Mar 7, 2021 · Mar 6, 2021
diff --git a/src/string/prefix-function.md b/src/string/prefix-function.md
 So let us compute the value of the prefix function $\pi$ for $i + 1$.
 If $s[i+1] = s[\pi[i]]$, then we can say with certainty that $\pi[i+1] = \pi[i] + 1$, since we already know that the suffix at position $i$ of length $\pi[i]$ is equal to the prefix of length $\pi[i]$.
 This is illustrated again with an example.
 $$\underbrace{\overbrace{s_0 ~ s_1 ~ s_2}^{\pi[i]} ~ \overbrace{s_3}^{s_3 = s_{i+1}}}\_{\pi[i+1] = \pi[i] + 1} ~ \dots ~ \underbrace{\overbrace{s_{i-2} ~ s_{i-1} ~ s_{i}}^{\pi[i]} ~ \overbrace{s_{i+1}}^{s_3 =s_i + 1}}\_{\pi[i+1] = \pi[i] + 1}$$
 $$\underbrace{\overbrace{s_0 ~ s_1 ~ s_2}^{\pi[i]} ~ \overbrace{s_3}^{s_3 = s_{i+1}}}\_{\pi[i+1] = \pi[i] + 1} ~ \dots ~ \underbrace{\overbrace{s_{i-2} ~ s_{i-1} ~ s_{i}}^{\pi[i]} ~ \overbrace{s_{i+1}}^{s_3 =s_{i + 1}}}\_{\pi[i+1] = \pi[i] + 1}$$

 If this is not the case, $s[i+1] \neq s[\pi[i]]$, then we need to try a shorter string.
 In order to speed things up, we would like to immediately move to the longest length $j \lt \pi[i]$, such that the prefix property in the position $i$ holds, i.e. $s[0 \dots j-1] = s[i-j+1 \dots i]$:
Original file line number	Diff line number	Diff line change
Expand Up		@@ -63,7 +63,7 @@ To accomplish this, we have to use all the information computed in the previous
		So let us compute the value of the prefix function $\pi$ for $i + 1$.
		If $s[i+1] = s[\pi[i]]$, then we can say with certainty that $\pi[i+1] = \pi[i] + 1$, since we already know that the suffix at position $i$ of length $\pi[i]$ is equal to the prefix of length $\pi[i]$.
		This is illustrated again with an example.
		$$\underbrace{\overbrace{s_0 ~ s_1 ~ s_2}^{\pi[i]} ~ \overbrace{s_3}^{s_3 = s_{i+1}}}\_{\pi[i+1] = \pi[i] + 1} ~ \dots ~ \underbrace{\overbrace{s_{i-2} ~ s_{i-1} ~ s_{i}}^{\pi[i]} ~ \overbrace{s_{i+1}}^{s_3 =s_i + 1}}\_{\pi[i+1] = \pi[i] + 1}$$
		$$\underbrace{\overbrace{s_0 ~ s_1 ~ s_2}^{\pi[i]} ~ \overbrace{s_3}^{s_3 = s_{i+1}}}\_{\pi[i+1] = \pi[i] + 1} ~ \dots ~ \underbrace{\overbrace{s_{i-2} ~ s_{i-1} ~ s_{i}}^{\pi[i]} ~ \overbrace{s_{i+1}}^{s_3 =s_{i + 1}}}\_{\pi[i+1] = \pi[i] + 1}$$

		If this is not the case, $s[i+1] \neq s[\pi[i]]$, then we need to try a shorter string.
		In order to speed things up, we would like to immediately move to the longest length $j \lt \pi[i]$, such that the prefix property in the position $i$ holds, i.e. $s[0 \dots j-1] = s[i-j+1 \dots i]$:
Expand Down