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chapter26.tex
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@ -9,17 +9,17 @@ time.
\index{pattern matching}
For example, a fundamental problem related to strings
is the \key{pattern matching} problem:
For example, a fundamental string processing
problem is the \key{pattern matching} problem:
given a string of length $n$ and a pattern of length $m$,
our task is to find the positions where the pattern
occurs in the string.
our task is to find the occurrences of the pattern
in the string.
For example, the pattern \texttt{ABC} occurs two
times in the string \texttt{ABABCBABC}.
The pattern matching problem is easy to solve
The pattern matching problem can be easily solved
in $O(nm)$ time by a brute force algorithm that
goes through all positions where the pattern may
tests all positions where the pattern may
occur in the string.
However, in this chapter, we will see that there
are more efficient algorithms that require only
@ -31,8 +31,13 @@ $O(n+m)$ time.
\index{alphabet}
An \key{alphabet} is a set of characters
that may appear in strings.
Throughout the chapter, we assume that
zero-based indexing is used in strings.
Thus, a string \texttt{s} of length $n$
consists of characters
$\texttt{s}[0],\texttt{s}[1],\ldots,\texttt{s}[n-1]$.
The set of characters that may appear
in strings is called an \key{alphabet}.
For example, the alphabet
$\{\texttt{A},\texttt{B},\ldots,\texttt{Z}\}$
consists of the capital letters of English.
@ -40,9 +45,12 @@ consists of the capital letters of English.
\index{substring}
A \key{substring} is a sequence of consecutive
characters of a string.
The number of substrings of a string is $n(n+1)/2$.
For example, the substrings of the string
characters in a string.
We use the notation $\texttt{s}[a \ldots b]$
to refer to a substring of \texttt{s}
that begins at position $a$ and ends at position $b$.
A string of length $n$ has $n(n+1)/2$ substrings.
For example, the substrings of
\texttt{ABCD} are
\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D},
\texttt{AB}, \texttt{BC}, \texttt{CD},
@ -52,9 +60,9 @@ For example, the substrings of the string
A \key{subsequence} is a sequence of
(not necessarily consecutive) characters
of a string in their original order.
The number of subsequences of a string is $2^n-1$.
For example, the subsequences of the string
in a string in their original order.
A string of length $n$ has $2^n-1$ subsequences.
For example, the subsequences of
\texttt{ABCD} are
\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D},
\texttt{AB}, \texttt{AC}, \texttt{AD},
@ -69,19 +77,18 @@ A \key{prefix} is a subtring that starts at the beginning
of a string,
and a \key{suffix} is a substring that ends at the end
of a string.
For example, for the string \texttt{ABCD},
the prefixes are
\texttt{A}, \texttt{AB}, \texttt{ABC} and \texttt{ABCD}
and the suffixes are
For example,
the prefixes of \texttt{ABCD} are
\texttt{A}, \texttt{AB}, \texttt{ABC} and \texttt{ABCD},
and the suffixes of \texttt{ABCD} are
\texttt{D}, \texttt{CD}, \texttt{BCD} and \texttt{ABCD}.
\index{rotation}
A \key{rotation} can be generated by moving
characters one by one from the beginning
to the end of a string (or vice versa).
For example, the rotations of the string
\texttt{ABCD} are
the characters of a string one by one from the beginning
to the end (or vice versa).
For example, the rotations of \texttt{ABCD} are
\texttt{ABCD}, \texttt{BCDA}, \texttt{CDAB} and \texttt{DABC}.
\index{period}
@ -97,13 +104,13 @@ For example, the shortest period of
A \key{border} is a string that is both
a prefix and a suffix of a string.
For example, the borders of the string \texttt{ABACABA}
For example, the borders of \texttt{ABACABA}
are \texttt{A}, \texttt{ABA} and \texttt{ABACABA}.
\index{lexicographical order}
Strings are usually compared using the \key{lexicographical order}
that corresponds to the alphabetical order.
Strings are compared using the \key{lexicographical order}
(which corresponds to the alphabetical order).
It means that $x<y$ if either $x \neq y$ and $x$ is a prefix of $y$,
or there is a position $k$ such that
$x[i]=y[i]$ when $i<k$ and $x[k]<y[k]$.
@ -112,11 +119,10 @@ $x[i]=y[i]$ when $i<k$ and $x[k]<y[k]$.
\index{trie}
A \key{trie} is a tree structure that
A \key{trie} is a rooted tree that
maintains a set of strings.
Each string is stored as
a chain of characters starting at
the root node.
Each string in the set is stored as
a chain of characters that starts at the root.
If two strings have a common prefix,
they also have a common chain in the tree.
@ -156,38 +162,39 @@ For example, consider the following trie:
This trie corresponds to the set
$\{\texttt{CANAL},\texttt{CANDY},\texttt{THE},\texttt{THERE}\}$.
