就像玩元组的多米诺骨牌（棋盘游戏）一样-如何优化？

Question

我想订购数百个2-15元组的子集。想象一个元组就像一块多米诺骨牌。两个数字固有地合为一体。没有定义哪个是第一个，哪个是第二个数字。这样就可以翻转了。

查找链中最多碎片（多元组）的多米诺骨牌（元组）的顺序。形成链的条件就像在多米诺骨牌中一样：第n个元组的第二个值==第n + 1个元组的第一个值。

其他注释：元组不一定形成循环

通过检查每个组合得到的技巧，强行强加它。一个由6个元组组成的链已经具有45360个可能性（带有翻转）

#brute forcing it by trying every possibility

#one subset
MN=[(3,4), (4,5), (8,5), (17,3), (34,56), (4,34)]
n=len[MN]

#initializing some variables for counting, etc.
count=0 #count the possibilities
score_max=0
best_sequence=None

from itertools import permutations
#helper functions
def calculate_optimization_score(MN):
    n=len(MN)
    score=0
    for i in range(n-1):
        if MN[i][1]==MN[i+1][0]:
            score=score+1
    return score

def flip_MN(MN, bool_ls):
    MN_new=list(MN) #to be safe, make a copy
    for i,e in enumerate(MN_new):
        if bool_ls[i]==True:
            MN_new[i]=tuple(reversed(MN_new[i]))
    return MN_new


for nb_flip_MN in range(n):
    #    make True False array with nb_flip_MN True-values
    bool_ls=[]
    for i in range(nb_flip_MN):
        bool_ls.append(True)
    for i in range(n-nb_flip_MN):
        bool_ls.append(False)
    perm = set(list(permutations(bool_ls))) # and all permutations of that

    for i,e in enumerate(perm):
        #make the sequence with the flipped MN according to perm
        MN_with_flip=flip_MN(MN, e)

        for j,sequence in enumerate(list(permutations(MN_with_flip))):
            score=calculate_optimization_score(sequence)
            count=count+1

            if score>score_max:
                score_max=score
                best_sequence=sequence