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If you only care about who started the fight, it can be done in If you want to keep the questions simple, it can be done in First: Method using only simple questions: Shorter method, using more complex questions:


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I was going to post a different answer, only to realize it was the same as Magma's. So I had to find a new one:


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Note: Answer:


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The drink is: Because the other things are: Smooth dance: Ball sport: Country's Capital: Month: Which are:


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This is probably not the minimum, but it is an improvement to the answer from @Sriotchilism O'Zaic: with the sizes In 9 parts: In 8 parts: In 7 parts: In 6 parts: In 5 parts: I used the $18$ pieces from the linked puzzle as my starting point and


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The problem might be similar to this problem: Given a multiset M of rational numbers where the sum of M equals 1, this is "n-distributable" for an n n∈N, if there exists a partition M1⊔...⊔Mn of M, that the sum of each subset Xi equals 1/n. We would like to find for a fixed k (k∈N) the minimal possible cardinality of a (a1,a2,..,ak) distributable multiset. ...


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Answer: Reasoning:


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Solution: Why?


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2 and 1 3 and 8 4 and 7 5 and 6 with the picture


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Partial currently:


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I took the practical approach and came up with a strategy that could be optimal (I have not tried to prove it). In the worst case scenario it takes: I made a drawing to show my strategy. The yellow squares are the rooms the police officer looks into that night, the black squares are rooms in which the thief logically cannot be.


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This is a very nice 2-dimensional variation of this question.


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I don't know how to do the exact maths (maybe someone else can help me out) but:


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This answer's a little eccentric, but bear with me. (Warning: this answer would fit a lateral-thinking puzzle, but since it is not explicitly a non-lateral-thinking puzzle, just treat this as light-hearted entertainment.) There is no killer.


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There is only one person is the killer.


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Therefore:


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As others have said, the person telling the truth is Here's another approach to figuring this out: From this we conclude (since we know there is only one truth-teller) that:


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The person telling the truth is because Since there can only be one person telling the truth, and we know that is telling the truth, and is therefore the killer.


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As an alternate line of thinking: Therefore This means


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I think the answer is because


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Who tells the truth is: because


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The person telling the truth is: because: and consequently, the killer is:


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I would like to throw in some math this might make it easier to comprehend for some people. Lets introduce some notation: Agents $a_i,$ in our case $i \in \{1,..,K\}$, where $K$ is the number of agents. $t^n_i = i + K (n - 1)$ is the amount of money $i-$th agent takes during $n-$th turn. $S^n_i = \sum_{k=1}^n t^k_i = ni + K\frac{n(n-1)}{2}$ is the sum an ...


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The completed map: Method:


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From @Retro's work Rule 9 gives From Retro Therefore


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PARTIAL ANSWER First off, this is a great puzzle. I really liked spending time on it. I think it's best to start this problem by finding all possibilities. I also abbreviated each business name. Next up is all conditionals. Give me a bit to continue this. I want to make sure everything I've written is correct.


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Each node has 4 edges, so is visited twice. We start and end on node A, but it must be visited some time in between as well. Let's split this up into cases depending on how many other nodes are visited before we come back to A the first time. This gives a total of: More explicitly, the routes are:


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Edit : This is an answer to the question: How many different paths exist from A to A that go through each other point exactly once ? However, per the comments, it seems the OP is looking for something else. I cannot figure out what exactly yet. There are paths that qualify. Indeed,


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I managed to find the 8 in So we are trying to find the 5th largest card in a bunch of 12, by measuring them in batches of four. Here's my strategy: Now we have identified, for certain, a couple of cards we can exclude: A1 and A2 (both have at least 8 cards smaller than them), and B4, C2, C3 and C4 (all have at least 5 cards bigger than them). We also have ...


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I can do it in Note: Continuing... Assume the worst case. Any other results would obviously be easier to solve. For any of these options: Thus, we have completed the process and identified card 8 in at most Note, that as a side-effect of identifying card 8, we also identified cards 12, 11, 10, & 9. There might be a more efficient option to ...


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This answer only assumes that the guards and the keys are taking up physical space, i.e. each guard has a door to which they are closest, and each door has a guard to which it is closest (i.e. a one-to-one relationship). Edit: I realized I need to update my answer to prevent the lying guard from pointing to the non-working key:


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Now you have one lock open and have eliminated a key, so ask the same question to the other guard, pointing at the remaining unopened lock. If the question needs to be asked simultaneously of both guards, then we need to add a little more information, so it becomes:


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Further clarification on the rules suggests that the problem is solvable. I will leave this answer as I find it humorous and interesting, but it was only true according to my original interpretation of the rules. And I will delete my other answer.


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Hmm..my head is hurting Since And


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I can do it in:


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If the population size is N, then you need at most 2N questions.


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It is false when you say it, but true after. Thus they let your pup in, then let you in.


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Totally using Deusovi's logic here, so credit to him: But I'm stuck if each guard doesn't have their own lock.


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Your statement is: Because On the other hand, Comment


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My dog is in the city Initially it will be outside so that will be false and they will let the dog in. Then it will be true and you will be let in. It does depend on assessing them in that order. I guess for the original order one could do: I am outside the city with my dog (true) I get let in I am outside the city with my dog (now false) My dog gets ...


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The results depend on whether the children care at all about one another's survival. I shall assume that each would rather have more land than more surviving siblings. It's also not clear what happens to "the ducal lands" if there is no survivor other than the new monarch; I assume the monarch gets them. If a strict majority is required So, in this ...


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Not entirely sure if this counts as "using a paradox", but you might try If he does, you have told a true statement, so you get to go in, too. If he doesn't, he is violating his own rules, since you have just told a false sentence. So he cannot do that. This is all based on the observation that so letting both in isn't against the guard's rules as stated....


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You could say Reasoning


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I can figure out the fitting key for one lock specifically with the question: If I asked you which key you need for lock A, to which key would you point? Because: Obviously the truth teller points to the right key. The liar has to also point to the right key because if he pointed to one of the wrong keys he would answer the question correctly and tell ...


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As long as the two locks are distinguishable somehow, this can easily be solved with a variation on the 'standard' trick: This works because: (If the two locks are indistinguishable, there is of course no way to determine which key unlocks which lock.)


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You should agree with the citizens that Then with the following strategy, everyone can stay in the village:


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Further efficiency:


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