|
|
| Line 4: |
Line 4: |
| http://planeta.rambler.ru/users/hajgas/ ����� ���� ����������� | | http://planeta.rambler.ru/users/hajgas/ ����� ���� ����������� |
|
| |
|
| ==A possible definition of quasirandom subsets of <math>[3]^n</math>==
| | http://www.justin.tv/angelinajolienaked/profile angelina jolie naked |
| | |
| As with all the examples above, it is more convenient to give a definition for quasirandom functions. However, in this case it is not quite so obvious what should be meant by a balanced function.
| |
| | |
| Here, first, is a possible definition of a quasirandom function from <math>[2]^n\times [2]^n</math> to <math>[-1,1].</math> We say that f is c-quasirandom if <math>\mathbb{E}_{A,A',B,B'}f(A,B)f(A,B')f(A',B)f(A',B')\leq c.</math> However, the expectation is not with respect to the uniform distribution over all quadruples (A,A',B,B') of subsets of <math>[n].</math> Rather, we choose them as follows. (Several variants of what we write here are possible: it is not clear in advance what precise definition will be the most convenient to use.) First we randomly permute <math>[n]</math> using a permutation <math>\pi</math>. Then we let A, A', B and B' be four random intervals in <math>\pi([n]),</math> where we allow our intervals to wrap around mod n. (So, for example, a possible set A is <math>\{\pi(n-2),\pi(n-1),\pi(n),\pi(1),\pi(2)\}.</math>)
| |
| | |
| As ever, it is easy to prove positivity. To apply this definition to subsets <math>\mathcal{A}</math> of <math>[3]^n,</math> define f(A,B) to be 0 if A and B intersect, <math>1-\delta</math> if they are disjoint and the sequence x that is 1 on A, 2 on B and 3 elsewhere belongs to <math>\mathcal{A},</math> and <math>-\delta</math> otherwise. Here, <math>\delta</math> is the probability that (A,B) belongs to <math>\mathcal{A}</math> if we choose (A,B) randomly by taking two random intervals in a random permutation of <math>[n]</math> (in other words, we take the marginal distribution of (A,B) from the distribution of the quadruple (A,A',B,B') above) and condition on their being disjoint. It follows from this definition that <math>\mathbb{E}f=0</math> (since the expectation conditional on A and B being disjoint is 0 and f is zero whenever A and B intersect).
| |
| | |
| Nothing that one would really like to know about this definition has yet been fully established, though an argument that looks as though it might work has been proposed to show that if f is quasirandom in this sense then the expectation <math>\mathbb{E}f(A,B)f(A\cup D,B)f(A,B\cup D)</math> is small (if the distribution on these "set-theoretic corners" is appropriately defined).
| |
