# Thread: A Fun puzzle

1. I am just back from a funeral (nobody close, thank heavens), so I want to have some FUN.

A friend posted the following puzzle on another forum to which I subscribe, so I copy/paste without permission or attribution. This is clearly against the rules here, but hey! I'm the boss, and, as a friend here pointed out, I can do as I please. So I quote:

A star system contains an odd number of planets. One (and only one) astronomer on each planet gazes at some other planet in this star system. Each astronomer looks at the planet closest to the planet on which the astronomer lives. Assuming the distances between pairs of planets are unique, show that at least one planet is not gazed upon by any of the astronomers.

I offered a slightly frivolous argument, using no math, which was nonetheless based on logic as I saw it. It was semi-rejected on the grounds that it was a non-constructivist argument, in that it used the well ordering principle.

So in the name of FUN, I invite you all to have a go using verbal logic (no math), then I'll show you my frivolity on the subject.

PS I did also attempt a math proof, which I can show if desired......  2.

3. This one's simple. Since there are an odd number of planets and a unique distance for every possible pair, then there must be at least one pair of planets that gaze upon each other, in that the distance between that pair is the smallest distance among pairs for both planets. There must also be a planet that has the "largest" distance to the closest planet to it, being that every distance is unique.

Now, the first planet will gaze upon the closest planet to it, which means that there must be a planet that is closer to the second planet for the second planet to gaze upon, and a closer planet for the third, and this will continue until you get to the last planet, which will have to turn around and gaze back at the planet behind it, being that it will be the closest planet to it, otherwise some other planet would have been gazing upon the last planet. Assuming a pairing pattern, since there are an odd number of planets, the planets can't perfectly pair off either, since there will always be one left over, which will be ungazed upon.

The planet with the largest closest distance, being that there are an odd number of planets, will always be ungazed upon.  4. Originally Posted by Arcane_Mathematician
Since there are an odd number of planets and a unique distance for every possible pair, then there must be at least one pair of planets that gaze upon each other, in that the distance between that pair is the smallest distance among pairs for both planets.
Yes, this is an application of the well-ordering principle. A good place to start, and where I did too.
There must also be a planet that has the "largest" distance to the closest planet to it,
I don't understand this phrase.

Assuming a pairing pattern, since there are an odd number of planets, the planets can't perfectly pair off either, since there will always be one left over, which will be ungazed upon.
I THINK I agree, but I cannot tell, since I have no idea what this means

The planet with the largest closest distance,
Anyway, here was my jocular (and I still think correct) reasoning.....

Since the spacing of planets is unique, there must a pair that are closer to each other than each is to any other planet, and they will thus be observing each other mutually. Nuke them both. There may be other such pairs, of course, nuke them too, and invite the remaining astronomers to retrain their telescopes accordingly to the same rule.

Once again, there will be at least one pair who are mutually closest, and therefore mutually observing. Blast them away, and once again invite the now-terrified astronomers to retrain their telescopes, also using the same rule. Continue to terrorize the system in this way.

Note that, as we are destroying pairs, or multiples of pairs of planets, if the starting number of planets is even, we will be left with two mutually observing planets - KABOOM, now there are none.

If, on the other hand, the starting number is odd, we will be left with one planet that is observed by no-one. (The fact that it has nothing to look at is immaterial to the puzzle)

Now some math.....

Suppose I let denote the number of planets in our system.

First note that the puzzle does NOT insist that, for even, there will be NO unobserved planets. But for simplicity I shall assume this.

Now note that no astronomer can observe himself, and therefore has planets in his "line if sight". Since he may (and must) observe only one of these, for this guy, let's call him , there will be planets un-observed by him.

Let's call the number of planets unobserved by as ; since the the number of observers is the same as the umber of planets, let's write the total number of planets unobserved by astronomers as Suppose that . Then clearly this planet is un-observed; Now let . Then these must be mutually observing, and .

