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Intermediate
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Figure 12-2 Building an index
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If you were looking for the words SQL Server, the query would scan the root page. The value of O would be found as well as T. Because S comes before T, the page that is needed to find the data would be on O. The query would then move to the interme diate-level page that O pointed to. Note that this single operation has immediately eliminated three-fourths of the possible pages in a single step by scanning a small sub set of values. The query would then scan the intermediate-level page and find the value S. It would then jump to the page to which this entry points. At this point, the query has scanned exactly two pages in the index to find the data that was requested. No matter which letter you choose, it would require scanning exactly two pages to locate the page that contained the words that started with that letter. This is what it means to have a balanced tree. Every search that is performed always transits the same number of levels in the index as well as the same number of pages in the index to locate the piece of data you are interested in.
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12
Designing the Physical Database
Index Levels
The number of levels in an index and the number of pages within each level of an index are determined by simple mathematics. A data page in SQL Server is 8192 bytes in size, which can be used to store up to 8060 bytes of actual user data. If you build an index on a char(60) column, each row in the table requires 60 bytes of storage, which also means 60 bytes of storage for each row within the index. If there are only 100 rows of data in the table, you need 6,000 bytes of storage. Because all the entries fit on a single page of data, the index has a single page that is the root page as well as the leaf page. In fact, you can store 134 rows in the table and still allo cate only a single page to the index. As soon as you add the 135th row, all the entries can no longer fit on a single page, so two additional pages are created. The first page contains the first half of the entries, the second page contains the second half of the entries, and the root page has two rows of data. This process creates an index with a root page and two leaf-level pages. This index does not need an intermediate level created because the root page can con tain all the values at the beginning of the leaf-level pages. At this point, locating any row in the table requires scanning exactly two pages in the index. You can continue to add rows to the table without affecting the number of levels in the index until you reach 17,957 rows. You then have 134 leaf level pages with 134 entries each. The root page has 134 entries corresponding to the first row on each of the leaflevel pages. When the 17,957th row of data is added to the table, another page needs to be allocated to the index at the leaf level, but the root page cannot hold 135 entries because it would exceed the 8,060 bytes that are allowed. So an intermediate level is introduced that contains two pages. The first page contains the initial entry for the first half of the leaf-level pages, and the second page contains the initial entry for the second half of the leaf-level pages. The root page contains two rows at this point, cor responding to the initial value for each of the two intermediate-level pages. The next time an intermediate level needs to be introduced is when the 2,406,105th row of data is added to the table. As you can see, this type of structure enables SQL Server to locate rows in extremely large tables very quickly. In this example, finding a row in the table with nearly 2.5 million rows requires SQL Server to scan only three pages of data, and the table could grow to more than 300 million rows before it would require SQL Server to read four pages to find any row.
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