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We must still rearrange the array a The simplest way is to declare a second array of Comparable, which we call copy We can write the correct sorted order into copy and then write from copy back into a The cost of doing so is an extra array and a total of 2N Comparable copies

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The name is simply a string and can be any name you choose The ID is a Uniform Resource Name (URN) that uniquely identi es the template part As you can see from Figure 1017, the URN contains the name of the template part, the date, and the time, which helps to uniquely identify the template part If you want to create a new template part, you must change this URN For example, you could change the name part of the URN and the date and time by hand (You ll notice that when you change the name of the template part, the ID changes as well, but not vice versa) If you decide that you want to regenerate the template part ID based on the new name you entered, just click the Reset button Once you change the URN and save, you have a completely new template part (Of course, since it s possible to manually edit the template part ID in the Template Part Properties dialog, it is also possible to enter the ID of an existing template part This is only useful if you want to overwrite an existing part and not create a new one)





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The algorithm in Section 991 has a potentially important problem By using copy, we doubled the space requirement We can assume that N is large (otherwise, we would use insertion sort) and that the size of a Comparable object is large (otherwise, we would not bother using a pointer implementation) Thus we can reasonably expect that we are operating near the memory limits of our machine Although we can expect to use an extra v e c t o r of pointers, we cannot necessarily expect an extra v e c t o r of Comparable objects to be available Thus we need to rearrange the array a in place, without resorting to an extra array A second consequence of our decision to use copy is that a total of 2N C o m ~ a r a b l e o ~ i e s used Although this situation is an im~rovement c are u over the original algorithm, we can improve the algorithm even more In particular, in the following improvement, we never use more than 3 N l 2 Comparable copies, and on almost all inputs we use only a few more than N Not only do we save space, but also we save time Before stepping through the code, let us get a general idea of what needs to be done Surprisingly, we have already done it before To get an idea of what we have to do, let us start with i = 2 With p [ 2 I pointing at a [ 4 1 , we know that we need to move a [ 41 to a [ 2 I First, we must save a [ 2 1, or we will not be able to place it correctly later Thus we havetmp=a[2] andthena[2]=a[4]Whena[4] hasbeenmovedtoa[21, we can move something into a [ 4 I, which is essentially vacant By examining p [ 41 , we see that the correct statement is a [ 4 ] =a [ 3 ] Now we need to move something into a [ 3 1 With p [ 3 ] pointing at a [ 2 I , we know that we want to move a [ 2 1 there But a [ 2 1 has been overwritten at the start of this rearrangement; as its original value is in tmp, we finish with a [ 3] =tmp This process shows that, by starting with i equal to 2 and following the p array, we form a cyclic sequence 2,4, 3, 2, which corresponds to.

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