# Tag Info

39

Solved in 10 moves including "Add a comment" and "Add Comment" buttons at the start and end I was muddling around in the "inspect element" for way too long trying to find something I could use in the html.

34

The answer should be Here is how you do it.

27

This is hugely inspired by @rhavelka's answer, so make sure to upvote him/her as well! Just found an alternative of where the phrases can come from:

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It might be possible to do a little better, but I think I can get Explanation:

24

The trick is this:

20

Not sure if minimal, but here is a way to do it in Method Also, as Jaap Scherphuis points out in the comments

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Probably not optimal, but here's an approach that saves at least

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If the person can stay in those 2 minutes, the problem will be unsolvable because he can only move if and only if we check his room. Otherwise, here is a solution with exactly 15 14 (thanks @JaapScherphuis!) checkings, just in time!

15

This is my first answer on the puzzling stack exchange and I'm really not used to explaining things like this so it's probably going to be rather convoluted. I'll probably come back and edit it later when I can figure out how to make this clearer. Solution for a maximum of 4 weightings: First divide the coins into 4 equally sized piles. There are now 3 ...

15

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The same idea as APrough's answer, but presented visually instead of with words: 📅 🏠 🏜️ 🏜️ 🏜️ 🏜️ 🏜️ 🏁 0 🤠🍗🍗🍗🍗🙂🍗🍗🍗🍗🙂🍗🍗🍗🍗 1 🤠🍗🍗🍗🙂🍗🍗🍗🙂🍗🍗🍗 1 🤠🍗🍗🍗🍗🙂🍗🍗🍗🍗🙂🍗 2 🙂 🤠🍗🍗🍗🙂🍗🍗🍗 2 🙂 🤠🍗🍗🍗🍗🙂🍗🍗 3 🙂 🙂🍗 🤠🍗🍗🍗 4 🙂🙂 🤠🍗🍗 5 🙂🙂 🤠🍗 6 🙂🙂 🤠

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Alternative solution: Then here's how that works Here is the total number of rations and helpers required

13

A complementary result of tehtmi's answer: Using the same strategy, is the maximum number of coins you can get, and +1 is impossible. Proof:

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The optimal strategy takes Here's why this is necessary: And here's a way to do it: Fun fact:

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They win with probability

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You can solve the problem via integer linear programming as follows. Let $n$ be the number of coins, so we need at most $n$ bags. For $b \in \{1,\dots,n\}$, let nonnegative integer decision variable $x_b$ be the number of coins in bag $b$, with $x_b$ nonincreasing. Let $z$ represent $\max_b \{b\cdot x_b\}$, which is the number of coins the king will take. ...

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If you are unsure about the puzzle, you are probably misquoting a classical puzzle as Rand Al'Thor mentions. If you cannot distinguish the type of the coins, regardless of how you make two piles, there is always a way to swap a heads-up and a tails-up coin between the two piles, changing the heads-up counts. So you cannot guarantee the counts are the same. ...

10

Can't you just do ?

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Their best strategy is Their chances with this strategy are Optimality

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Here is a proof that hexomino's answer of is best possible. Here's how. The alternative is (This can be expressed more concisely using the language of graph theory, but it seemed better not to.)

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We can limit the loss to Bags are as follows: Examples of results for different sizes the knight could pick: This seems even closer to optimal because Some notes on technique and strategy: I put together a spreadsheet which In the final configuration shown above, it looks like this: Some explanation: The spreadsheet can also automatically calculate ...

8

Assumption: Shifty coin only switches if weighed. I believe this is optimal solution; if (case 1) otherwise (case 2) If it is Case 1; if (Case 1.1 - They are equal) if (Case 1.2 - They are not equal) then if Case 2 happens (no balance at first weighing); if there is balance (Case 2.1), (also meaning it was heavier at first) if no balance again as ...

8

I think I can do it in How to achieve it: Now, why is this optimal?

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If the first player is the one who says either 2nd January or 1st February, then the winning player is Proof: The first player The strategy is to Note: I assumed that the game only goes over one year. If players can go on to the next January, then

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Pretty much all Nim games can be solved by starting at the end and working backwards. Let's first solve the basic Nim, with only one pile, and simple actions (take 1-4). Let's enumerate the endgame situations by how many sticks are left, and see what kind of pattern emerges: Rules: 2 players, one pile, take 1-4, taking last wins 0: loss, cannot take 1: ...

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Phew finally got it in 9 I believe. Took a while to find, with help from other posters (see their post for better explanation of some shared steps)

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@StephenTG's answer is the best. Here's a proof: Since there are 100 possible numbers, they must encode into 100 "states". In other words,

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This solution was derived independently but the result is equivalent to the answer already given by Mark Tilford and the confident tone of a suggestion by Greg Martin led to the realization that this does provide the highest possible probability of winning the car. Imagining a repeating cycle of doors, . . . Left - Middle - Right - Left - Middle - Right - ...

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The poor robot might get slightly less dizzy with

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