Computes the number of possible orderings of n objects under the relations < and?

The code Let f(n,k) be the number of possible orderings with exactly k inequality relations (and (n - 1 - k) quality relations) Easy to see, that f(n,0) = 1, f(n,n-1) = n! For any n Now we want to compute f(n,k) for any k. Imagine, you moved away one number (any), so there are now (n-1) numbers.

There are two possabilities: 1) There are (k-1) inequalities in those (n-1) numbers, and there are (k+1)(f(n-1,k-1)) ways to add n'th number so that new inequality added 2) There are k inequlaities in those (n-1) numbers, and there are (k+1)(f(n-1,k)) ways to add n'th number with no additional inequality So, f(n,k) = (k+1)(f(n-1,k-1) + f(n-1,k)) The final answer F(n) = f(n,0) + f(n,1) + ... + f(n,n-1).

The code Let f(n,k) be the number of possible orderings with exactly k inequality relations (and (n - 1 - k) quality relations). Easy to see, that f(n,0) = 1, f(n,n-1) = n! For any n.

Now we want to compute f(n,k) for any k. Imagine, you moved away one number (any), so there are now (n-1) numbers. There are two possabilities: 1) There are (k-1) inequalities in those (n-1) numbers, and there are (k+1)(f(n-1,k-1)) ways to add n'th number so that new inequality added.2) There are k inequlaities in those (n-1) numbers, and there are (k+1)(f(n-1,k)) ways to add n'th number with no additional inequality.So, f(n,k) = (k+1)(f(n-1,k-1) + f(n-1,k)).

The final answer F(n) = f(n,0) + f(n,1) + ... + f(n,n-1).

If you just have For example, let's do a set of size 3 (like your example). We can have 1, 2 or 3 equivalence classes. There's only 1 way to have 1 equivalence class, 3 ways to have 2 equivalence classes and 1 way to have 3 equivalence classes (see later for how to compute these).

This gives 1 * 1! + 3 * 2! + 1 * 3!

= 13 total orderings. This breaks the original problem into a simpler one: counting ways to partition a set into k parts. If you write a function S(N, k) to do this, you can get the answer to the original problem by computing: sum(k!

* S(N, k) for k in 1, 2, ..., N). You can use these recurrence relations to count partitions (see en.wikipedia.org/wiki/Stirling_number_of...): S(N, k) = S(N - 1, k - 1) + k * S(N - 1, k) S(N, 1) = 1 S(N, N) = 1 And you can use dynamic programming to compute this function.

You can have recursive function for this: F(1) = 1; F(2) = 3; F(n) = F(n-1) * n + F(n-2) * n*(n-1)/2 + F(n-3) * n*(n-1)(n-2)/3! +.... + F(1) * n + 1; for example F(2) = 1 * 2 +1 = 3; F(3) = F(2) * 3 + F(1) * 3 + 1 = 9 + 3 + 1 = 13; Why? Think you know F(n-1) now you want insert An in your sequences, An can be in the 0,1,2,...nth place in the sequence, no matter, counting the number of sequences: A0,...An-1 F(n-2) * n * (n-1)/ 2 why?... (think yourself it's easy) how many sequences end with "K" "=" sing?

F(n-k+1) * n * ...*(n-k+1)/ k! And there is one sequence full = this can be calculated polynomially simply.

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.