Now, we may assume that, as the number of observers increases from 1 to N, the total number of un-observed planets decreases. So, since we may not assume that each planet is gazed upon by only one other, I claim there must be some number , less than , such that denotes the total number of planets unobserved by observers, so that for and for However, if is even, then I may always find some such that, say, .

But if is even, then is odd, and there is no number such that .  5. Using an argument similar to Arcane's one could say there is one most remote planet, i.e. whose nearest neighbor is farther away than anybody else's nearest neighbor. This planet would not be observed by any other planet, because every other planet has a nearer neighbor. But this would apply for an even number of planets as well.  6. With an even number of planets there's a chance that two planets will share the farthest nearest neighbor distance.  7. I don't get this at all: what exactly is meant by the " farthest nearest ....distance"?

Arcane said something similar. Please explain what this means  8. Originally Posted by Harold14370
But this would apply for an even number of planets as well.
At first glance I thought this would be true even for an even number of planets, but then I realized that with an even number of planets, you could just "pair" the planets so that the distance between a pair of planets is much smaller than the distance between one pair to the next pair. Then each planet would look at its pair and that planet would look back at it.  9. Originally Posted by Guitarist
I don't get this at all: what exactly is meant by the " farthest nearest ....distance"?

Arcane said something similar. Please explain what this means
You're right. Forget what I said. Two planets could have the same farthest nearest, if their nearest is to each other.  10. Originally Posted by Guitarist
I don't get this at all: what exactly is meant by the " farthest nearest ....distance"?

Arcane said something similar. Please explain what this means
Define x_p as the distance to p's nearest neighbor, then define X as the largest x_p. That's the farthest nearest neighbor, or whatever you want to call it. The planet with the largest distance between it and its nearest neighbor.  11. Magimaster said it quite well, sorry for my vague terms.  12. Originally Posted by Guitarist
I am just back from a funeral (nobody close, thank heavens), so I want to have some FUN.

A friend posted the following puzzle on another forum to which I subscribe, so I copy/paste without permission or attribution. This is clearly against the rules here, but hey! I'm the boss, and, as a friend here pointed out, I can do as I please. So I quote:

A star system contains an odd number of planets. One (and only one) astronomer on each planet gazes at some other planet in this star system. Each astronomer looks at the planet closest to the planet on which the astronomer lives. Assuming the distances between pairs of planets are unique, show that at least one planet is not gazed upon by any of the astronomers.

I offered a slightly frivolous argument, using no math, which was nonetheless based on logic as I saw it. It was semi-rejected on the grounds that it was a non-constructivist argument, in that it used the well ordering principle.

So in the name of FUN, I invite you all to have a go using verbal logic (no math), then I'll show you my frivolity on the subject.

PS I did also attempt a math proof, which I can show if desired......
There are 2n +1 planets for n =0,1,2,3,... with (2n+1)!/2!(2n-1)! different pairwise distance between planets. An astronomer on each planet observes his nearest neighbor. It is to be shown that at least one planet is not observed.

Proceed by induction. If n=0 the theorem is clear.

Assume inductively that there are 2n+1, n>0, planets and the theorem is true for 2k+1 planets whenever k<n.

Since the pairwise distances are unique there is a unique pair for which the distance is a minimum. Relabeling if necessary, we take this pair to corespond to planets 0 and 1. So astronomers on planets 0 and 1 are known to observer each other.

Case 1). No observers in planets 2,3,..., 2n+1 observer planet 0 or planet 1
In this case we consider the system of planet 2,3,...2n+1 and apply the inductive hypothesis to conclude that some planet is not observed.

Case 2) At least one observer on planets 2,3,..., 2n+1 observes planet 0 or planet 1.

Again relabeling if necessary we assume that planet 0 is observed by planet 2.

Each observer on planets the 2n planets 2,3, ...2n+1 observes only one planet.
We know that planet 2 observes planet 0. So there are 2n-1 observers available in the observer pool and since each observer observes only one planet, at most 2n-1 of the 2n planets are observed.

QED  13. Ya, well, the only "slight" difference between your proof and mine is that yours is rigourous.

PS Thank you for sharing that  Bookmarks
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