数字列表上所有可能的操作组合，以查找特定数字

Question

I know that there are other questions like mine but the only problem is that they get all the combinations for all of the varibales in list but I want it so that the user enters the number that the want and the numbers they need to make the desired number. This is the code that I have:

numbers = []
operators = ['+', '*', '-', '/']
desire = int(input("Enter the number you want: "))
num1 = int(input("Enter First number: "))
num2 = int(input("Enter Second number: "))
num3 = int(input("Enter Third number: "))
num4 = int(input("Enter Fourth number: "))
numbers.append(num1)
numbers.append(num2)
numbers.append(num3)
numbers.append(num4)

But I have no idea how to expand on this

This is an example on what the code should do:

Say that the number they want made is 24 and

Say the numbers that they enter are 1, 9, 8, 2

output should be this:

9 - 1 + 8 * 2 = 24

etc...

All possible solutions need to be listed

All suggestions will be greatly appriciated

Answer 1

You could use permutations from the itertools module to arrange numbers and operators in all possible ways into a string formula. Then use eval() to compute the result.

For example:

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)

[UPDATE1] If you are allowed to use the same operator more than once (as suggested by your comment on 1+1+1*8=24), you will need to use combinations_with_replacement to generate more operator patterns:

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)

Essentially, this only differs from the previous example by the insertion of the for operCombo in ... loop.

Note: The combinations will generate formulas that look exactly the same so you will want to avoid printing solutions that have already been seen (as I did here). Duplications would also occur in the previous example if any numbers were repeated in the input.

Also note that in order for 9-1+8*2 to result in 24, the multiplication must be performed before additions and subtractions (ie under precedence rules) otherwise 9-1+8*2=32. You would need to support parentheses to cover different orders of operation.

[UPDATE2] Supporting parentheses is a bit more involved depending on how many numbers you want to allow. For 4 numbers, there are 11 patterns:

• No parentheses: A+B+C+D
• A+B group: (A+B)+C+D
• B+C group: A+(B+C)+D
• C+D group: A+B+(C+D)
• A+B and C+D groups: (A+B)+(C+D)
• A+B+C group: (A+B+C)+D
• B+C+D group: A+(B+C+D)
• A+B group + C: ((A+B)+C)+D
• A + group B+C: (A+(B+C))+D
• B+C group + D: A+((B+C)+D)
• B + group C+D: A+(B+(C+D))

If you have more than 4 numbers there will be more patterns of parentheses grouping.

Here's an example (for 4 numbers):

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 # ignore variations on parentheses alone
            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

Groupings could result in divisions by zero, so a try:except block had to be added.

This produces the following result:

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

To generate the parentheses grouping patterns for more numbers, you can use this function:

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 # recursion
    patterns = [] # final, add "+" between X value placeholders or groups
    for pat in sorted(set(result),key=lambda x:len(x)):
        pat = re.sub("X(?=X)", r"X+",  pat)  # XX --> X+X
        pat = re.sub("X\(",    r"X+(", pat)  # X( --> X+(
        pat = re.sub("\)X",    r")+X", pat)  # )X --> )+X
        pat = re.sub("\)\(",   r")+(", pat)  # )( --> )+(
        patterns.append(pat)
    return patterns

And then replace groups = ["X+X+X+X",... with groups = groupPatterns(len(numbers)) in the previous example.

OR, create a completely generic function for any number of values, with or without grouping and operator reuse:

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 # ignore variations on parentheses alone
                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)

Answer 2

Here is what I came up with using eval() to mathematically evaluate the string of math operators (note this is not a very secure method and malicious users could potentially hack your program through it. If you are deploying then maybe look at Evaluating a mathematical expression in a string )

numbers = []
operators = ['+', '*', '-', '/']
desire = int(input("Enter the number you want: "))
num1 = input("Enter First number: ")
num2 = input("Enter Second number: ")
num3 = input("Enter Third number: ")
num4 = input("Enter Fourth number: ")
numbers.append(num1)
numbers.append(num2)
numbers.append(num3)
numbers.append(num4)

for operator1 in operators:
    for operator2 in operators:
        for operator3 in operators:
            problem = numbers[0] + operator1 + numbers[1] + operator2 + numbers[2] + operator3 + numbers[3]
            result = int(eval(problem))
            if result == desire:
                print("{} = {}".format(problem, result))

My first simple test

Enter the number you want: 40
Enter First number: 10
Enter Second number: 10
Enter Third number: 10
Enter Fourth number: 10

Yields

10+10+10+10 = 40

A more complex test

Enter the number you want: 18
Enter First number: 6
Enter Second number: 3
Enter Third number: 4
Enter Fourth number: 4

Yeilds:

6*3+4-4 = 18

6*3*4/4 = 18

6*3-4+4 = 18

6*3/4*4 = 18

6/3+4*4 = 18

It should be noted though that this solution does not consider the various orders for your numbers. I'll see if I can craft something more devious

