# Solve this logic puzzle **URL:** \ **Category:** The Game Room\ **Created:** [November 12, 2008, 3:20pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351 "2008-11-12T15:20:05Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [November 13, 2008, 3:25pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351/21 "2008-11-13T15:25:32Z")
> [@Quercus](#): > > Going to 3-D, at a minimum you could add two more (straight up and straight down), so we know that at least six unscathed is possible in 3-D. I suspect more are possible, but don’t know the exact solution. In 3-D Euclidean geometry, if you have a star close to\* each of the vertexes of a cube, and one at the centre of the cube, then all the other stars attack the one in the centre, and the one in the centre attacks one of the others. That gives 7 unattacked. There may be a way to get 8 unattacked. I think it would involve having 8 on the surface of a sphere, and (using latitude & longitude) having 3 on the equator at 0, -120 and +120 degrees, then 3 each on circles in the northern and southern hemispheres at -60, +60 and 180 degrees. I don’t know enough spherical trigonometry to know if you could do that so that each is more than 60 degrees from the other, so that the centre point is closer. \*It has to be close to, rather than at, so that all the distances are different. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [November 13, 2008, 3:30pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351/22 "2008-11-13T15:30:11Z")
> [@Giles](#): > > There may be a way to get 8 unattacked. I think it would involve having 8 on the surface of a sphere, and (using latitude & longitude) having 3 on the equator at 0, -120 and +120 degrees, then 3 each on circles in the northern and southern hemispheres at -60, +60 and 180 degrees. I don’t know enough spherical trigonometry to know if you could do that so that each is more than 60 degrees from the other, so that the centre point is closer. Sorry, of course you can! (1) 31N, 60E (2) 31N, 60W (3) 31N, 180 (4) 0, 0 (5) 0, 120E (6) 0, 120W (7) 31S, 60E (8) 31S, 60W (9) 31S, 180 (10) centre of the sphere. ---
**Author:** ![Balance](https://avatars.discourse-cdn.com/v4/letter/b/ccd318/32.png) [@Balance](https://boards.straightdope.com/u/Balance)\ **Post date:** [November 13, 2008, 7:02pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351/23 "2008-11-13T19:02:12Z")
> [@Giles](#): > > Sorry, of course you can! > (1) 31N, 60E > (2) 31N, 60W > (3) 31N, 180 > (4) 0, 0 > (5) 0, 120E > (6) 0, 120W > (7) 31S, 60E > (8) 31S, 60W > (9) 31S, 180 > (10) centre of the sphere. Doesn’t this violate the no-equidistant-pairs restriction? Is there enough leeway for offsets to avoid it? Interesting extrapolation, regardless–I was only playing with the original 7-system scenario. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [November 13, 2008, 9:57pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351/24 "2008-11-13T21:57:20Z")
> [@Balance](#): > > Doesn’t this violate the no-equidistant-pairs restriction? Is there enough leeway for offsets to avoid it? Interesting extrapolation, regardless–I was only playing with the original 7-system scenario. I was assuming very small offsets to allow for the non-equidistance. I think there’s at least 1 degree to play with there. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [November 13, 2008, 11:19pm UTC](https://boards.straightdope.com/t/solve-this-logic-puzzle/472351/25 "2008-11-13T23:19:17Z")
> [@](#): > > That argument works for three stars, but you’ve got to deal with more cases with a larger number. For example, if they were in order on a straight line, you could have A closest to B, B closest to C, C closest to D, D closest to E, E closest to F, and F closest to G, so there’s no pairing off. Now, if you bend that line into a circle, what stops G being closest to A? Obviously, it can’t be, but you have more possible cases to deal with than just pairing off. You need some sort of a inductive argument, where you take the stars out of contention two at a time, without assuming any pairing off in the original arrangement. And the beauty of the inductive argument is that it works for any odd number, so you could restate the puzzle as involving 1,000,000,001 stars, and it still works. > [@](#): > > Yes, but Little Nemo’s post assumed that you only have to deal with pairs and one triangle left over. To give a complete argument, you need to deal with other possible polygons (which is not too bad in the case of 7, but becomes difficult with a large number) or you have to have an explicit method of taking out stars two at a time. I’ll admit I worded my post very poorly. I don’t think that every even-numbered system will break down into pairs. I just figured it was possible so an even-numbered solution is possible. But an odd-numbered solution is not. As I wrote, an odd-numbered system would require at least one group of three stars where A is closer to B than C, B is closer to C than A, and C is closer to A than B. And such a triangle cannot be constructed. For such a triangle to exist AB\