How To Open Abandoned Crates: The Multitool Way

[Pckables]

So you're sitting around Cargo doing your usual thing, when suddenly a miner walks in with this weird locked crate that they don't know how to open. With it's 4-digit lock, how could anyone hope to find out what goodies are inside?!
There are two ways to open these Deca-lock abandoned crates. One way is to blow it open; either with a gun or explosion. This destroys some of the crate's contents, which could be REALLY valuable (or not).

The other method, is actually solving the lock.

What You Need to Know:

Abandoned Crates have a randomized 4 digit code, digits 0-9, and you have 10 tries to guess the correct one before it (harmlessly) explodes, leaving nothing behind.
Hitting the crate with an ID, PDA, or melee weapon will also cause it to destruct instantly, so don't do that by accident.

There are no repeat digits.
Every digit in the code is unique; you will never have two of the same number in a crate's code. Use this knowledge to your advantage.

You can back out of a code input attempt by submitting an empty guess.
If you accidentally click on the crate, prompting a code entry, just leave it blank and submit, and you will instead leave the crate alone. Every attempt is precious.

You need a multitool.
When you hit an abandoned crate with a multitool, it tells you two things: How many attempts left before the crate is destroyed, and how close your last code input was to the real code.
You are only shown the MOST RECENT attempt though, so do not forget to use the multitool after every input, or else you learn nothing.

Write things down.
Use your IC notes, your PDA (make sure you don't smack the crate after taking notes), or even just a paper and pencil, but write down the results of every input (shortened preferably) or else you'll forget.

What To Do:

Let's get crackin'. First order of business is to throw away that first crate the miners gave you because they probably used up 4 of the 10 guesses on random things.
Once you get your hands on a crate with at least 9 attempts intact, as well as a multitool and some sort of note-taker, find a nice spot to sit down in for the next 5 minutes or so.

What you want to be taking note of is the code you inputted, how many correct numbers were in the incorrect place, and how many correct numbers were in the correct place of every attempt.
For simplicity's sake, you should write it down as this:
[input] [#]c [#]i  - Input being your guess, c being the number of correct digits in the correct place, and i being the number of correct digits in the incorrect place. Remember that c and i will add up to the total amount of correct digits you have in that attempt. 1c and 1i means 2 correct digits, just that one is in the correct spot and one isn't.

The method detailed in this guide is what I call the "Snake" method, because you slide up one number at a time and use the head and tail to find out which digits are part of the code.

Start with the code 0123. The crate will tell you it's wrong unless you're the luckiest person in the world. Then use the multitool to found out how many digits were in the correct and incorrect places, write it down, and move on to the code 1234. Repeat to 2345, 3456, all the way to 5678 (You can usually skip 6789, as you can deduce when theres a 9 in the code when you've only found 3 digits.)

Once you have everything written down, it's time for the Deduction phase. See where the amount of digits increases or decreases, and that will tell you what numbers are in the code.
Then you can go back and see when these numbers are in the correct position, and narrow it down to 1 or 2 possible answers.

Here is an easy example:

 

0123 0c 2i
1234 1c 2i
2345 0c 2i
3456 0c 3i
4567 1c 1i
5678 1c 0i

First things first: Lets find the numbers.
We can see that the total of correct digits increased with 1234, meaning 4 has to be a correct digit, as it is the only new digit introduced.
We see a decrease in 2345, meaning 1 has to be correct digit, as the total decreased as it left the code.
Another increase in 3456 tells us 6 is one of the correct digits
And a decrease in 4567 tells us that 3 is our last missing part of the correct code.
Our digits are 1,3,4, and 6, but not in that order. Let's now try to deduce the positions.

We can see that three of the four digits were in the correct place at some point in the code, as seen by the total amount of c's across all attempts.
Let's start with the easiest. We can see that in 5678, 6 is the only correct digit, and its also in the correct place, meaning 6 is the second digit.
Then, knowing where 6 is supposed to go, that means that the correct digit in the correct place in 4567 must be 4. This is the first digit.
Now the tricky part: We have three digits in 1234, and only one of them is in the correct spot. We know that it can't be 4, as that goes into the first spot. It can't be 1, as we know that 4 is supposed to be in the first spot. This only leaves use with 3, being in the third spot.
Now we know the code is 4-6-3-?. As we have gathered, the last number can only be 1.

The correct code is 4631

Alright, now let's go for a much harder example, this one from an actual abandoned crate.

0123 0c 2i
1234 0c 2i
2345 1c 1i
3456 0c 2i
4567 0c 2i
5678 2c 0i
6789 0c 1i

Note that I decided to check 6789. This is not checking for 9, as it can be deduced that there are none, but to check if 5 was correct.
The difficulty here is that there is no increase or decrease in totals except in 6789. This is because there are correct digits that are 4 numbers apart; be wary of these.

