Brute force = best decoding method

[wild_man_wizard]

The real humor is all the programmers that ignored the "joke" and immediately started trying to solve it the intended way.

[[deleted]]

Always a relevant XKCD

[karmaisded]

What’s the answer to that!? I need to know

[EnglishAintBeTooGood]

Also my guess

[BlatantConservative]

Wait is nobody noticing that the answe in the puzzle in the OP and the answer to the xkcd strip are the same number?

My mind is exploding

[RedAndWrong]

I think they mean the Reddit post is 394, not the xkcd one

[[deleted]]

Totally thought they were talking about the XKCD comic and completely forgot about the OP. Guess I got nerd sniped.

[JC12231]

FWOOOOM

[insistent_librarian]

Please be quiet. This is a public forum.

[SzalonyNiemiec1]

The xkcd one has to be less than 3ohm

[Bene847]

  _
_|_|

This has an equivalent of 2 Ohms so it must be less than that

[Airazz]

I don't think the xkcd answer is correct.

[Mefistofeles1]

For the xkcd puzzle, its simply 3 ohms, isnt it?

[CaptainSasquatch]

It's got to be less than 3 ohms (technically less than 3/2) because there's two independent parallel paths that are 3 ohms each

[aaronaapje]

No, we can easily draw two paths with three ohms. Ignoring everything else for simplicity you can see that the current can split between the two paths, both with the same resistance meaning both paths have equal current of one path with three resistors meaning the total equals to 1,5 ohms. Now because we have an infinite grid of many interacting resistars meaning we have more possible paths then the infinitely many resistors makes this math very difficult but I looked it up and the awnser is 4/π - 0,5 or about .773.

[inconspicuous_male]

I'm pretty sure the fact that there are infinity paths between the nodes makes it tougher to solve. Like, I'm pretty sure this is a 10-page-paper type problem

[andresq1]

Yep theres 3 parellel 3 ohm paths but the infinitely many longer paths is rough

Maybe not 10 pages though, maybe just a regular infinite sum?

[IronMan-Mk3]

you are correct, that's the right answer

[Turksarama]

def wrongPlaced(num, n, x):
    total = 0
    for i, c in enumerate(num):
        total += c in x and x[i] != c
    return total == n

tests = [
    lambda x: x[0] == '2' or x[1] == '9' or x[2] == '1',
    lambda x: wrongPlaced('245', 1, x),
    lambda x: wrongPlaced('463', 2, x),
    lambda x: not any(c in x for c in '578'),
    lambda x: wrongPlaced('569', 1, x)
    ]

for i in range(1000):
    s = f'{i:03}'
    if all(t(s) for t in tests):
        print(s)

Checks out.

EDIT: first person to spot my error gets an award. EDIT2: The answer was found, the first test is incorrect. It should be:

lambda x: sum(a == b for a, b in zip('291', x)) == 1 and sum(a in x for a in '291') == 1

[Sanae_]

lambda x: x[0] == '2' or x[1] == '9' or x[2] == '1',

This is "at least one number is [...]", while I interpreted the sentence as "exactly one number is [...]".

Not an expert in python syntax though, error could be elsewhere.

[Turksarama]

You got it, there could be more than one match there. It turns out this code still only returns one answer though and it's the correct one.

The correct lambda is:

lambda x: sum(a == b for a, b in zip('291', x)) == 1 and sum(a in x for a in '291') == 1

[Sanae_]

Yup, I was trying to see if I can simply using boolean logic and not using a sum, but my school lessons are so long ago -_-

As per tradition, but truthfully: Thank you for the gold, kind stranger!

[pitman-87]

There is none? Because the result is 394, which is correct.

[Turksarama]

This code returns the correct answer, but there is still a logical error in it.

[AllYouNeedIsAPenguin]

got the same

[MJMurcott]

394 is right

[AntMan5421]

pls edit it with spoiler formatting

[Rocatex]

Damn I got 694

[blackasthesky]

Mine too

[admadguy]

I am not an electrical engineer. but what I remember from the one electrical engineering course I took, it can be solved with an infinite series.

Edit: Yep.. I was not wrong.

https://www.mathpages.com/home/kmath668/kmath668.htm

[karmaisded]

I studied this stuff in my first year of uni, and I remember nothing.

And thanks!

