我有一个代表数独游戏的9x9多维数组。我需要将它分解为9个3x3的许多组件。怎么做？我完全不知道从哪里开始。
game = [
[1, 3, 2, 5, 7, 9, 4, 6, 8],
[4, 9, 8, 2, 6, 1, 3, 7, 5],
[7, 5, 6, 3, 8, 4, 2, 1, 9],
[6, 4, 3, 1, 5, 8, 7, 9, 2],
[5, 2, 1, 7, 9, 3, 8, 4, 6],
[9, 8, 7, 4, 2, 6, 5, 3, 1],
[2, 1, 4, 9, 3, 5, 6, 8, 7],
[3, 6, 5, 8, 1, 7, 9, 2, 4],
[8, 7, 9, 6, 4, 2, 1, 5, 3]
]
分成块，它变成
chunk_1 = [
[1, 3, 2],
[4, 9, 8],
[7, 5, 6]
]
chunk_2 = [
[5, 7, 9],
[2, 6, 1],
[3, 8, 4]
]
...and so on
答案 0 :(得分：5)
方法Matrix#minor是为此量身定制的：
require 'matrix'
def sub3x3(game, i, j)
Matrix[*game].minor(3*i, 3, 3*j, 3).to_a
end
chunk1 = sub3x3(game, 0, 0)
#=> [[1, 3, 2], [4, 9, 8], [7, 5, 6]]
chunk2 = sub3x3(game, 0, 1)
#=> [[5, 7, 9], [2, 6, 1], [3, 8, 4]]
chunk3 = sub3x3(game, 0, 2)
#=> [[4, 6, 8], [3, 7, 5], [2, 1, 9]]
chunk4 = sub3x3(game, 1, 0)
#=> [[6, 4, 3], [5, 2, 1], [9, 8, 7]]
...
chunk9 = sub3x3(game, 2, 2)
#=> [[6, 8, 7], [9, 2, 4], [1, 5, 3]]
Ruby没有＆＃34;行＆＃34;的概念。和＆＃34;列＆＃34;数组。因此，为方便起见，我将参考3x3＆＃34;子阵列＆＃34; game的偏移i和j i = 0,1,2，j = 0,1,2），作为m = Matrix[*game]的3x3子矩阵，其左上角值为行偏移3*i和3*j的列偏移m，转换为数组。
这是相对低效的，因为为每个＆＃34; chunk＆＃34;的计算创建了一个新矩阵。考虑到阵列的大小，这不是一个问题，而是提高效率，而不是重新考虑整体设计。创建九个局部变量（而不是一个九个数组的数组）是不可能的。
这里建议在所有打开的单元格填满后检查game（使用上面的方法sub3x3）的有效性。请注意，我使用了Wiki description of the game，其中唯一有效的条目是数字1-9，我假设当玩家在单元格中输入值时，代码会强制执行该要求。
def invalid_vector_index(game)
game.index { |vector| vector.uniq.size < 9 }
end
def sub3x3_invalid?(game, i, j)
sub3x3(game, i, j).flatten.uniq.size < 9
end
def valid?(game)
i = invalid_vector_index(game)
return [:ROW_ERR, i] if i
j = invalid_vector_index(game.transpose)
return [:COL_ERR, j] if j
m = Matrix[*game]
(0..2).each do |i|
(0..2).each do |j|
return [:SUB_ERR, i, j] if sub3x3_invalid?(game, i, j)
end
end
true
end
valid?(game)
#=> true
请注意，这会返回true，意味着game有效，或者数组既表示解决方案无效，又包含可用于通知玩家原因的信息。< / p>
现在尝试
game[5], game[6] = game[6], game[5]
所以
game
#=> [[1, 3, 2, 5, 7, 9, 4, 6, 8],
# [4, 9, 8, 2, 6, 1, 3, 7, 5],
# [7, 5, 6, 3, 8, 4, 2, 1, 9],
# [6, 4, 3, 1, 5, 8, 7, 9, 2],
# [5, 2, 1, 7, 9, 3, 8, 4, 6],
# [2, 1, 4, 9, 3, 5, 6, 8, 7],
# [9, 8, 7, 4, 2, 6, 5, 3, 1],
# [3, 6, 5, 8, 1, 7, 9, 2, 4],
# [8, 7, 9, 6, 4, 2, 1, 5, 3]]
valid?(game)
#=> [:SUB_ERR, 1, 0]
行和列显然仍然有效，但此返回值表示至少有一个3x3子数组无效且数组
[[6, 4, 3],
[5, 2, 1],
[2, 1, 4]]
是第一个发现无效的。
答案 1 :(得分：4)
这是一项有趣的练习！
game.each_slice(3).map{|stripe| stripe.transpose.each_slice(3).map{|chunk| chunk.transpose}}.flatten(1)
这很麻烦，不需要定义每个chunk_1, chunk_2, ...。
如果您需要chunk_2，可以使用extract_chunks(game)[1]
它输出[chunk_1, chunk_2, chunk_3, ..., chunk_9]，所以它是一个数组数组：
1 3 2
4 9 8
7 5 6
5 7 9
2 6 1
3 8 4
4 6 8
3 7 5
2 1 9
6 4 3
5 2 1
...
您可以定义一种方法来检查此网格是否有效（它是）：
def extract_chunks(game)
game.each_slice(3).map{|stripe| stripe.transpose.each_slice(3).map{|chunk| chunk.transpose}}.flatten(1)
end
class Array # NOTE: Use refinements if you don't want to patch Array
def has_nine_unique_elements?
self.flatten(1).uniq.size == 9
end
end
def valid?(game)
game.has_nine_unique_elements? &&
game.all?{|row| row.has_nine_unique_elements? } &&
game.all?{|column| column.has_nine_unique_elements? } &&
extract_chunks(game).all?{|chunk| chunk.has_nine_unique_elements? }
end
puts valid?(game) #=> true
each_slice，flatten(1)删除一个级别并返回一行单元格行数组。 ：）答案 2 :(得分：0)
您可以创建一个从给定索引生成单个3X3块的方法。由于数独板的长度为9，因此将为您生成9个3X3块。见下文。
#steps
#you'll loop through each index of the board
#to get the x value
#you divide the index by 3 and multiply by 3
#to get the y value
#you divide the index by 3, take remainder and multiply by 3
#for each x value, you can get 3 y values
#this will give you a single 3X3 box from one index so
def three_by3(index, sudoku)
#to get x value
x=(index/3)*3
#to get y value
y=(index%3)*3
(x...x+3).each_with_object([]) do |x,arr|
(y...y+3).each do |y|
arr<<sudoku[x][y]
end
end
end
sudoku = [ [1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2],
[1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2],
[1,2,3,4,5,6,7,8,9],
[2,3,4,5,6,7,8,9,1],
[3,4,5,6,7,8,9,1,2]]
p (0...sudoku.length).map {|i| three_by3(i,sudoku)}
#output:
#[[1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2],
# [1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2],
# [1, 2, 3, 2, 3, 4, 3, 4, 5],
# [4, 5, 6, 5, 6, 7, 6, 7, 8],
# [7, 8, 9, 8, 9, 1, 9, 1, 2]]