The character * in a node means that
one of the strings in the set ends at the node.
a string in the set ends at the node.
Such a character is needed, because a string
may be a prefix of another string.
For example, in the above trie, \texttt{THE}
is a prefix of \texttt{THERE}.
We can check if a trie contains a string
in $O(n)$ time where $n$ is the length of the string,
We can check in $O(n)$ time whether a trie
contains a string of length $n$,
because we can follow the chain that starts at the root node.
We can also add a new string to the trie
in $O(n)$ time using a similar idea.
If needed, new nodes will be added to the trie.
We can also add a string of length $n$ to the trie
in $O(n)$ time by first following the chain
and then adding new nodes to the trie if necessary.
Using a trie, we can also find
for a given string the longest prefix
that belongs to the set.
In addition, by storing additional information
Using a trie, we can find
the longest prefix of a given string
such that the prefix belongs to the set.
Moreover, by storing additional information
in each node,
it is possible to calculate the number of
strings that have a given prefix.
we can calculate the number of
strings that belong to the set and have a
given string as a prefix.
A trie can be stored in an array
\begin{lstlisting}
int t[N][A];
int trie[N][A];
\end{lstlisting}
where $N$ is the maximum number of nodes
(the maximum total length of the strings in the set)
and $A$ is the size of the alphabet.
The nodes of a trie are numbered
$1,2,3,\ldots$ so that the number of the root is 1,
and $\texttt{t}[s][c]$ is the next node in the chain
from node $s$ using character $c$.
$0,1,2,\ldots$ so that the number of the root is 0,
and $\texttt{trie}[s][c]$ is the next node in the chain
when we move from node $s$ using character $c$.
\section{String hashing}
@ -199,7 +206,7 @@ allows us to efficiently check whether two
strings are equal\footnote{The technique
was popularized by the Karp–Rabin pattern matching
algorithm \cite{kar87}.}.
The idea is to compare the hash values of the
The idea in string hashing is to compare hash values of
strings instead of their individual characters.
\subsubsection*{Calculating hash values}
@ -217,15 +224,15 @@ based on their hash values.
A usual way to implement string hashing
is \key{polynomial hashing}, which means
that the hash value is calculated using the formula
that the hash value of a string \texttt{s}
of length $n$ is
\[(\texttt{s}[0] A^{n-1} + \texttt{s}[1] A^{n-2} + \cdots + \texttt{s}[n-1] A^0) \bmod B ,\]
where \texttt{s} is a string of length $n$
(so $s[0],s[1],\ldots,s[n-1]$
are the codes of the characters),
where $s[0],s[1],\ldots,s[n-1]$
are interpreted as the codes of the characters of \texttt{s},
and $A$ and $B$ are pre-chosen constants.
For example, the codes of the characters
in the string \texttt{ALLEY} are:
of \texttt{ALLEY} are:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) grid (5,2);
@ -246,41 +253,37 @@ in the string \texttt{ALLEY} are:
\end{center}
Thus, if $A=3$ and $B=97$, the hash value
of the string \texttt{ALLEY} is
of \texttt{ALLEY} is
\[(65 \cdot 3^4 + 76 \cdot 3^3 + 76 \cdot 3^2 + 69 \cdot 3^1 + 89 \cdot 3^0) \bmod 97 = 52.\]
\subsubsection*{Preprocessing}
It turns out that using polynomial hashing,
we can calculate the hash value of any substring
of a string
in $O(1)$ time after an $O(n)$ time preprocessing.
The idea is to construct an array $h$ such that
$h[k]$ contains the hash value of the prefix
of the string that ends at position $k$.
Using polynomial hashing, we can calculate the hash value of any substring
of a string \texttt{s} in $O(1)$ time after an $O(n)$ time preprocessing.
The idea is to construct an array \texttt{h} such that
$\texttt{h}[k]$ contains the hash value of the prefix $\texttt{s}[0 \ldots k]$.
The array values can be recursively calculated as follows:
\[
\begin{array}{lcl}
h[0] & = & \texttt{s}[0] \\
h[k] & = & (h[k-1] A + \texttt{s}[k]) \bmod B \\
\texttt{h}[0] & = & \texttt{s}[0] \\
\texttt{h}[k] & = & (\texttt{h}[k-1] A + \texttt{s}[k]) \bmod B \\
\end{array}
\]
In addition, we construct an array $p$
where $p[k]=A^k \bmod B$:
In addition, we construct an array $\texttt{p}$
where $\texttt{p}[k]=A^k \bmod B$:
\[
\begin{array}{lcl}
p[0] & = & 1 \\
p[k] & = & (p[k-1] A) \bmod B. \\
\texttt{p}[0] & = & 1 \\
\texttt{p}[k] & = & (\texttt{p}[k-1] A) \bmod B. \\
\end{array}
\]
Constructing these arrays takes $O(n)$ time.
