Let's play tag!
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I have a short and bitter-sweet riddle for the community this evening. As always, I hope you enjoy it; good luck!
I want to play a game of tag; however, there are some very important rules:
- We both start at the same point.
- You cannot tag until the third turn or greater.
- Each turn you take one step.
- Each turn I double the number of steps I take.
- Our stride is exactly the same.
- We travel in a straight line.
- We both take one initial step.
- Our game is played on an infinitely flat plane
- All steps are forwards, never backwards.
- Assume we are infinitely traveling in a straight line.
Can you tag me?
Please explain your answer.
Hint:
Your step is on top, mine on bottom. Please note the
math
tag.
$frac{1}{1}, frac{1}{2}, frac{1}{4}, frac{1}{8}, frac{1}{16}, frac{1}{32}...$
riddle mathematics knowledge lateral-thinking
$endgroup$
|
show 6 more comments
$begingroup$
I have a short and bitter-sweet riddle for the community this evening. As always, I hope you enjoy it; good luck!
I want to play a game of tag; however, there are some very important rules:
- We both start at the same point.
- You cannot tag until the third turn or greater.
- Each turn you take one step.
- Each turn I double the number of steps I take.
- Our stride is exactly the same.
- We travel in a straight line.
- We both take one initial step.
- Our game is played on an infinitely flat plane
- All steps are forwards, never backwards.
- Assume we are infinitely traveling in a straight line.
Can you tag me?
Please explain your answer.
Hint:
Your step is on top, mine on bottom. Please note the
math
tag.
$frac{1}{1}, frac{1}{2}, frac{1}{4}, frac{1}{8}, frac{1}{16}, frac{1}{32}...$
riddle mathematics knowledge lateral-thinking
$endgroup$
1
$begingroup$
Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
$endgroup$
– Krad Cigol
Feb 10 at 3:09
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@KradCigol I've updated my post to clarify this question and another.
$endgroup$
– PerpetualJ
Feb 10 at 3:14
1
$begingroup$
Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
$endgroup$
– Dr Xorile
Feb 10 at 6:22
1
$begingroup$
I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
$endgroup$
– Steve Bennett
Feb 10 at 11:01
1
$begingroup$
@Nautilus TAG YOU'RE IT!
$endgroup$
– Riddler
Feb 14 at 1:21
|
show 6 more comments
$begingroup$
I have a short and bitter-sweet riddle for the community this evening. As always, I hope you enjoy it; good luck!
I want to play a game of tag; however, there are some very important rules:
- We both start at the same point.
- You cannot tag until the third turn or greater.
- Each turn you take one step.
- Each turn I double the number of steps I take.
- Our stride is exactly the same.
- We travel in a straight line.
- We both take one initial step.
- Our game is played on an infinitely flat plane
- All steps are forwards, never backwards.
- Assume we are infinitely traveling in a straight line.
Can you tag me?
Please explain your answer.
Hint:
Your step is on top, mine on bottom. Please note the
math
tag.
$frac{1}{1}, frac{1}{2}, frac{1}{4}, frac{1}{8}, frac{1}{16}, frac{1}{32}...$
riddle mathematics knowledge lateral-thinking
$endgroup$
I have a short and bitter-sweet riddle for the community this evening. As always, I hope you enjoy it; good luck!
I want to play a game of tag; however, there are some very important rules:
- We both start at the same point.
- You cannot tag until the third turn or greater.
- Each turn you take one step.
- Each turn I double the number of steps I take.
- Our stride is exactly the same.
- We travel in a straight line.
- We both take one initial step.
- Our game is played on an infinitely flat plane
- All steps are forwards, never backwards.
- Assume we are infinitely traveling in a straight line.
Can you tag me?
Please explain your answer.
Hint:
Your step is on top, mine on bottom. Please note the
math
tag.
