我如何生成一个嵌套数组，其中包含所有可能的4位数排列6个数字与重复值?鲁比(人名)

Question

给定一个数n = 1,2,3,4,5,6

我想生成一个嵌套数组S，它将包含n的所有可能的4位数排列。因为6^4 = 1296，所以有1296种可能的排列。

的例子:S =[(1, 1, 1, 1),(1, 1, 1, 2)[1, 1、2、2]…[6、6、6、6]]

我从第一个值为[1,1,1,1]的索引开始嵌套循环然后使用范围为0..的for in循环并试图将S[i]的值结转到S[i+1]然后增加S[i+1][x]的值，其中x总是从3开始，然后递减，直到达到0，然后再次变为3。我的过程的问题是，当我试图增加S[i+1][x]， S[i]也增加它的S[i][x]。在下面的代码中，S被称为'all_possible_combination '

all_possible_combinations = Array.new(1296) {Array.new(4)}
all_possible_combinations[0] = [1, 1 ,1 ,1]
x = 3
for i in 0..1295
if i + 1 == 1296
break
else
all_possible_combinations[i+1] = all_possible_combinations[i]
all_possible_combinations[i+1][x] += 1
x -= 1
if x == 0
x = 3
end
end
end

[附图显示调试过程，其中Si][x]也被增加

Answer 1

您可以按如下方式计算该数组。

a = [1, 2, 3, 4, 5, 6]

b = a.repeated_permutation(4).to_a
#=> [[1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 1, 3], [1, 1, 1, 4], [1, 1, 1, 5],
#    [1, 1, 1, 6], [1, 1, 2, 1], [1, 1, 2, 2], [1, 1, 2, 3], [1, 1, 2, 4],
#    ...
#    [6, 6, 5, 3], [6, 6, 5, 4], [6, 6, 5, 5], [6, 6, 5, 6], [6, 6, 6, 1],
#    [6, 6, 6, 2], [6, 6, 6, 3], [6, 6, 6, 4], [6, 6, 6, 5], [6, 6, 6, 6]]

b.size
#=> 1296

看到数组# repeated_permutation

如果数组a可能包含重复的排列，并且您希望删除重复的排列，您可能希望添加array# uniq。

a = [1, 1, 3, 1, 1, 6]

b = a.repeated_permutation(4).to_a.uniq
#=> [[1, 1, 1, 1], [1, 1, 1, 3], [1, 1, 1, 6], [1, 1, 3, 1],
#    [1, 1, 3, 3], [1, 1, 3, 6], [1, 1, 6, 1], [1, 1, 6, 3], 
#    [1, 1, 6, 6], [1, 3, 1, 1], [1, 3, 1, 3], [1, 3, 1, 6],       
#    [1, 3, 3, 1], [1, 3, 3, 3], [1, 3, 3, 6], [1, 3, 6, 1],
#    [1, 3, 6, 3], [1, 3, 6, 6], [1, 6, 1, 1], [1, 6, 1, 3],
#    [1, 6, 1, 6], [1, 6, 3, 1], [1, 6, 3, 3], [1, 6, 3, 6],
#    [1, 6, 6, 1], [1, 6, 6, 3], [1, 6, 6, 6], [3, 1, 1, 1],
#    [3, 1, 1, 3], [3, 1, 1, 6], [3, 1, 3, 1], [3, 1, 3, 3],
#    [3, 1, 3, 6], [3, 1, 6, 1], [3, 1, 6, 3], [3, 1, 6, 6],
#    [3, 3, 1, 1], [3, 3, 1, 3], [3, 3, 1, 6], [3, 3, 3, 1],
#    [3, 3, 3, 3], [3, 3, 3, 6], [3, 3, 6, 1], [3, 3, 6, 3],
#    [3, 3, 6, 6], [3, 6, 1, 1], [3, 6, 1, 3], [3, 6, 1, 6],
#    [3, 6, 3, 1], [3, 6, 3, 3], [3, 6, 3, 6], [3, 6, 6, 1],
#    [3, 6, 6, 3], [3, 6, 6, 6], [6, 1, 1, 1], [6, 1, 1, 3],
#    [6, 1, 1, 6], [6, 1, 3, 1], [6, 1, 3, 3], [6, 1, 3, 6],
#    [6, 1, 6, 1], [6, 1, 6, 3], [6, 1, 6, 6], [6, 3, 1, 1],
#    [6, 3, 1, 3], [6, 3, 1, 6], [6, 3, 3, 1], [6, 3, 3, 3],
#    [6, 3, 3, 6], [6, 3, 6, 1], [6, 3, 6, 3], [6, 3, 6, 6],
#    [6, 6, 1, 1], [6, 6, 1, 3], [6, 6, 1, 6], [6, 6, 3, 1],
#    [6, 6, 3, 3], [6, 6, 3, 6], [6, 6, 6, 1], [6, 6, 6, 3],
#    [6, 6, 6, 6]]

b.size
#=> 81

Answer 2

要创建一个序列，其中每个元素都是基于前一个元素生成的，有Enumerator.produce，例如:

enum = Enumerator.produce([1, 1, 1, 1]) do |a, b, c, d|
d += 1                # ^^^^^^^^^^^^
# initial value
if d > 6
d = 1
c += 1
end
if c > 6
c = 1
b += 1
end
if b > 6
b = 1
a += 1
end
if a > 6
raise StopIteration # <- stops enumeration
end
[a, b, c, d]          # <- return value = next value
end

我故意使示例保持简单，为四位数字中的每一位使用一个显式变量。当然，你也可以有一个数组，并使用一个小循环来处理自增/进位。

以上给出:

enum.count        #=> 1296
enum.first(3)     #=> [[1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 1, 3]]
enum.to_a.last(3) #=> [[6, 6, 6, 4], [6, 6, 6, 5], [6, 6, 6, 6]]