27
votes
Accepted
15
votes
Accepted
Drowning Squirrel
The optimal solution has length
So, can the squirrel escape when T=30
seconds?
This problem is called Bellman's lost in a forest problem, and for an equilateral triangle, the optimal solution was ...
- 4,238
14
votes
Accepted
12
votes
3 blindfolded people dividing 9 coins
Another strategy:
And it works because...
PS: And I realize Dmitry already gave that solution in a crypted comment.
- 25.1k
11
votes
Accepted
Can Alice form a unit square?
Spoiler alert: I won't bother spoiler tagging the following wall of text: proceed at your own risk.
The answer is that Alice wins for any finite $N$. In fact, the more general version holds:
Let $S$ ...
- 18.3k
10
votes
Accepted
9
votes
Beyond earth and countries fighting for land
I'm presuming here that each player's primary goal is to win (i.e. claim, or otherwise be rewarded with) as much land as possible, and that a secondary, tie-breaking goal is to minimize the amount of ...
- 2,655
8
votes
Accepted
Think in a Odd way
The first person can't do better than 50%, as they have no prior information to act on, so the ceiling on survival is 9.5 prisoners.
In fact, we can reach that ceiling using the following strategy
- 19.4k
7
votes
Accepted
7
votes
7
votes
Accepted
7
votes
6
votes
Accepted
6
votes
Accepted
A bit complicated 20 card games v2
The player always has the four options:
Adding them all together gives
This gives the lower bound:
Next, we show that we can indeed prevent the player from doing better.
With the arrangement
The ...
- 17.1k
6
votes
Accepted
Equilibrium of differences
(NB: I've chosen to interpret "the four numbers Alice chooses must be non-repeatable" as "Alice cannot choose the same digit more than once". There are other possible ...
- 72.4k
5
votes
Accepted
An overconfident grandmaster 2
In this particular position, having an extra free move with the restrictions given (i.e., no Rc1xc6xc7) does not seem enough for White to win the game:
Two different pieces must be played and not the ...
- 5,551
5
votes
5
votes
5
votes
3 blindfolded people dividing 9 coins
This can be solved in no more than 4 passes independent of the number of coins that everyone has. (I.E. Start with 201 coins with each person having 67 at the end, after 4 passes.)
- 151
5
votes
Decision Paralysis
The one with the winning strategy is quite surprisingly
First, let us note that
First options
Case 1
Case 2
Case 2.1
- 8,038
5
votes
Accepted
Dethy Mafia: Solved on the first day
At any point:
For example:
These results prove that
Thus, to prove Eve the Goon:
Similar logic can also be done in reverse:
Edit: changed a word to remove self-targeting cases, since the question ...
- 2,757
5
votes
Accepted
Ball weighing - second heaviest
The least number of (two-pan) weighings required is
The flowchart below details how to perform the weighings; an unlabelled poset represents all knowledge gathered of the relative weights of the ...
- 6,834
5
votes
Accepted
5
votes
How to Divide 1,987 Gold Coins Among 40 Pirates?
If there is an alliance formation (there can be several such cases), then in the worst case the rest of the pirates form an alliance and go against $P_1$. In this case, $P_1$ can ensure:
Thus, after $...
- 1,018
4
votes
Accepted
Conclusively solving the extreme gerrymandering puzzle
The answer, in short:
Now let's take a look at why:
Now, we can show that
Thus,
- 1,922
4
votes
Attacking Hyenas
We can slightly rephrase this question as "For a given number of attacked hyenas, what is the maximum number of unattacked hyenas we can have?"
Particularly, since hyenas can be divided into ...
- 7,182
4
votes
How to Divide 1,987 Gold Coins Among 40 Pirates?
The first pirate would love to end up with one pile of 1947 coins and 40 piles of 1 coin. That would be the best for that pirate. So the first pirate definitely splits into and 1986 and 1.
The last ...
- 5,588
3
votes
More Genuine and Fake Coins
Somewhat based off the previous answer but I think we can do better than $8$ weightings.
- 2,424
3
votes
Dethy Mafia: Solved on the first day
Braegh's solution describes a situation in which the goon could be determined on the first day. However, it makes some assumptions about Eve's play:
a) Eve reveals her "result" first, with ...
- 983
3
votes
Accepted
The mower's challenge
Edit: The error in my proof should be fixed, I still claim the same speed.
I have a solution for when the speed is strictly greater than:
Step 1:
Step 2:
Step 3:
- 146
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