For context, I am an amateur bridge player, and in many cases, it helps to sort my hand in 13 cards in alternating colors from greatest to least so I can see what cards I am working with, so that basically inspired this problem.

Suppose you have a list of integer tuples (a_1, b_1), (a_2, b_2), ..., (a_n, b_n). You wish to arrange the list in a certain order that meets the following three criteria:

All tuples with first element a_i are grouped together. That is, you shouldn’t have a spare a_i anywhere else. 1. Within a grouping with first element a_i, the group is ordered in decreasing order of the b_i’s. 1. Two adjacent groupings identified by elements a_i != a_j must have a_i and a_j differ in parity IF POSSIBLE. That is, if a_i is even, then all adjacent groupings must have a_j as odd, and vice versa. If all elements have a_i’s of a single parity, then only rules 1 and 2 apply. A move consists of moving any tuple to any index i. Any element that was already at index i now moves to index i-1.

For example, if we are given {(1, 7), (3, 8), (2, 7), (2, 9), (1, 10)}

We can move (1, 7) to index 4, getting {(3, 8), (2, 7), (2, 9), (1, 10), (1, 7)}.

Now we can move (2, 7) to index 2, getting {(3, 8), (2, 9), (2, 7), (1, 10), (1, 7)}.

Thus this list required 2 moves to transform it into a list that satisfies all three conditions.

Is there an algorithm/procedure that finds the fastest way to do this, or the optimal number of moves?

EDIT: Added clarification rule 3. It may be the case that some lists have only one parity in their first element, i.e. {(2, 6), (2, 5), (4, 3), (4, 7), (4, 5)}. In this case, the third rule does not apply, but the first two rules do apply. So we would need one move to turn this list into a valid list: {(2, 6), (2, 5), (4, 7), (4, 5), (4, 3)}.