无马AntiKnight

来源:本站 作者:Halcyon 发布时间:2023-03-22 19:22:37

无马 Anti Knight

 

无马数独是一种全盘限制类的题目,它看起来像是标准数独。

Anti Knight is a variant with all-grid limitted. It looks like Classic.

标准数独基础上,如果一格是A,从这个格起走一步马步能到的单元格内不能是A。下图展示了当C3=A时,诸如A2或B5这样的格子不能是A。

Classic rules applied. Additionally, if a cell is A, cells have Knight Step with this cell cannot be A. The next picture shows cells like A2 or B5 cannot be A when C3=A.

马步是象棋步法的一种,所以无马数独是一种无国际象棋类变型(无缘也是其中一种,它也被称为无王数独)。

Knight Step is one of the Chess Step, so Anti Knight is a kind of the Anti Chess variants (Untouch is another kind of them, which is also called Anti King).

 

和无缘数独类似,特殊的区块和删减非常常见。但无马数独中,特殊区块的删减域更广,因为马步经常比国王步更远一些。

Similar to Untouch, special Blocks and deletings are common. But deletings own wider domain in Anti Knight, for Knight Steps  are usually further than King Steps.

 

六阶潜规则Hidden Rules in 6x6

涂色区域有三种数字,白色区域有另外三种数字。

In 6x6 puzzles, 3 kinds of numbers in colored cells, 3 kinds of numbers in other cells.

符合同位数独规则。Disjoint rules are obeyed.

 

九阶潜规则 Hidden Rules in 9x9

① 如果图中8格只有1个重复数字A,那么E5=A. 

If 8 cells like the yellow structure have just one repeated number A, then E5=A.

证明:如果E5=X,那么X必须在黄色区域中出现两次。

Proof: If E5=X, X must appear twice in yellow cells.

② 如果四角上这四组2x2方格都不是A,那么E5=A。

If these four 2x2 squares on corner cannot be A, then E5=A.

证明:红色区域没有A意味着紫色区域也没有A(利用互补)。但A必须出现9次,所以每个2x5的区域里有两个A,剩下E5=A。

Proof: Red cells cannot own A means purple cells cannot own A either (by complementariness). But A must be on the grid for 9 times, so each 2x5 part contains 2 A, and E5=A.

 

练习 Practice

① 六阶无马数独中,B2有多少个马步格?

How many cell has a Knight Step with B2 in Anti Knight 6x6?

② 哪些格不能是5或6?Which cells cannot be 5 or 6?

答案 Answer

① 4  B1, D5, E3