寻找真正的数独难题 [英] Looking for really hard sudoku puzzles

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本文介绍了寻找真正的数独难题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

有没有人有一些超级,超级难的数独谜题?


今年2月,我有足够的时间用Java编写数独的解码器。现在,我正在寻找可解决的难题,但我的程序无法解决。然而,到目前为止,报纸上的大多数难题都在2-3个周期内得到解决。


我在谷歌上发现的那些据说很难,在3个周期内得到解决。


任何帮助都会非常感激!

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该程序适用于每个单元格的可能值。

1.确定每个单元格中的可能值。

2.检查行。如果有任何值只有1个空单元格可以取该值,则将空单元格设置为该值。

3.列。

4. Nonets,(3X3 Squares 。)

5.单个单元格。

Does anyone have some super, super hard Sudoku puzzles?

Back in February this year, I had enough time to finally program a Sudoku solver in Java. Right now, I''m looking for solvable puzzles, but ones that my program cannot solve. However, most hard puzzles in the newspapers are solved within 2-3 cycles so far.

The ones I''ve found on google which are said to be hard, get solved in 3 cycles.

Any help would be very much appreciated!
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The program works on the possible values for each cell.
1. Determine the possible values in each cell.
2. Check rows. If there is any value in which only 1 empty cell can take that value, set empty cell to that value.
3. Columns.
4. Nonets, (3X3 Squares.)
5. Indiviual cells.

推荐答案

好吧,我的手机上有一些解算器可能无法解决的问题,所以这里有几个:
Well, I have some on my mobile which that solver might not be able to solve, so here are a few:
展开 | 选择 | Wrap | < span class =codeLinkonclick =LineNumbers(this);>行号


感谢您的谜题;我刚检查过:我的老求解器瞬间解决了这些问题。 (见关于数独的旧文章)。


亲切的问候,


Jos


ps 。解决时间分别为14,2和9毫秒,但包括类加载等。这些都是非常不准确的时间。
Thanks for those puzzles; I just checked: my old solver solves them all instantaneously. (see that old article about Sudoku).

kind regards,

Jos

ps. solving took 14, 2 and 9 ms respectively but that includes the class loading etc. Those are highly inaccurate timings.


由于您的输入,管理以查找程序中的错误。在再次运行这3个问题之后,我得到了。

1.解决了。 5个周期。

2.没有足够的数据或无法解决的难题。

3.数据不足或无法解决的难题。


将不得不寻找改进算法的方法,以便它可以解决最后的2。
Managed to find a bug in the program thanks to your inputs. After running those 3 problems through again, I got.
1. Solved. 5 cycles.
2. Not enough data, or unsolvable puzzle.
3. Not enough data, or unsolvable puzzle.

Will have to look up ways to refine the algorithm, so that it can solve the last 2.


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