updating chapter 4 and rebuilding book.pdf

chapter04.tex

@@ -755,40 +755,22 @@ \subsubsection{Algorithm 3}
 so the total time complexity is $O(n \log n)$.
 
 \subsubsection{Efficiency comparison}
-
 The following table shows how efficient
 the above algorithms are when $n$ varies and
 the elements of the lists are random
-integers between $1 \ldots 10^9$:
+integers between $1 \ldots 2 \cdot 10^7$:
 
 \begin{center}
 \begin{tabular}{rrrr}
 $n$ & Algorithm 1 & Algorithm 2 & Algorithm 3 \\
 \hline
-$10^6$ & $1.5$ s & $0.3$ s & $0.2$ s \\
-$2 \cdot 10^6$ & $3.7$ s & $0.8$ s & $0.3$ s \\
-$3 \cdot 10^6$ & $5.7$ s & $1.3$ s & $0.5$ s \\
-$4 \cdot 10^6$ & $7.7$ s & $1.7$ s & $0.7$ s \\
-$5 \cdot 10^6$ & $10.0$ s & $2.3$ s & $0.9$ s \\
+$10^6$ & $2.8$ s & $2.2$ s & $2.2$ s \\
+$2 \cdot 10^6$ & $5.7$ s & $4.6$ s & $5$ s \\
+$3 \cdot 10^6$ & $8.9$ s & $7.4$ s & $8.6$ s \\
+$4 \cdot 10^6$ & $12.3$ s & $10$ s & $12.1$ s \\
+$5 \cdot 10^6$ & $15.7$ s & $12.1$ s & $15$ s \\
+$10^7$ & $31.7$ s & $24.5$ s & $32.5$ s \\
 \end{tabular}
 \end{center}
 
-Algorithms 1 and 2 are equal except that
-they use different set structures.
-In this problem, this choice has an important effect on
-the running time, because Algorithm 2
-is 4–5 times faster than Algorithm 1.
-
-However, the most efficient algorithm is Algorithm 3
-which uses sorting.
-It only uses half the time compared to Algorithm 2.
-Interestingly, the time complexity of both
-Algorithm 1 and Algorithm 3 is $O(n \log n)$,
-but despite this, Algorithm 3 is ten times faster.
-This can be explained by the fact that
-sorting is a simple procedure and it is done
-only once at the beginning of Algorithm 3,
-and the rest of the algorithm works in linear time.
-On the other hand,
-Algorithm 1 maintains a complex balanced binary tree
-during the whole algorithm.
+Algorithms 1 and 3 are equal and Algorithm 2 is the fastest. which is intuitive as the time complexity of Algorithm 1 and Algorithm 3 are O(nlogn) which is bigger than O(n) when n tends to be very large(like $10^7$).