19.2 View Serializability. Recall our discussion in Section?? of how our true goal in the design of a

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1 View Serializability Recall our discussion in Section?? of how our true goal in the design of a scheduler is to allow only schedules that are serializable. We also saw how differences in what operations transactions apply to the data can aæect whether or not a given schedule is serializable. We also learned in Section?? that schedulers normally enforce ëconæict serializability," which guarantees serializability regardless of what the transactions do with their data. However, there are weaker conditions than conæict-serializability that also guarantee serializability. In this section we shall consider one such condition, called ëview-serializability." Intuitively, view-serializability considers all the connections between transactions and U such that writes a database element whose value U reads. he key diæerence between view- and conæictserializability appears when a transaction writes a value that no other transaction reads èbecause some other transaction later writes its own value for è. In that case, the w èè action can be placed in certain other positions of the schedule èwhere is likewise never readè that would not be permitted under the deænition of conæict-serializability. In this section, we shall deæne view-serializability precisely and give a test for it View Equivalence Suppose we have two schedules S 1 and S 2 of the same set of transactions. Imagine that there is a hypothetical transaction 0 that wrote initial values for each database element read by any transaction in the schedules, and another hypothetical transaction f that reads every element written by one or more transactions after each schedule ends. hen for every read action r i èè in one of the schedules, we can ænd the write action w j èè that most closely preceded the read in question. 1 Wesay j is the source of the read action r i èè. Note that transaction j could be the hypothetical initial transaction 0, and i could be f. If for every read action in one of the schedules, its source is the same in the other schedule, we say that S 1 and S 2 are view-equivalent. Surely, viewequivalent schedules are truly equivalent; they each do the same when executed on any one database state. If a schedule S is view-equivalent to a serial schedule, we say S is view-serializable. Example 19.1 : Consider the schedule S deæned by: 1 : r 1 èè w 1 èbè 2 : r 2 èbè w 2 èè w 2 èbè 3 : r 3 èè w 3 èbè 1 While we have not previously prevented a transaction from writing an element twice, there is generally no need for it to do so, and in this study it is useful to assume that a transaction only writes a given element once.

2 2 Notice that we have separated the actions of each transaction vertically, to indicate better which transaction does what; you should read the schedule from left-to-right, as usual. In S, both 1 and 2 write values of B that are lost; only the value of B written by 3 survives to the end of the schedule and is ëread" by the hypothetical transaction f. S is not conæict-serializable. o see why, ærst note that 2 writes before 1 reads, so 2 must precede 1 inahypothetical conæict-equivalent serial schedule. he fact that the action w 1 èbè precedes w 2 èbè also forces 1 to precede 2 in any conæict-equivalent serial schedule. Yet neither w 1 èbè nor w 2 èbè has any long-term aæect on the database. It is these sorts of irrelevant writes that view-serializability is able to ignore, when determining the true constraints on an equivalent serial schedule. More precisely, let us consider the sources of all the reads in S: 1. he source of r 2 èbè is 0, since there is no prior write of B in S. 2. he source of r 1 èè is 2, since 2 most recently wrote before the read. 3. Likewise, the source of r 3 èè is he source of the hypothetical read of by f is he source of the hypothetical read of B by f is 3, the last writer of B. Of course, 0 appears before all real transactions in any schedule, and f appears after all transactions. If we order the real transactions è 2 ; 1 ; 3 è, then the sources of all reads are the same as in schedule S. hat is, 2 reads B, and surely 0 is the previous ëwriter." 1 reads, but 2 already wrote, sothe source of r 1 èè is 2,asinS. 3 also reads, but since the prior 2 wrote, that is the source of r 3 èè, as in S. Finally, the hypothetical f reads and B, but the last writers of and B in the schedule è 2 ; 1 ; 3 è are 2 and 3 respectively, also as in S. We conclude that S is a view-serializable schedule, and the schedule represented by the order è 2 ; 1 ; 3 è is a view-equivalent schedule Polygraphs and the est for View-Serializability here is a generalization of the precedence graph, which we used to test conæict serializability in Section??, that reæects all the precedence constraints required by the deænition of view serializability. We deæne the polygraph for a schedule to consist of the following: 1. node for each transaction and additional nodes for the hypothetical transactions 0 and f. 2. For each action r i èxè with source j, place an arc from j to i.

