vjg vjg - 1 month ago 8
Python Question

an unknown number of cycles

If

n=1
I want to write:

for a1 in range(-10,10,1):
if 13*a1 == 1:
print('Success')


if
n=2
I want to write:

for a1 in range(-10,10,1):
for a2 in range(-10,10,1):
if 13*a1+27*a2==1:
print('Success')


if
n=3
I want to write:

for a1 in range(-10,10,1):
for a2 in range(-10,10,1):
for a3 in range(-10,10,1):
if 13*a1+27*a2+37*a3==1:
print('Success')


etc.
n
can be 1 or 2 or 3 .. 20.
n
is a random number from 1 to 20
How can I write the next code:

for a1 in range(-10, 10, 1):
for a2 in range(-10,10,1):
....
for an in range(-10,10,1): #n is a random number from 1 to 20
if some_number_1*a1+...+some_number_n*an == 1:
print('Success')


My attempt to find a solution:


  1. Maybe we can use recursion

    def recur(a):
    for a in range(-10,10,1):
    recur(a)


    this method will not work

  2. Maybe we can use lists.
    For example,

    list[1]=-10,..., 10
    list[2]=-10,...,10
    ...
    list[n]=-10,...,10


    So we have matrix(n,21). But I don't see how I can use this matrix to solve my problem.

  3. Maybe we need to use trees but I've never worked with them

  4. Maybe I can use a library to solve the equation
    a1*x1+a2*x2+...+an*xn=1
    but it's not interesting. I want to know how I can solve this task myself.


Answer

Use itertools.product:

def func(a):
    for i in itertools.product(range(-10, 10), repeat=len(a)):
        if sum(x*y for x, y in zip(i, a)) == 1:
            print("success")