Ordering compositions of integers, where cyclic shifts are not considered distinct A composition of an integer $n$ of length $k$, is a tuple (ordered list) of $k$ positive integers whose elements sum to $n$, (sometimes also called integer composition, ordered partition or ordered integer partition). For instance, $[1,1,2,4], [1,2,1,4], [1,2,4,1], [4,4],$ and $[1,1,1,3,1,1]$ are examples of distinct compositions of $8$. These have lengths $4,4,4,2$, and $6$, respectively.

My question is: if I consider standard lexicographical ordering of all compositions of a fixed length $k$ of an integer $n$ (see example below for $k=4$, $n=8$), which is the most efficient way to check whether a given composition is the smallest (with respect to this ordering) among all its cyclic shifts?

I'm demonstrating with an example: Below I enumerate the partitions of 8 of length 4. For example, the composition $[1,1,5,1]$ has a smaller cyclic shift composition $[1,1,1,5]$; the composition $[2,3,2,1]$ has a smaller cyclic shift $[1,2,3,2]$; $[2,2,2,2]$, and $[1,3,2,2]$ are the smallest cyclic shifts of such a composition. In the case of $[2,3,2,1]$, I observed that there's a $1$ in there, so I shifted the composition until I got $[1,2,3,2]$, but I'm not sure how I do a similar check in the general case.

\begin{array}{llll}
    1 &amp; 1 &amp; 1 &amp; 5 \\
    1 &amp; 1 &amp; 2 &amp; 4 \\
    1 &amp; 1 &amp; 3 &amp; 3 \\
    1 &amp; 1 &amp; 4 &amp; 2 \\
    1 &amp; 1 &amp; 5 &amp; 1 \\
    1 &amp; 2 &amp; 1 &amp; 4 \\
    1 &amp; 2 &amp; 2 &amp; 3 \\
    1 &amp; 2 &amp; 3 &amp; 2 \\
    1 &amp; 2 &amp; 4 &amp; 1 \\
    1 &amp; 3 &amp; 1 &amp; 3 \\
    1 &amp; 3 &amp; 2 &amp; 2 \\
    1 &amp; 3 &amp; 3 &amp; 1 \\
    1 &amp; 4 &amp; 1 &amp; 2 \\
    1 &amp; 4 &amp; 2 &amp; 1 \\
    1 &amp; 5 &amp; 1 &amp; 1 \\
    2 &amp; 1 &amp; 1 &amp; 4 \\
    2 &amp; 1 &amp; 2 &amp; 3 \\
    2 &amp; 1 &amp; 3 &amp; 2 \\
    2 &amp; 1 &amp; 4 &amp; 1 \\
    2 &amp; 2 &amp; 1 &amp; 3 \\
    2 &amp; 2 &amp; 2 &amp; 2 \\
    2 &amp; 2 &amp; 3 &amp; 1 \\
    2 &amp; 3 &amp; 1 &amp; 2 \\
    2 &amp; 3 &amp; 2 &amp; 1 \\
    2 &amp; 4 &amp; 1 &amp; 1 \\
    3 &amp; 1 &amp; 1 &amp; 3 \\
    3 &amp; 1 &amp; 2 &amp; 2 \\
    3 &amp; 1 &amp; 3 &amp; 1 \\
    3 &amp; 2 &amp; 1 &amp; 2 \\
    3 &amp; 2 &amp; 2 &amp; 1 \\
    3 &amp; 3 &amp; 1 &amp; 1 \\
    4 &amp; 1 &amp; 1 &amp; 2 \\
    4 &amp; 1 &amp; 2 &amp; 1 \\
    4 &amp; 2 &amp; 1 &amp; 1 \\
    5 &amp; 1 &amp; 1 &amp; 1 \\
\end{array}
You have to start with the smallest part. If there is only one of that size, you are done. Otherwise find the smallest number following one of the smallest parts and so on.
A rule for the order of all compositions of a positive integer n is derived in the paper, “Relationships between Ordered Compositions and Fibonacci Numbers” published in the Journal of Mathematics research $(Vol-7, No-3, 2015)$ under Canadian Center of Science and Education.
        ‘Rule for Significant Order of Compositions of n’ or ‘SOC(n)’ : Under SOC(n), the summands of the 1st C(n) are all 1; the last C(n) is n itself; and  for n ≥ r ≥ 2,  if any kth C(n) is: $x_1 + …  + x_r$  then $(k + 1)$-th  C(n) is: $x_1 + …+ x_{r – 2}  + (x_{r – 1 + 1)} +$ the  sum of $x_{r – 1}$ summands which are all 1 such that if $r \ge 3$ then the first $r – 2$ summands of k-th C(n) appear also in (k + 1)th C(n) in the same order, but if r = 2 then such common summands of k-th C(n) and (k + 1)-th C(n) cannot exist. The number of summands of k-th C(n) and (k + 1)th C(n) under SOC(n) are $r$ and $r + x_{r – 2}$ respectively.
Example: SOC(4)  ≡ $4, 1 + 1 + 1 + 1,   1 + 1 + 2,   1 + 2 + 1,   1 + 3,   2 + 1 + 1,   2 + 2,   3 + 1,   4$.
        In the connection of your enquiry, you can download and read the paper.