How do I simplify these nested for loops? - python-3.x

Question

I was given this riddle :
My first idea was to solve it using python. (Later I realized algebra would have been much easier).
Here was my initial code:
a = 0
b = 0
c = 0
d = 0
for a in range(100):
for b in range(100):
for c in range(100):
for d in range(100):
if float(a + b)/4 == 8.0:
if float(a + c)/4 == 13.0:
if float(c - d)/4 == 6.0:
if float(b + d)/4 == 8.0:
print(float(a + b)/4, end=" ")
print(float(a + c)/4, end=" ")
print(float(c + -d)/4, end=" ")
print(float(b + d)/4, end=" and values are:")
print(a/4, end=" ")
print(b/4, end=" ")
print(c/4, end=" ")
print(d/4)
It worked, but obviously this is not very readable.
Looking on stack overflow leads to the solution being recursive functions, however I do not know how exactly to apply this because there are four different mathematical calculations.
For example this is what I tried at first:
val = [0, 0, 0, 0]
def recursive_solution(section):
for val[section] in range(100):
for val[section+1] in range(100):
if section == 0:
if float(val[section])/4 + float(val[section + 1])/4 == 8:
recursive_solution(1)
recursive_solution(0)
But this obviously does not work, as the second index is added to the fourth index, and one of the operations is a subtraction, and you can't add the fourth index to the fifth one because the fifth one doesn't exist.

Answer 1

Here's what I would do from your code: use a cartesian product and and statements:
from collections import product
for a, b, c, d in product(*([range(100)] * 4))
if (
float(a + b) / 4 == 8.0
and float(a + c) / 4 == 13.0
and float(c - d) / 4 == 6.0
and float(b + d) / 4 == 8.0
):
print(float(a + b) / 4, end=" ")
print(float(a + c) / 4, end=" ")
print(float(c + -d) / 4, end=" ")
print(float(b + d) / 4, end=" and values are:")
print(a / 4, end=" ")
print(b / 4, end=" ")
print(c / 4, end=" ")
print(d / 4)