After this, the hash value of a substring
that begins at position $a$ and ends at position $b$
After this, the hash value of any substring
$\texttt{s}[a \ldots b]$
can be calculated in $O(1)$ time using the formula
\[(h[b]-h[a-1] p[b-a+1]) \bmod B\]
\[(\texttt{h}[b]-\texttt{h}[a-1] \texttt{p}[b-a+1]) \bmod B\]
assuming that $a>0$.
If $a=0$, the hash value is simply $h[b]$.
If $a=0$, the hash value is simply $\texttt{h}[b]$.
\subsubsection*{Using hash values}
@ -364,7 +367,7 @@ The probability of one or more collisions is
\[1-(1-\frac{1}{B})^n.\]
\textit{Scenario 3:} Strings $x_1,x_2,\ldots,x_n$
\textit{Scenario 3:} All pairs of strings $x_1,x_2,\ldots,x_n$
are compared with each other.
The probability of one or more collisions is
\[ 1 - \frac{B \cdot (B-1) \cdot (B-2) \cdots (B-n+1)}{B^n}.\]
@ -398,7 +401,7 @@ $B \approx 10^9$.
The phenomenon in scenario 3 is known as the
\key{birthday paradox}: if there are $n$ people
in a room, the probability that some two people
in a room, the probability that \emph{some} two people
have the same birthday is large even if $n$ is quite small.
In hashing, correspondingly, when all hash values are compared
with each other, the probability that some two
@ -417,7 +420,7 @@ which makes the probability of a collision very small.
Some people use constants $B=2^{32}$ and $B=2^{64}$,
which is convenient, because operations with 32 and 64
bit integers are calculated modulo $2^{32}$ and $2^{64}$.
However, this is not a good choice, because it is possible
However, this is \emph{not} a good choice, because it is possible
to construct inputs that always generate collisions when
constants of the form $2^x$ are used \cite{pac13}.
@ -426,17 +429,16 @@ constants of the form $2^x$ are used \cite{pac13}.
\index{Z-algorithm}
\index{Z-array}
The \key{Z-array} of a string
contains for each position of the string
the length of the longest substring
that begins at that position and is a prefix of the string.
Such an array can be efficiently constructed
using the \key{Z-algorithm}\footnote{The Z-algorithm
was presented in \cite{gus97} as the simplest known
method for linear-time pattern matching, and the original idea
was attributed to \cite{mai84}.}.
The \key{Z-array} \texttt{z} of a string \texttt{s}
of length $n$ contains for each $k=0,1,\ldots,n-1$
the length of the longest substring of \texttt{s}
that begins at position $k$ and is a prefix of \texttt{s}.
Thus, $\texttt{z}[k]=p$ tells us that
$\texttt{s}[0 \ldots p-1]$ equals $\texttt{s}[k \ldots k+p-1]$.
Many string processing problems can be efficiently solved
using the Z-array.
For example, the Z-array of the string
For example, the Z-array of
\texttt{ACBACDACBACBACDA} is as follows:
\begin{center}
@ -498,48 +500,45 @@ For example, the Z-array of the string
\end{tikzpicture}
\end{center}
For example, the value at position 6 of the
above Z-array is 5,
In this case, for example, $\texttt{z}[6]=5$,
because the substring \texttt{ACBAC} of length 5
is a prefix of the string,
is a prefix of \texttt{s},
but the substring \texttt{ACBACB} of length 6
is not a prefix of the string.
It is often a matter of taste whether to use
string hashing or the Z-algorithm.
Unlike hashing, the Z-algorithm always works
and there is no risk for collisions.
On the other hand, the Z-algorithm is more difficult
to implement and some problems can only be solved
using hashing.
is not a prefix of \texttt{s}.
\subsubsection*{Algorithm description}
The Z-algorithm scans the string from left
to right, and calculates for each position
the length of the longest substring that
is a prefix of the string.
A straightforward algorithm
would have a time complexity of $O(n^2)$,
but the Z-algorithm has an important
optimization which ensures that the time complexity
is only $O(n)$.
Next we describe an algorithm,
called the \key{Z-algorithm}\footnote{The Z-algorithm
was presented in \cite{gus97} as the simplest known
method for linear-time pattern matching, and the original idea
was attributed to \cite{mai84}.},
that efficiently constructs the Z-array in $O(n)$ time.
The algorithm calculates the Z-array values
from left to right by both using information
already stored in the Z-array and comparing substrings
character by character.
The idea is to maintain a range $[x,y]$ such that
the substring from $x$ to $y$ is a prefix of
the string and $y$ is as large as possible.