$frac{1}{1}, frac{1}{2}, frac{1}{4}, frac{1}{8}, frac{1}{16}, frac{1}{32}...$
riddle mathematics knowledge lateral-thinking
riddle mathematics knowledge lateral-thinking
edited Feb 13 at 17:02
PerpetualJ
asked Feb 10 at 1:35
PerpetualJPerpetualJ
4,092547
4,092547
1
$begingroup$
Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
$endgroup$
– Krad Cigol
Feb 10 at 3:09
$begingroup$
@KradCigol I've updated my post to clarify this question and another.
$endgroup$
– PerpetualJ
Feb 10 at 3:14
1
$begingroup$
Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
$endgroup$
– Dr Xorile
Feb 10 at 6:22
1
$begingroup$
I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
$endgroup$
– Steve Bennett
Feb 10 at 11:01
1
$begingroup$
@Nautilus TAG YOU'RE IT!
$endgroup$
– Riddler
Feb 14 at 1:21
|
show 6 more comments
1
$begingroup$
Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
$endgroup$
– Krad Cigol
Feb 10 at 3:09
$begingroup$
@KradCigol I've updated my post to clarify this question and another.
$endgroup$
– PerpetualJ
Feb 10 at 3:14
1
$begingroup$
Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
$endgroup$
– Dr Xorile
Feb 10 at 6:22
1
$begingroup$
I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
$endgroup$
– Steve Bennett
Feb 10 at 11:01
1
$begingroup$
@Nautilus TAG YOU'RE IT!
$endgroup$
– Riddler
Feb 14 at 1:21
1
1
$begingroup$
Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
$endgroup$
– Krad Cigol
Feb 10 at 3:09
$begingroup$
Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
$endgroup$
– Krad Cigol
Feb 10 at 3:09
$begingroup$
@KradCigol I've updated my post to clarify this question and another.
$endgroup$
– PerpetualJ
Feb 10 at 3:14
$begingroup$
@KradCigol I've updated my post to clarify this question and another.
$endgroup$
– PerpetualJ
Feb 10 at 3:14
1
1
$begingroup$
Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
$endgroup$
– Dr Xorile
Feb 10 at 6:22
$begingroup$
Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
$endgroup$
– Dr Xorile
Feb 10 at 6:22
1
1
$begingroup$
I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
$endgroup$
– Steve Bennett
Feb 10 at 11:01
$begingroup$
I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
$endgroup$
– Steve Bennett
Feb 10 at 11:01
1
1
$begingroup$
@Nautilus TAG YOU'RE IT!
$endgroup$
– Riddler
Feb 14 at 1:21
$begingroup$
@Nautilus TAG YOU'RE IT!
$endgroup$
– Riddler
Feb 14 at 1:21
|
show 6 more comments
8 Answers
8
active
oldest
votes
$begingroup$
So, if we both go
infinitely many times
then your position will be
$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac{1}{2}$
and my position will be
$1+1+1+1+1+dots= -frac{1}{2}$, shown by Dirichlet regularization
hence
I tag you.
$endgroup$
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
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– Kevin
Feb 13 at 16:32
add a comment |
$begingroup$
The answer is
Yes.
We can do this because:
Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.
$endgroup$
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
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@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
add a comment |
$begingroup$
The answer is:
Yes. Either A tags B before the game starts,
or
A's steps are twice as long as B's, and so tag occurs at the end of round 1.
$endgroup$
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
add a comment |
$begingroup$
The answer is
Yes
Why?
I have really long arms
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add a comment |
$begingroup$
It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply
No.
The distance $d_{you}$ you've traveled after $n$ turns is $d_{you} = n^2$, and the distance $d_{me}$ I've traveled is $d_{me} = n$.
Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.
One important thing to note that the accepted answer fails to account for:
While $sum_{n=1}^{infty} n^2$ and $sum_{n=1}^{infty} n$ are both equal to $-frac{1}{2}$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_{n=1}^{infty} n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_{n=1}^{infty} n = infty$, since its sequence of partial sums increases monotonically without bound."
The reason this is the more appropriate way to approach the puzzle is
At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.
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$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
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– Kevin
Feb 14 at 18:21
add a comment |
$begingroup$
Kinda lame, but
Yes, if you have low stamina and takes some breaks?
