背景
我们有一个家庭传统，我和我的兄弟姐妹的圣诞礼物是用一个代码来识别的，这个代码只能使用与我们相关的数字来解决。例如，代码可以是出生月份*年龄+毕业年份（这是一个简单的）。如果数字是 8 * 22 + 2020 = 2196，那么 2196 将写在我所有的圣诞礼物上。
我已经创建了一个 Python 类来解决具有某些约束的代码，但我想知道是否可以递归地执行它。
当前代码
第一个函数为在 target_values 中产生值的所有可能的数字和操作组合返回一个结果集
#Master algorithm (Get the result set of all combinations of numbers and cartesian products of operations that reach a target_value, using only the number_of_numbers_in_solution)
#Example: sibling1.results[1] = [(3, 22, 4), (<built-in function add>, <built-in function add>), 29]. This means that 3 + 22 + 4 = 29, and 29 is in target_values
import operator
from itertools import product
from itertools import combinations
NUMBER_OF_OPERATIONS_IN_SOLUTION = 2 #Total numbers involved is this plus 1
NUMBER_OF_NUMBERS_IN_SOLUTION = NUMBER_OF_OPERATIONS_IN_SOLUTION + 1
TARGET_VALUES = {22,27,29,38,39}
def getresults( list ):
#Add the cartesian product of all possible operations to a variable ops
ops = []
opslist = [operator.add, operator.sub, operator.mul, operator.truediv]
for val in product(opslist, repeat=NUMBER_OF_OPERATIONS_IN_SOLUTION):
ops.append(val)
#Get the result set of all combinations of numbers and cartesian products of operations that reach a target_value
results = []
for x in combinations(list, NUMBER_OF_NUMBERS_IN_SOLUTION):
for y in ops:
result = 0
for z in range(len(y)):
#On the first iteration, do the operation on the first two numbers (x[z] and x[z+1])
if (z == 0):
#print(y[z], x[z], x[z+1])
result = y[z](x[z], x[z+1])
#For all other iterations, do the operation on the current result and x[z+1])
else:
#print(y[z], result, x[z+1])
result = y[z](result, x[z+1])
if result in TARGET_VALUES:
results.append([x, y, result])
#print (x, y)
print(len(results))
return results
然后是一个类，它接受每个人的个人参数并获取结果集
def getalpha( str, inverse ):
"Converts string to alphanumeric array of chars"
array = []
for i in range(0, len(str)):
alpha = ord(str[i]) - 96
if inverse:
array.append(27 - alpha)
else:
array.append(alpha)
return array;
class Person:
def __init__(self, name, middlename, birthmonth, birthday, birthyear, age, orderofbirth, gradyear, state, zip, workzip, cityfirst3):
#final list
self.listofnums = []
self.listofnums.extend((birthmonth, birthday, birthyear, birthyear - 1900, age, orderofbirth, gradyear, gradyear - 2000, zip, workzip))
self.listofnums.extend(getalpha(cityfirst3, False))
self.results = getresults(self.listofnums)
最后，一个“解决代码”方法从结果集中获取并找到产生完整目标值列表的任何可能组合。
#Compares the values of two sets
def compare(l1, l2):
result = all(map(lambda x, y: x == y, l1, l2))
return result and len(l1) == len(l2)
#Check every result in sibling2 with a different result target_value and equal operation sets
def comparetwosiblings(current_values, sibling1, sibling2, a, b):
if sibling2.results[b][2] not in current_values and compare(sibling1.results[a][1], sibling2.results[b][1]):
okay = True
#If the indexes aren't alphanumeric, ensure they're the same before adding to new result set
for c in range(0, NUMBER_OF_NUMBERS_IN_SOLUTION):
indexintersection = set([index for index, value in enumerate(sibling1.listofnums) if value == sibling1.results[a][0][c]]) & set([index for index, value in enumerate(sibling2.listofnums) if value == sibling2.results[b][0][c]])
if len(indexintersection) > 0:
okay = True
else:
okay = False
break
else:
okay = False
return okay
#For every result, we start by adding the result number to the current_values list for sibling1, then cycle through each person and see if a matching operator list leads to a different result number. (Matching indices as well)
#If there's a result set for everyone that leads to five different numbers in the code, the values will be added to the newresult set
def solvecode( sibling1, sibling2, sibling3, sibling4, sibling5 ):
newresults = []
current_values = []
#For every result in sibling1
for a in range(len(sibling1.results)):
current_values = []
current_values.append(sibling1.results[a][2])
for b in range(len(sibling2.results)):
if comparetwosiblings(current_values, sibling1, sibling2, a, b):
current_values.append(sibling2.results[b][2])
for c in range(len(sibling3.results)):
if comparetwosiblings(current_values, sibling1, sibling3, a, c):
current_values.append(sibling3.results[c][2])
for d in range(len(sibling4.results)):
if comparetwosiblings(current_values, sibling1, sibling4, a, d):
current_values.append(sibling4.results[d][2])
for e in range(len(sibling5.results)):
if comparetwosiblings(current_values, sibling1, sibling5, a, e):
newresults.append([sibling1.results[a][0], sibling2.results[b][0], sibling3.results[c][0], sibling4.results[d][0], sibling5.results[e][0], sibling1.results[a][1]])
current_values.remove(sibling4.results[d][2])
current_values.remove(sibling3.results[c][2])
current_values.remove(sibling2.results[b][2])
print(len(newresults))
print(newresults)
这是我想知道是否可以优化并制成递归算法的最后一个“solvecode”方法。在某些情况下，添加或删除一个兄弟姐妹可能会有所帮助，这会递归看起来不错（我妈妈有时会在一个兄弟姐妹上犯错，或者我们有一个新的兄弟/嫂子）
感谢您的任何帮助！我希望你至少能从我奇怪的家庭传统中得到一笑。
编辑：如果您想测试算法，这里有一组兄弟姐妹的示例，它们会产生一个正确的解决方案
#ALL PERSONAL INFO CHANGED FOR STACKOVERFLOW
sibling1 = Person("sibling1", "horatio", 7, 8, 1998, 22, 5, 2020, "ma", 11111, 11111, "red")
sibling2 = Person("sibling2", "liem", 2, 21, 1995, 25, 4, 2018, "ma", 11111, 11111, "pho")
sibling3 = Person("sibling3", "kyle", 4, 21, 1993, 26, 3, 2016, "ma", 11111, 11111, "okl")
sibling4 = Person("sibling4", "jamal", 4, 7, 1991, 29, 2, 2014, "ma", 11111, 11111, "pla")
sibling5 = Person("sibling5", "roberto", 9, 23, 1990, 30, 1, 2012, "ma", 11111, 11111, "boe")