A flag is defined via 3 colors (ordering doesn't matter)with nine colors what is the maximum number of flags one can have given that any two flags can share at most 1 color?

<t>Thank you for your answers.Indeed Gauss-Patterson generalises in someway Gauss-Kronrod. These two methods allows for some error measure.This "property" helps one make a quadrature adaptive.In some cases one needs higher dimension rules. for eg a 255 points Gauss Patterson rule. Those weights and ...

Hi,Do you have a documentation about the Gauss-Patterson integration method? (how to generate the quadrature points)All i can find in internet is 127 points nothing more!Thx

Hi all,Does anyone have this article : "some remarks on the supermodular order" by Muller and Scarsini?Thank you very much

Hi all,Does anyone know an efficient way to invert a bi-dimensional gaussian? Find x such that N(x,y,rho) = z, N being the cumulative density function of a bi-dimensional gaussianThank you

A formulae that allows to generalise the result to any dimension is : (n being the number of points and d the dimension of our vectorial space)P(n,d) = 0.5 * (P(n-1,d) + P(n-1,d-1)

the problem is that you make your reasonning given an hemisphere, while it's otherwise : given two or three points you can find an hemisphere...

There are effectively approximations but also exact formulaes and for a d-dimensionnal sphere.for a sphere it's n²-n+2 * 2^(1-n)

<r>You got it right zarnywhoop in the circle case.An other way of doing it is dividing the perimeter into 2p equal arcs, then only considering the semi-circles composed of the arcs. One can apply theinclusion-exclusion principle and only the let k go to infinity.This reasonning, as well as a discuss...

How stupid am i?you're right!I think that the sphere thing is a little hard. Try the same exercise with a circle instead of a sphere and a semicircle instead of an hemisphere.start with 3 points

Maybe i am wreong : Let's consider a sphere of radius one. For me a great circle is a circle of radius one. Two points A and B can't define one great circle unless the distance AB = 2.

Farmer : for one two or three points, they always are on the same hemisphere.adgyboy is right because aym is not considering all the cases. It's just a lower bound. The first two don't determine a great circle in all cases.

Maybe it's a classical one!You have n points on a sphere, what is the probability that all of them are on the same hemisphere?Sorry if it has already been posted

Very good zarnywhoop.In general, for a sequence of lenght n>0, the maximum monotonous sub-sequence is of lenght : 1 + FLOOR[ SQRT[n-1]].

The result is the same as mine and i'm sure it's right but i don't understand the reasonning (i mean the word : qed and the construction of S_i : ?starting from h_i if h_i < h_j for for i > j?Text)