Here is a algorithms/riddle i just solved that was kind of fun. So heres the setup. You have a circular table that spins with four evenly spaces opaque boxes on it around the outside. In each box is a coin that is either heads or tails. You get a maximum of 5 rounds, between each round you are blindfolded and the table ia spun randomly. During each round you can pick a box to open, look, the pick a second box to open and look. You can then flip none, any, or all, of the coins with open boxes. Then the boxes are closed before the next round. If at any point all four coins are the same then you win the game.
You must come up with a strategy that **guarantees** in 5 rounds or less that all 4 coins will be the same.
Tommorow i will take a picture of the flow chart i came up with the answer. In the meantime please give hints, answer, or thoughts in the comments but be sure to CW them. I will reply tomorrow with a picture of the answer and CW that as well. Good luck!
PS for a slightly harder version all the same rules apply except you need to pick both boxes to open simultaneously rather than one after thr other. With this harder version it is still possible in 5 rounds or less.
#riddle #algorithm #computerscience #programming #algorithms #puzzle
@freemo No. If you know they're all the same, you can just leave them for the remaining turns. Or if you got them all the same without realising it, you can flip them out, flip them back, whatever you like until your N turns are up. I'm just removing the assumption that the fact the game hasn't ended implies all four are not the same.
@khird as far as i know the smallest N for a garunteed win is 5, you can sometimes win sooner but 5 is the smallest garunteed.
@freemo I don't believe you can guarantee a win in five without the instant-win condition in the original problem.
@freemo unless you can logically deduce from the the coins you've seen that it must be solved or must not be solved, you cannot make the assumption in either direction.
@khird without an oracle telling you that you've won i dont think its possible to do because of the random element.
Mathematically proving there is no solution to a conjecture is often itself an extremely hard task. That said i may be able to articulate why it isnt solvable if i give it some thought.
Any solution relies on first getting the coins to a state where you have 3 coins the same and the state of the 4th is unknown. The problem is i dont see how you can ever deduce the unknown state without an oracle since you can spin it as many times as you want and may always get the same choices every time. Flip it all you want and you can never find the unknown state without an oracle.
So proof of impossiblity is this.
Imagine you are so unlucky that every time you pick two cups, even if you alternate adjacent and opposite picks, you always keep picking the 2 out of the same 3 cups. In other words your so unlucky you never pick one specific cup.
No matter what you do the state of that one cup is always unknown to you and therefore you can never garuntee you will end any one round in a winning state.
For similar reasons, even under the original instant-win rules, a version with five boxes is unsolvable. No matter how many times you open any arrangement of two boxes, fate can always allow there to be two boxes that don't match and you never see.
I think five boxes, opening three per turn, should work, but I haven't figured out what the target number of moves should be.
With 4 boxes you have two modes of selecting, adjacent or across, with 5 boxes you also have 2 modes of operation, adjacent or skip one, and you cant do opposite. But with 6 boxes you get 3 modes, adjacent, skip one, or across. I need to think about this some more but i suspect if a certain number of boxes is solvable depends on the number of modes. Compared to number of boxes. It may be (i need to ponder) that 6 boxes is solvable where 5 is not.
@freemo only opening two boxes per turn, you can't do five or six. I think there might be a solution if you allow the player to open three per turn, at least for five.
For five boxes, choosing two: whenever you choose adjacent, suppose you choose boxes one and two; whenever you choose skip-one, suppose you choose one and three. Box four contains a heads, and box five contains a tails. You'll never be able to get all five the same.
For six, choosing two: suppose your adjacent choices are one and two, your skip-one are one and three, and your opposite are one and four. You'll never be able to get five to match six.
I think if you select 3 you can solve 5 or 6 boxes, you dont need 4 to solve 6 just 3. So its not number of boxes minus 2.
@freemo I don't think three flips solves six boxes. Your patterns are 1-2-3, 1-2-5, 1-3-5, 2-3-5, unless I'm forgetting one. If box four doesn't match box six you're out of luck.
Ill work through my logic on 6 boxes tomorrow. I think i see how to solve it but i need to get it on paper, see if there is a hole in my logic. Also will be helpful to nail down a workable notation.
@freemo some further observations:
For any number of boxes:
If you're allowed to choose all the boxes, a win is trivially possible: flip all to heads.
If you're allowed to choose all but one of the boxes, a win is trivially possible: flip all to heads; if you then see a tails turn it to heads, if not turn all coins to tails.
For five boxes:
There's no solution with three choices, because there's no penultimate state from which you can guarantee that you'll choose all the remaining heads or all the remaining tails. So five boxes has only the trivial solutions.
For six boxes:
I have a solution for four choices in five moves, which may not be optimal (I haven't searched exhaustively for other strategies).
The number of patterns for given numbers of boxes and choices is from the OEIS: https://oeis.org/A047996, under "example". I don't understand the notation they use for the generating formula, but they call it a "circular binomial coefficient" which makes sense.
@khird I actually created a notation/process where you can determine if a configuration is solvable. I didnt have a chance to try it out on various combos. Works like this.
Represent selection modes with binary numbers. So selecting one out of 4 boxes might look like 0001, 2 selected out of 4 would be 0011 or 0101.
To see if two selection modes are equivalent reduce them to what im calling "standard form". What that means is you left or right shift the binary number (taking any digits that roll off and replacing it on the other end) and do this until the number is in its smallest representation. So 1010 would have a standard form of 0101 (most significant digits on the left). If the standard form of two selection modes are the same then they are equivelant.
Next compute all possible selection modes in standard form for the given parameters. For the original puzzle this would be:
No other selection modes are possible.
Next do an OR operation on all possible selection modes, here we get:
If the resulting binary number has one or fewer 0s then a solution is possible. If it has greater then a solution is impossible.
So new problem, given a certain number of boxes, and a number of flips per round, how can we determine if it is solvable, and the number of rounds needed to garuntee a solve, assuming an oracle.
@freemo jumping right to the general case!
I think it's going to be exponential difficulty in at least flips-per-round, and probably also in boxes.
My algorithm is this:
Go through all known cases: if I were told the state of the boxes, for each pattern I could choose, what possible combinations would I find?
For each state-choice-combination triplet, and each possible combination I could set the coins to, what state (or superposition of states) could I leave it in?
Go through the unknown states the same way, by superposing the known ones, building all the way up to completely unknown (all states possible).
Make the decision tree by minmaxing, like in two-player game theory. If the starting state of "completely unknown" has a path to either all-heads or all-tails, count the steps in the path. If there isn't a path linking the starting state to the winning states, it's unsolvable.
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@freemo Let's extend the riddle by saying you have to set the coins the same after exactly N moves. You're allowed to have all four coins the same before then, but you won't win unless they're still the same (or again the same) at exactly N moves.
What's the smallest N for which there's a guaranteed winning strategy?