Distinct relative ordering of four consecutive numbers Find a permutation of the set $\{1,2,\dots, 24\}$ with the property that each consecutive group of four integers (including around the corners) has a distinct relative order. 
When we are looking at the set $\{1,2,3,4,5,6\}$, I found that the permutation $146253$ satisfies that each consecutive group of three integers (including around the corner) has distinct relative order. I do not see how to generalize this. I have found this useful site which on page 10 describes a possible method to use; but note that this does not guarantee that we use all $24$ numbers. How can I solve this question?
 A: The linked document gave an example for a universal cycle that required only five symbols:$$1,2,3,4,1,5,3,4,2,1,5,4,2,1,3,5,4,1,3,5,2,4,3,5$$
The diagram of inequalities, presented here without the arrows, is:
$$\begin{array}{ccccc}
a&b&c&d&f\\
e& &g&h& \\
 &i& & &k\\
j& & &l& \\
 &m&o& &p\\
n& & &q& \\
 & &s& &t\\
r&u& &v& \\
 & &w& &x\\
\hline
1&2&3&4&5\\
\end{array}$$
Note that this universal cycle only requires a minimum of five symbols. We can write this cycle with $24$ numbers by assigning unique values to each variable. We only need to ensure that the values in the first column are less than the values in the second, and those in the second are less than those in the third, and so on. (This is not a strict requirement, but the easiest way to avoid breaking the cycle). Taking these columns as sets, we have:
$$\begin{array}{rrcl}
\text{Set 1}&\{a,e,j,n,r\}&=&\{1,2,3,4,5\}\\
\text{Set 2}&\{b,i,m,u\}&=&\{6,7,8,9\}\\
\text{Set 3}&\{c,g,o,s,w\}&=&\{10,11,12,13,14\}\\
\text{Set 4}&\{d,h,l,q,v\}&=&\{15,16,17,18,19\}\\
\text{Set 5}&\{f,k,p,t,x\}&=&\{20,21,22,23,24\}
\end{array}$$
And then one possible cycle is:
$$1,6,10,15,2,20,11,16,7,3,21,17,8,4,12,22,18,5,13,23,9,19,14,24$$
