8.1 Straight Insertion and Shell's Method 331 void piksr2(int n,float arr[],float brr[]) Sorts an array arr [1..n]into ascending numerical order,by straight insertion,while making the corresponding rearrangement of the array brr[1..n]. int i,j; float a,b; for (j=2;j<=n;j++){ Pick out each element in turn asarr[i]; b=brr[j]; 1=j-1; while (i>0&&arr[i]a){ Look for the place to insert it. arr[i+1]=arr[i]; brr[i+1]=brr[i]; 1--； 83g d granted for 19881992 arr[i+1]=a; Insert it. brr[i+1]=b; 11800 from NUMERICAL RECIPES I For the case of rearranging a larger number of arrays by sorting on one of ⊙ them,see 88.4. Shell's Method 豆e3d Americ computer, This is actually a variant on straight insertion,but a very powerful variant indeed. The rough idea,e.g.,for the case of sorting 16 numbers n1...n16,is this:First sort, 9 by straight insertion,each of the 8 groups of 2 (n1,ng),(n2,n1o),...,(ns,n6). Next,sort each of the 4 groups of 4 (n1,n5,ng,n13),...,(n4,ns,n12,n16).Next SCIENTIFIC sort the 2 groups of 8 records,beginning with (n1,n3,n5,n7,ng,n11,n13,n15). Finally,sort the whole list of 16 numbers. Of course,only the last sort is necessary for putting the numbers into order.So what is the purpose of the previous partial sorts?The answer is that the previous sorts allow numbers efficiently to filter up or down to positions close to their final resting places.Therefore,the straight insertion passes on the final sort rarely have to go past more than a"few"elements before finding the right place.(Think of sorting Numerical 10-621 a hand of cards that are already almost in order. The spacings between the numbers sorted on each pass through the data(8,4,2,1 43106 in the above example)are called the increments,and a Shell sort is sometimes Recipes called a diminishing increment sort.There has been a lot of research into how to choose a good set of increments,but the optimum choice is not known.The set ...8,4,2,1 is in fact not a good choice,especially for N a power of 2.A much North better choice is the sequence (3-1)/2,..,40,13,4,1 (8.1.1) which can be generated by the recurrence i1=1, ik+1=3k+1,k=1,2,. (8.1.2) It can be shown (see 11])that for this sequence of increments the number of operations required in all is of order N3/2 for the worst possible ordering of the original data.8.1 Straight Insertion and Shell’s Method 331 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machine￾readable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). void piksr2(int n, float arr[], float brr[]) Sorts an array arr[1..n] into ascending numerical order, by straight insertion, while making the corresponding rearrangement of the array brr[1..n]. { int i,j; float a,b; for (j=2;j<=n;j++) { Pick out each element in turn. a=arr[j]; b=brr[j]; i=j-1; while (i > 0 && arr[i] > a) { Look for the place to insert it. arr[i+1]=arr[i]; brr[i+1]=brr[i]; i--; } arr[i+1]=a; Insert it. brr[i+1]=b; } } For the case of rearranging a larger number of arrays by sorting on one of them, see §8.4. Shell’s Method This is actually a variant on straight insertion, but a very powerful variant indeed. The rough idea, e.g., for the case of sorting 16 numbers n 1 ...n16, is this: First sort, by straight insertion, each of the 8 groups of 2 (n1, n9), (n2, n10), ... , (n8, n16). Next, sort each of the 4 groups of 4 (n1, n5, n9, n13), ... , (n4, n8, n12, n16). Next sort the 2 groups of 8 records, beginning with (n1, n3, n5, n7, n9, n11, n13, n15). Finally, sort the whole list of 16 numbers. Of course, only the last sort is necessary for putting the numbers into order. So what is the purpose of the previous partial sorts? The answer is that the previous sorts allow numbers efficiently to filter up or down to positions close to their final resting places. Therefore, the straight insertion passes on the final sort rarely have to go past more than a “few” elements before finding the right place. (Think of sorting a hand of cards that are already almost in order.) The spacings between the numbers sorted on each pass through the data (8,4,2,1 in the above example) are called the increments, and a Shell sort is sometimes called a diminishing increment sort. There has been a lot of research into how to choose a good set of increments, but the optimum choice is not known. The set ..., 8, 4, 2, 1 is in fact not a good choice, especially for N a power of 2. A much better choice is the sequence (3k − 1)/2,..., 40, 13, 4, 1 (8.1.1) which can be generated by the recurrence i1 = 1, ik+1 = 3ik + 1, k = 1, 2,... (8.1.2) It can be shown (see [1]) that for this sequence of increments the number of operations required in all is of order N 3/2 for the worst possible ordering of the original data