从一组2元组生成3元组

时间:2019-05-23 19:12:23

标签: python algorithm tuples

在先前的问题中:

Generating maximum number of 3-tuples from a list of 2-tuples

我从@AChampion得到了一个答案,如果2元组的数目可被3整除,这似乎是可行的。但是,例如,如果我们有10个2元组,则解决方案将失败。在摸索了一段时间之后,我觉得不可能找到一个完美的解决方案:

(1,2)(1,3),(1,4),(2,3),(2,4),(3,4)

因此,我有兴趣找到一种解决方案,以最大程度地减少剩余元组的数量。在上面的示例中,结果可能是:

(1,2,3)           # derived from (1,2), (1,3), (2,3)
(1,4),(2,4),(3,4) # remainder tuples 

从3个2元组生成3元组的规则是:

(a,b), (b,c), (c,a) -> (a, b, c)

即2元组是一个长度为3的循环。3元组中元素的顺序并不重要,即:

(a,b,c) == (c,a,b)

我实际上对我们有一个数字n的情况很感兴趣:

for x in range(1,n+1):
    for y in range(1,n+1):
        if x!=y:
            a.append((x,y))

# a = [ (1,2),...,(1,n), (2,1),(2,3),...,(2,n),...(n,1),...,(n,n-1) ]

从a中,最小化生成3元组时剩余的2元组的数量。每个2元组只能使用一次。

我花了好几个小时来解决这个问题,但是对于一般情况,我似乎无法想出一个优雅的解决方案(嗯,我也没有找到一个丑陋的解决方案:-)。有什么想法吗?

1 个答案:

答案 0 :(得分:1)

为此,您需要创建用于替换的组合数量。然后遍历您的数据以获取包含以上任意组合的3个项目并替换它们。 我已经完成了好几个步骤。

from itertools import combinations

# create replacements elements
number_combinations_raw = list(combinations(range(1, 5), 3))

# create proper number combinations
number_combinations = []
for item in number_combinations_raw:
    if (item[0] + 1 == item[1]) and (item[1] + 1 == item[2]):
        number_combinations.append(item)

# create test data
data = [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4)]

# reduce data
reduce_data = []
for number_set in number_combinations:
    count = 0
    merged_data = []
    for item in data:
        if (number_set[0] in item and number_set[1] in item) or (number_set[1] in item and number_set[2] in item) \
            or (number_set[0] in item and number_set[2] in item):
            merged_data.append(item)
            count += 1
    if count == 3:
        reduce_data.append((number_set, merged_data))

# delete merged elements from data list and add replacement
for item in data:
    for reduce_item in reduce_data:
        for element in reduce_item[1]:
            if element in data:
                data.remove(element)

        data = [reduce_item[0]] + data

# remove duplicated replaced elements
final_list = list(dict.fromkeys(data))

输出:

[(1, 2, 3), (1, 4), (2, 4)]