昨天研究了下卡诺图, 写了 [python] 卡诺图化简, 结果发现还有重复的问题, 于是今天花了半天时间加了这段逻辑供大家参考.
## ts 是
# 1. 根据分组建立数字和组的对应关系
kt = {}
for i,t in enumerate(ts):
for s in t:
if s not in kt: kt[s] = []
kt[s].append(i)
# 2. 先将必选组拿出来, 数字只在该组出现过.
mteam = []
for n in kt:
if len(kt[n]) == 1 and kt[n][0] not in mteam:
mteam.append(kt[n][0])
# 3. 确定已覆盖有的数字
over_n = []
for m in mteam:
for n in ts[m]:
if n not in over_n: over_n.append(n)
# 4. 找到没有覆盖的数字和它在分组的关系
no_over_t = []
for n in kt:
if n in over_n: continue
no_over_t.append(kt[n])
# 5. 剩余的部分, 分组没有被选,那就排列组合一下
# ps: 我这里没有足够的时间测试和样例研究和测试, 直接排列组合一下, 不知道有没有问题.
import itertools
tp = itertools.product(*no_over_t)
est_all = [list(t)+mteam for t in tp] # 得到所有分组
# 打印看看
for i, t in enumerate(est_all):
print(f'分组{i+1} :',end = '')
for e in t:
print(ts[e], end = ', ')
print()
# 输入:
t = [2,3,7,9,10,11,12,13,18,19,22,23,26,27,30,31]
# 输出:
分组1 :{9, 11}, {12, 13}, {19, 3, 23, 7}, {2, 3, 10, 11, 18, 19, 26, 27}, {18, 19, 22, 23, 26, 27, 30, 31},
分组2 :{9, 13}, {12, 13}, {19, 3, 23, 7}, {2, 3, 10, 11, 18, 19, 26, 27}, {18, 19, 22, 23, 26, 27, 30, 31},
现在加入这段逻辑后, 输出还算正常了, 欢迎大家评论区讨论.
