你可以使用itertools模块中的排列函数将数字和运算符按照所有可能的顺序组成字符串公式。然后使用eval()函数计算结果。
例如:
from itertools import permutations
numbers = ["1","9","8","2"]
target = 24
operators = ["+","-","*","/"]
for values in permutations(numbers,len(numbers)):
for oper in permutations(operators,len(numbers)-1):
formula = "".join(o+v for o,v in zip([""]+list(oper),values))
if eval(formula) == target: print(formula,"=",target)
[更新1] 如果您可以多次使用相同的运算符(根据您在 1+1+1*8=24 上的评论建议),您需要使用 combinations_with_replacement 来生成更多的运算符模式:
from itertools import permutations,combinations_with_replacement
numbers = ["1","1","1","8"]
target = 10
operators = ["+","-","*","/"]
seen = set()
for values in permutations(numbers,len(numbers)):
for operCombo in combinations_with_replacement(operators,len(numbers)-1):
for oper in permutations(operCombo,len(numbers)-1):
formula = "".join(o+v for o,v in zip([""]+list(oper),values))
if formula not in seen and eval(formula) == target:
print(formula,"=",target)
seen.add(formula)
实际上,这与先前的示例唯一不同之处在于插入了for operCombo in ...
循环。
注意:生成的组合公式看起来完全相同,因此您应该避免打印已经出现过的解决方案(就像我这里做的那样)。如果输入中有任何数字重复,则在先前的示例中也会发生重复。
还要注意,为了使9-1+8*2的结果为24,必须在加减法之前执行乘法(即按照优先级规则),否则9-1+8*2=32。您需要支持括号以覆盖不同的运算顺序。
[更新2] 支持括号取决于您想要允许多少个数字。对于4个数字,有11种模式:
- 没有括号:A+B+C+D
- A+B组合:(A+B)+C+D
- B+C组合:A+(B+C)+D
- C+D组合:A+B+(C+D)
- A+B和C+D组合:(A+B)+(C+D)
- A+B+C组合:(A+B+C)+D
- B+C+D组合:A+(B+C+D)
- A+B组合 + C:((A+B)+C)+D
- A + 组合 B+C:(A+(B+C))+D
- B+C组合 + D:A+((B+C)+D)
- B + 组合 C+D:A+(B+(C+D))
如果您有超过4个数字,则会出现更多的括号组合模式。
这里是一个示例(针对4个数字):
from itertools import permutations,combinations_with_replacement
numbers = ["9","8","1","2"]
target = 24
operators = ["+","-","*","/"]
groups = ['X+X+X+X', 'X+X+(X+X)', 'X+(X+X)+X', '(X+X+X)+X', '(X+X)+X+X', 'X+(X+X+X)', '((X+X)+X)+X', 'X+(X+(X+X))', 'X+((X+X)+X)', '(X+X)+(X+X)', '(X+(X+X))+X']
seen = set()
for values in permutations(numbers,len(numbers)):
for operCombo in combinations_with_replacement(operators,len(numbers)-1):
for oper in permutations(operCombo,len(numbers)-1):
formulaKey = "".join(oper+values)
if formulaKey in seen: continue
for pattern in groups:
formula = "".join(o+p for o,p in zip([""]+list(oper), pattern.split("+")))
formula = "".join(v+p for v,p in zip([""]+list(values),formula.split("X")))
try:
if eval(formula) == target:
print(formula,"=",target)
seen.add(formulaKey)
break
except: pass
分组可能导致除以零,因此必须添加 try:except 块。
这会产生以下结果:
9*8/(1+2) = 24
9+8*2-1 = 24
9*8/(2+1) = 24
9-1+8*2 = 24
9-(1-8*2) = 24
9-1+2*8 = 24
(9-1)*2+8 = 24
9/(1+2)*8 = 24
9/((1+2)/8) = 24
9-(1-2*8) = 24
9+2*8-1 = 24
9/(2+1)*8 = 24
9/((2+1)/8) = 24
8+(9-1)*2 = 24
8*9/(1+2) = 24
8*9/(2+1) = 24
8-(1-9)*2 = 24
8/(1+2)*9 = 24
8/((1+2)/9) = 24
8+2*(9-1) = 24
8*2+9-1 = 24
8*2-1+9 = 24
8/(2+1)*9 = 24
8/((2+1)/9) = 24
8-2*(1-9) = 24
8*2-(1-9) = 24
2*(9-1)+8 = 24
2*8+9-1 = 24
2*8-1+9 = 24
2*8-(1-9) = 24
要生成更多数字的括号分组模式,您可以使用此函数:
from itertools import product
import re
def groupPatterns(count,pattern=None):
arr = pattern or "X"*count
if len(arr) < 2 : return [arr]
result = []
for mid in range(1,len(arr)):
leftPattern = groupPatterns(count,arr[:mid])
rightPattern = groupPatterns(count,arr[mid:])
for left,right in product(leftPattern,rightPattern):
result += [left + right]
if len(left) > 1 : result += ["(" + left + ")" + right]
if len(right) > 1 : result += [left + "(" + right + ")"]
if len(left) > 1 and len(right) > 1:
result += ["(" + left + ")(" + right + ")"]
if pattern: return result
patterns = []
for pat in sorted(set(result),key=lambda x:len(x)):
pat = re.sub("X(?=X)", r"X+", pat)
pat = re.sub("X\(", r"X+(", pat)
pat = re.sub("\)X", r")+X", pat)
pat = re.sub("\)\(", r")+(", pat)
patterns.append(pat)
return patterns
然后在前一个示例中,使用groups = groupPatterns(len(numbers))
替换groups = ["X+X+X+X",...
。
或者,为任意数量的值创建完全通用的函数,无论是否具有分组和运算符重用:
from itertools import permutations,combinations_with_replacement
def numbersToTarget(numbers,target,reuseOper=True,allowGroups=True,operators=["+","-","*","/"]):
groups = groupPatterns(len(numbers)) if allowGroups else [ "+".join("X"*len(numbers)) ]
seen = set()
for values in permutations(numbers,len(numbers)):
for operCombo in combinations_with_replacement(operators,len(numbers)-1) if reuseOper else [operators]:
for opers in permutations(operCombo,len(numbers)-1):
formulaKey = str(opers)+str(values)
if formulaKey in seen: continue
for pattern in groups:
formula = "".join(o+p for o,p in zip([""]+list(opers), pattern.split("+")))
formula = "".join(str(v)+p for v,p in zip([""]+list(values),formula.split("X")))
try:
if eval(formula) == target:
seen.add(formulaKey)
yield formula
break
except: pass
for formula in numbersToTarget([9,8,1,2],24):
print("24 =",formula)
for formula in numbersToTarget([9,8,1,2,5],0,allowGroups=False):
print("0 =",formula)