This is a question from CLRS book. (Chapter 32, string matching, the question is the problem for the whole chapter, it's in the end of the chapter)
Let $y^i$ denote the concatenation of string y with itself $i$ times. For example, $(ab)^{3}$ = $ababab$. We say that a string $x\in X$ has repetition factor $r$ if x = $y^{r}$ for some string $y\in X$ and some $r > 0$. Let $p(x)$ denote the largest $r$ such that $x$ has repetition factor $r$. Give an efficient algorithm that takes as input a pattern $P[1..m]$ and computes the value $p(P_i)$ for $i = 1, 2,\dots,m$. What is the running time of your algorithm?
I found an answer like this: First compute Prefix function (based on the prefix function from the book), so we return $π$. Then, Suppose that $π[i] = i −k$. If $k|i$, we know that $k$ is the length of the primitive root, so, the word has a repetition factor of $\frac{i}{k}$. We also know that there is no smaller repetition factor $i$. Now, suppose that we have $k$ not dividing $i$. We will show that we can only have the trivial repetition factor of 1. Suppose we had some repetition $y^{r} = \Pi$. Then, we know that $π[i] ≥ y^{r}−1$. However, if we have it strictly greater than this, this means that we can write the $y$’s themselves as powers because we have them aligning with themselves.
COMPUTE-PREFIX-FUNCTION (P)
1 m = P.length
2 let π be a new array
3 π[1]= 0
4 k = 0
5 for q = 2 to m
6 while k > 0 and P[k+1] != P[q]
7 k = π[q]
8 if P[k+1] != P[q]
9 k = k + 1
10 π[q] = k
11 return π
The prefix function is a part of KMP algorithm
KMP-MATCHER (T,P)
1 n= T.length
2 m =P.length
3 π=COMPUTE-PREFIX-FUNCTION (P)
4 q= 0 // number of characters matched
5 for i = 1 to n // scan the text from left to right
6 while q > 0 and P[q+1] != T[i]
7 q= π[q] // next character does not match
8 if P[q+1] == T[i+1]
9 q = q + 1 // next character matches
10 if q == m // is all of P matched?
11 print “Pattern occurs with shift” i - m
12 q=π[q] // look for the next match
I can't still really understand completely the answer. Why should we suppose $k$ not dividing $i$? And the explanation for the case of $k$ not dividing $i$, the repetition factor is 1, is confusing to me.
