那你看看 30 楼的实现吧，我认为我这个答案是满足你的需要的。

如果不做限制的话，那这题那就是简单的取几个数排列组合，再去重就是了。不过做题之前是应该向出题者弄清楚规则需求。

我理解的九宫格规则是：数字只能连相邻的数字，且不能出现重复。如果我没理解错的话，那就没有那么复杂吧。我前面也写了一个实现。

# -*- coding: utf-8 -*-
"""
主要实现思路
主要是根据手势密码锁来定义基本操作规则（只能连接临近的数字，不能越过数字进行连接），再运用类比法构造实现（不能出现重复的数字）。
"""
#定义基本规则
key=[[[1,2],[1,4],[1,5]],[[2,1],[2,3],[2,4],[2,5],[2,6]],[[3,2],[3,5],[3,6]],[[4,1],[4,2],[4,5],[4,7],[4,8]],[[5,1],[5,2],[5,3],[5,4],[5,6],[5,7],[5,8],[5,9]],[[6,3],[6,5],[6,8],[6,9],[6,2]],[[7,4],[7,5],[7,8]],[[8,7],[8,9],[8,4],[8,5,],[8,6]],[[9,6],[9,5],[9,8]]]

#1个
def rank1(key):
    skey=[]
    for i in range(1,10):
         skey.append(key[i-1][0][0])
    return skey
#2个
def rank2(key):
    skey=[]
    key1=rank1(key)
    for key2 in key1:
       for dkey in key[key2-1]:
            skey.append(dkey)
    return skey
#3个
def rank3(key):
    skey=[]
    key2=rank2(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<3):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey
#4个
def rank4(key):
    skey=[]
    key2=rank3(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<4):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey
#5个
def rank5(key):
    skey=[]
    key2=rank4(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<5):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey

#6个
def rank6(key):
    skey=[]
    key2=rank5(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<5):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey
#7个
def rank7(key):
    skey=[]
    key2=rank6(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<5):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey
 #8个
def rank8(key):
    skey=[]
    key2=rank7(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<5):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey
#9个
def rank9(key):
    skey=[]
    key2=rank8(key)
    for dkey in key2:
        for key3 in key[dkey[-1]-1]:
            tim=[]
            if (key3[1] not in dkey) and (len(tim)<5):
                for okey in dkey:
                        tim.append(okey)
                tim.append(key3[1])
                skey.append(tim)
    return skey


print(rank4(key))