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Faster Lock Combination

Difficulty:

Description:

Over on Steampunk Island, Bow Ninecandle is having trouble opening a padlock. Do some research and see if you can help open it!

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Solution

Solution

For this particular lock, there is a method to finding the correct combination. The algorithm is a bit confusing at first. I do recommend watching the video recommended by the hint:

[198] Close Up On How To Decode A Dial Combination Lock In 8 Attempts Or Less

I definitely found the youtube video a bit confusing as it wasn't explained all that clearly. I found a PDF that is way better at explaining it for me. I could not find who made this PDF to give them thanks but here it is:

How_To_Decode_Dial_Combo_Locks.pdf

Once you have the sticky number and the guess numbers you can narrow down the guesses to 8 guesses maximum.

I will not repeat the algorithm here as its clearly laid out in the PDF, but if you follow the algorithm it will allow you to guess the combo very quickly.

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Script It!

Script It!

For the very lazy, I created a python script to do the math for you. You just have to enter the sticky number and the 2 guess numbers and it will spit out the possible combos for you.

#Find the absolute shortest distance between 2 numbers with a modulus
def abs_distance(n1, n2, mod):
    d = abs((n1+mod)-(n2+mod))%mod
    return min(d, mod-d)

sticky_number = int(input("Enter the sticky number:"))
guess_number1 = int(input("Enter the first guess number:"))
guess_number2 = int(input("Enter the second guess number:"))

#first number is sticky number + 5
first_number = sticky_number + 5
sticky_remainder = first_number % 4
third_possibilities = [guess_number1, guess_number2]

for i in range(10,40,10):
    n = (guess_number1+i)%40
    third_possibilities.append(n)
      
for i in range(10,40,10):
    n = (guess_number2+i)%40
    third_possibilities.append(n)

third_numbers = []
for possibility in third_possibilities:
    if possibility % 4 == sticky_remainder:
        third_numbers.append(possibility)

input_str = "Which 3rd possibility [" + ','.join(str(x) for x in third_numbers) + "] is correct (looser)? " 
third_number = int(input(input_str))
sticky_remainder = sticky_remainder + 2
second_possibilities = [sticky_remainder]

for i in range(8,40,8):
    n = (sticky_remainder+i)%40
    second_possibilities.append(n)
    
sticky_remainder = sticky_remainder + 4
second_possibilities.append(sticky_remainder)

for i in range(8,40,8):
    n = (sticky_remainder+i)%40
    second_possibilities.append(n)

print(second_possibilities)
second_numbers = []

for possibility in second_possibilities:
    if abs_distance(third_number,possibility, 40) > 2:
        second_numbers.append(possibility)

print("Possible Combinations:")
for n in second_numbers:
    print(f"({first_number},{n},{third_number})")

Enter the sticky number:17
Enter the first guess number:8
Enter the second guess number:1
Which 3rd possibility [18,38] is correct (looser)? 18

Possible Combinations:
(22,4,18)
(22,12,18)
(22,28,18)
(22,36,18)
(22,8,18)
(22,24,18)
(22,32,18)
(22,0,18)

The script output the 8 possibilities. Turns out it was the 5th guess (22,8,18).

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Cheating

Cheating

If you look in the javascript console, there is a variable called "lock_numbers". The correct combination is in there.

In the above example, enter the numbers 34, 28, 14, then pull the lock down to unlock it.