# Tag Info

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The volume filled is:

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There is an easy way to solve it in the minimum number of moves. The reason this works is:

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Let's number the hallways (doors) from $1$ to $2001$ from west to east. One important observation is that: This is because: Now, another important observation is that, if a dead-end room connects hallway $i$ and $i+1$: Combining both observations: How long will it take for us to survive? This is a visual help:

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We can work backwards to figure this out: Conclusion:

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

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One way to do this is: This could be modified to fix one minor concern:

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I think the following strategy works for any $N$. Note that it works for any kind of hat-labelling, as long as those labels are sortable and all distinct, so not just positive integers. This is a bit complicated, so here is a small example with $N=5$. Instead of numbers on the hats, I'll use some random letters instead, otherwise it could get a bit ...

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This is However

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I would: Then after the final number came in:

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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!

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I have found a set where the weight of the heaviest box is Here are the weights of the boxes General Strategy Minimality confirmed by Oray using computation.

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I don't know why this works, but I tried my old strategy from other similar puzzles and it has worked in several cases so far: Then, One final note - since I don't know why it works, I also can't prove that it always works, unfortunately. But in my many years of solving puzzles in this genre, that strategy always seems to eventually land on a solution. ...

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I can do it in This is optimal because: More detailed argument for symmetry:

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Kevin Ferland and I have a paper on this: Maximal Crossword Grids, Journal of Combinatorial Mathematics and Combinatorial Computing (Feb. 2019) See also OEIS A243826, which has a link to Ferland's earlier paper. Here are a summary table and selected examples from Illustrations of solutions for 3 ≤ n ≤ 50 PDF at OEIS A243826.  \small\begin{array}{rccc|rccc|...

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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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My strategy would be to Then

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Each person gets Then they will Once that is done, they trade This leaves them with So from there, each person It is possible to break this system by This answer is

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You can build a structure something along the lines of a da Vinci bridge, where the weight of the structure itself keeps it from falling apart: It's easiest to build on the tabletop; you can then pick it up by the bottom nail, watch how gravity locks the structure into place, and then balance the whole on the vertical nail.

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They win with probability

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Improvement on the 2-explorer case: First attempt at 3-explorer case, using the same strategy and formatting:

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Or, create a new blank board, and click on it all the positions of the 'on' switches from the original board. Take this new pattern back to the original board, and click every 'off' position from the new board onto the original board. All lights are on!

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I believe that the most intuitive answer is that

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Suppose Thelma asks her mother on the first day. I originally misread the question, thinking that the rule was that if you ask one parent on two consecutive days, they won't say yes twice. This is a pretty interesting question too, so I'll keep the answer for that question below: Clearly she must alternate which parent she asks, because if she asks the same ...

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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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Source: I played this game a lot while distracted and this seems to work pretty well. [Sort of unrelated, but 2048 is one of the first C++ programs I wrote. Code is here: link to code.]

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As an easy-to-remember simple implementation of @athin's excellent answer: Starting before the first door, and until you walk out the exit After each exit, As mentioned by @FlorianF, this With an expected time (assuming random positioning and ignoring E-W movement) of

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Can't you just do ?

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This can be done in Steps

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