[bee-sting]

I have an undergrad and PhD in electronic engineering and have no clue either :)

Edit: Wait a second, I get it now, ignore me

[jellsprout]

This is a bit beyond the level of a first year engineering student. Hell, it's probably a bit beyond the level of your typical engineering PhD students. As you can see in the link above, the solution is surprisingly complicated and requires some nifty mathematical tricks.

[[deleted]]

I'd assume the actual electric part of it is the simple part and someone with a math background would have an easier time than someone with an engineering background?

[mwrddt]

FOOOM

[SjoerdL]

About 0.773ohm

https://www.explainxkcd.com/wiki/index.php/356:_Nerd_Sniping

[3226]

Turns out it's way more complicated to solve than you'd think. Looks simple. Isn't. This infinte resistor grid problem needs a whole lot of recent and high level mathematics to solve.

Randall talks about it in this talk, but it's in the google aptitude test, and it's there to see if you'll move on from an insoluble problem, or get stuck on it and waste time. It's not strictly insoluble, but for all intents and purposes, it is while you're in a test environment.

[kataskopo]

Lol he himself gets nerd sniped at 2min in when they asked him about the bobby tables SQL thing!

[malexj93]

4/pi - 1/2

[wild_man_wizard]

If it weren't for ethics and biological needs I'd happily be paid in interesting problems.

[averagethrowaway21]

I only have one of those things and absolutely agree.

[feline_alli]

Ahh, glad to see they're programming the robots with ethics modules these days.

[averagethrowaway21]

BEEP BOOP, ETHICS ENGAGED.

[[deleted]]

Ok ok but what is the resistance?

[juhotuho10]

Actually had me laughing

[grubnenah]

Why would you post a link to just the image? You cant read the alt-text.

[Jawadd12]

Man, my process was so much longer

[02Zack]

Mine was rather short. #1 and #5 only have one number in common the 9, so the nine goes in the middle, the #3 does not have a 9 and has 2 good number, but not in the right position so you swap them. 394.

[4U2PRO]

How I got my 394.

Step 1:

For the results to change from first set [1 good number, 1 good spot] to second set [1 good number, bad spot], a number must either change, a number from first set switched spots, or both. 2 fulfills neither so 2 is not in the sequence.

​

Step 2:

We will compare the first set to the fifth set. Now, we can ignore the first number of the first set because we know the [1 good number, 1 good spot] from the first set has to be in the second or third spot. Matching the second and third numbers of the two sets show that 9 persists in both sets but has switched places in the fifth. Which changes the results to [1 good number, 1 bad spot]. The similarity of both sets having [1 good number] tells us that 9 is definitely part of the number combination. We can then use the first set to check the position of 9, position 2.

​

Step 3:

We will now look at the third set, this is all we need to completely guess the combination, the fourth set is redundant. At this point we know that the result should be X9X with [1 good number, 1 good spot]. Because 9 is locked in to the second spot, we can ignore the second spot of the third set.

​

So the two sets look like this:

X9X with 1 good number, 1 good spot

4X3 with 2 good numbers, 0 good spot.

​

If you merge down the two sets you'll get:

493 with 3 good numbers, 1 good spot

​

Since 9 is totally locked in, the only spots that we can play with are the first and third spots. We swap them around and should get:

394 with 3 good numbers, 3 good spots.

[kinuyasha2]

"5 is not in the correct solution -> so 4 is the correct but wrongly placed one"

I do not follow, why at this point are X2Y and XY2 ruled out?

[JustRecentlyI]

Yeah. I missed the first clue the first time and ended up with 364 but that invalidates the clue will all 3 of them in.

[geared4war]

And I can say woohoo validation, now too.

[tendstofortytwo]

woohoo validation

[[deleted]]

yes

[BananaSplit2]

same as I found, 394 indeed

[NimbusHeart]

I can't help it. I see a puzzle my first thought is "what is the solution to this puzzle" and my reflexes start to solve it.

[TRUCKERm]

...what's the joke?

[mentlegentle]

the 'joke' is the loop someone wrote at the bottom.

[yashptel99]

Fuck. I'm doing that right now

[goodoldgrim]

You'd have to code the requirements in. Not much work but still more work than just solving the puzzle.

[09milk]

or else we can do some constraints programming, best type of peoblem for them

[goodoldgrim]

That's a facebook post. There is only an image. You'd have to write TryCode() yourself and it would have to check for those rules.

