Problem Link:

Difficulty:

easy-medium

Pre-requisites:

Sperner's Theorem

Problem:

Given an integer N, find the size of smallest set S such that there exists an antichain of size N among the subsets of S.

Explanation:

The relation between groups of cities A and B served by two spoons can be equivalently stated as: A-B is not empty and B-A is not empty, which is same as A ⊄ B and B ⊄ A.

If we consider the poset of all subsets of a set S, then the two elements A and B above are incomparable, therefore all the N sub-sets form an antichain.

People could obtain the solution by

guess work, or,
evaluating the first few few terms and then using oies to get the required sequence, or,
reinventing proof of Sperner's Theorem.

The statement and proof of Sperner's Theorem is given below in very simple language.

Given the Sperner's Theorem, we would still need to find smallest M such that M choose ⌊ M/2 ⌋ is at least N. For N ≤ 10¹⁸, M ≤ 64, and therefore, one could compute these values offline(or online in his code itself), and do a binary search to lookup the best value. Time complexity O(T).

Proof of Sperner's Theorem

Given a set S of size M, you want to pick a family F of subsets of S (i.e. F ⊂ 2^S) such that no set in the family is subset of another set in the family (i.e. x, y ∈ F ⇒ x ⊄ y). The maximum possible size of such a family of subsets is M choose ⌊ M/2 ⌋

It is clear that if we choose all subsets in the family F to be of same size, k, then the family would satisfy the above property, and its size will be M choose k. This is maximum when k=⌊ M/2 ⌋. Therefore, the bound suggested above can be attained by picking all sbsets of size ⌊ M/2 ⌋. Now it remains to show that this is also the best possible bound.

Lets say we have got a family F of subsets satisfying above property, and it has a₁ subsets of size 1, a₂ subsets of size 2 ... a_M subsets of size M.

Then we can generate a permutation of S by selecting a set S' in F and concatenating a permutation of the elements of S' with a permutation of the non-members. If |S'| = i, it will be associated in this way with i!(n − i)! permutations. Each permutation can only be associated with at most one set in F. This is because otherwise two different prefixes of a single permutation would be associated with different subsets(say S₁ and S₂) in F, in which case, one(out of S₁ and S₂) would be a subset of the other. Therefore, the number of permutations that can be generated by this procedure is

sum [i=1 to M] a_i * i! (M-i)!

And the above number is no more than M!. For k=⌊ M/2 ⌋ we already have

sum [i=1 to M] a_i * k! (M-k)! ≤ sum [i=1 to M] a_i * i! (M-i)! ≤ M!

⇒ sum [i=1 to M] a_i≤ M!/ (k! * (M-k)!)

Setter's Solution:

Can be found here

Tester's Solution:

Mahbub's code
Sergey's code

Editorialist's Solution:

Can be found here, and code used to generate values offline is here