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Pondering about a very real mosquito in my room I came up with this puzzle:

You are in a room of dimensions $10$ m $\times$ $10$ m $\times$ $3$ m. There is also a (point-shaped) mosquito in the room but you do not know where (the mosquito is invisible to you).

You have an electric flyswatter that you can pass through the room and that will kill the mosquito if the mosquito is inside a volume of air swiped by the flyswatter.

Your quest to kill the mosquito proceeds as follows:

  1. The mosquito moves any distance up to 1 m from its current position
  2. You swipe any (not necessarily continuous) volume of air up to $V_{max}$ with the flyswatter

The question is: What is the minimum value of $V_{max}$ that guarantees that there is a strategy that allows you to kill the mosquito in a finite number of tries?

It is trivial to show that $V_{max}$ is strictly less than the volume of the room for any room of finite dimensions and any mosquito speed less than the maximum dimension of the room, but I have not managed to find a method to establish the minimum.

Does this puzzle reduce to another (existing) puzzle with a known answer maybe?

Edit: Proof that $V_{max} \le \frac{1}{2} V_{room}$:

Starting from any wall, swipe the "first half" of the room. If you did not kill the mosquito, swipe the other half of the room. If you still did not kill the mosquito, you know that the mosquito must have just crossed from one half to the other, which means it is now at most 1 m from the plane separating the two halves of the room. You can now easily swipe a volume that is guaranteed to include the mosquito.

Edit 2: Proof that $V_{max} \le (1 + \epsilon) m × 10 m × 3m$:

As noted by @ccorn, just start from the smallest face of the room with a plane $(1+\epsilon)$ m thick and proceed to the opposite wall of the room in increments of $\epsilon$.

This appears to be where the obvious strategies that work by "confining" the mosquito to a decreasing amount of space end. The remaining question is: Is $V_{max}$ less than the bound proven in Edit 2?

Edit 3: Proof that $V_{max} \gt \frac{4}{3} \pi m^3$:

If $V_{max} \le \frac{4}{3} \pi m^3$ (the volume of a sphere with radius 1 m), every point that you swipe will have an unswiped point at most 1 m away. That means by swiping you do not gain any information about the next position of the mosquito.

Community
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    I take it the mosquito is invisible? Or can we see it on some sort of time-delay? – Chris Eagle Jul 28 '13 at 19:55
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    The mosquito is invisible. You will know when you killed it from the popping sound. –  Jul 28 '13 at 19:56
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    Also, there is no restriction on the shape of the volume of air that you swipe (obviously, it has to be measurable, else it would be difficult to say it is less than $V_{max}$). –  Jul 28 '13 at 19:57
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    (Very) superficially related problem: http://en.wikipedia.org/wiki/Angel_problem. –  Jul 28 '13 at 19:59
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    More closely related: http://en.wikipedia.org/wiki/Princess_and_monster_game. –  Jul 28 '13 at 20:04
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    I'm guessing that the mosquito is supposed to be a single point in space? (If the mosquito's body has nonempty interior, we can kill it in one turn by swiping the set of points with rational coordinates, so $\min V_\max=0$.) – Chris Culter Jul 28 '13 at 20:05
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    @ChrisCulter: Correct, the mosquito is point-shaped (edited to clarify) – which immediately explains why it is invisible. –  Jul 28 '13 at 20:08
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    In addtion to the proof in the edit: $2$m times the smallest face is enough - just start from one wall and move to the opposite wall in $1$m steps. – Hagen von Eitzen Jul 28 '13 at 20:25
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    $V_{\text{max}}\leq V_{\text{c}}(\epsilon)$ for any $\epsilon>0$ where $V_{\text{c}}(\epsilon) = 1\text{m}(1+\epsilon)\times 10\text{m}\times3\text{m}$. Beginning at one side, just swipe the room with increments of $\epsilon\text{m}$ along the first dimension. – ccorn Jul 28 '13 at 20:25
  • True! But is there a volume less than $V_{c}(\epsilon)$ that suffices? –  Jul 28 '13 at 20:29
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    Alternate formulation: Let $E(A)$ denote the erosion operator (1 m radius) applied to a region $A$ inside the room. The goal is to construct a sequence $C_{n}= E(S_n \cup C_{n-1})$ with $S_n$ (swiped region in iteration $n$) having fixed volume $V$, and with $C_{n}$ ("covered" region, where we know for sure the mosquito is not) increasing, and eventually filling the room. – leonbloy Aug 03 '13 at 20:48
  • Considering my comment above, the bound of comment 2 should be tight. That strategy is what gives less "shell thickness" to be eroded, and hence is optimal. Any starting region with less than that volume disappears after erosion, hence we cannot make any progress. – leonbloy Aug 03 '13 at 23:16

1 Answers1

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Not really an answer, but it can serve as a starting point.

As noted in the comments (Hagen von Eitzen and ccorn) -and included in the edited question- a swipe volume $V > V_0= x \times y \times d$ (where $x,y$ are the smallest sides of the box and $d$ is the maximum distance travelled by the mosquito; here $x \times y \times d = 3 m \times 10 m \times 1 m=30 m^3$) is enough. One starts swiping a slice adjacent to the smallest face of the box.

In general: assume that before swipe $n$ we knew for sure that the mosquito was not in the excluded volume $C_{n}$, and assume that we swipe (without success) the region $S_n$. Now, we know that the mosquito is not in the region $S_n \cup C_n$; but in the next iteration it can be in any place a distance $d$, hence all we know is that $C_{n+1} = \psi_d(S_n \cup C_n)$, where $\psi_d(X)$ is the erosion morphological operator (it substracts from region $X$ all points that are at distance less than $d$ from its complement $\bar X$).

The goal is to construct an increasing sequence $C_n$ that eventually fills the box. That's possible with the above construction.

Now, suppose that $V<V_0$. Then, it's clear that the above approach (start with a slice on the smaller face and try to make it thicker) does not work: the erosion removes all the excluded volume and we cannot make progress. What remains (what seems difficult) is to show that there is no better strategy, in the long run. For example, one could try starting with a corner (tethraedron) instead of a slice, so that the first erosion leaves us with some non-empty excluded volume, but soon (at least, when we've covered half of the box) we loose that advantage and we cannot progress.

Specifically, it would be enough to prove this (seemingly true) conjecture: Let $A$ be any region inside the box, with volume greater than half of the box (and less than the full box). Then, the "eroded volume" (points with distance less than $d$ from $\bar A$) is at least that of the corresponding slice parallel to the smaller face: $V(A) - V(\psi_d(A)) \ge x\times y\times d $

leonbloy
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  • the conjecture can be shown at least for the special case that the lengths of $(A-S_n) \cap ({x}\times{y}\times[0,10])$ are between $1$ and $9$, using a Fubini type argument. This is sort of reducing to the 1-d version of the problem, where you can see that if the room is not completely full, yet volume is greater than 1, then you can erode at least 1 unit of length. – Louis Aug 06 '13 at 15:58
  • oops, meant $(A\cup S_n)$ in the comment above – Louis Aug 06 '13 at 17:29