all 20 comments

sorted by: hot top controversial new old
[–] 50 points 10 months ago

She didn't ask for the expiration date

  • source
  • [–] 27 points 10 months ago (3 children)

    This is funny and all but first of all the guards could just say I don't know and they wouldn't even be lying.

  • source
  • hideshow 3 child comments
  • [–] 21 points 10 months ago (2 children)

    Look at this guy, not even spending 20 minutes to memorize his credit card numbers to impulse buy even easier!

  • source
  • parent
  • hideshow 2 child comments
  • [–] 7 points 10 months ago

    Ugh my bank keeps calling me and making me validate my identity by rattling off the card number and security code, to warn me that hackers have charged my card absurd amounts! It would be nice if they called before the hackers steal my money. That's how I memorised it.

  • source
  • parent
  • [–] 69 points 10 months ago

    Attention all Fortnite gamers: John Wick is in great danger and he needs your help to wipe out the squads in the Tilted Towers, but to do this he needs a golden scar and a couple of chug jugs. To help him, all he needs is your credit card number, the three digits on the back, and the expiration month and year. But, you gotta be quick, so John Wick can secure the bag and achieve the epic Victory Royal!

  • source
  • [–] 22 points 10 months ago (4 children)

    Workaround: interpret "and" as "plus", and say a single number which is the two summed together.

  • source
  • hideshow 4 child comments
  • [–] 15 points 10 months ago (2 children)

    Nah, the lowest digit of a credit card is a checksum, and the first few digits identify the credit card company, so it's pretty trivial to narrow down the possible options

  • source
  • parent
  • hideshow 2 child comments
  • [–] 7 points 10 months ago (1 child)

    Standard compliance rules generally allow showing the first six and last four digits on a statement or receipt or whatever. That leaves 6 digits, and since you know it has to match the cksum, 1/100,000 odds on straight guess.

    You may be able to knock down two more digits if you know which bank and card type (debit, prepaid, etc) it is. The first 6 digits used to encapsulate that data but there's a transition to 8 now.

  • source
  • parent
  • hideshow 1 child comment
  • [–] 3 points 10 months ago

    Sure, but in this case we have a credit card number summed with a 3-4 digit number (which you don't have to guess the length bc of the first digits of the processor identificatoon). You can try every permutation of 3-4 digit number taken out, and eliminate all the ones that don't have a valid checksum

  • source
  • parent
  • [–] 21 points 10 months ago (2 children)

    In the first frame which guard is lying? Seems like neither.

  • source
  • hideshow 2 child comments
  • [–] 15 points 10 months ago* (last edited 10 months ago) (2 children)

    For those interested, ignoring the contradictory presentation of the riddle (as the knights themselves would not say the riddle since one always lies and one always tells the truth), the solution is simple. Ask the knights what the other knight would answer when asked what door is correct, and they will both say which path not to go to. Thus you pick the path that neither Knight says!

    Logic:

    Liar: Will say the wrong option, as they're being asked which door the truth telling knight would say (and they will lie about what the truth-teller would say)
    ------------------------------------------
    Truth-teller: Will say the wrong option, as they're being asked which door the liar would say (and they'll tell the truth about that)
    

    NOTE: This can be expanded to a case with n doors by asking the knights to provide all the options that the other knight could say, and each will provide n-1 options, so you'd pick the one option that neither knight says. It is possible the liar may not list all options, but the truth-teller would, so the problem could still be worked out regardless (and you'd know which knight is the liar in that case).

  • source
  • hideshow 2 child comments
  • [–] 4 points 10 months ago (1 child)
  • [–] 5 points 10 months ago* (last edited 10 months ago)

    The fun is in discrete logic! There is a whole field of mathematics (discrete mathematics) that analyzes the logic of situations that involve binary (true/false) statements, such as the solution to this riddle.

    The riddle can be summed up as the following:

    Let Y represent a question asked of the knights.
    Let A(x) be a boolean function representing the answer of the truth-teller.
    Let B(x) be a boolean function representing the answer of the liar.
    
    B(x)≡¬A(x)
    
    An answer can only be determined if the knights do not contradict, otherwise it could not be determined which knight was lying, for instance when Y≡(A(Y)∧B(Y))∨(¬A(Y)∧¬B(Y))≡T.
    A trivial question whose answer is logically true would result in the following form:
    
    Y≡(A(Y)∧B(Y))∨(¬A(Y)∧¬B(Y))
    A(Y)≡T
    B(Y)≡F
    Y≡(T∧F)∨(F∧T)≡F∨F≡F
    ∴Y≡F
    
    Such a question would allow identification of which knight is lying, but would not provide information on which door is true.
    