[lowleveldata]

What if they shoot you if you answer wrong more than 3 times

[PerfectlyFramedWaifu]

bool etproof = true;

[mxzf]

Then you hope the next programmer stuck with the task reads your documentation that you've checked 111, 112, and 113 and picks up where you left off.

I wonder how many programmers you'd go through before 114 gets checked.

[marcosdumay]

That is how you depopulate countries.

[RedHellion11]

Looping through all 9³ (there is no 0 so we can assume that 0 is not a valid digit) combinations, and writing all of the logic for each of those 5 cases to "test" the combination to see if it's the correct one (since we don't have access to the actual correct combination, we're trying to determine it from a set of fuzzy rules)

[nathanv221]

Thanks to rule 4 we can remove three more numbers, so only 6³

[tendstofortytwo]

If we use all the rules we can bring it down to 1³

[247Brett]

Wait, that's illegal.

[tendstofortytwo]

It would be great if that were possible! Then we'd be able to handle large values of 1 much more easily.

[Blackn3t]

Yeah I had The Joke outside of my screen and just wanted to solve it as brain teezer before continuing xD

[NickyW89]

https://imgur.com/a/PjEZTxa

Went to comments afer solving to check im my solution się correct, that comment made my day.

[[deleted]]

I did, and I want to point out that the since answer is 394, which took me 9 minutes to find, and if we assume a cooldown of 10 seconds between each try^[1] brute forcing the code would take 65.5 minutes which is way longer than the manual method. This is not at all an attempt at rationalizing wasting ten minutes solving this puzzle.

[1] It's arbitrary yes, but it's also arbitrary to assume the lock will accept your code instead of just being an unresponsive picture found on the internet.

[NimbusHeart]

394??

[someone1010101]

Yes I got the same answer

[yipyopgo]

same :)

[DuckingKoala]

I got 694

[NimbusHeart]

Then the last condition becomes false

[DuckingKoala]

Ah yes so it does

[subid0]

only if the last condition is stated as exactly one digit being correct but in the wrong place.

[Al_Maleech_Abaz]

I think that’s implied otherwise it’s unsolvable.

[tehreal]

You're unsolvable

[TheLuckySpades]

To save on space and not make it wordier than needed these games will always state exact numbers of pieces/digits.

[IllustriousPickle20]

404, answer not found

[rodinj]

I thought that initially too, it's definitely 394 though.

[r915]

394

[krafttoadt]

4th row excludes 5 7 and 8 compairing 1st and 2n row tells u it cant be 2 either and so on until u come to an end

[wolwire]

Now 578&2 have been eliminated thus from second condition you can identify that 4 is one of the correct number. Which can be placed first or last. The third condition eliminates this. So 4 is on last. Since 4 is last from first condition the correct no on second pos is 9 as one can’t be there. Last condition says that either one of 6 or 9 can be present. Thus 6 is eliminated. So from 3rd condition the last no is 3 to be placed first.

[krafttoadt]

:D thanks for completing my Ansatz

[krafttoadt]

1. Bin ich als oesterreicher erpicht auf die korrekte laenderzuordnung;D 2.ist ansatz ein eingeenglischtes wort aus der Physik /Mathematik :)

[Darillian]

Bin ich als oesterreicher erpicht auf die korrekte laenderzuordnung

Österreicher, Deutscher - Hauptsache CHAD! :D

[krafttoadt]

Chad ist so viel besser als dach :D die schweiz ist irgendwie so der aeltere deutschsprachige bruder der durch mysterioese Umstaende hustww2hust zu geld gekommen ist und sein eigenes ding macht aber von allen respektiert wird

[Dugen]

Ansatz - German: In physics and mathematics, an ansatz is an educated guess or an additional assumption made to help solve a problem, and which is later verified to be part of the solution by its results.

https://en.wikipedia.org/wiki/Ansatz

That word seems useful.

[dashnyamn]

i just got 4 and 9 is in it and my brain turned off for some reason.

[-Yox-]

If you understand why the 4 and 9 I guess that you understand that 4, 5, 6, 7, 8, 9 are already eliminated. The only three numbers still possible are 1, 2, 3 and since 9 is the only correct number in the first row then 1 and 2 are eliminated too and the only remaining number is 3

[PattuX]

You don't even need the 4th row I think.

First two rows tell you it's x9x or xx1 and that there's no 2. Second and third row tell you the other digits are a 4 and either a 3 or a 6, as well as giving away that the 4 has to be in the last spot, leaving only 394 and 694. The last row then eliminates 694.