$endgroup$
add a comment |
$begingroup$
You say it's an infinite plane so:
Let's take a well known graph that stretches off to infinity; $y = frac{1}{x}$
As the values in x reach positive infinity, they become closer and closer to y = 0, until eventually it flips over to negative y, and the x becomes negative infinity, where the x values then increase to near y = 0.
One could liken this graph to your problem, I believe that as the number of steps taken doubles every time, it would go from a positively infinite number back to negatively infinite, and eventually catch back up again with the other player who only moves 1 step at a time.
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add a comment |
$begingroup$
The answer is
yes
Explanation:
you did not mention what type of geometric plane so it might be curved plane where one meet again
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add a comment |
Your Answer
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So, if we both go
infinitely many times
then your position will be
$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac{1}{2}$
and my position will be
$1+1+1+1+1+dots= -frac{1}{2}$, shown by Dirichlet regularization
hence
I tag you.
$endgroup$
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
add a comment |
$begingroup$
So, if we both go
infinitely many times
then your position will be
$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac{1}{2}$
and my position will be
$1+1+1+1+1+dots= -frac{1}{2}$, shown by Dirichlet regularization
hence
I tag you.
$endgroup$
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
add a comment |
$begingroup$
So, if we both go
infinitely many times
then your position will be
$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac{1}{2}$
and my position will be
$1+1+1+1+1+dots= -frac{1}{2}$, shown by Dirichlet regularization
hence
I tag you.
$endgroup$
So, if we both go
infinitely many times
then your position will be
$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac{1}{2}$
and my position will be
$1+1+1+1+1+dots= -frac{1}{2}$, shown by Dirichlet regularization
hence
I tag you.
answered Feb 11 at 15:22
athinathin
8,58722776
8,58722776
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
add a comment |
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
5
5
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
$begingroup$
-1 As I argue in my answer, the naive approaches that give $sum_{x}^{infty} n^2$ and $sum_{x}^{infty} n$ equal to $infty$ are still correct in some contexts, and those contexts are actually the more reasonable approaches for this puzzle.
$endgroup$
– Kevin
Feb 13 at 16:32
add a comment |
$begingroup$
The answer is
Yes.
We can do this because:
Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.
$endgroup$
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
add a comment |
$begingroup$
The answer is
Yes.
We can do this because:
Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.
$endgroup$
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
add a comment |
$begingroup$
The answer is
Yes.
We can do this because:
Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.
$endgroup$
The answer is
Yes.
We can do this because:
Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.
edited Feb 10 at 3:17
answered Feb 10 at 3:10
Krad CigolKrad Cigol
1,056210
1,056210
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
add a comment |
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
Ahhh, nice adjustment; you're close with your thinking, serves me right for not thinking about lateral solutions. The real question is, would I escape, even if the plane is infinite? This is definitely a suitable answer to the way the question was originally phrased so don't remove the edit, just edit with a new guess. I'll be awarding $+50$ bounty for thinking outside the box with your spheroid solution.
$endgroup$
– PerpetualJ
Feb 10 at 3:37
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
$begingroup$
@PerpetualJ: Thanks! In the infinite plane - straight line case, it does seem like you’ll escape. I’ll see if I can figure it out.
$endgroup$
– Krad Cigol
Feb 10 at 4:07
add a comment |
$begingroup$
The answer is:
Yes. Either A tags B before the game starts,
or
A's steps are twice as long as B's, and so tag occurs at the end of round 1.
$endgroup$
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
add a comment |
$begingroup$
The answer is:
Yes. Either A tags B before the game starts,
or
A's steps are twice as long as B's, and so tag occurs at the end of round 1.
$endgroup$
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
add a comment |
$begingroup$
The answer is:
Yes. Either A tags B before the game starts,
or
A's steps are twice as long as B's, and so tag occurs at the end of round 1.
$endgroup$
The answer is:
Yes. Either A tags B before the game starts,
or
A's steps are twice as long as B's, and so tag occurs at the end of round 1.
answered Feb 10 at 1:40
JonMark PerryJonMark Perry
20.7k64099
20.7k64099
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
add a comment |
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
$endgroup$
– PerpetualJ
Feb 10 at 1:41
$begingroup$
Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA
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– PerpetualJ
Feb 10 at 1:41
add a comment |
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The answer is
Yes
Why?
