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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.

$\frac{1}{1}, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}...$

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    $\begingroup$ 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, 2019 at 3:09
  • $\begingroup$ @KradCigol I've updated my post to clarify this question and another. $\endgroup$
    – Hazel へいぜる
    Feb 10, 2019 at 3:14
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    $\begingroup$ 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, 2019 at 6:22
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    $\begingroup$ 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, 2019 at 11:01
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    $\begingroup$ @Nautilus TAG YOU'RE IT! $\endgroup$
    – Riddler
    Feb 14, 2019 at 1:21

8 Answers 8

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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.

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    $\begingroup$ -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, 2019 at 16:32
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Hazel へいぜる
    Feb 10, 2019 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, 2019 at 4:07
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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.

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  • $\begingroup$ Very good attempt, but note the knowledge tag. :) +1 for taking the same route my daughter tries every time! HAHA $\endgroup$
    – Hazel へいぜる
    Feb 10, 2019 at 1:41
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The answer is

Yes

Why?

I have really long arms

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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. $1, 1.5, 1.75, 1.875, 1.9375...$ $\endgroup$
    – Hazel へいぜる
    Feb 13, 2019 at 17:01
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    $\begingroup$ @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, 2019 at 18:21
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Kinda lame, but

Yes, if you have low stamina and takes some breaks?

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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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You say it's an infinite plane so:

Let's take a well known graph that stretches off to infinity; $y = \frac{1}{x}$ enter image description here
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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