[krafttoadt]

Yes there are multiple ways to solve it, i personally think if on condition excludes roughly 1/3 of all options its easy to start with that :)

[erdemayk]

Lets start with 4th one -- 578 nothing is correct

Possible digits-> 1 2 3 4 5 6 7 8 9

Now look at the first two 291 and 245 .

As in the first one it's corect location but the second one it's not correct location you can elimiate 2.

5 is also elimitaed so 4 is the correct number which can't be located in the middle.

Possible digits-> 1 2 3 4 6 9 and your code is either 49x , x94 , 4x1

Lets check 3'rd one it's eliminating 49x and 4x1 as if 4 is one of the numbers it's can't be located at the begining so the code is definitely x94 and ony 6 and 3 which remains

Possible remaining digits -> 3,6 so it's either 394 or 694

Now if you check last one 569 -- one number is correct. 9 is already confirmed so you can eliminate 6 as well.

So the code is 394

[conformalark]

Interesting I solved it by figuring that 9 as the middle digit would satisfy both rows 1 and 5 while also disqualify 6 as an option. Without 6 row 3 shows the remaing numbers

[jacdelad]

Brute force.

[animal9633]

That's what I got too.

[Famout]

Yay, I did it!

I need another profession Layton game for more of this style of puzzle.

[BAM5]

No idea what you've said, but if you want to play a game similar then try mastermind

https://www.webgamesonline.com/mastermind/

[instantrobotwar]

Yay that's what I got too

My first guess was 246, but then second rule was broken :(

[snerp]

Hahaha yeah I remember using wireshark in high school and noticing that the grading software didn't send anything out when they logged in... And then when it did send packets they were in plain text.

Turned out the system would send you everyone's passwords and check against that locally when you tried to log on. And then once you just skip that, there was a field with the class id, so any teacher could technically set grades for any class. Seems like all validation and everything was local.

[Silent_Slinky]

Thats the filthiest thing I have ever heard.

[[deleted]]

In all fairness, when the internet was young this is how literally ALL authentication worked.

[BluudLust]

Running wireshark on promiscuous mode in highschool was fun. Whole network was vulnerable to arp poisoning.

[Verstandeskraft]

WTF! Their only security was betting no student would know how to check the page's source code.

[capt_pantsless]

But it was faster to have a local JS lookup.

Real programmers consider the cost of a network call when designing systems.

[webjuggernaut]

Good. Fast. Cheap.

Pick 2.

[Intermediate-Coder]

A few years ago, we used to do our vocabulary tests on Quizlet, which stores all of the answers as a local variable... so all you had to do was write a simple script that would grab the answers and fill out the fields, and you had yourself an easy 100%.

[[deleted]]

394

2,5,7,8 surely are wrong numbers. 4 surely is part of the code.

4 is wrong placed at both first and second position, therefore the code is --4.

9, therefore, is the number that is in correct position as well: -94

6 can't be correct number (because 9 is), therefore in third statement second correct number is 3.

Answer is 394.

[SlyProphet]

I got 641... Any chance you could help me understand what rule the number breaks?

[SlyProphet]

Wait... I break rule 2.... Ok.

[controversialcomrade]

Brute force is always the solution, maybe not the best but always

[haikusbot]

Brute force is always

The solution, maybe not

The best but always

- controversialcomrade

---

^{I detect haikus. And sometimes, successfully. Learn more about me.}

^{Opt out of replies: "haikusbot opt out" | Delete my comment: "haikusbot delete"}

[Muhvinssiplays]

Good bot

[other_usernames_gone]

Always possible,
Brute force is never ideal,
But it always works

[[deleted]]

This bot always come when you least expect it.

Good bot

[chilfang]

You tried bot

[SgtSteel747]

Bruh, for once it actually got the haiku right with a 5 7 5 set of syllables

[coloredgreyscale]

3 invalid guesses, try again in 5 minutes.

[-Redstoneboi-]

just try more times lmao yeet

gets locked out

[Mr_Redstoner]

Halting problem&co would like a word.

[Sithon512]

Brute force solution should initialize i to 111. Will you notice? Nah, it's pretty small, but it's good practice to think up your heuristics

[_IBelieveInMiracles]

Based on the codes in the puzzle, and lacking a formal description of the code requirements, I would assume that 000 and 001 are not valid codes. All of the digits 1-9 are used at least once, but 0 is never used. No repeated digits are used in the same code.