I have really long arms
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add a comment |
$begingroup$
The answer is
Yes
Why?
I have really long arms
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add a comment |
$begingroup$
The answer is
Yes
Why?
I have really long arms
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The answer is
Yes
Why?
I have really long arms
answered Feb 11 at 13:53
CubemasterCubemaster
1,677334
1,677334
add a comment |
add a comment |
$begingroup$
It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply
No.
The distance $d_{you}$ you've traveled after $n$ turns is $d_{you} = n^2$, and the distance $d_{me}$ I've traveled is $d_{me} = n$.
Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.
One important thing to note that the accepted answer fails to account for:
While $sum_{n=1}^{infty} n^2$ and $sum_{n=1}^{infty} n$ are both equal to $-frac{1}{2}$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_{n=1}^{infty} n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_{n=1}^{infty} n = infty$, since its sequence of partial sums increases monotonically without bound."
The reason this is the more appropriate way to approach the puzzle is
At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.
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This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
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– PerpetualJ
Feb 13 at 17:01
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@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
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– Kevin
Feb 14 at 18:21
add a comment |
$begingroup$
It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply
No.
The distance $d_{you}$ you've traveled after $n$ turns is $d_{you} = n^2$, and the distance $d_{me}$ I've traveled is $d_{me} = n$.
Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.
One important thing to note that the accepted answer fails to account for:
While $sum_{n=1}^{infty} n^2$ and $sum_{n=1}^{infty} n$ are both equal to $-frac{1}{2}$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_{n=1}^{infty} n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_{n=1}^{infty} n = infty$, since its sequence of partial sums increases monotonically without bound."
The reason this is the more appropriate way to approach the puzzle is
At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.
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$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
$endgroup$
– Kevin
Feb 14 at 18:21
add a comment |
$begingroup$
It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply
No.
The distance $d_{you}$ you've traveled after $n$ turns is $d_{you} = n^2$, and the distance $d_{me}$ I've traveled is $d_{me} = n$.
Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.
One important thing to note that the accepted answer fails to account for:
While $sum_{n=1}^{infty} n^2$ and $sum_{n=1}^{infty} n$ are both equal to $-frac{1}{2}$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_{n=1}^{infty} n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_{n=1}^{infty} n = infty$, since its sequence of partial sums increases monotonically without bound."
The reason this is the more appropriate way to approach the puzzle is
At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.
$endgroup$
It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply
No.
The distance $d_{you}$ you've traveled after $n$ turns is $d_{you} = n^2$, and the distance $d_{me}$ I've traveled is $d_{me} = n$.
Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.
One important thing to note that the accepted answer fails to account for:
While $sum_{n=1}^{infty} n^2$ and $sum_{n=1}^{infty} n$ are both equal to $-frac{1}{2}$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_{n=1}^{infty} n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_{n=1}^{infty} n = infty$, since its sequence of partial sums increases monotonically without bound."
The reason this is the more appropriate way to approach the puzzle is
At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.
edited Feb 14 at 21:27
answered Feb 13 at 16:29
KevinKevin
1,0231821
1,0231821
$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
$endgroup$
– Kevin
Feb 14 at 18:21
add a comment |
$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
$endgroup$
– Kevin
Feb 14 at 18:21
$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
This is a very thorough answer! My puzzle was technically focused on convergence of the reciprocals of powers of two where starting from 1, as $n$ steps approaches infinity, we converge on 2. $1, 1.5, 1.75, 1.875, 1.9375...$
$endgroup$
– PerpetualJ
Feb 13 at 17:01
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
$endgroup$
– Kevin
Feb 14 at 18:21
$begingroup$
@PerpetualJ I don't think that convergence has any useful meaning in your puzzle. The fraction $frac{1}{n^2}$ is the fraction of the distance you travel that I travel on turn $n$. It makes no meaningful sense to sum up those fractions. What would the sum represent?
