For the four people puzzle. Assign every suit a number 0-3. All four people guess differently what the sum of all the cards is modulo 4, meaning exactly one person will get it right. Using the sum and the view of the other cards the one who guessed right can work out their own card.
Can you crack puzzles devised by the brilliant cryptological minds at the NSA? Find out by logging in to our totally legit, completely innocuous, not in any way a honeypot that collects extensive info on you, puzzle site!
> Can you crack puzzles devised by the brilliant cryptological minds at the NSA?
almost obligatory snark:
is this the same NSA which allowed/enabled one Edward Snowden to walk out with a gigantic treasure trove of super ultra mega top secret documents, vital to national security, on a USB thumb drive (or whatever it was)?
sincere question
because this point is among the many data points I now have in my mind, as a US citizen and adult voter, when I get to decide exactly how much power I'd like those "smart" folks to have, as they build/run a global 24x7 passive data snoop system on, effectively, all of humanity.
equations may be elegant, perfect, ideal
executable software: less so
human procedures & rules & everyday happenstance & one-off oopsies: far less
"The players may not communicate in any way" - It seems to me that all the answers are a form of communication. In fact the dictionary defines communicate as, "share or exchange information."
The solution to the 4 person problem is actually pretty simple.
Let's assume that we know the answer to the 2 person problem.
We have 4 persons; A, B, C and D.
At first the persons guess their colour (without revealing it of course) based on their direct neighbor.
So for instance:
A <-----> B
C <-----> D
A and B being neighbors, they guess their colours based on the solution of the 2 persons problem. Same goes for C and D.
Without loss of generality, let's assume C guesses correctly his card colour.
Since he can see the cards of A and B, he knows who will guess their colour correctly.
For instance, if they have the same colour, A will guess correctly, if not, then B will guess correctly.
Without loss of generality, let's assume A guesses correctly his card colour. He then sees that C will guess correctly.
The participants agreed beforehand that clubs and hearts are associated with 0 while diamonds and spades are associated with 1.
A and C now apply the solution to the 2 person problem to guess their shape group (either (diamonds or spades) or (clubs or hearts)):
A B
|
|
C D
It is guaranteed that at least one has guessed the shape group correctly. Since they both guessed correctly the color, it is guaranteed that one of them will guess the shape correctly.
"If you think you know the answer, write it on the back of your classified one time pad key and send it to: NSA, Box 35 Fort Meade (our fair city) Maryland".
Some of these are poorly worded. Why does the answer to the May one keep talking about 30 seconds passing when the only time mentioned in the problem is 30 minutes? 30 seconds gives Nick at least 10 seconds per possible solution, he should have tried them all by then.
It should talk about the number of tries so far (one each), the length of time it took is irrelevant. But why are they even taking turns when they have separate padlocks and could easily brute-force it? I get the concept they're going for but the premise doesn't fit and just confuses things.
It said no communication during but you could devise a strategy before play. What I would do on the first challenge is say I would write down the opposite color of what I see and tell the other person the write down the same color they see so you will definitely at least win $50 "or 100$ if you have the same color". For the suite challenge do the same strategy but split 2 people write down the same suite and have a plan for the other 2 to pick what they don't see.
Coincidentally enough, this was a submitted puzzler to the NPR syndicated show "Car Talk," except there were 10 men in a line wearing black or white hats. The goal was to determine the optimal strategy for guessing hat colors.
I'm not sure how orienting the card that is on your own head would help. You don't know what it is.
That being said the method of "Bob always tells the truth, Alice lies" is sort of a card orientation way for two people. Alice puts her card up sideways and Bob knows that to mean he should tell the truth. Likewise Bob doesn't rotate his and Alice knows to lie.
However that only works with two people and they have opposite spins.
Maybe it can work with more than two and you have two rounds of guessing. Also you need to know that you have differing spins.
There are so many potential hidden channels!
If 2nd can see what the 1st one guessed -- that's also easy
Or if they just agree on the timing of the guess..
1. The strategy is that they should agree to write down the same colour(either red or black), irrespective of what they see on the opposite persons forehead
2. Same as above. This time all of them should agree to write any single suit.
By probability, at least one of them will get it right.
1 They should write their own color, so this is not a winning strategy.
[Spoiler] As stated by other the winning strategy is that one says always the other person's color and the other say the opposite. That's because or the color are the same or are different.
The June one too. A simple three if then statements can solve the problem easily even increasing the 1000 coins to a number physically possible by the ship to hold.
Make Bruce guess opposite of what he sees on Ava and Ava guess same color as what she sees on Bruce. This covers atleast one correct guess for all for Cases BB, RR, BR, RB.
