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October 21

dinner party problem

This problem was relayed to me: I want to schedule 14 dinners, one dinner for each week for 14 weeks. Each dinner will have one host, and two guests. There are 14 people in the pool, so dinner one will be hosted by person one, and so on. I want to guarantee that no dinner has the same three people. Once I gave up trying to remember what I've read about the Fano plane and the like, a solution quickly came to me: Guests[i] = (i+1) mod 14, (i+3) mod 14. No two participants ever meet twice. Now here's my question: I suspect that all solutions with this property are isomorphic; is there an easy way to confirm or deny that? —Tamfang (talk) 01:34, 21 October 2008 (UTC)[reply]

I'm not entirely sure how you mean isomorphic in this case, but if you mean something like you can always reindex the participants in such a way that the resulting structure looks the same as what you have described, then no. Break the 14 people into two groups of 7. Perform your same trick modulo 7 on each group. Now you have a different solution that partitions all dinner into two non-overlapping sets. In which case it is trivially true that no reindexing of participants could regenerate your map. Dragons flight (talk) 02:45, 21 October 2008 (UTC)[reply]

Thanks. —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

I thought of another counterexample before reading the above, by considering star polygons. My solution can be drawn as a 14/3 star inscribed in a 14-gon. If the second guest is (i+4) rather than (i+3), the 14/3 star becomes two 7/2 stars. If two graphs are isomorphic, their adjacency matrices ought to have the same eigenvalues, and these two do not. —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

So then I generated the eigenvalues of the adjacency matrices of all pair-avoiding solutions of the form (i+p) mod 14, (i+q) mod 14 and found that the three mentioned above are the only ones! —Tamfang (talk) 05:02, 21 October 2008 (UTC)[reply]

Points on a plane

On a two-dimensional plane, is the average distance between any one point and all the others a constant function relative to the area of the plane? Nadando (talk) 03:54, 21 October 2008 (UTC)[reply]

A plane is infinite. Do you mean a region in the plane? —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

I guess. Yes. Nadando (talk) 04:04, 21 October 2008 (UTC)[reply]

I would say that it also depend on the shape of the region. Think of the averge distance between two persons on a train, compared with the average distance of two persons in the hall of the station, assuming the area is the same...--79.38.22.37 (talk) 07:19, 21 October 2008 (UTC)And the driver of the train is in average twice as far from all the other people as the inspector in the middle is from the others guys, that's why he stays there. --79.38.22.37 (talk) 08:30, 21 October 2008 (UTC)[reply]

Let's compare two different shapes. If we take a disk of radius r then the average distance of any point in the disk from the centre of the disk is

{\frac {1}{\pi r^{2}}}\int \limits _{0}^{r}2\pi x^{2}\,dx={\frac {2r}{3}}

On the other hand, if we take a long, thin rectangle with length a and width b where a>>b, then the average distance of any point in the rectangle from the centre of the rectangle is approximately a/4.

Now by making b equal to πr²/a, we can create a long thin rectangle with the same area as our disk, but with an average distance from centre as large as we like. So average distance is not related to area in a simple way. Gandalf61 (talk) 09:02, 21 October 2008 (UTC)[reply]

Significant figures

I understand the trailing zeros bit, what i don't understand is leading-zeroes. The article doesn't explain 'why' you would ignore leading-zeroes. If i've got 2 pieces of data - one that is 0.0213 and another which is 0.0000213 then are they both 213 when expressed to 3 significant figures? If so, how does this overcome the issue that one number is (to me) quite markedly larger than the other? Does the significant figures thing only work/is only relevant when all results showing as expected to be within a given range of each other? I.e. if everytime you do experient X you get a result that is 0.0000# (where # is a number) then is that when significant figures serve their purpose? 194.221.133.226 (talk) 15:25, 21 October 2008 (UTC)[reply]

To 3 significant figures these numbers are 0.0213 and 0.0000213. Is our article not clear on this? Algebraist 16:11, 21 October 2008 (UTC)[reply]

The concept of "significant figures" does not describe the magnitude of a number; it merely describes the precision. The number 0.0213 expressed to three significant figures is 0.0213; expressed to two significant figures, it's 0.021; and expressed to one significant figure, it's 0.02. The idea is that a measurement of "2.5 cm" is not more precise than a measurement of "2.5 miles"—the difference between 2.5 cm and 2.6 cm, for an object which would be measured at that scale, is equivalent to the difference between 2.5 miles and 2.6 miles, for an object which would be measured at that scale. —Bkell (talk) 16:15, 21 October 2008 (UTC)[reply]

I find that significant figures are much easier to determine if you write the number in scientific notation. 2.13*10^-2 and 2.13*10^-5 clearly both have three significant figures, as does 2.13*10^2. --Carnildo (talk) 20:48, 21 October 2008 (UTC)[reply]