Hi, I am one of the puzzle authors of the Festo Coding Challenge 2020.
Here you can find the code I used to compute the solution codes myself. You can also find the solutions to the logic puzzles and the mystery puzzles here.
All my code is written in python.
def is_valid(n):
s = str(n)
return ('5' in s) != ('7' in s)
l = []
i = 0
while len(l) < 1000:
if is_valid(i):
l.append(i)
i += 1
print('Solution for puzzle 1.1: {}'.format(l[-1]))Solution for puzzle 1.1: 2379
import numpy as np
import itertools
d = np.genfromtxt('1_2_christmas_cards.csv',
dtype = int,
delimiter = ',',
missing_values = '-',
filling_values = -1)
friendlist = np.arange(1,9)
best_length = np.inf
best_order = None
for order in itertools.permutations(friendlist):
l = [0] + list(order) + [9]
length = np.sum(d[l[:-1],l[1:]])
if length < best_length:
best_length = length
best_order = l
chars = ' abcdefgh '
out = ''.join([chars[i] for i in best_order[1:-1]])
print('Solution for puzzle 1.2: {}'.format(out))Solution for puzzle 1.2: hcdfbgea
Solution strategies: This puzzle type is called "Star Battle" (German: Doppelstern). On google you find many more puzzles of this type and some strategy guides.
Solution for puzzle 1.3: 1135113445
How to find it: In the last post in the blog, there are links to four images. One link is pointing to a youtube-video instead of the actual image. You can find the correct link by using one of the other images' links and replacing the filename by screen.jpg, correct link is https://coding-challenge.festo.com/blog/images/screen.jpg. Now you can read the two parts of the note on the screen, pointing to the github account named fin-der-lohn. On github.com/fin-der-lohn you find a repository. The solution code is in there, if you look at the older commit.
Solution for puzzle 1.4: 1337potato
import re
prog = re.compile('.*2.*0.*2.*0.*')
l = []
i = 1
while len(l) < 1000000:
if not prog.match(str(i)):
l.append(i)
i +=1
print('Solution for puzzle 2.1: {}'.format(l[-1]))Solution for puzzle 2.1: 1001270
import numpy as np
import itertools
d = np.genfromtxt('2_2_christmas_shopping.csv',
dtype = int,
delimiter = ',',
missing_values = '-',
filling_values = -1)
shop_types = ' sspphhddtt '
S = [1,2]
P = [3,4]
H = [5,6]
D = [7,8]
T = [9,10]
best_length = np.inf
best_order = None
for shop_selection in itertools.product(S,P,H,D,T):
for order in itertools.permutations(shop_selection):
l = [0] + list(order) + [11]
length = np.sum(d[l[:-1],l[1:]])
if length < best_length:
best_length = length
best_order = l
shoplist = ['-', 's1', 's2', 'p1', 'p2', 'h1', 'h2', 'd1', 'd2', 't1','t2', '-']
out = ''.join([shoplist[i] for i in best_order[1:-1]])
print('Solution for puzzle 2.2: {}'.format(out))Solution for puzzle 2.2: s1d2p2h2t1
The clues on the sides are regular expressions.
Solution for puzzle 2.3: RCFKLIZER
The poem is source code in the language Rockstar.
Remove the first three words "A Rockstar knows," to obtain valid rockstar code. An online interpreter can be found here.
Solution for puzzle 2.4: 4.1925
def rooms():
s = set([1])
while True:
a = min(s)
yield a
s.remove(a)
s.update(a*x for x in [7,11,13])
l = []
for n in rooms():
l.append(n)
if len(l) >= 200:
break
print('Solution for puzzle 3.1: {}'.format(l[-1]))An alternative approach is to generate numbers of the form 7^a * 11^b * 13^c for many combinations of a, b and c and then sorting these numbers.
A third approach (not recommended) is to compute the prime factorization of each number 1,..., and check the condition.
This number sequence is similar to the sequence of regular numbers, also called Hamming numbers.
Solution for puzzle 3.1: 1185367183
import numpy as np
import itertools
d = np.genfromtxt('3_2_delivery_service.csv',
dtype = int,
delimiter = ',',
missing_values = '-',
filling_values = -1)
stops = [1,1,2,2,3,3,4,4,5,5]
s = set()
for order in itertools.permutations(stops):
s.add(order)
def process(raw_order):
is_loaded = np.zeros(5,bool)
order = []
for i in raw_order:
if not is_loaded[i-1]:
order.append(i)
else:
order.append(i+5)
is_loaded[i-1] = not is_loaded[i-1]
load = sum(is_loaded)
if load > 3:
return False, None
return True, tuple(order)
s2 = set()
for x in s:
success, o = process(x)
if success:
s2.add(o)
best_length = np.inf
best_order = None
for order in s2:
l = [0] + list(order) + [11]
length = np.sum(d[l[:-1],l[1:]])
if length < best_length:
best_length = length
best_order = l
stoplist = ['-', 'r1', 'r2', 'r3', 'r4', 'r5', 'c1', 'c2', 'c3', 'c4', 'c5', '-']
out = ''.join([stoplist[i] for i in best_order[1:-1]])
print('Solution for puzzle 3.2: {}'.format(out))Solution for puzzle 3.2: r2c2r1r5r4c4r3c5c1c3
(Cable lengths: 78, 38 and 11)
Solution strategies: This puzzle is a variant of a "Slither link" puzzle (German: Rundweg). Wikipedia has quiet a list of solution strategies. In this puzzle, though, we cannot use any techniques that avoid closed loops. Instead at some points, we have to make sure the right cable goes into the right plug.
Solution for puzzle 3.3: 783811
The text refers to the image on the challenge's 404-page, for example, here: (https://coding-challenge.festo.com/anything). You can access it, for example, by reading through the css-stylesheet. Its url is https://coding-challenge.festo.com/images/cybersecurity.png
If you check the alpha-channel of the image, you will find some text that tells you to check the frequencies (fft2). Computing the fft2 of the image, reveals the solution code.
import numpy as np
import matplotlib.pyplot as plt
img = plt.imread('cybersecurity.png')
img_gray = img.mean(axis=2)
img_freq = np.abs(np.fft.fft2(img_gray))
plt.imshow(np.log(img_freq), cmap='Blues_r')This will reveal this image:
Instead of taking the mean over the four channels (r,g,b,alpha), any single channel can also be used.
Solution for puzzle 3.4: Top!