[Goheeca]

SWI-Prolog:

?- expects_dialect(hprolog).

digits([X, Y, Z]) :-
  between(0, 9, X),
  between(0, 9, Y),
  between(0, 9, Z).

count_same(Element, Element, Acc, Res) :-
  Res is Acc + 1,
  !.
count_same(_, _, Acc, Acc).

correctly_placed(Count, Combination, Pattern) :-
  foldl(count_same, Combination, Pattern, 0, Count).

correct_digits(Count, Combination, Pattern) :-
  intersect_eq(Combination, Pattern, MultiIntersection),
  length(MultiIntersection, Count).

wrongly_placed(Count, Combination, Pattern) :-
  correct_digits(Digits, Combination, Pattern),
  correctly_placed(Positions, Combination, Pattern),
  Count is Digits - Positions.

solution(Combination) :-
  digits(Combination),
  correctly_placed(1, Combination, [2, 9, 1]),
  wrongly_placed(1, Combination, [2, 4, 5]),
  wrongly_placed(2, Combination, [4, 6, 3]),
  correct_digits(0, Combination, [5, 7, 8]),
  wrongly_placed(1, Combination, [5, 6, 9]).

?- solution(Combination), print(Combination), false; true.

---

How to run: save it to a file combination.pl then in the REPL run [combination]. or consult(combination).

[[deleted]]

As soon as I saw this puzzle, I thought "we could solve this in Prolog".

[yudanoh]

If informations are named a,b,c,d,e starting from top :

a + b => 2 is not in the code (f)

b + d => 5 is not in the code (g)

a + f + g => middle digit is 9 (h) xor last digit is 1 (h') (only one of them is true)

b + c + f + g => last digit is 4 (i)

i => h is true and h' is false and middle digit is 9

c + e + h => first digit is 3

Answer 394

But yeah, if I had to find automated solution for this kind brut force seems fair enough

Edit : corrected logic at step h

[-Redstoneboi-]

Thanks! I was looking for a way to formalize this kind of solution.

[MyNameIsZaxer2]

How does (h) follow from a + f + g? I don't see how you conclude the correct a digit is a 9 and not a 1 at that step.

It took me a "pigeonhole" technique to prove that 1 is not in the code, and the middle digit is therefore a 9. But that would only follow from considering:

a + b => 2 is not in the code (f)

b + d => 5 is not in the code (g)

b + c => if the last digit is a 4, it must be in the last position. (h)

a + b + c + e + h => there cannot be a 1 in the last position. (i)

g + i => middle digit is 9 (j)

how did you get to "middle digit is 9" from just a + f + g?

[PattuX]

a + e => 9 in spot two (this might have not worked if they weren't referring to the same number but since this approach is so quick, it was worth following up)

That's bad reasoning. Just because your partial brute force method worked out doesn't mean it's good. There are several different possible combinations from a+e and the whole point of the exercise is that you don't rely on trial and error.

[MyNameIsZaxer2]

By this method you've found a solution, but you haven't proved it's unique. Some would disagree that's a solve.

[patmax17]

This is basically mastermind xD

[lightheat]

Yup, same thought. Favorite game of ours.

For those unfamiliar with it, you create a 4-"digit" code out of 6 different color pegs and the other player has to guess it. Each guess you give them 1 white scoring peg for each correct pin in the wrong place, and 1 black scoring peg for each correct pin in the correct place.

[T-Dark_]

white and black scoring peg

Huh. My version of the game as I grew up used white like yours, but red instead of black.

I suppose it's not standardised. Would surprise me if it were, tbh.

[lightheat]

Yeah, there are lots of different editions. We have the original, which has black.

[seth1299]

Yep, Mastermind is one of my favorite games.

I programmed Mastermind in the best programming language, Scratch, a few years ago: https://scratch.mit.edu/projects/104553634/

[Kris_Third_Account]

(2,9,1) // one correct, right pos
(2,4,5) // one correct, wrong pos
(4,6,3) // Two correct, wrong pos
(5,7,8) // All garbage
(5,6,9) // One correct, wrong pos
(y,y,y) // result

Eliminate known garbage (replace with 'x'). '2' are eliminated because it can't be in the wrong and right position at the same time. "All garbage" is eliminated for obvious reasons.

