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+#+title: A lemma on suffix array searching
+#+filetags: @results suffix-array
+#+OPTIONS: ^:{} num: num:t
+#+hugo_front_matter_key_replace: author>authors
+#+toc: headlines 3
+#+date: <2024-10-05 Sat>
+
+
+We'll prove that using the "faster" binary search algorithm (see [[#faster-search]]) that tracks the LCP
+with the left and right boundary of the remaining search interval has amortized
+runtime
+
+$$
+O\Big(\lg_2(n) + |P| + \lg_2(Occ(P))\cdot |P|\Big),
+$$
+when $P$ is a randomly sampled fixed-length pattern from the text and $Occ(P)$ counts the number of occurrences of $P$ in the text.
+
+Thus, when searching for patterns that follow the same distribution
+as the input text, the performance of the faster search method is nearly as good as the
+LCP-based $O(|P| + \lg_2 n)$ method when patterns have only few matches.
+
+First some background.
+
+* Suffix arrays
+
+Suffix arrays were introduced by [cite/t:@suffix-arrays-manber-myers-90].
+A suffix array $S$ of a text $T$ of length $n = |T|$ is a permutation of
+$[n]=\{0, \dots, n-1\}$ such that $T[S[i]..] < T[S[i+1]..]$, that is, the suffixes
+are sorted by the order $S$.
+
+Given the suffix array, one can search for a pattern $P$ of length $|P|$,
+which returns the position $p$ of the /first/ suffix $T[S[p]..]\geq P$.
+
+In general, we are usually interested in the entire /interval/ of suffixes
+$S[l..r]$ that start with the given pattern. This can simply be done by using
+two independent searches, as so does not affect the theoretical complexity. But
+of course, in practice it's more efficient to find both boundaries in one go.
+
+* Searching methods
+
+[cite/author/bc:@suffix-arrays-manber-myers-90] introduce three methods to
+search the suffix array, all based on binary search.
+
+** Naive $O(|P|\cdot \lg_2 n)$ search
+
+The naive method works using a simple binary search on the suffix array.
+
+#+caption: Python code for searching a suffix array the naive way.
+#+begin_src py
+def search(T, S, P):
+    n = len(T)
+    l, r = 0, n
+    while l < r:
+        m = (l+r)//2
+        M = T[S[m]:]
+        if M < P:
+            l = m + 1
+        else:
+            r = m
+    return S[l]
+#+end_src
+
+The binary search needs at most $\lceil \lg_2(n+1)\rceil$ iterations, and in
+each iteration the slowest operation is the comparison of the suffix $M=T[S[m]..]$
+with the pattern =P=, which in the worst case compares $|P|$ characters in $O(|P|)$ time.
+Thus, the overall runtime is $O(|P| \cdot \lg_2 n)$.
+
+** Faster $O(|P|\cdot \lg_2 n)$ search
+:PROPERTIES:
+:CUSTOM_ID: faster-search
+:END:
+One of the bad cases of the naive method above is that in each iteration, the
+entire pattern $P$ is matched. As the remaining search interval $[l, r]$
+narrows, the remaining suffixes get more and more similar to $P$. In fact, once
+we know that the first $c$ characters of $L=S[T[l]..]$ match the first $c$
+characters of $R=S[T[r]..]$, every remaining string in the interval also starts
+with these same first $c$ characters, and so does $P$.
+Thus, in the comparison of $M=T[S[m]..]$ and $P$, we can skip the first $c$ characters.
+
+In practice, we implement this by keeping the lengths $c_l$ and $c_r$ longest common prefix of $P$ with
+$L$ and $R$, and skipping the first $\min(c_l, c_r)$ character comparisons.
+
+#+caption: Python code for searching a suffix array the faster way.
+#+begin_src py
+def search(T, S, P):
+    n = len(T)
+    l, r = 0, n
+    c_l, c_r = 0, 0
+    while l < r:
+        m = (l+r)//2
+        M = T[S[m]:]
+        c = min(c_l, c_r)
+        # The lcp and the comparison can be computed at the same time.
+        c_m = c + lcp(M[c:], P[c:])
+        if M[c:] < P[c:]:
+            l = m + 1
+            c_l = c_m
+        else:
+            r = m
+            c_r = c_m
+    return S[l]
+#+end_src
+
+In the worst case, this method is still $O(|P| \cdot \lg_2 n)$, but it's widely
+known that this method performs very well in practice, and can even be faster than
+the $O(|P|+\lg_2 n)$ method below.
+
+** LCP-based $O(|P| + \lg_2 n)$ search
+
+Now suppose that for every step in the binary search, we have already precomputed values
+$x_l=LCP(L, M)$ and $x_r=LCP(M, R)$.
+(This is possible in $O(n)$ time and space using a bottom-up approach on the
+implicit binary search tree that contains around $n$ nodes.)
+
+Now assume that in a step of the binary search, we have $l\geq r$.
+Again following the original paper, there are three cases, as they nicely
+illustrated.
+
+#+caption: Three cases of the binary search, taken from [cite/t:@suffix-arrays-manber-myers-90].
+#+attr_html: :class inset large
+[[file:lcp-cases.png]]
+
+The black bars indicate the length of the LCP of each suffix with $P$, and $l$
+and $r$ are the length of the LCP of the left and right of the interval with
+$P$. Assume that $l\geq r$ (the symmetric case is equivalent).
