }

里面用到的 Item 结构以及 sameCandidates 接口为：

struct Item {
    u8 sameSum;
    u8 celIdxs[9];
    Item() {sameSum = 0;}
};

bool CQuizDealer::sameCandidates(u8 cel1, u8 cel2)
  {
      u8 sum = m_seqCell[cel1].candidates[0];
      if (sum != m_seqCell[cel2].candidates[0])
          return false;
      for (u8 idx = 1; idx <= sum; ++idx)
          if (m_seqCell[cel1].candidates[idx] != m_seqCell[cel2].candidates[idx])
              return false;
      return true;
  }

filterRowCandidatesEx 接口实现小修改

u8 CQuizDealer::filterRowCandidatesEx(u8 row)
{
    ...
    for (u8 idx = 0; idx < 3; ++idx) {
        ret = filterRowByPolicy1(row, idx, vals, blkTakens);
        if (IsDone(ret))
            return ret;
    }
    for (u8 idx = 0; idx < 3; ++idx) {
        ret = filterRowByPolicy2(row, idx, vals, blkTakens);
        if (IsDone(ret))
            return ret;
    }
    return ret;
}

filterColCandidatesEx 接口的修改类似。

shrinkRowCandidatesP1 接口实现小修改

u8 CQuizDealer::shrinkRowCandidatesP1(u8* pBlk, u8* pRow, u8 zeroSum, u8 row, u8 colBase)
{
    ...
    u8 shrunken = 0;
    for (u8 col = colBase; col < colBase + 3; ++col) {
        ...
        if (ret != RET_PENDING) {
            printf(" worked by row-ply1.\n");
            if (ret == RET_OK)
                shrunken = 1; // complete
            else if (ret == RET_SHRUNKEN && shrunken == 0)
                shrunken = 2; // incomplete
        }
    }
    if (shrunken == 1) {
        ret = RET_OK;
        printf("%u) row %d shrunken ply-1 by blk %d\n", m_steps, (u8)row + 1, (u8)colBase / 3 + 1);
    }
    else if (shrunken == 2)
        ret = RET_SHRUNKEN;
return ret;
}

shrinkRowCandidatesP2 接口以及 shrinkColCandidatesP1 和 shrinkColCandidatesP2 接口的修改类似。

其他小修改

// 1.0 2021/9/20
// 2.0 2021/10/2
// 2.1 2021/10/4
#define STR_VER "Sudoku Solver 2.2 2021/10/10 by readalps\n\n"

void showOrderList()
{
    ...
    printf("onlygrp: only using group shrink\n");
    printf("step: step forward\n");
    ...
}

void dealOrder(std::string& strOrder)
{
    ...
    else if ("onlygrp" == strOrder)
        CQuizDealer::instance()->setOnlyGrpMode();
    else if ("step" == strOrder)
    ...

实例分析

继续以 SudokuSolver 1.0：用C++实现的数独解题程序 【二】 里试验过的“最难”数独题为例做分析。在第一次 run 命令的输出中有如下信息：

506) Forward guess [1,4] level 11 at 2 out of 2
[2,1]: 1 4 5 9 shrunken to  5 9 worked by row-ply2.
[2,8]: 4 5 9 shrunken to  5 9 worked by row-ply2.
509) Guess [1,3] level 12 at 1 out of 2

第 2 行有两个空位发生了不完全收缩，走的不是第一类收缩算法（by grp），而是第二类收缩算法（by ply）。重新用 runtil 506 命令进到当时的上下文做深入分析：

...
506) Forward guess [1,4] level 11 at 2 out of 2
820 703 601
003 600 807
070 090 203

050 007 104
000 045 706
007 100 935

001 009 568
008 500 319
090 000 472

Steps:506
Candidates:
[1,3]: 4 5 9    [1,5]: 5    [1,8]: 4 5 9
[2,1]: 1 4 5 9    [2,2]: 1 4    [2,5]: 1 2 5
[2,6]: 1 2 4    [2,8]: 4 5 9    [3,1]: 1 4 5 6
[3,3]: 4 5 6    [3,4]: 4 8    [3,6]: 1 4 8
[3,8]: 4 5    [4,1]: 2 3 6 9    [4,3]: 2 6 9
[4,4]: 2 3 8 9    [4,5]: 2 3 6 8    [4,8]: 2 8
[5,1]: 1 2 3 9    [5,2]: 1 3 8    [5,3]: 2 9
[5,4]: 2 3 8 9    [5,8]: 2 8    [6,1]: 2 4 6
[6,2]: 4 6 8    [6,5]: 2 6 8    [6,6]: 2 6 8
[7,1]: 2 3 4 7    [7,2]: 3 4    [7,4]: 2 3 4
[7,5]: 2 3 7    [8,1]: 2 4 6 7    [8,2]: 4 6
[8,5]: 2 6 7    [8,6]: 2 4 6    [9,1]: 3 5 6
[9,3]: 5 6    [9,4]: 3 8    [9,5]: 1 3 6 8
[9,6]: 1 6 8
The foremost cell with 1 candidate(s) at [1,5]

At guess level 11 [1,4] 2
Run time: 326 milliseconds; steps: 506, solution sum: 0.

Order please:

由 The foremost cell with 1 candidate(s) at [1,5] 可知当时 [1,5] 位置可以直接填值。先走完这一步：

Order please:
step
820 753 601
003 600 807
070 090 203

050 007 104
000 045 706
007 100 935

001 009 568
008 500 319
090 000 472

Steps:507
Candidates:
[1,3]: 4 9    [1,8]: 4 9    [2,1]: 1 4 5 9
[2,2]: 1 4    [2,5]: 1 2    [2,6]: 1 2 4
[2,8]: 4 5 9    [3,1]: 1 4 5 6    [3,3]: 4 5 6
[3,4]: 4 8    [3,6]: 1 4 8    [3,8]: 4 5
[4,1]: 2 3 6 9    [4,3]: 2 6 9    [4,4]: 2 3 8 9
[4,5]: 2 3 6 8    [4,8]: 2 8    [5,1]: 1 2 3 9
[5,2]: 1 3 8    [5,3]: 2 9    [5,4]: 2 3 8 9
[5,8]: 2 8    [6,1]: 2 4 6    [6,2]: 4 6 8
[6,5]: 2 6 8    [6,6]: 2 6 8    [7,1]: 2 3 4 7
[7,2]: 3 4    [7,4]: 2 3 4    [7,5]: 2 3 7
[8,1]: 2 4 6 7    [8,2]: 4 6    [8,5]: 2 6 7
[8,6]: 2 4 6    [9,1]: 3 5 6    [9,3]: 5 6
[9,4]: 3 8    [9,5]: 1 3 6 8    [9,6]: 1 6 8
The foremost cell with 2 candidate(s) at [1,3]

At guess level 11 [1,4] 2

Order please:

走完 507 步时，第 2 行的空位候选值分布情况为：

[2,1]: 1 4 5 9
[2,2]: 1 4    [2,5]: 1 2    [2,6]: 1 2 4
[2,8]: 4 5 9

这里，5 和 9 只能填入两个空位，即 [2,1] 和 [2,8] 中，因此，这两个空位只能是一个填 5 另一个填 9。于是，这两个空位里的除 5 和 9 之外的其他候选值都可以排除掉。这和上一篇的实例分析部分发现的 grp 算法可改进之处还不一样，这是新发现的一处可以改进 grp 算法的地方。