    Let D represent a set of m doors where there is exactly 1 correct door.
    
    ∃!n∈ℕ:D(n)≡T∧n≤m
    
    The player ideally wants to discover which value n is correct so they can choose a door accordingly.
    Let R represent the following question: "What are all the doors the other knight could say are correct?".
    Let S represent all possible answers to the question "What is the correct door?" for each knight respectively.
    Let P be the set representing the value n where D(n)≡T, and Q be the set representing all values n where D(n)≡F,
    
    P⊂D
    P={n}
    Q⊂D
    Q=D-P
    
    Thus the following properties are true:
    
    P∩Q=∅
    P∪Q=D
    
    Consider that the answer S would be either t∈P if a knight is telling the truth or t∈Q if a knight is lying.
    Thus when answering question R:
    
    A(R)≡B(S)≡t∈Q
    B(R)≡¬A(S)≡t∈Q
    
    Since question R asks for all possible answers, both knights would provide an answer equivalent to set Q.
    
    ∴ the correct answer is D-Q; the one door that neither knight says.
    

    I'm a little rusty on my discrete mathematics, as I haven't done it in 3+ years, but hopefully my answer is logically sound! Feel free to correct me.

    EDIT: This original puzzle is Knights and Knaves, so to be true to the original, the "truth-telling knight" is just the knight, and the "liar knight" is a knave. I just presented it this way because there was no mention of it in this comic, so I didn't want to be confusing.

  • source
  • parent
  • [–] 6 points 10 months ago

    Alternative answer: What's in your browser history?

  • source
  • [+] 4 points 10 months ago (1 child)
    15 comments in c/femcelmemes@lemmy.blahaj.zone view that post
    [–] 34 points 10 months ago (6 children)

    Oooh! This is a great solution to this problem! Just write down both numbers and perform a luhn check:

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

    The one that lies only has a 1 in 10 chance of making up a number that will pass the check! Cool shortcut: If it starts with a 1, 2, 7, 8, or 9, they're lying. Those are reserved for things that aren't credit cards 👍

  • source
  • hideshow 6 child comments
  • [–] 19 points 10 months ago (2 children)

    But you only can ask one question and asking for their CC information won't help you find the exit.

  • source
  • parent
  • hideshow 2 child comments
  • [–] 2 points 10 months ago (2 children)

    I thought credit card numbers were just for credit cards, what are the "things that aren't credit cards"? Genuine question I'm just curious.

  • source
  • parent
  • hideshow 2 child comments
  • [–] 28 points 10 months ago (1 child)

    This kinda falls apart with the knights talking at the start. Either there is 1 that only tells truth with the other not being bound by those rules or one that only tells lies with the other one doing whatever since one of those 2 is telling a lie.

  • source
  • hideshow 1 child comment
  • [–] 11 points 10 months ago* (last edited 10 months ago) (1 child)

    For those interested, ignoring the contradictory presentation of the riddle (as the knights themselves would not say the riddle since one always lies and one always tells the truth), the solution is simple. Ask the knights what the other knight would answer when asked what door is correct, and they will both say which path not to go to. Thus you pick the path that neither Knight says!

    Logic:

    Liar: Will say the wrong option, as they're being asked which door the truth telling knight would say (and they will lie about what the truth-teller would say)
    ------------------------------------------
    Truth-teller: Will say the wrong option, as they're being asked which door the liar would say (and they'll tell the truth about that)
    

    NOTE: This can be expanded to a case with n doors by asking the knights to provide all the options that the other knight could say, and each will provide n-1 options, so you'd pick the one option that neither knight says. It is possible the liar may not list all options, but the truth-teller would, so the problem could still be worked out regardless (and you'd know which knight is the liar in that case).

  • source
  • hideshow 1 child comment
  • [–] 3 points 10 months ago

    Am I mistaken for recalling that in the original riddle, the question should have a boolean outcome (yes/no)?

    Having a question with a big domain makes it a lot easier to guess what is a false and true outcome.

  • source
  • [–] 3 points 10 months ago (2 children)

    The easiest solution is to ask what is the color of their capes that they are wearing

  • source
  • hideshow 2 child comments