Since the characters in the ranges $[0,y-x]$
and $[x,y]$ are the same,
we can use this information to calculate
the Z-array values in the range $[x,y]$.
To efficiently calculate the Z-array values,
the algorithm maintains a range $[x,y]$ such that
$\texttt{s}[x \ldots y]$ is a prefix of \texttt{s}
and $y$ is as large as possible.
Since we know that $\texttt{s}[0 \ldots y-x]$
and $\texttt{s}[x \ldots y]$ are equal,
we can use this information when calculating
Z-values for positions $x+1,x+2,\ldots,y$.
The time complexity of the Z-algorithm is $O(n)$,
because the algorithm only compares strings
character by character starting at position $y+1$.
If the characters match, the value of $y$ increases,
and it is not needed to compare the character at
position $y$ again
but the information in the Z-array can be used.
At each position $k$, we first
check the value of $\texttt{z}[k-x]$.
If $k+\texttt{z}[k-x]<y$, we know that $\texttt{z}[k]=\texttt{z}[k-x]$.
However, if $k+\texttt{z}[k-x] \ge y$,
$\texttt{s}[0 \ldots y-k]$ equals
$\texttt{s}[k \ldots y]$, and to determine the
value of $\texttt{z}[k]$ we need to compare
the substrings character by character.
Still, the algorithm works in $O(n)$ time,
because we start comparing at positions
$y-k+1$ and $y+1$.
For example, let us construct the following Z-array:
@ -602,10 +601,8 @@ For example, let us construct the following Z-array:
\end{tikzpicture}
\end{center}
The first interesting position is 6 where the
length of the common prefix is 5.
After calculating this value,
the current $[x,y]$ range will be $[6,10]$:
After calculating the value $\texttt{z}[6]=5$,
the current $[x,y]$ range is $[6,10]$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -673,17 +670,15 @@ the current $[x,y]$ range will be $[6,10]$:
\end{tikzpicture}
\end{center}
Now, it is possible to calculate the
subsequent values of the Z-array
more efficiently,
Now we can calculate
subsequent Z-array values
efficiently,
because we know that
the ranges $[0,4]$ and $[6,10]$
contain the same characters.
First, since the values at
positions 1 and 2 are 0,
we immediately know that
the values at positions 7 and 8
are also 0:
$\texttt{s}[0 \ldots 4]$ and
$\texttt{s}[6 \ldots 10]$ are equal.
First, since $\texttt{z}[1] = \texttt{z}[2] = 0$,
we immediately know that also
$\texttt{z}[7] = \texttt{z}[8] = 0$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -755,9 +750,7 @@ are also 0:
\end{tikzpicture}
\end{center}
After this, we know that the value
at position 9 will be at least 2,
because the value at position 3 is 2:
Then, since $\texttt{z}[3]=2$, we know that $\texttt{z}[9] \ge 2$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -826,8 +819,8 @@ because the value at position 3 is 2:
\end{tikzpicture}
\end{center}
Since we have no information about the characters
after position 10, we have to begin to compare the strings
However, we have no information about the string
after position 10, so we need to compare the substrings
character by character:
\begin{center}
@ -901,10 +894,8 @@ character by character:
\end{tikzpicture}
\end{center}
It turns out that the length of the common
prefix at position 9 is 7,
and thus the new range $[x,y]$ is $[9,15]$:
It turns out that $\texttt{z}[9]=7$,
so the new $[x,y]$ range is $[9,15]$:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -973,10 +964,9 @@ and thus the new range $[x,y]$ is $[9,15]$:
\end{tikzpicture}
\end{center}
After this, all subsequent values of the Z-array
can be calculated using the values already
stored in the array. All the remaining values can be
directly retrieved from the beginning of the Z-array:
After this, all the remaining Z-array values
can be determined by using the information
already stored in the Z-array:
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -1045,10 +1035,18 @@ directly retrieved from the beginning of the Z-array:
\subsubsection{Using the Z-array}
As an example, let us consider again
It is often a matter of taste whether to use
string hashing or the Z-algorithm.
Unlike hashing, the Z-algorithm always works
and there is no risk for collisions.
On the other hand, the Z-algorithm is more difficult
to implement and some problems can only be solved
using hashing.
As an example, consider again
the pattern matching problem,
where our task is to find the positions
where a pattern $p$ occurs in a string $s$.
where our task is to find the occurrences
of a pattern $p$ in a string $s$.
We already solved this problem efficiently
using string hashing, but the Z-algorithm
provides another way to solve the problem.
@ -1123,7 +1121,7 @@ the Z-array is as follows:
The positions 5 and 10 contain the value 3,
which means that the pattern \texttt{ATT}
occurs in the corresponding positions
in the string \texttt{HATTIVATTI}.
of \texttt{HATTIVATTI}.
The time complexity of the resulting algorithm
is linear, because it suffices to construct