The only thing we can deduce at the start here is that 5 is a correct digit, as the total decreased with 6789.
Knowing this, and the fact there is no increase or decrease in total anywhere else, another correct digit must be exactly four less: So 1 is also a correct digit.
Now we have to find the other two numbers with very minimal clues.
Based on the notes, our options are 0,2,3 and 6,7,8,9. We can deduce that the two missing digits also have a difference of four, else we would see a change in total somewhere. This rules out 0, 8, and 9, as they do not have partners with a difference of 4 that would work in this instance. (To test numbers, imagine the code in your head and go through ALL your attempts, seeing if the totals line up).
So our numbers are either 1,5,2,6 or 1,5,3,7.
How do we figure out which one is correct? The secret is in the correct placements.

Say we assume the correct numbers are 1,5,2,6 but not in that order. 5678 would tell us that 5 would have to be the first number, and 6 the second number.
HOWEVER: Our attempt on 2345 would contradict this, saying that either 2 is correct in the first position, or 5 is correct in the last.
They can't both be correct, meaning that this isn't the correct combination of numbers.
We are only left with 1,5,3,7.

5678 tells us that 5 is first and 7 is third.
2345 tells us that 3 must be second, as we already know where the 5 goes.
The 1 has only one spot left to take.

The correct code is 5371

Opening abandoned crates only takes 3-5 minutes on some of the easier codes. Tougher ones can really twist your brain, but I think that makes it more fun.

Sure you can blow them open, but you never know when you destroy that sweet energy cutlass or compact defibrillator that could have been gotten by finding the code.

Edited March 24, 2018 by Pckables
Minor Cleanup

[IK3I]

I tend to use a different route for my solutions. It's based off of reducing the space of possible digits, then testing conditionals

 

1. check 0123 and 4567

This splits the space into up to three groups. We know how many from each set by how many show up in each answer

2. if more than one group is used, check its proportion of the total against other groups

so if you have 2 from 0-3, 1 from 4-7 and 1 from 8-9, then you should pick two from the first group, and one from the other groups

3. This divides your groups into grids of potential digits for each position and group. You should strive to make each continued guess eliminate as many items from the grid as possible

 

This can get rather complex, so I'll just show an example of how can you put this into use

CODE TO SOLVE:  1376

A. 0123 0 2
B. 4567 0 2

• Code contains 2 from first set, two from second set, and neither 8 or 9

C. 1045 1 0

• We now know that 3 of the four correct digits are on the right side of the initial guess group
• We can determine which is the correct one by working through permutations moving positions and numbers.
• Any guess from here on will have a minimum of 2 correct digits

D. 3026 1 1

• Because we only have two correct digits, we know that the digit from the first group must be wrong. This follows from the fact that the second group can only ever contribute one wrong digit, which is has.
  • Therefore, we should work our way through the first set, and pick another permutation of the second set maintaining one position from the previous guess.
• We can deduce something further about the fourth digit in particular here.
  • It has 5 possible digits, as C tells us the only first group answer is 5, we can already see that neither 3 or seven are correct in the second group bringing us to the following three possibilities: 2,5,6
    • If 2 is the correct digit, then it follows that 3 must be in the correct position in D and 6 must not be in the code.
    • If 5 is the correct digit, then it follows that 3 is in the correct position in D due to A and either 2 or 6 is not in the code.
    • If 6 is the correct digit, then it follows that either 2 or 3 is not in the code and either 1 or 4 is correct
• We should now seek to eliminate each of these options by testing as many as possible at once

E. 1635 1 2

• We've just tested all of our statements with this sequence.
• If 2 is correct, then 3 would have to be in position one. We know that either the 1 or 5 must be correct as we have three correct digits with two selections from the first group If the one is correct, then 2 is not in the last position. If the five is correct, then 2 is not in the last postion.
  • Therefore 2 is not in the last position
• If 5 is correct, then 3 must be in position 1.
  • We know that 3 and 6 are also in the code.
    • We can see that 6 has been in all positions except one, but has never been in the correct position.
    • Therefore, it must be in position one.
  • Two numbers can not be in position one, therefore, five is incorrect
• If 6 is correct, then no issues arise and we can deduce the following:
  • 1 is correct
  • 6 is in the fourth position
  • 3 is in the second position as its the only position is has not been in while also never being in a correct position
  • 7 must be in position three as we know that the 2 isn't in the code from D

F. 1376
Done

Edited March 24, 2018 by IK3I

[Pckables]

I encourage everyone with their own working strategies to post them in here, so that others looking for help can choose a method that clicks well with them.

Those looking for practice: You can always take four numbers from a random number generator (making sure you don't get repeat numbers) and work with that as the hidden code. You'll have to do a bit of pretending, acting like you don't know the real code, but the goal is figuring out the process rather than getting to the end.

[IK3I]

well this is just 4x10 mastermind, so I'm sure you can find an inbrowser clone if you want to test yourself

[Taaketa]

I get a KA fully upgraded and blast the crate for 40 seconds.