20
It's undefined although a slight tightening up of the question would make it defined.
What the question lacks is a specification of what happens when a player reaches the limit of the universe. To be more specific:
Suppose each player executes their optimum strategy of moving at maximum speed in the negative direction. What exactly happens at $t=1/3$ when ...
15
Fundamentally, the issue is that
For every fixed cop strategy, there exists a thief strategy that beats it.
For every fixed thief strategy, there exists a cop strategy that beats it.
For an easier picture, this is effectively what the game looks like:
Cop chooses a real number $C\in(0,1)$ (basically $\liminf_{t\in\mathbb{R}^+} C(t)$ of their position)
...
13
Base question
Addition: Bonus question
11
Cleaned up answer:
To make things concrete, let's say that the cop's strategy is represented as a function c(t), which is dependent on the thief's trajectory w(t), and vice-versa.
The cop's strategy is c(t) = max(2/3-t, w(t)). The thief's strategy is w(t) = c(t)/2.
Now we can observe for the cop's strategy that
We can also observe for the thief's strategy ...
11
is right, basically because
The problem with one of the two suggested approaches is that
The space in which they're moving is sort of like a 1-dimensional hyperbolic space: they can get infinitesimally close to the boundary but they can never actually reach it.
TL;DR: it seems that, on an open line segment, Zeno's paradox is actually true.
10
I think that
Proof
8
If the largest number is $N$, then there are
possibilities for the two smaller numbers. Summing the possibilities for the smallest two numbers when the largest is each of 11,12,13,14, and 15 gives a total of
possible solutions for all three numbers. An efficient means of choosing a solution from a number of possibilities is to discard approximately half ...
7
The usual heuristic for this sort of thing is that at any given point there's some number of possible solutions -- e.g., for the puzzle discussed here there are 5!=120 possible arrangements of the pieces -- and each question will, once answered, reduce that number (to the number of configurations for which that answer is correct), and you want to reduce as ...
6
The number line’s origin’s being excluded means that . . .
It also means that the maximum speeds of both cop and thief . . .
6
EDIT: I don't know!
Let's call the thief's position as a function of time $\Theta(t)$, and the cop's position as a function of time $C(t)$. We know each function is continuous. We don't know if they're differentiable, but based on the maximum speed restriction we can say $|\Theta(b)-\Theta(a)| < b-a$ and $|C(b)-C(a)| < b-a$ for any two times $a<b$.
...
4
Expanding on obscuran's answer: https://puzzling.stackexchange.com/a/112389/11569
There is no answer because the game has no equilibrium, as obscuran explains.
Claim 1. For every fixed thief strategy (i.e. continuous path $f(t)$), there exists a cop strategy that beats it.
Claim 2. For every fixed cop strategy (i.e. continuous path $g(t)$), there exists a ...
4
So it looks like the answer is
Reasoning
4
This is a question about puzzles, so it's not inherently off-topic for the site. But there's not much of a meaningful answer we can give to some of these questions.
1. According to what strategies could one ask the two types of questions?
2. Where to begin? Does it make sense to ask Questions A first for squares on the rim, as opposed to central squares ...
3
No matter what,
2
The number of combinations of 3 different digits among 15 digits is 3 choose 15.
The number of combinations where all digits are smaller than 10 is 3 choose 10.
So number of combinations where at least 1 number is greater than 10 is:
2
The thief will be caught.
The cop simply moves at his maximum speed towards 0 until he catches the thief.
The thief is constantly going to retreat, certainly, and has an infinite number of positions to move to before 0 - but sometime prior to, and almost exactly, t=2/3, the cop will have checked every infinitely possible position and has to have found the ...
2
For this answer I'm going to assume that neither the thief nor the cop can get outside of the line.
If we consider a small variation to the question where the line have ends at 0 and 1 it's clear that
So what happens when we remove the 0 and 1 points? There is an infinitesimal small reduction in length to travel. That means that
But what I would like to ...
2
I'm going to do something rather weird here, and create a second answer. That's because I want to answer a slightly different variant of the question, and propose a controversial result.
My other answer considers the formulation of the question, and also addresses the case where the endpoints of the universe are "allowed" but are outside the ...
2
The optimal strategy for both the cop and the thief is to move in the negative direction at full speed. For these strategies, we have functions that describe the position of the thief and cop for time t:
$ d_t(t) = 1/3 - t $
$ d_c(t) = 2/3 - t $
However, $d_t(t)$ is not defined for $t>=1/3$. For all $t<1/3$ (the range over which the problem is ...
2
We don't know because the problem is not well defined. We know the cop moves with velocity $-1$ on the time interval $[0,\frac 23)$. We know the thief moves with velocity $-\frac 12$ on the same interval. Before $t=\frac 23$ the cop has not caught the thief. We don't know what to do at the limit because neither one is allowed to move to $0$. The stated ...
2
1
Certain connectivity rules must always apply. Every corner piece must be 1-connected (not 3-connected), since it could at most be 2-connected (which is illegal). Also, every 1-connected dot connects to a 3-connected dot because if it paired with a 1-connected dot, the two would form a disconnected/unreachable "island."
Next: Fill in lines you ...
1
If we do not require all the captured pawns to remain until the end of the jump sequence, then there is even a solution with 22 captures:
But I don't even think this is the maximum. Anyway, since this is not an answer to the question as stated, let me know if you think it should be deleted.
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