UPDATE

I've crafted a way to consider all permutations of numbers as well

def make_order_combinations():
    number_orders = []
    for i in range(4):
        for j in range(4):
            for k in range(4):
                for z in range(4):
                    if i != j and i != k and i != z and j != k and j != z and k != z:
                        number_orders.append((i, j, k, z))
    return number_orders


def solve_given_number_order(number_order):
    for operator1 in operators:
        for operator2 in operators:
            for operator3 in operators:
                problem = numbers[number_order[0]] + operator1 + numbers[number_order[1]] + operator2 + numbers[number_order[2]] + operator3 + numbers[number_order[3]]
                # print(problem)
                result = eval(problem)
                # print(result)
                if result == desire:
                    print("{} = {}".format(problem, result))

numbers = []
operators = ['+', '*', '-', '/']
desire = int(input("Enter the number you want: "))
num1 = input("Enter First number: ")
num2 = input("Enter Second number: ")
num3 = input("Enter Third number: ")
num4 = input("Enter Fourth number: ")
numbers.append(num1)
numbers.append(num2)
numbers.append(num3)
numbers.append(num4)

list_of_orders = make_order_combinations()
for order in list_of_orders:
    solve_given_number_order(order)

Now the test

Enter the number you want: 67
Enter First number: 15
Enter Second number: 4
Enter Third number: 7
Enter Fourth number: 1

Yields

15*4+7*1 = 67
15*4+7/1 = 67.0
15*4+1*7 = 67
15*4*1+7 = 67
15*4/1+7 = 67.0
15*1*4+7 = 67
15/1*4+7 = 67.0
4*15+7*1 = 67
4*15+7/1 = 67.0
4*15+1*7 = 67
4*15*1+7 = 67
4*15/1+7 = 67.0
4*1*15+7 = 67
4/1*15+7 = 67.0
7+15*4*1 = 67
7+15*4/1 = 67.0
7+15*1*4 = 67
7+15/1*4 = 67.0
7+4*15*1 = 67
7+4*15/1 = 67.0
7+4*1*15 = 67
7+4/1*15 = 67.0
7+1*15*4 = 67
7*1+15*4 = 67
7/1+15*4 = 67.0
7+1*4*15 = 67
7*1+4*15 = 67
7/1+4*15 = 67.0
1*15*4+7 = 67
1*4*15+7 = 67
1*7+15*4 = 67
1*7+4*15 = 67

Where you can see it does consider all possible rearrangement of the numbers as well. Order of operations does still apply though and therefore the output:

1*7+4*15 = 67

Should be read as (1*7)+(4*15) = 67

Answer 3

this is really not well tested (and probably does too much work) but may get you started:

from operator import mul, add, sub, truediv
from itertools import permutations, combinations_with_replacement

operators = (mul, add, sub, truediv)
desire = 24

numbers = [1, 9, 8, 2]

OP2SYM = {mul: '*', add: '+', sub: '-', truediv: '/'}

for op0, op1, op2 in combinations_with_replacement((mul, add, sub, truediv), 3):
    for n0, n1, n2, n3 in permutations(numbers, 4):
        # print(op0, op1, op2)
        # print(n0, n1, n2, n3)
        if op0(n0, op1(n1, op2(n2, n3))) == desire:
            print('{} {} ({} {} ({} {} {}))'.format(
                n0, OP2SYM[op0], n1, OP2SYM[op1], n2, OP2SYM[op2], n3))
        if op0(op1(n0, n1), op2(n2, n3)) == desire:
            print('({} {} {}) {} ({} {} {})'.format(
                n0, OP2SYM[op0], n1, OP2SYM[op1], n2, OP2SYM[op2], n3))
        if op2(op1(op0(n0, n1), n2), n3) == desire:
            print('(({} {} {}) {} {}) {} {}'.format(
                n0, OP2SYM[op0], n1, OP2SYM[op1], n2, OP2SYM[op2], n3))

it outputs

((8 * 2) + 9) - 1
((2 * 8) + 9) - 1

a simpler idea would be to constuct the strings of the form '6*3-4+4' and use ast.literal_eval to evaluate them

Answer 4

You can try with the permutations module inside itertools

from itertools import permutations, combinations
numbers = ""
solutions = []
operators = "+*-/"
desire = int(input("Enter the number you want: "))
num1 = input("Enter First number: ")
num2 = input("Enter Second number: ")
num3 = input("Enter Third number: ")
num4 = input("Enter Fourth number: ")
#concatenate the input
numbers = num1 + num2 + num3 + num4    
#generate all possible permutations of this characters
num_permutations = [p for p in permutations(numbers)]
op_combinations = [p for p in combinations(operators,3)]

for n_perm in num_permutations:
   for op_perm in op_combinations:
      cur_expression = ""
      for i in range(3):
         cur_expression += n_perm[i] + op_perm[i]
      cur_expression += n_perm[3]
      tmp_solution = eval(cur_expression)
      if desire == tmp_solution:
         solutions.append(tmp_solution)