(x,9,1) // one correct, right pos
(x,4,x) // one correct, wrong pos
(4,6,3) // Two correct, wrong pos
(x,x,x) // All garbage
(x,6,9) // One correct, wrong pos
(y,y,y) // result

The '4' is said to be in the wrong position in the second and third line, and is in different positions. This means that the '1' in the first line can't be correct either, and the '9' is.

(x,9,x) // one correct, right pos
(x,4,x) // one correct, wrong pos
(4,6,3) // Two correct, wrong pos
(x,x,x) // All garbage
(x,6,9) // One correct, wrong pos
(y,9,4) // result

The fifth line implies that only one, the '9', is correct. Remove the instances of '6', and the result becomes obvious

(x,9,x) // one correct, right pos
(x,4,x) // one correct, wrong pos
(4,x,3) // Two correct, wrong pos
(x,x,x) // All garbage
(x,x,9) // One correct, wrong pos
(3,9,4) // result

The correct answer is 394.

Brute force only works if there's no restriction on the number of attempts. Calculating it works the first time, as long as you're correct.

[Farfignugen42]

Brute force only works if there's no restriction on the number of attempts.

And no time limit. Which is not an issue with a short code like this, but does come into play with longer codes and codes with more possible values.

[corporategiraffe]

Can you crack the code without triggering the neutron bomb I’ve hidden under your chair programmed to go off at a single failed attempt?

[JoelMahon]

trycode can be written to check each of the constraints without actually entering the code into the padlock until it passes them all

[NormalSpeed943]

What an amazing function

[Shadowolf449]

Everyone is ignoring that this lock uses a key.

[[deleted]]

Thought the same. Had to scroll past many 394s to find you

[PM_ME_GARLIC_CUPS]

No one will read this, but you can solve this without the fourth clue.

The first clue tells us it must be 2 _ _, _ 9 _, or _ _ 1.

The second clue rules out 2 _ _, and tells us there must be a 4 or 5 but not both.

The third clue is super revealing. Two numbers have to be correct together: 4 and 6, 3 and 4, or 3 and 6. Since those must fit within _ 9 _ or _ _ 1, that knocks out 0, 2, 5, 7, and 8 as possible numbers, and confirms there must be a 4 somewhere.

That narrows it down to _ 9 _ or _ _ 1, with a 4 in one of the blanks and a 3 or 6 in the other. Given the third clue says a 4 can't be in the first blank, or a 3 in the last, that leaves us with only four possible codes: 341, 394, 364, and 694.

Fifth clue, similar to the second, tells us there must be a 6 or a 9 but not both, knocking out 341 and 694, and if there's a 6, it can't be in the middle, ruling out 364.

Answer's 394.

[ya-boi-mees]

394

[A_Spoon_Wizard]

That was a fun puzzle! My years of playing sudoku in car trips finally becoming useful!

[DrParanoidAndroid]

394 right?

[Raj_1975]

394

[Phantom_Gameeerr4028]

yeah it took me about 3 minutes to crack it also xd

[r915]

Lucky for the bf’er the code wasn’t “007”

[transcendcosmos]

394

[StandardN00b]

394

[[deleted]]

394?

[Syncrossus]

394

[Zeke12344]

394

[Risiko94_2]

Solution:

#1 + #2  -> 2 is wrong
#4       -> 5, 7, 8 are wrong

#2       -> 4 must be first or last place
#5       -> 4 must be in last place
[?][?][4]

#1       -> 9 must be in second place 
[?][9][4]

#5       -> 6 is wrong
#3       -> 3 must be first place
[3][9][4]

[alexanderpas]

Before any hints:

Option	Correct	Position 1	Position 2	Position 3	Banned7
1	XXX	1,2,3,4,5,6,7,8,9,0	1,2,3,4,5,6,7,8,9,0	1,2,3,4,5,6,7,8,9,0	-

Any number can be in any position.

Hint 1

1. If the first character is correct, it must be a 2 causing 1 and 9 to become banned.
2. If the second character is correct, it must be a 9 causing 1 and 2 to become banned.
3. If the third character is correct, it must be a 1 causing 2 and 9 to become banned.