$endgroup$
– Kevin
Feb 14 at 18:21
add a comment |
$begingroup$
Kinda lame, but
Yes, if you have low stamina and takes some breaks?
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add a comment |
$begingroup$
Kinda lame, but
Yes, if you have low stamina and takes some breaks?
$endgroup$
add a comment |
$begingroup$
Kinda lame, but
Yes, if you have low stamina and takes some breaks?
$endgroup$
Kinda lame, but
Yes, if you have low stamina and takes some breaks?
answered Feb 11 at 12:21
Henrique ValleHenrique Valle
3665
3665
add a comment |
add a comment |
$begingroup$
You say it's an infinite plane so:
Let's take a well known graph that stretches off to infinity; $y = frac{1}{x}$
As the values in x reach positive infinity, they become closer and closer to y = 0, until eventually it flips over to negative y, and the x becomes negative infinity, where the x values then increase to near y = 0.
One could liken this graph to your problem, I believe that as the number of steps taken doubles every time, it would go from a positively infinite number back to negatively infinite, and eventually catch back up again with the other player who only moves 1 step at a time.
$endgroup$
add a comment |
$begingroup$
You say it's an infinite plane so:
Let's take a well known graph that stretches off to infinity; $y = frac{1}{x}$
As the values in x reach positive infinity, they become closer and closer to y = 0, until eventually it flips over to negative y, and the x becomes negative infinity, where the x values then increase to near y = 0.
One could liken this graph to your problem, I believe that as the number of steps taken doubles every time, it would go from a positively infinite number back to negatively infinite, and eventually catch back up again with the other player who only moves 1 step at a time.
$endgroup$
add a comment |
$begingroup$
You say it's an infinite plane so:
Let's take a well known graph that stretches off to infinity; $y = frac{1}{x}$
As the values in x reach positive infinity, they become closer and closer to y = 0, until eventually it flips over to negative y, and the x becomes negative infinity, where the x values then increase to near y = 0.
One could liken this graph to your problem, I believe that as the number of steps taken doubles every time, it would go from a positively infinite number back to negatively infinite, and eventually catch back up again with the other player who only moves 1 step at a time.
$endgroup$
You say it's an infinite plane so:
Let's take a well known graph that stretches off to infinity; $y = frac{1}{x}$
As the values in x reach positive infinity, they become closer and closer to y = 0, until eventually it flips over to negative y, and the x becomes negative infinity, where the x values then increase to near y = 0.
One could liken this graph to your problem, I believe that as the number of steps taken doubles every time, it would go from a positively infinite number back to negatively infinite, and eventually catch back up again with the other player who only moves 1 step at a time.
answered Feb 11 at 12:48
AHKieranAHKieran
5,5551144
5,5551144
add a comment |
add a comment |
$begingroup$
The answer is
yes
Explanation:
you did not mention what type of geometric plane so it might be curved plane where one meet again
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add a comment |
$begingroup$
The answer is
yes
Explanation:
you did not mention what type of geometric plane so it might be curved plane where one meet again
$endgroup$
add a comment |
$begingroup$
The answer is
yes
Explanation:
you did not mention what type of geometric plane so it might be curved plane where one meet again
$endgroup$
The answer is
yes
Explanation:
you did not mention what type of geometric plane so it might be curved plane where one meet again
edited Feb 22 at 6:21
athin
8,58722776
8,58722776
answered Feb 10 at 13:19
user56760
add a comment |
add a comment |
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1
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Can paths be different? Will you try your best not to get caught, or will you take a path where you can get caught?
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– Krad Cigol
Feb 10 at 3:09
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@KradCigol I've updated my post to clarify this question and another.
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– PerpetualJ
Feb 10 at 3:14
1
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Are all steps the same length? Are the turns sequential? Each turn is a straight line, but can you reorient between turns?
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– Dr Xorile
Feb 10 at 6:22
1
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I'm confused by both parties starting at the same point. If we're next to each other, hasn't the tagger already won?
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– Steve Bennett
Feb 10 at 11:01
1
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@Nautilus TAG YOU'RE IT!
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– Riddler
Feb 14 at 1:21