哈斯克尔：获得数独解算器块 [英] Haskell: Get block for Sudoku solver

问题描述

我试图为我正在开发的一个项目构建一个Sudoku解算器。我有一个9x9网格，每个位置编号为0..80，从左到右，从上到下

例如：

0 1 2 3 4 5 6 7 8

9 10 ...

我在尝试返回表示位置落入的3x3网格的Int列表。例如，对于网格（0,0）中的位置1，它将返回[0,1,2,9,10,11,18， 19,20]和位于网格（0,2）的位置8，它将返回[6,7,8,15,16,17,24,25,26]。

我写了一个函数，它返回所需的3x3网格，在一个9x9的Sudoku中：

  getBlock :: Int  - > Int  - > [（Int]] 
 getBlock xy = [[（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq + 0），（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq +（y * sudokuSizeSq + 0）*（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq +（sudokuSizeSq-1））]，[（（x * sudokuSizeSq + 1）* sudokuSize） （（x * sudokuSizeSq + 1）* sudokuSize）+（y * sudokuSizeSq + 1）...（（x * sudokuSizeSq + 1）* sudokuSize）+（y * sudokuSizeSq +（sudokuSizeSq-1））] （（x * sudokuSizeSq + 2）* sudokuSize）+（y * sudokuSizeSq + 1）...（（x * sudokuSizeSq + 2）* sudokuSize） +（y * sudokuSizeSq +（sudokuSizeSq-1））]]] 
  

其中：
sudokuSizeSq是宽度的平方（3）
sudokuSize是宽度（9）

x和y代表可能的网格。

这有效，但是我想增加它，因此它可以适用于更大的网格。我改变了我的代码到后面，但它似乎不会加载函数。

  getBlock :: Int  - > Int  - > [（Int]] 
 getBlock xy = [[（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq + 0），（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq +（y * sudokuSizeSq + 0）*（（x * sudokuSizeSq + 0）* sudokuSize）+（y * sudokuSizeSq +（sudokuSizeSq-1））]，[（（x * sudokuSizeSq + 1）* sudokuSize） （（x * sudokuSizeSq + 1）* sudokuSize）+（y * sudokuSizeSq + 1）...（（x * sudokuSizeSq + 1）* sudokuSize）+（y * sudokuSizeSq +（sudokuSizeSq-1））] （x * sudokuSizeSq +（sudokuSizeSq-1））* sudokuSize）+（y * sudokuSizeSq + 1）...（（x * sudokuSizeSq +（sudokuSizeSq-1））* sudokuSize）+（y * sudokuSizeSq + 0） x * sudokuSizeSq +（sudokuSizeSq-1））* sudokuSize）+（y * sudokuSizeSq +（sudokuSizeSq-1））]] 
  

我这样做的方式似乎很麻烦，有没有更好的方法来做到这一点或解决我遇到的问题。谢谢。

解决方案

  [a + b 
 | a<  - 取sudokuSizeSq $ map（sudokuSize *）[x * sudokuSizeSq ..] 
，b<  - 取sudokuSizeSq [y * sudokuSizeSq ..]] 
  

I am trying to build a Sudoku solver for a project I am working on. I have a 9x9 grid with each position numbered 0..80, left to right, top to bottom

e.g.:

0 1 2 3 4 5 6 7 8

9 10 ...

I am trying to return a list of Int's that represent the 3x3 grid that a position falls in. For example for position 1, which is in grid (0,0) it would return [0,1,2,9,10,11,18,19,20] and for position 8, which is in grid (0,2) it would return [6,7,8,15,16,17,24,25,26].

I have written a function which returns the required 3x3 grid, on a 9x9 Sudoku:

getBlock :: Int -> Int -> [[Int]]
getBlock x y = [[((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))],[((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))],[((x * sudokuSizeSq + 2) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + 2) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + 2) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))]]

Where: sudokuSizeSq is the square of the width (3) sudokuSize is the width (9)

x and y represent the possible grids.

This works, however I would like to increase it so it will work for larger grids. I changed my code to the follow but it seems then wont load the function.

getBlock :: Int -> Int -> [[Int]]
getBlock x y = [[((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + 0) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))],[((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + 1) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))]..[((x * sudokuSizeSq + (sudokuSizeSq-1)) * sudokuSize) + (y * sudokuSizeSq + 0),((x * sudokuSizeSq + (sudokuSizeSq-1)) * sudokuSize) + (y * sudokuSizeSq + 1)..((x * sudokuSizeSq + (sudokuSizeSq-1)) * sudokuSize) + (y * sudokuSizeSq + (sudokuSizeSq-1))]]

The way I'm doing it seems very cumbersome, is there a better way to do this or a way to fix the problem I'm having. Thanks.

解决方案

[ a + b
| a <- take sudokuSizeSq $ map (sudokuSize *) [x * sudokuSizeSq .. ]
, b <- take sudokuSizeSq [y * sudokuSizeSq ..] ]

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