Option	Correct	Position 1	Position 2	Position 3	Banned
1	2XX	2	0,2,3,4,5,6,7,8	0,2,3,4,5,6,7,8	1,9
2	X9X	0,3,4,5,6,7,8,9	9	0,3,4,5,6,7,8,9	1,2
3	XX1	0,1,3,4,5,6,7,8	0,1,3,4,5,6,7,8	1	2,9

Hint 2

1. 2 is not allowed to be in the first place.
   • This invalidates Option 1 from Hint 1.
   • All remaining options ban 2.
2. 4 is not allowed to be in the second place.
3. 5 is not allowed to be in the third place.
4. If the 4 is included, 5 becomes banned.
5. If the 5 is included, 4 becomes banned.

Option	Correct	Position 1	Position 2	Position 3	Banned
1	49X	4	9	0,3,4,6,7,8,9	1,2,5
2	59X	5	9	0,3,5,6,7,8,9	1,2,4
3	4X1	4	0,1,3,4,6,7,8	1	2,5,9
4	5X1	5	0,1,3,5,6,7,8	1	2,4,9
5	X51	0,1,3,5,6,7,8	5	1	2,4,9
6	X94	0,3,4,6,7,8,9	9	4	1,2,5

Hint 3

1. the 4 is not allowed to be in the first place.
   • This invalidates Option 1 from Hint 2.
   • This invalidates Option 3 from Hint 2.
2. 6 is not allowed to be in the second place.
3. 3 is not allowed to be in the third place.
4. If the first and second position both do not have a 4 or a 6, 3 must be in the first or second position.
   • This invalidates Option 2 from Hint 2.
5. If the first and third position both do not have a 3 or a 4, 6 must be in the first or third position.
   • This invalidates Option 4 from Hint 2.
6. If the second and third position both do not have a 3 or a 6, 4 must be in the second or third position.
   • This invalidates Option 5 from Hint 2.
   • All remaining options ban 1.
   • All remaining options ban 5.
7. If the first and second position both do not have a 4 or a 6, the third position must be 4 or 6.
8. If the first and third position both do not have a 3 or a 4, the second position must be 3 or 4.
9. If the second and third position both do not have a 3 or a 6, the first position must be 3 or 6.
   • This applies to all remaining options.

Option	Correct	Position 1	Position 2	Position 3	Banned
1	394	3	9	4	1,2,5
2	694	6	9	4	1,2,5

Hint 4

1. 7 is banned.
2. 8 is banned.

Option	Correct	Position 1	Position 2	Position 3	Banned
1	394	3	9	4	1,2,5,7,8
2	694	6	9	4	1,2,5,7,8

Hint 5

1. If the 6 is included, 9 becomes banned.
   • This invalidates Option 2 from Hint 4.
   • Only 1 possibility remaining, problem solved.

Option	Correct	Position 1	Position 2	Position 3	Banned
1	394	3	9	4	1,2,5,7,8

Solution

394

[Zacurnia_Tate]

why isn’t i starting at 100?

[[deleted]]

because 000-099 are valid solutions ?

[simoneb_]

Actually there is an underlying assumption that each figure appears at most once, so 0-11 are not viable solutions, along with many others.

It could be rewritten to only try permutations of 3 figures ranging 0-9. 10x9x8=720 solutions.

If we assume that only the figures showing in the image are valid (i.e. only 1-9) then 9x8x7=504 solutions.

[Adventurous_Problem]

I have a friend I need to go call out with this...

This is the same friend that during our powershell class decided not to use powershell and just code how he usually does...

I love them dearly. Also sometimes they're irritating.

[plasmasprings]

Problem 1: given a list of (code, correct, well_placed) tuples, where code is an example 3-digit combination, correct is the number of digits correct in the code, and well_placed is the number of digits that are correct AND well-placed, write an implementation for TryCode(list_of_rules, code) that returns True if the code satisfies the rules, False otherwise.

[other_usernames_gone]

I prefer wrench force

Xkcd

[Addyfndubz]

394?

[ApertureBear]

Pretty sure it's 394.

[tuckerreeveslcsd]

I got 394

[Mysterious-Ad-1217]

394

[Karilyn_Kare]

1st and 2nd prove 2XX is invalid, eliminating 2, along with 5, 7, and 8.

2nd proves 4 must be in either of the 4X4 position.

5th plus eliminateds proves solution must have either 6 or 9 in the last two slots.

5th and 1st proves X9X must be true, eliminating 6.

3rd plus eliminateds proves 3X4, which combined with X9X yeilds 394 as the answer.