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Use 0-indexing with C++ arrays

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Antti H S Laaksonen 2017-04-17 17:59:27 +03:00
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chapter03.tex
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@ -49,15 +49,15 @@ For example, the array
\node at (6.5,0.5) {$5$};
\node at (7.5,0.5) {$6$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
will be as follows after sorting:
@ -73,15 +73,15 @@ will be as follows after sorting:
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -97,18 +97,17 @@ A famous $O(n^2)$ time sorting algorithm
is \key{bubble sort} where the elements
''bubble'' in the array according to their values.
Bubble sort consists of $n-1$ rounds.
Bubble sort consists of $n$ rounds.
On each round, the algorithm iterates through
the elements of the array.
Whenever two consecutive elements are found
that are not in correct order,
the algorithm swaps them.
The algorithm can be implemented as follows
for an array
$\texttt{t}[1],\texttt{t}[2],\ldots,\texttt{t}[n]$:
for an array \texttt{x}:
\begin{lstlisting}
for (int i = 1; i <= n-1; i++) {
for (int j = 1; j <= n-i; j++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n-1; j++) {
if (t[j] > t[j+1]) swap(t[j],t[j+1]);
}
}
@ -118,7 +117,7 @@ After the first round of the algorithm,
the largest element will be in the correct position,
and in general, after $k$ rounds, the $k$ largest
elements will be in the correct positions.
Thus, after $n-1$ rounds, the whole array
Thus, after $n$ rounds, the whole array
will be sorted.
For example, in the array
@ -135,16 +134,16 @@ For example, in the array
\node at (5.5,0.5) {$2$};
\node at (6.5,0.5) {$5$};
\node at (7.5,0.5) {$6$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -165,16 +164,16 @@ as follows:
\node at (7.5,0.5) {$6$};
\draw[thick,<->] (3.5,-0.25) .. controls (3.25,-1.00) and (2.75,-1.00) .. (2.5,-0.25);
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -191,17 +190,16 @@ as follows:
\node at (7.5,0.5) {$6$};
\draw[thick,<->] (5.5,-0.25) .. controls (5.25,-1.00) and (4.75,-1.00) .. (4.5,-0.25);
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -218,17 +216,16 @@ as follows:
\node at (7.5,0.5) {$6$};
\draw[thick,<->] (6.5,-0.25) .. controls (6.25,-1.00) and (5.75,-1.00) .. (5.5,-0.25);
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -245,16 +242,16 @@ as follows:
\node at (7.5,0.5) {$9$};
\draw[thick,<->] (7.5,-0.25) .. controls (7.25,-1.00) and (6.75,-1.00) .. (6.5,-0.25);
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -289,16 +286,16 @@ For example, in the array
\node at (5.5,0.5) {$5$};
\node at (6.5,0.5) {$9$};
\node at (7.5,0.5) {$8$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
the inversions are $(6,3)$, $(6,5)$ and $(9,8)$.
@ -360,16 +357,16 @@ For example, consider sorting the following array:
\node at (5.5,0.5) {$2$};
\node at (6.5,0.5) {$5$};
\node at (7.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -389,16 +386,16 @@ as follows:
\node at (6.5,0.5) {$2$};
\node at (7.5,0.5) {$5$};
\node at (8.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (5.5,1.4) {$5$};
\node at (6.5,1.4) {$6$};
\node at (7.5,1.4) {$7$};
\node at (8.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (5.5,1.4) {$5$};
% \node at (6.5,1.4) {$6$};
% \node at (7.5,1.4) {$7$};
% \node at (8.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -419,16 +416,16 @@ as follows:
\node at (6.5,0.5) {$5$};
\node at (7.5,0.5) {$8$};
\node at (8.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (5.5,1.4) {$5$};
\node at (6.5,1.4) {$6$};
\node at (7.5,1.4) {$7$};
\node at (8.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (5.5,1.4) {$5$};
% \node at (6.5,1.4) {$6$};
% \node at (7.5,1.4) {$7$};
% \node at (8.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -445,16 +442,16 @@ subarrays and creates the final sorted array:
\node at (5.5,0.5) {$6$};
\node at (6.5,0.5) {$8$};
\node at (7.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
@ -546,6 +543,7 @@ whose indices are elements in the original array.
The algorithm iterates through the original array
and calculates how many times each element
appears in the array.
\newpage
For example, the array
\begin{center}
@ -559,16 +557,16 @@ For example, the array
\node at (5.5,0.5) {$3$};
\node at (6.5,0.5) {$5$};
\node at (7.5,0.5) {$9$};
\footnotesize
\node at (0.5,1.4) {$1$};
\node at (1.5,1.4) {$2$};
\node at (2.5,1.4) {$3$};
\node at (3.5,1.4) {$4$};
\node at (4.5,1.4) {$5$};
\node at (5.5,1.4) {$6$};
\node at (6.5,1.4) {$7$};
\node at (7.5,1.4) {$8$};
%
% \footnotesize
% \node at (0.5,1.4) {$1$};
% \node at (1.5,1.4) {$2$};
% \node at (2.5,1.4) {$3$};
% \node at (3.5,1.4) {$4$};
% \node at (4.5,1.4) {$5$};
% \node at (5.5,1.4) {$6$};
% \node at (6.5,1.4) {$7$};
% \node at (7.5,1.4) {$8$};
\end{tikzpicture}
\end{center}
corresponds to the following bookkeeping array:
@ -658,6 +656,7 @@ int n = 7; // array size
int t[] = {4,2,5,3,5,8,3};
sort(t,t+n);
\end{lstlisting}
\newpage
The following code sorts the string \texttt{s}:
\begin{lstlisting}
string s = "monkey";
@ -776,7 +775,7 @@ For example, the following code searches for
an element $x$ in the array \texttt{t}:
\begin{lstlisting}
for (int i = 1; i <= n; i++) {
for (int i = 0; i < n; i++) {
if (t[i] == x) {} // x found at index i
}
\end{lstlisting}
@ -815,7 +814,7 @@ depending on the value of the middle element.
The above idea can be implemented as follows:
\begin{lstlisting}
int a = 1, b = n;
int a = 0, b = n-1;
while (a <= b) {
int k = (a+b)/2;
if (t[k] == x) {} // x found at index k
@ -848,9 +847,9 @@ been found or we know that it does not appear in the array.
The following code implements the above idea:
\begin{lstlisting}
int k = 1;
int k = 0;
for (int b = n/2; b >= 1; b /= 2) {
while (k+b <= n && t[k+b] <= x) k += b;
while (k+b < n && t[k+b] <= x) k += b;
}
if (t[k] == x) {} // x was found at index k
\end{lstlisting}