First Sort II

Consider the following algorithm for sorting a list:

• 1. Starting from the beginning of the list, check each pair of adjacent elements in turn.
• 2. If the elements are out of order:
  • a. Move the smallest element of the pair at the beginning of the list.
  • b. Restart the process from step 1.
• 3. If all pairs are in order, stop.

For example, the list { 4 1 3 2 } is sorted as follows:

• 4 1 3 2 (4 and 1 are out of order so move 1 to the front of the list)
• 1 4 3 2 (4 and 3 are out of order so move 3 to the front of the list)
• 3 1 4 2 (3 and 1 are out of order so move 1 to the front of the list)
• 1 3 4 2 (4 and 2 are out of order so move 2 to the front of the list)
• 2 1 3 4 (2 and 1 are out of order so move 1 to the front of the list)
• 1 2 3 4 (The list is now sorted)

Let F(L) be the number of times step 2a is executed to sort list L. For example, F({ 4 1 3 2 }) = 5.

We can list all permutations P of the integers {1, 2, ..., n} in lexicographical order, and assign to each permutation an index I_n(P) from 1 to n! corresponding to its position in the list.

Let Q(n, k) = min(I_n(P)) for F(P) = k, the index of the first permutation requiring exactly k steps to sort with First Sort. If there is no permutation for which F(P) = k, then Q(n, k) is undefined.

For n = 4 we have:

Let R(k) = min(Q(n, k)) over all n for which Q(n, k) is defined.

Find R(12¹²).

Written by gamwe6 who lives and works in San Francisco building useful things. You should follow him on Twitter