If you're on a Linux system, cat file.txt | tr 'A-Za-z' 'N-ZA-Mn-za-m' should do the trick, where the text you want to translate is in file.txt. At 04:47 AM 3/31/2016, Tom Rokicki wrote:

It's rot13. The name should tell you everything. Most good mail readers let you ungarble it with a single keystroke.

On Thu, Mar 31, 2016 at 7:35 AM, Simon Plouffe <simon.plouffe@gmail.com> wrote:

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Hello, what language is this : V jnf unccl jvgu guvf fbyhgvba, hagvy V abgvprq gung jura A vf guerr, guvf tvirf frira qvssrerag beqrevatf, naq gurer ner bayl fvk gbgny beqrevatf bs guerr guvatf! ????

Google translate cannot even recognize what it is,

Ok, I see, you have something big in your mouth and this is english, but unable to pronounce it correctly ??

Have a nice day.

Simon Plouffe

2016-03-31 13:17 GMT+02:00 Andy Latto <andy.latto@pobox.com>:

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On Wed, Mar 30, 2016 at 11:35 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:

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But now I'm wondering about the more general case. For instance the number of sort orders of distances to N random points in a plane, as measured from any point in the plane. I think it would not be a constant.

Here's a fallacious proof I came up with for the wrong answer. I had fun figuring out the fallacy, so maybe some other funsters will too. There's a hint as to what's wrong in rot13 at the bottom; I'll post the fallacy, together with the correct answer, in a day or two.

For any two of the points, A, B, draw the perpendicular bisector of AB. If we measure from a point on one side of this bisector, A will come before B in the ordering, while if we measure from a point on the other side, B will come before A. So with N points, we have N(N-1)/2 bisectors dividing up the plane into regions, and we'll get a different ordering for each region. If we have K lines in the plane in general position, they divide the plane into K(K + 1)/2 + 1 regions; this is easily seen by induction; when we add the K'th line, it intersects the existing K-1 lines in K points, dividing it into K pieces (K-2 segments and 2 rays); each piece divides one planar region in 2, so we have added K regions, so K lines give us the 1 region we start with plus 1 + 2 + 3 + .... + K more.

So with N points, we have N(N-1)/2 perpendicular bisectors, so they divide the plane into

1 + (N(N-1)/2)((N(N-1)/2) + 1) / 2 = (N^4 - 2 N^3 + 3 N^2 - 2N + 8)/8 possible orderings.

V jnf unccl jvgu guvf fbyhgvba, hagvy V abgvprq gung jura A vf guerr, guvf tvirf frira qvssrerag beqrevatf, naq gurer ner bayl fvk gbgny beqrevatf bs guerr guvatf!

Andy -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --