bibin
/
cphb
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

Use 0-indexing also in strings [closes #27]

This commit is contained in:
Antti H S Laaksonen 2017-04-17 18:20:20 +03:00
parent 35d53d39d4
commit b9a237181b
1 changed files with 163 additions and 160 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

323
chapter26.tex
View File

@ -108,7 +108,7 @@ 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 or there is a position $k$ such that
$x[i]=y[i]$ when $i<k$ and $x[k]<y[k]$. $x[i]=y[i]$ when $i<k$ and $x[k]<y[k]$.
\section{Trie structures} \section{Trie structure}
\index{trie} \index{trie}
@ -218,9 +218,10 @@ based on their hash values.
A usual way to implement string hashing A usual way to implement string hashing
is polynomial hashing, which means is polynomial hashing, which means
that the hash value is calculated using the formula that the hash value is calculated using the formula
\[(c[1] A^{n-1} + c[2] A^{n-2} + \cdots + c[n] A^0) \bmod B ,\] \[(\texttt{s}[0] A^{n-1} + \texttt{s}[1] A^{n-2} + \cdots + \texttt{s}[n-1] A^0) \bmod B ,\]
where $c[1],c[2],\ldots,c[n]$ where \texttt{s} is a string of length $n$
are the codes of the characters in the string, (so $s[0],s[1],\ldots,s[n-1]$
are the codes of the characters),
and $A$ and $B$ are pre-chosen constants. and $A$ and $B$ are pre-chosen constants.
For example, the codes of the characters For example, the codes of the characters
@ -263,8 +264,8 @@ of the string that ends at position $k$.
The array values can be recursively calculated as follows: The array values can be recursively calculated as follows:
\[ \[
\begin{array}{lcl} \begin{array}{lcl}
h[0] & = & 0 \\ h[0] & = & \texttt{s}[0] \\
h[k] & = & (h[k-1] A + c[k]) \bmod B \\ h[k] & = & (h[k-1] A + \texttt{s}[k]) \bmod B \\
\end{array} \end{array}
\] \]
In addition, we construct an array $p$ In addition, we construct an array $p$
@ -279,7 +280,9 @@ Constructing these arrays takes $O(n)$ time.
After this, the hash value of a substring After this, the hash value of a substring
that begins at position $a$ and ends at position $b$ that begins at position $a$ and ends at position $b$
can be calculated in $O(1)$ time using the formula can be calculated in $O(1)$ time using the formula
\[(h[b]-h[a-1] p[b-a+1]) \bmod B.\] \[(h[b]-h[a-1] p[b-a+1]) \bmod B\]
assuming that $a>0$.
If $a=0$, the hash value is simply $h[b]$.
\subsubsection*{Using hash values} \subsubsection*{Using hash values}
@ -477,22 +480,22 @@ For example, the Z-array for the string
\node at (15.5, 0.5) {1}; \node at (15.5, 0.5) {1};
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -581,30 +584,30 @@ For example, let us construct the following Z-array:
\node at (15.5, 0.5) {?}; \node at (15.5, 0.5) {?};
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The first interesting position is 7 where the The first interesting position is 6 where the
length of the common prefix is 5. length of the common prefix is 5.
After calculating this value, After calculating this value,
the current $[x,y]$ range will be $[7,11]$: the current $[x,y]$ range will be $[6,10]$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -652,22 +655,22 @@ the current $[x,y]$ range will be $[7,11]$:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -676,12 +679,12 @@ Now, it is possible to calculate the
subsequent values of the Z-array subsequent values of the Z-array
more efficiently, more efficiently,
because we know that because we know that
the ranges $[1,5]$ and $[7,11]$ the ranges $[0,4]$ and $[6,10]$
contain the same characters. contain the same characters.
First, since the values at First, since the values at
positions 2 and 3 are 0, positions 1 and 2 are 0,
