how to loop over the elementary arithmetic symbols

Question

I want to check if 7 digits can get to 100 when putting elementary arithmetic symbols between them.

def is_hundred(n1,n2,n3,n4,n5,n6,n7):
    p = [+,-,*,/]
    for p1 in p:
        for p2 in p:
            for p3 in p:
                for p4 in p:
                    for p5 in p:
                        for p6 in p:
                            if n1 p1 n2 p2 n3 p3 n4 p4 n5 p5 n6 p6 n7 == 100:
                                return "success"

how can i replace the variables with the arithmetic symbols that are in the list?

Answer 1

In case you didn't hear it: using eval is evil. So that's definitely not code that I'd put in production. That said, this exercise is also something that I will probably never have to deploy to prod...

The idea: use the arguments and "brute-force" all the possibilities of adding operations between them. Construct these options as strings, and for each one of them, when done, evaluate the total (using eval) and if the total is 100 - print it to stdout:

def is_hundred(args, res=''):
    if res == '':
        res = args[0]
        args = args[1:]
    if not args:
        if eval(res) == 100:
            print(res)
    else:
        first = args[0]
        for op in ['+','-','*','/']:
            is_hundred(args[1:], "{}{}{}".format(res, op, first))


# you can pass any number of arguments in a list: 7, 8, 15...
is_hundred([2,3,4,8,10,11])  # 2+3+4+8*10+11

Answer 2

Here is a sample code which search a pair of operators. the i in iadd means inplace add operator. You can simply switch to add. product creates duplicate permutation of operators.

from operator import iadd, isub, imul, itruediv
from itertools import product

operators = (iadd, isub, imul, itruediv)

def is_hundred(values):
    n_operators = len(values) - 1

    for ops in product(operators, repeat=n_operators):
        value = values[0]
        for op, v in zip(ops, values[1:]):
            value = op(value, v)
        if value == 100:
            print(list(ops))
            return True
    else:
        return False

print(is_hundred([99,1,1,1,1,1]))
print(is_hundred([1,1,1,1]))