Use 0-based array indexing

This commit is contained in:
Antti H S Laaksonen 2017-04-19 21:12:31 +03:00
parent a456fedeff
commit ee7ede62d5
1 changed files with 32 additions and 32 deletions

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@ -774,21 +774,21 @@ The following array corresponds to the above tree:
\node at (14.5,0.5) {$1$}; \node at (14.5,0.5) {$1$};
\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) {$14$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
@ -846,27 +846,27 @@ can be found as follows:
\node at (14.5,0.5) {$1$}; \node at (14.5,0.5) {$1$};
\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) {$14$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
Node 5 is at position 3, node 8 is at position 6, Node 5 is at position 2, node 8 is at position 5,
and the node with lowest level between and the node with lowest level between
positions $3 \ldots 6$ is node 2 at position 4 positions $2 \ldots 5$ is node 2 at position 3
whose level is 2. whose level is 2.
Thus, the lowest common ancestor of Thus, the lowest common ancestor of
nodes 5 and 8 is node 2. nodes 5 and 8 is node 2.