we immediately know that we immediately know that
the values at positions 8 and 9 the values at positions 7 and 8
are also 0: are also 0:
\begin{center} \begin{center}
@ -731,22 +734,22 @@ are also 0:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\draw[thick,<->] (7.5,-0.25) .. controls (7,-1.25) and (2,-1.25) .. (1.5,-0.25); \draw[thick,<->] (7.5,-0.25) .. controls (7,-1.25) and (2,-1.25) .. (1.5,-0.25);
@ -755,8 +758,8 @@ are also 0:
\end{center} \end{center}
After this, we know that the value After this, we know that the value
at position 10 will be at least 2, at position 9 will be at least 2,
because the value at position 4 is 2: because the value at position 3 is 2:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -804,29 +807,29 @@ because the value at position 4 is 2:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25); \draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
Since we have no information about the characters Since we have no information about the characters
after position 11, we have to begin to compare the strings after position 10, we have to begin to compare the strings
character by character: character by character:
\begin{center} \begin{center}
@ -879,22 +882,22 @@ character by character:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
%\draw[thick,<->] (11.5,-0.25) .. controls (11,-1.25) and (3,-1.25) .. (2.5,-0.25); %\draw[thick,<->] (11.5,-0.25) .. controls (11,-1.25) and (3,-1.25) .. (2.5,-0.25);
\end{tikzpicture} \end{tikzpicture}
@ -902,8 +905,8 @@ character by character:
It turns out that the length of the common It turns out that the length of the common
prefix at position 10 is 7, prefix at position 9 is 7,
and thus the new range $[x,y]$ is $[10,16]$: and thus the new range $[x,y]$ is $[9,15]$:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.7] \begin{tikzpicture}[scale=0.7]
@ -951,22 +954,22 @@ and thus the new range $[x,y]$ is $[10,16]$:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
% \draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25); % \draw[thick,<->] (9.5,-0.25) .. controls (9,-1.25) and (4,-1.25) .. (3.5,-0.25);
\end{tikzpicture} \end{tikzpicture}
@ -1022,22 +1025,22 @@ directly retrieved from the beginning of the Z-array:
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\node at (14.5, 2.5) {15}; \node at (14.5, 2.5) {14};
\node at (15.5, 2.5) {16}; \node at (15.5, 2.5) {15};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -1102,24 +1105,24 @@ the Z-array is as follows:
\node at (13.5, 0.5) {0}; \node at (13.5, 0.5) {0};
\footnotesize \footnotesize
\node at (0.5, 2.5) {1}; \node at (0.5, 2.5) {0};
\node at (1.5, 2.5) {2}; \node at (1.5, 2.5) {1};
\node at (2.5, 2.5) {3}; \node at (2.5, 2.5) {2};
\node at (3.5, 2.5) {4}; \node at (3.5, 2.5) {3};
\node at (4.5, 2.5) {5}; \node at (4.5, 2.5) {4};
\node at (5.5, 2.5) {6}; \node at (5.5, 2.5) {5};
\node at (6.5, 2.5) {7}; \node at (6.5, 2.5) {6};
\node at (7.5, 2.5) {8}; \node at (7.5, 2.5) {7};
\node at (8.5, 2.5) {9}; \node at (8.5, 2.5) {8};
\node at (9.5, 2.5) {10}; \node at (9.5, 2.5) {9};
\node at (10.5, 2.5) {11}; \node at (10.5, 2.5) {10};
\node at (11.5, 2.5) {12}; \node at (11.5, 2.5) {11};
\node at (12.5, 2.5) {13}; \node at (12.5, 2.5) {12};
\node at (13.5, 2.5) {14}; \node at (13.5, 2.5) {13};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The positions 6 and 11 contain the value 3, The positions 5 and 10 contain the value 3,
which means that the pattern \texttt{ATT} which means that the pattern \texttt{ATT}
occurs in the corresponding positions occurs in the corresponding positions
in the string \texttt{HATTIVATTI}. in the string \texttt{HATTIVATTI}.