New answers tagged strategy
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Solution or Origin of logic puzzle
The constraints are rigorous; the dwarfs can not meet at either side, nor anywhere on the track.
The puzzle suggests that they are able to communicate using buckets and rocks. Failing that... ...
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Accepted
Find the best plan for White
White can win with the move
Then, if black plays d2, white plays
If black plays e2 first, then white plays
If black plays f3, then white plays
The general strategy is to
If black plays Ra8, white ...
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The mower's challenge
Here's a strategy with speed
I believe the optimal speed is an infimum rather than a minimum.
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The mower's challenge
Better solution:
It can be done when the mower is:
A slower speed than this is definitely still possible.
We'll say the left intersection has $LU$ and $LD$ of grass heading up and down respectively, ...
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Catch the invisible and omniscient thief
Fun fact (partially related to the puzzle). What if the thief has infinite speed, but there is more than one cop (cops communicate with each other and work together)?
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Catch the invisible and omniscient thief
As Florian F established, if the thief is able to reach the center of an intersection that has three or more edges, the cop cannot ever catch the thief as the thief can always dance around the center, ...
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Catch the invisible and omniscient thief
Lets invert numbers and say thief speed and set to 1, and v is cop speed. Easier for me that way.
Vertices are labeled A1-D4, roads are labeled BC3 meaning road ...
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Catch the invisible and omniscient thief
I believe the answer
if $S_t$
if $S_t$
I believe the more original question would be
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Catch the invisible and omniscient thief
Just to show that dodging around the intersection isn't enough, consider the simpler city when there is just one intersection, four roads that go a unit distance away from it, and $S_c > 7*S_t$.
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Catch the invisible and omniscient thief
I have to remove my post about the idea of escaping the cop by hanging around an intesection. Despite concluding that it is in fact not valid, people still refer to it as valid
Instead, as explained ...
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