A322750 Number of directed graphs of 2*n vertices each having an in-degree and out-degree of 1 such that the graph specifies a pairwise connected gift exchange with no reciprocal gifts.

0, 0, 2, 48, 2640, 250368, 34110720, 6347520000

Offset

0,3

Comments

The sequence is the number of unique arrangements of directed graphs connecting 2*n vertices, where vertices occur in pairs, and meeting the following requirements:

1. Each vertex has an out-degree and in-degree of 1.

2. No edge connects vertices that are paired.

3. Starting with any pair, following the edges of paired vertices connects all vertices.

4. There are no closed walks of two vertices (i.e., no reciprocal connections).

The requirements were chosen to yield a nice gift exchange between a set of couples. Acknowledgement to the additional members of the initial, inspirational gift exchange group: Cat, Brad, Kim, Ada, Graham, Nolan, and Leah.

The fraction of graphs meeting the requirements is approximately 0.07. Starting with n=2, the fractions are (0.083333333, 0.066666667, 0.06547619, 0.068994709, 0.071212121, 0.072810787). Is there a way to compute the percentage of graphs satisfying the condition in the limit as n approaches infinity?

Links

Table of n, a(n) for n=0..7.

Example

For n = 0, there are no pairs; a(0) = 0 since no edges exist.
For n = 1, there is one pair; a(1) = 0 since requirements 1 and 2 can't be satisfied.
For n = 2, there are two pairs; a(2) = 2 graphs given by these edge destinations:
    ((2, 3), (1, 0))
    ((3, 2), (0, 1))
while ((2, 3), (0, 1)) is not allowed because the first and third edges form a 2-vertex walk.

Prog

(Python)
from itertools import permutations as perm
def num_connected_by_pairs(assigned, here=0, seen=None):
    seen = (seen, set())[seen is None]
    for proposed in [(here - 1, here), (here, here + 1)][(here % 2) == 0]:
        if proposed not in seen:
            seen.add(proposed)
            num_connected_by_pairs(assigned, assigned[proposed], seen)
    return len(seen)
def valid(assigned, pairs):
    self_give = [assigned[i] == i for i in range(len(assigned))]
    is_reciprocal = [assigned[a] == i for i, a in enumerate(assigned)]
    same_pair = [assigned[i] == i + 1 or assigned[i+1] == i for i in range(0, 2*pairs, 2)]
    if pairs == 0 or True in self_give + is_reciprocal + same_pair:
        return False
    num_connected = [num_connected_by_pairs(assigned, here) for here in range(2, 2*pairs, 2)]
    return min(num_connected) == 2*pairs
print([len([x for x in perm(range(2*pairs)) if valid(x, pairs)]) for pairs in range(0, 6)])

Crossrefs

A322751 allows reciprocal connections.

A010050 is the number of graphs (2n)!.

Sequence in context: A186416 A210723 A346454 * A367537 A346019 A087085

Adjacent sequences: A322747 A322748 A322749 * A322751 A322752 A322753

Keyword

nonn,more

Author

Russell Y. Webb, Dec 25 2018

Status

approved