dimension of graph $K_n$ as defined in Proofs from the book

Question

Let $N = {1, \ldots, n}$, $n \geq 3$, and consider $m$ permutations $\pi_1, . . . , \pi_m$ of $N$. We say that the permutations $\pi_i$ represent $K_n$ if to every three distinct numbers $i, j, k$ there exists a permutation $\pi$ in which $k$ comes after both $i$ and $j$. The dimension of $K_n$ is then the smallest $m$ for which a representation $\pi_1, . . . , \pi_m$ exists.

The above definition states the dimension of complete graphs $K_n$.I could easily see why $\dim(K_3)=3$ since we can take permutations $(1,2,3),(3,1,2),(2,3,1)$. I saw that $\dim(K_n) = 4$ for $n\leq 12$, but I am having a hard time showing this for $K_5,K_6,K_7$. Could anyone tell me if the following are correct? I feel like each of them is wrong.

$\dim(K_5) = 4$ since taking $(1,2,3,4,5),(5,1,2,4,3),(3,5,2,4,1),(3,5,1,4,2)$ we have each pair in front of every number.

$\dim(K_6) = 4$ since we can take $(1,2,3,4,5,6),(6,5,4,3,2,1),(4,1,6,5,3,2),(2,4,5,1,6,3).$

$\dim(K_7) = 4$ since we can take $(7,6,5,4,3,2,1),(1,5,4,3,2,7,6),(3,4,5,1,2,7,6),(2,7,6,5,4,3)$.

Answer 1

The one for $K_5$ looks right to me.

The one for $K_6$ is certainly wrong, because there is no permutation where $4$ comes after both $3$ and $6$.

The one for $K_7$ is certainly wrong, because there is no permutation where $2$ comes after both $1$ and $7$.

Here's a python script:

from itertools import combinations
from collections import defaultdict

n = 6
pi = [(1,2,3,4,5,6), (6,5,3,4,2,1), (4,1,6,5,3,2), (2,4,5,1,6,3)]

pairs = defaultdict(set)

for p in pi:
    for k in range(2,n):
        for c in combinations(p[:k], 2):
            if c[0]< c[1]: 
                pairs[p[k]].add(c)
            else:
                pairs[p[k]].add((c[1], c[0]))

for c in combinations(range(1,n+1), 2):
    for k in pairs:
        if k not in c and c not in pairs[k]:
            print("%d doesn't follow %s"%(k, c))

This produces the output:

4 doesn't follow (1, 5)
4 doesn't follow (1, 6)
4 doesn't follow (2, 5)
4 doesn't follow (2, 6)
5 doesn't follow (2, 6)
5 doesn't follow (3, 6)

I didn't check $K_7$ but you can easily modify the script to check it yourself.

EDIT

$$(1,2,3,4,5,6),(6,2,3,5,4,1),(6,1,3,4,5,2),(6,1,2,5,4,3)$$

works for $K_6$. We can find this quickly. WLOG we have $(1,2,3,4,5,6).$ Then $6$ is taken care of, so we can put it first in the other $3$ permutations, and $1,2,3$ present the most difficulty, so we put then last. That leaves only $4$ and $5$ to deal with, and since $5$ has come after $4$ once already, we put $4$ after $5$ twice and $5$ after $4$ once. Then everything falls into place.

I also found $$(1,2,3,4,5,6,7),(7,2,3,5,6,4,1),(7,1,3,4,6,5,2),(7,1,2,6,5,4,3)$$ for $K_7$ by a similar approach. It took a little fiddling around with $4,5,6$ this time. The first couple of things I tried didn't work, but I checked them with the script, and figured it out.