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multiple of 45 solution in Creative category for Hubspot Amulet by Sim0000
from itertools import product
def checkio(m):
for x in product(range(-180,180,45), repeat = 3): # 45 = gcd(225, 315)
t = [sum(m[j][i]*x[j] for j in range(3)) % 360 for i in range(3)]
if t == [0, 225, 315]: return list(x)
# Answer is only multiple of 45.
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
def check_it(func, matrix):
result = func(matrix)
if not all(-180 <= el <= 180 for el in result):
print("The angles must be in range from -180 to 180 inclusively.")
return False
f, s, t = result
temp = [0, 0, 0]
temp[0] += f
temp[1] += matrix[0][1] * f
temp[2] += matrix[0][2] * f
temp[0] += matrix[1][0] * s
temp[1] += s
temp[2] += matrix[1][2] * s
temp[0] += matrix[2][0] * t
temp[1] += matrix[2][1] * t
temp[2] += t
temp = [n % 360 for n in temp]
if temp == [0, 225, 315]:
return True
else:
print("This is the wrong final position {0}.".format(temp))
return False
assert check_it(checkio,
[[1, 2, 3],
[3, 1, 2],
[2, 3, 1]]), "1st example"
assert check_it(checkio,
[[1, 4, 2],
[2, 1, 2],
[2, 2, 1]]), "2nd example"
assert check_it(checkio,
[[1, 2, 5],
[2, 1, 1],
[2, 5, 1]]), "3rd example"
March 15, 2014
