4 ms·
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 per
by im4w1l 10y ago
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.
- ulrikrasmussen 10y agoI don't think the players can see the other players' answers. The puzzle reads like an instance of the 100 hats puzzle, but instead of maximizing the number of correct answers, you just have to guarantee at least one. Edit: I should've read your answer more carefully.
- josefdlange 10y agoGist of this, since formatting got screwy: https://gist.github.com/josefdlange/04c9eb14a9ddbdfae8eb2e34b91d56df https://gist.github.com/josefdlange/04c9eb14a9ddbdfae8eb2e34... Seeing the others' answers I don't think matters in this solution. The strategy -- given that it's allowable to discuss strategy before drawing cards -- is each person has a different modulo result (x) they have to compute the number they must add [to the sum of the other three], (m) in the following algebra. Solve for m to get the suit your should guess. x = ([sum of others] + m) % 4 Suits: 0: Diamond 1: Heart 2: Club 3: Spade Example: [PLAYER]: (SUIT THEY HAVE), (CHOSEN MODULO RESULT) A: (0) (0) B: (1) (1) C: (3) (2) D: (3) (3) For A's m: ((1 + 3 + 3) + m) % 4 == 0, therefore m = 1 For B's m: ((0 + 3 + 3) + m) % 4 == 1, therefore m = 3 For C's m: ((0 + 1 + 3) + m) % 4 == 2, therefore m = 2 For D's m: ((0 + 1 + 3) + m) % 4 == 3, therefore m = 3 D guesses correct. There are many permutations of the drawn cards, but I think this one does solve it. All props to the person up this thread who initially answered the 4-player problem; I'm just explaining their methodology as I understood it.
- ulrikrasmussen 10y agoThanks! I misread the answer and took it for the solution to a different but related puzzle
- pliny 10y agoIf they have (A,B,C,D) = (0,1,0,1): A sees 1+0+1 -> guesses (2+0)%4 = 2 (wrong) B sees 0+0+1 -> guesses (1+1)%4 = 2 (wrong) C sees 1+0+1 -> guesses (2+2)%4 = 0 (wrong) C sees 0+1+0 -> guesses (1+3)%4 = 0 (wrong) EDIT: This is twice incorrect, as indicated below.
- im4w1l 10y ago1. You got your signs messed up. They should subtract the observation from the total. 2. C correctly guesses 0 in your example.
- nonsequitur 10y agoThere's no assignment to the „m“ variables that would make your algorithm work in the general case. im4w1l's strategy is just: Player i returns (i - [Sum of the other players' suits]) % 4 (With i ranging from 0 to 3)
- josefdlange 10y ago`m` is the result; this scratch here is meant to resemble algebra, not programming. You know, solve for `m`, etc.
- josefdlange 10y agoBut the subtraction is simpler. I didn't want to confuse anyone with moduloing a negative number.
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- javitury 10y agoTake a look at: https://gist.github.com/javiertury/bf71d37ff26643eca939c664264346ea https://gist.github.com/javiertury/bf71d37ff26643eca939c6642... I've tried my best to illustrate why this is the correct solution in the url above. Long story short: Call S to the sum of all the numbers on the hats of every player. The sum of all the numbers a player can see will be less than S but the difference with S will be less than "n - 1". Also the difference between the largest guess(for the sum of all the hats) and S won't be larger than "n - 1". There is now a problem, we want a set of guesses with a size no larger than "n" but in reality it can be as large as size "2(n - 1) + 1 = 2n - 1". Taking the set of possible rational guesses modulo "n" creates a collision, now some guesses will share the same remainder modulo n. However because no guess can be farther than "n - 1" with respect to S, the set of guesses modulo "n" won't create that collision for "S mod n". Any player that was told to make the remainder of the total sum mod n equal to "S mod n" will always have the correct answer. If you want to think visually imagine the modular clock for "n" which has "S mod n" as the reference point. Because no other point can't be farther apart than "n - 1" distance from S, only S can give the remainder "S mod n". If there are "n" players and we assign each one a different remainder "mod n", one of them will have the correct answer for sure.
- josefdlange 10y agoThat is very, very clever.
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- philbarr 10y agoIt seems that we could expand this for n people with n choices. So in the first case where there are only 2 people and 2 colours, they assign, say 0 to black and 1 to red. Then they each sum up the other choices they can see (which happens to be only one other choice in this case) plus their own assigned number (0 or 1) and answer that modulo 2. Er..I think...?
- brian_cloutier 10y agoYeah, that sounds right. Through that lens, the first problem is a special case of the second problem.
- brian_cloutier 10y agoTo give some intuition as to why these answers are "obviously" correct; for the 2-person case you have 4 possible results: (Ava, Bob) -> (B, B), (B, R), (R, B), (R, R) It's easy to see that there are only two alternatives: they either both have the same suit or they have different suits. Luckily, there are two players, and so they can try both options simultaneously. The same thing works in the 4-player case. You can try 4 strategies at once, is there any way to somehow reduce the state space to 4 options? If you give every suit a number from 0-3, then the sum of all the suits is obviously going to be somewhere between 0-3 modulo 4.