+The grey area is the length of the LCP of $L=T[S[l]..]$ and $M=T[S[m]..]$.
+
+Let $x = LCP(L, M)$. The three cases are, in order:
+- $x > l$ :: In this case, we know that $P$ is larger than $L$ in the
+   $l+1$'st character, and since $x>l$, $L$ and $M$ are equal in their $l+1$'st
+   character, so also $P$ is larger than $M$ in its $l+1$'st character and
+   $P>M$, so we branch right.
+- $x=l$ :: In this case, we know that $P$ shares the first $x$ characters with
+   $L$ and hence also with $M$. We now compare $P$ with $M$ starting at the
+   $l+1$'st character. Let's say that $h$ equal characters are compared.
+   If we branch left, the new value of $r$ is $l+h$, and if we branch right, the
+   new value of $l$ is $l+h$. Thus, $\max(l, r)$ always increases from $l$ to $l+h$.
+- $x<l$ :: We know that $P$ shares the first $l$ characters with $L$, and that
+   $L$ is less than $M$ in its first $x+1\leq l$ characters, so $P<M$, and we
+   branch left.
+
+We see that the first and last cases take constant time, and since $l\leq |P|$ and
+$r\leq |P|$, we have $\max(l, r)\leq m$. Since in the middle case the maximum is
+increased by $h$. the total number of characters compared in case b. is at most
+$m$. Thus, this method takes $O(|P| + \lg_2 n)$ time.
+
+
+* Analysing the faster search
+
+The main difference between the faster search and the LCP based search is that
+the faster search starts comparing characters between $P$ and $M$ at the
+$\min(l, r)+1$'st character, while the LCP version starts at the $\max(l,
+r)+1$'st character.
+Thus, if we can show that the total number of compared characters 'between'
+$\min(l, r)$ and $\max(l, r)$ is not too large in expectation (in particular, at
+most $|P|$), we recover the expected runtime of $(|P| + \lg_2 n)$.
+
+Let $0\leq i\leq n$ be a uniformly random position in the text, and let
+$P=T[i..i+m_P]$ be the pattern of length[fn::I'm avoiding using $m$ as the
+length of $P$ since it's also the position in the middle between $l$ and $r$
+already. Similarly I'm avoiding using $l$ for LCP length.] $m_P$ starting at position $i$. When
+$i>n-m_P$, the pattern is simply a bit shorter than $m_P$. In this case,
+we assume that $P$ includes a sentinel character at the end, and thus will have
+exactly $1$ occurrence in the text.
+
+Consider a step in the binary search, and assume again that $l \geq r$. We start
+comparing $P$ and $M$ at their $\min(l,r)+1=r+1$'st character, and let $y=LCP(P,
+M)$ be the computed length of the LCP of $M$ and $P$, which requires $h=y-r$
+comparisons of equal characters. We also still define $x=LCP(L, M)$
+although we do not explicitly compute this. There are a few cases:
+
+- $y < m_P$ ::
+  When $M$ does not start with $P$, we know that the pattern is larger or
+  smaller than $M$ with equal probability, since the pattern was randomly
+  sampled from the suffix array and thus each of the $m-l$ position left of $M$
+  and each of the $r-m$ positions right of $M$ has an equal probability of corresponding
+  to the chosen pattern. (In the case where $r-l$ is odd, we can randomly choose
+  between $m=\lfloor \frac{l+r}2\lfloor$ and $m=\lceil \frac{l+r}2\lceil$ to
+  truly equalize the probabilities.)
+
+  Thus, with probability $1/2$, the minimum of $l$ and $r$ increases to $y$, and
+  so in expectation, the sum of $l$ and $r$ increases by at least $(y-r)/2 =
+  h/2$. Since the sum of $l$ and $r$ is at most $2|P|$, the expected total number of
+  comparisons is at most $4|P|$.
+
+- $y = m_P$ ::
+  When $M$ starts with $P$, we trivially do at most $|P|$ comparisons.
+  When there are $Occ(P)$ occurrences of the pattern $P$, this situation
+  can happen at most $\lg_2 Occ(P)$ times, and so this incurs a total cost
+  bounded by $O(|P| \cdot \lg_2 Occ(P))$.
+
+We conclude that when we fix a length $m_P$ and uniformly random choose a pattern $P$ length $m_P$ random from the input
+text, the amortized cost of a search is
+$$
+O\Big(\lg_2(n) + m_P + \lg_2(Occ(P))\cdot m_P\Big).
+$$
+
+
+#+print_bibliography:
diff --git a/references.bib b/references.bib
index 5528e73..3fdfe61 100644
--- a/references.bib
+++ b/references.bib
@@ -3856,3 +3856,18 @@ @Article{suffix-arrays-with-a-twist
   url          = {http://dx.doi.org/10.31577/cai_2019_3_555},
   publisher    = {Central Library of the Slovak Academy of Sciences}
 }
+
+@Article{suffix-arrays-manber-myers-90,
+  author       = {Manber, Udi and Myers, Gene},
+  title        = {Suffix Arrays: A New Method for On-Line String Searches},
+  journal      = {SIAM Journal on Computing},
+  year         = 1993,
+  volume       = 22,
+  number       = 5,
+  month        = oct,
+  pages        = {935–948},
+  issn         = {1095-7111},
+  doi          = {10.1137/0222058},
+  url          = {http://dx.doi.org/10.1137/0222058},
+  publisher    = {Society for Industrial & Applied Mathematics (SIAM)}
+}