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Theorem 145 The insertion sort is correct See the solution to Problem 1423 on page 277 for a proof of this theorem Theorem 146 The insertion sort runs in O(n2) time See the solution to Problem 1424 on page 277 for a proof of this theorem Theorem 147 The insertion sort runs in O(n) time on a sorted sequence See the solution to Problem 1425 on page 283 for a proof of this theorem THE SHELL SORT Theorem 147 suggests that if the sequence is nearly sorted, then the insertion sort will run nearly in O(n) time That is true The shell sort exploits that fact to obtain an algorithm that in general runs in better than O(n15) time It applies the insertion sort repeatedly to skip subsequences such as {s0, s3, s6, s9, , sn 2} and {s1, s4, s7, s10, .

, sn 1} These are two of the three skip-3-subsequences Algorithm 144 The Shell Sort (Precondition: s = {s0 sn 1} is a sequence of n ordinal values) (Postcondition: The entire sequence s is sorted) 1 Set d = 1 2 Repeat step 3 until 9d > n 3 Set d = 3d + 1 4 Do steps 5 6 until d = 0 5 Apply the insertion sort to each of the d skip-d-subsequences of s 6 Set d = d/3 Suppose that s has n = 200 elements Then the loop at step 2 would iterate three times, increasing d from 1 to d = 4, 13, and 40 The first iteration of the loop at step 4 would apply the insertion sort to each of the 40 skip-40subsequences {s0, s40, s80, s120, s160}, {s1, s41, s81, s121, s161}, {s2, s42, s82, s122, s162}, .

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The SQL code is the same as would be issued interactively, but the SQL is stored in character strings inside the program, and then sent to the DBMS for execution In the case of JDBC, there is also a very first step required, which is to load the driver code for connecting to the database There are several mechanisms, or types of drivers for JDBC Some will connect by translating the calls into another standard called open database connection (ODBC); some will convert the JDBC calls into the native application programming interface (API) of the DBMS; some will convert the calls into an independent or vendor-specific network DBMS protocol One should check with the DBMS administrator for advice about which driver to load for the particular DBMS in use One can find information about JDBC drivers at: http://industryjavasuncom/products/jdbc/drivers.

, {s39, s79, s119, s159, s199} Then step 6 would reduce d to 13, and then the second iteration of the loop at step 4 would apply the insertion sort to each of the thirteen skip-13-subsequences {s0, s13, s26, s39, s52, s65, , s194}, {s1, s14, s27, s40, s53, s66, , s195}, , {s12, s25, s38, s51, s64, s77, , s193} Then step 6 would reduce d to 4, and the third iteration of the loop at step 4 would apply the insertion sort to each of the four skip-4-subsequences {s0, s4, s8, s12, , s196}, {s1, s5, s9, s13, , s197}, {s2, s6, s10, s14, , s198}, and {s3, s7, s11, s15, , s199} Then step 6 would reduce d to 1, and, and the fourth.

iteration of the loop at step 4 would apply the insertion sort to the entire sequence. This entire process would apply the insertion sort 58 times: 40 times to subsequences of size n1 = 5, 13 times to subsequences of size n2 = 15, 4 times to subsequences of size n3 = 50, and once to the entire sequence of size n4 = n = 200. At first glance, the repeated use of the insertion sort within the shell sort would seem to take longer than simply applying the insertion sort directly just once to the entire sequence. Indeed, a direct calculation of the total number of comparisons, using the complexity function n2, yields 40(n12) + 13(n22) + 4(n32) + 1(n42) = 40(52) + 13(152) + 4(502) + 1(2002) = 53,925 which is quite a bit worse than the single n2 = 2002 = 40,000 But after the first iteration of step 4, the subsequent subsequences are nearly sorted. So the actual number of comparisons needed there is closer to n. Thus, the actual number of comparisons is more like 40(n12) + 13(n2) + 4(n3) + 1(n4) = 40(52) + 13(15) + 4(50) + 1(200) = 1,595 which is quite a bit better than 40,000. Theorem 14.8 The shell sort runs in O(n1.5) time. Note that, for n = 200, n1.5 = 2001.5 = 2,829, which is a lot better than n2 = 2002 = 40,000. EXAMPLE 14.7 The Shell Sort

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