java数独解算器，而不是回溯解算器

Question

过去几周我一直在做一个数独游戏。 该游戏具有很多功能，如：玩游戏、打印数独、解数独

解算函数使用传统的回溯，但这不是问题所在，问题是我需要游戏能够生成一个人类可解的数独，为此我需要一种方法，能够像人类那样解数独

如果有人能帮我解决这个问题，我将不胜感激

Answer 1

大量针对人类玩家的数独解决策略在Andrew Stuart's Sudoku page上得到了很好的展示和解释：

**Show Possibles**         
1: Hidden Singles      
2: Naked Pairs/Triples     
3: Hidden Pairs/Triples    
4: Naked Quads     
5: Pointing Pairs      
6: Box/Line Reduction      
**Tough Strategies**  
7: X-Wing      
8: Simple Colouring        
9: Y-Wing      
10: Sword-Fish         
11: XYZ Wing       
**Diabolical Strategies** 
12: X-Cycles       
13: XY-Chain       
14: 3D Medusa      
15: Jelly-Fish         
16: Unique Rectangles      
17: Extended Unique Rect.      
18: Hidden Unique Rect's       
19: WXYZ Wing      
20: Aligned Pair Exclusion         
**Extreme Strategies**    
21: Grouped X-Cycles       
22: Empty Rectangles       
23: Finned X-Wing      
24: Finned Sword-Fish      
25: Altern. Inference Chains   
26: Sue-de-Coq         
27: Digit Forcing Chains       
28: Nishio Forcing Chains  
29: Cell Forcing Chains        
30: Unit Forcing Chains        
31: Almost Locked Sets         
32: Death Blossom      
33: Pattern Overlay Method         
34: Quad Forcing Chains        
**"Trial and Error"** 
35: Bowman's Bingo

作为一个相当频繁的玩家，我会将策略11之外的一切判断为“不再有趣”。但这可能是品味的问题

Answer 2

如果你只需要一个快速的随机数独游戏，你可以使用一种特殊的方法创建一个有效的数独模式，使用我刚才想出的以下算法：

You initialize an array with a randomized set of the numbers 1 to 9, 
technically it's easier if you initialize 3 arrays each with 3 length.
You can have these numbers be randomized, thus create a different sudoku.

[1 2 3] [4 5 6] [7 8 9]
Then you shift these:
[7 8 9] [1 2 3] [4 5 6]
[4 5 6] [7 8 9] [1 2 3]

Then you shift the numbers inside the arrays:

[3 1 2] [6 4 5] [9 7 8]
Then you shift the arrays themselves again:
[9 7 8] [3 1 2] [6 4 5]
[6 4 5] [9 7 8] [3 1 2]

Then you shift the numbers inside the arrays:

[2 3 1] [5 6 4] [8 9 7]
Then you shift the arrays again:
[8 9 7] [2 3 1] [5 6 4]
[5 6 4] [8 9 7] [2 3 1]

最后一组数独表：

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

这是有效的。之后，你可以拿出一些数字，然后用你已有的算法检查它是有一个解，还是有多个解。如果删除某个数字会产生多个，则可以撤消该数字并结束删除，或者尝试删除另一个数字