This is actually very similar to a question I was asked when interviewing to read Maths at Oxford. In the Oxford version, your brother and you are also playing a game. You have a 2-dimensional block of chocolate with a poison piece somewhere in the block. The poison piece is bright green to the eye. You each take it in turn to break off some chocolate (you have to follow the rows and columns of the chocolate, and the break must be a single break). To win you must avoid eating the poison piece and end the game by handing it to your brother. You have the choice of going first or second? When should you choose either?
[+] [-] IIAOPSW|9 years ago|reply
[+] [-] im4w1l|9 years ago|reply
[+] [-] sopooneo|9 years ago|reply
Sorry if your answer turns out to be obviously correct.
[+] [-] intended|9 years ago|reply
[+] [-] unknown|9 years ago|reply
[deleted]
[+] [-] bitwize|9 years ago|reply
[+] [-] Animats|9 years ago|reply
[1] https://web.archive.org/web/20130912120456/https://canyoufin...
[+] [-] sb8244|9 years ago|reply
[+] [-] syngrog66|9 years ago|reply
almost obligatory snark:
is this the same NSA which allowed/enabled one Edward Snowden to walk out with a gigantic treasure trove of super ultra mega top secret documents, vital to national security, on a USB thumb drive (or whatever it was)?
sincere question
because this point is among the many data points I now have in my mind, as a US citizen and adult voter, when I get to decide exactly how much power I'd like those "smart" folks to have, as they build/run a global 24x7 passive data snoop system on, effectively, all of humanity.
equations may be elegant, perfect, ideal
executable software: less so
human procedures & rules & everyday happenstance & one-off oopsies: far less
[+] [-] facepalm|9 years ago|reply
[+] [-] pokstad|9 years ago|reply
[+] [-] dangero|9 years ago|reply
[+] [-] fauria|9 years ago|reply
[+] [-] x0137294744532|9 years ago|reply
The participants agreed beforehand that clubs and hearts are associated with 0 while diamonds and spades are associated with 1. A and C now apply the solution to the 2 person problem to guess their shape group (either (diamonds or spades) or (clubs or hearts)):
It is guaranteed that at least one has guessed the shape group correctly. Since they both guessed correctly the color, it is guaranteed that one of them will guess the shape correctly.[+] [-] glu|9 years ago|reply
[+] [-] blueintegral|9 years ago|reply
[+] [-] goldenkey|9 years ago|reply
[+] [-] cwilkes|9 years ago|reply
[+] [-] tn13|9 years ago|reply
[+] [-] Kapow|9 years ago|reply
It should talk about the number of tries so far (one each), the length of time it took is irrelevant. But why are they even taking turns when they have separate padlocks and could easily brute-force it? I get the concept they're going for but the premise doesn't fit and just confuses things.
[+] [-] XeO3|9 years ago|reply
[+] [-] advisedwang|9 years ago|reply
[+] [-] cdevs|9 years ago|reply
[+] [-] cbgb|9 years ago|reply
[+] [-] flashman|9 years ago|reply
[+] [-] cwilkes|9 years ago|reply
That being said the method of "Bob always tells the truth, Alice lies" is sort of a card orientation way for two people. Alice puts her card up sideways and Bob knows that to mean he should tell the truth. Likewise Bob doesn't rotate his and Alice knows to lie.
However that only works with two people and they have opposite spins.
Maybe it can work with more than two and you have two rounds of guessing. Also you need to know that you have differing spins.
[+] [-] Fry-kun|9 years ago|reply
[+] [-] fractalb|9 years ago|reply
1. The strategy is that they should agree to write down the same colour(either red or black), irrespective of what they see on the opposite persons forehead
2. Same as above. This time all of them should agree to write any single suit.
By probability, at least one of them will get it right.
[Edit]: Didn't put it correctly before
[+] [-] faaabio1618|9 years ago|reply
[+] [-] LeartS|9 years ago|reply
[+] [-] malingo|9 years ago|reply
[+] [-] curiousgal|9 years ago|reply
[+] [-] ktta|9 years ago|reply
[+] [-] unknown|9 years ago|reply
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[+] [-] ForFreedom|9 years ago|reply
There I cracked that one :)
[+] [-] pcool_deathel|9 years ago|reply
[+] [-] zazaalaza|9 years ago|reply
see my solution here
[+] [-] unknown|9 years ago|reply
[deleted]
[+] [-] Cogito|9 years ago|reply
Always go first.
Assuming the table is symmetrical around only one point, the centre of the table, place the first coin on that point.
Every move after that place your coin in the location directly opposite the last coin played, the mirrored location around the point of symmetry.
Since you are always playing a symmetrical position, you know that there will always be a space available for you after the previous coin was played.
[edit] looks like the answer on the site is almost the exact wording as mine :)
[+] [-] datr|9 years ago|reply
[+] [-] teekert|9 years ago|reply
But where?