New answers tagged

-2

This is a scheduling problem. There is a huge literature on problems like this. This could be a puzzle, I suppose, if it were well formulated. As it stands, we don’t know the relative costs involved. What is the cost of a 10-second wait? What is the cost of an n-degree rotation? (Time and wear on the machine.) What is the cost of a flip? Also, we would ...


5

The maximum is 1 2 2 3 3 . 4 4 5 5 1 1 2 3 6 6 . 4 5 7 8 8 . 9 6 10 10 . 7 7 8 11 12 9 9 10 13 14 15 15 11 11 12 12 . 13 13 14 14 15 16 16 17 17 18 19 20 20 21 21 22 16 17 18 18 19 19 20 21 23 22 22 . 24 24 25 25 . 23 23 26 26 27 28 24 25 29 30 31 31 26 27 27 28 28 29 29 30 30 31 I obtained this solution via ...


3

To kickstart, But as we are given 7 empty squares, Which gives a total of


2

Trial answer with


3

The best strategy is to This will take steps, which is optimal: it breaks the least possible number of eggs, and if you were to use some other strategy to choose the floors, so the expected number of steps won't be any less for that other strategy, either. This happens because "infinity is really, really big", or more specifically: And so is, ...


2

Congratz to @Anon for finding the intended solution. I'm posting this just to show off a to scale picture I happened to have lying around:


4

Given an unbounded number of trials, we could simply test one egg over and over and over again until it breaks, and find our floor. "But that's too slow", you say, and you're right - so instead we The only information we need to solidify our method is...


3

Optimizing for number of throws t given a large amount of eggs, starting at throw n = 0 until success at throw t: This finds the max floor for arbitrarily high floor F in


9

TL;DR First, and easiest, we can do it in: Using the principle of the: The method is as follows: You then just need to rearrange your pieces. Why does this work? Now that we understand that method, how can we do it better? We now need to prove that we cannot do better. However, this is impossible. The following is a sketch of a proof:


8

The smallest number is going to be the one with the least digits. We need to make use of all the digits as efficiently as possible. Therefore, we should: Therefore, the solution is I’ll admit I used a computer (despite the no-computers tag) to generate the output, since a Bash command is much faster and more accurate than typing that up by hand. I ...


4

Let's start by removing the lower bound - say you can go into stamina-debt as far as you need to, and you keep playing until you escape. Then: Okay, but what does this say about the actual problem? In other words:


0

It can be done without programming or mathematics. If the set of coins is compact that means So the maximum number of coins is The minimum value is


2

Here's another 40 moves sequence that I found using the simulator: Note the brackets are [column, row]


2

Using the simulator, I was able to find 2 sequences requiring 41 moves each (see edit history for details, formatted T[y, x]). I was also able to find a single sequence requiring 40 moves (formatted t[x, y]):


2

My best shot: And for 10:


3

Here is one possible arrangement: Proof of optimality: In view of the answer by loopy walt, this also proves optimality for the case of decagon.


5

With the "Use as many and any black or white pieces (both kings included)" restriction, the theoretical maximum becomes Obviously, there can't be any legal position with that many white pieces, so we throw legality out of the window and start from here: And as long as we are careful to eat the bottom two pawn rows in good order, we are good to go ...


4

Variation 1. position must be legal: As there are at most 15 capturable white pieces this is a hard upper bound on the chain length. This bound can be achieved. Documentation: Variation 2. position need not be legal: I can do


0

You don't need any code for that. No spoilers as solution is quite obvious and it has been a day. 3x20 simplifies to 50+10. 20+20+10 simplifies to 50 and 20+20 > 20+10. Therefore, we cannot have 2x20. The same for 200. Anything else is obviously optimal when you have 1 coin of each type because 2 already merge to the next one. So, 6 coins with 385 total. ...


2



4

The maximum number of coins is with a minimum total value of achieved by The integer linear programming solution approach I used might be of interest. Let nonnegative integer decision variable $x_c$ be the number of coins of type $c$. The first problem is to maximize $\sum_c x_c$ subject to $\sum_c c\cdot x_c \le 495$ and "compactness" ...


4

proof-without-words (almost...)


11

Suppose we start with a full diagonal (A1-H8), and then see which queens we can remove. Removing any one queen is obviously fine. The other 7 queens cover all the other rows and columns, leaving only the removed queen's own square which is covered diagonally. If we remove two queens, If we remove more than two queens,


5

Lowest I was able to get it to (so far) was 5: Initial logic to this solution:


4

Converting a Disconnect Four puzzle to a SAT instance is very easy. Based on this


1

I think the answer is Trying to solve the puzzle from scratch doesn't give us much: However, case analysis on gives very interesting results. Note that Since we already forced a cell by case analysis, the total number of cells we need to force to get a unique solution is


5

If the number of people per table is $p$, then to cover each pair we must have $$\frac{600}{p} \binom{p}{2} \ge \binom{30}{2},$$ which implies that $p \ge \lceil 49/20 \rceil = 3$. The following set of $200$ triples of years covers every pair and contains each year exactly $20$ times: {{1,2,22},{1,3,5},{1,3,7},{1,4,20},{1,6,8},{1,7,15},{1,9,10},{1,11,12},{1,...


3

5 numbers are certainly enough to make it unsolvable. This is a simple example : You can never put one in the bottom left corner. Maybe someone has an optimization?


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