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CHAPTER 14 TREE-STRUCTURED DATA
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I must now quickly add that Tutorial D includes built-in support for the transitive closure operator, in accordance with which we could obtain the transitive closure of relation MM by means of the simpler expression TCLOSE ( MM ) The query Get components at all levels for part P1 (for example) can thus be formulated as follows: ( TCLOSE ( MM ) ) WHERE MAJOR_P# = P#('P1') Loosely speaking, the TRANCLO code shown earlier might be regarded as possible implementation code for the TCLOSE operator.
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At this point I d like to mention a couple of asides (which you might want to skip on a first reading). First, I remark that although it s certainly not relational, and it doesn t support transitive closure as such the SQL standard does allow us to write recursive expressions. (This feature was introduced with SQL:1999.) Thus, for example, the query Get components at all levels for part P1 might be formulated in SQL as follows: WITH RECURSIVE TC ( MAJOR_P#, MINOR_P# ) AS ( ( SELECT MM.MAJOR_P#, MM.MINOR_P# FROM MM ) UNION ( SELECT TC.MAJOR_P#, MM.MINOR_P# FROM TC, MM WHERE TC.MINOR_P# = MM.MAJOR_P# ) ) SELECT DISTINCT MINOR_P# FROM TC WHERE TC.MAJOR_P# = P#('P1') ; Lines 2 3 and lines 5 7 of this expression constitute the initial subquery and the recursive subquery, respectively; between them, they construct the transitive closure. Lines 8 10 constitute the final subquery. Please don t ask me to explain the example further. Second, I remark that the commercial Oracle product has had proprietary support for recursive queries for many years. For example: SELECT FROM CONNECT START LEVEL, MINOR_P# MM BY MAJOR_P# = PRIOR MINOR_P# WITH MAJOR_P# = P#('P1') ;
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Given the sample data of Figures 14-5 through 14-7, this expression yields the result shown in Figure 14-9 (LEVEL in that result shows the level below part P1 at which the component MINOR_P# appears in the tree see Figure 14-5). Note very carefully, however, that the result in question suffers from the serious defect that it s not a relation (which is why there s no primary key marked by double underlining in the figure). To be specific:
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CHAPTER 14 TREE-STRUCTURED DATA
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A point that s often forgotten in discussions of bill-of-materials and related matters is that the bill-of-materials problem per se is only one side of the coin; the other side is the where-used problem, also known as the part implosion problem, which is the task of determining, for any given part, what parts that part is a component of (again, regardless of level). The interesting point about the where-used problem is that trees or graphs like those of Figures 14-3, 14-5, and 14-6 are of no help with it! Why Because they re the wrong way up, as it were; equivalently, the edges go the wrong way. Clearly, what we need is the appropriate inverse structure (after all, the where-used problem is the inverse of the bill-of-materials problem, in a sense). By way of example, Figure 14-10 shows the inverse of the graph in Figure 14-6.5 Note that, since Figure 14-6 had three leaf nodes, Figure 14-10 has three root nodes; in other words, Figure 14-10
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5. Of course, we could obtain an inverse picture by simply repeating Figure 14-6 but reversing all of the arrows, but the result would violate our convention for drawing trees with the root at the top.
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CHAPTER 14 TREE-STRUCTURED DATA
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shows a set of trees, or what s sometimes called a forest (a forest of three trees, in this particular case if we agree for the moment to overlook the fact that they re not really trees at all, that is).
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