3 19.2. VIEW SERILIZBILIY 3 3. Suppose j is the source of a read r i èxè, and k is another writer of X. It is not allowed for k to intervene between j and i,soitmust appear either before j or after i. We represent this condition by an arc pair èshown dashedè from k to j and from i to k.intuitively, one or the other of an arc pair is ëreal," but we don't care which, and when we try to make the polygraph acyclic, we can pick whichever of the pair helps to make it acyclic. However, there are important special cases where the arc pair becomes a single arc: èaè If j is 0, then it is not possible for k to appear before j,sowe use an arc i! k in place of the arc pair. èbè If i is f, then k cannot follow i,sowe use an arc k! j in place of the arc pair B f B Figure 19.1: Beginning of polygraph for Example 19.2 Example 19.2 : Consider the schedule S from Example We show in Fig the beginning of the polygraph for S, where only the nodes and the arcs from rule è2è have been placed. We have also indicated the database element causing each arc. hat is, is passed from 2 to 1, 3, and f, while B is passed from 0 to 2 and from 3 to f. Now, we must consider what transactions might interfere with each of these æve connections by writing the same element between them. hese potential interferences are ruled out by the arc pairs from rule è3è, although as we shall see, in this example each of the arc pairs involves a special case and becomes a single arc. Consider the arc 2! 1 based on element. he only writers of are 0 and 2, and neither of them can get in the middle of this arc, since 0 cannot move its position, and 2 is already an end of the arc. hus, no additional arcs are needed. similar argument tells us no additional arcs are needed to keep writers of outside the arcs 2! 3 and 2! f. Now consider the arcs based on B. Note that 0, 1, 2, and 3 all write B. Consider the arc 0! 2 ærst. 1 and 3 are other writers of B; 0 and 2 also write B, but as we saw, the arc ends cannot cause interference, so we need not consider them. s we cannot place 1 between 0 and 2, in principle we need the arc pair è 1! 0 ; 2! 1 è. However, nothing can precede 0,so the option 1! 0 is not possible. We may in this special case just add the

4 4 arc 2! 1 to the polygraph. But this arc is already there because of, soin eæect, we makenochange to the polygraph to keep 1 outside the arc 0! 2. We also cannot place 3 between 0 and 2. Similar reasoning tells us to add the arc 2! 3, rather than an arc pair. However, this arc too is already in the polygraph because of, so we make no change. Next, consider the arc 3! f. Since 0, 1, and 2 are other writers of B, wemust keep them each outside the arc. 0 cannot be moved between 3 and f, but 1 or 2 could. Since neither could be moved after f,wemust constrain 1 and 2 to appear before 3. here is already an arc 2! 3, but we must add to the polygraph the arc 1! 3. his change is the only arc we must add to the polygraph, whose ænal set of arcs is shown in Fig f Figure 19.2: Complete polygraph for Example 19.2 Example 19.3 : In Example 19.2, all the arc pairs turned out to be single arcs as a special case. Figure 19.3 is an example of a schedule of four transactions where there is a true arc pair in the polygraph r 2 èè; r 1 èè; w 1 ècè; r 3 ècè; w 1 èbè; r 4 èbè; w 3 èè; r 4 ècè; w 2 èdè; r 2 èbè; w 4 èè; w 4 èbè; Figure 19.3: Example of transactions whose polygraph requires an arc pair Figure 19.4 shows the polygraph, with only the arcs that come from the source-to-reader connections. s in Fig we label each arc by the elementèsè that require it. We must then consider the possible ways that arc pairs could be added. s we saw in Example 19.2, there are several simpliæcations we can make. When avoiding interference with the arc j! i, the only transactions

5 19.2. VIEW SERILIZBILIY 5 that need be considered as k èthe transaction that cannot be in the middleè are: æ Writers of an element that caused this arc j! i, æ But not 0 or f, which can never be k, and æ Not i or j, the ends of the arc itself. With these rules in mind, let us consider the arcs due to database element, which is written by 0, 3, and 4. We need not consider 0 at all. 3 must not get between 4! f,sowe add arc 3! 4 ; remember that the other arc in the pair, f! 3 is not an option. Likewise, 3 must not get between 0! 1 or 0! 2, which results in the arcs 1! 3 and 2! 3. C B, C C B, B f D Figure 19.4: Beginning of polygraph for Example 19.3 Now, consider the fact that 4 also must not get in the middle of an arc due to. It is an end of 4! f, so that arc is irrelevant. 4 must not get between 0! 1 or 0! 2, which results in the arcs 1! 4 and 2! 4. Next, let us consider the arcs due to B, which is written by 0, 1, and 4. gain we need not consider 0. he only arcs due to B are 1! 2, 1! 4, and 4! f. 1 cannot get in the middle of the ærst two, but the third requires arc 1! 4. 4 can get in the middle of 1! 2 only. his arc has neither end at 0 or f, so it really requires an arc pair: è 4! 1 ; 2! 4 è. We show this arc pair, as well as all the other arcs added, in Fig Next, consider the writers of C: 0 and 1. s before, 0 cannot present a problem. lso, 1 is part of every arc due to C, so it cannot get in the middle. Similarly, D is written only by 0 and 2,sowe can determine that no more arcs are necessary. he ænal polygraph is thus the one in Fig esting for View-Serializability Since we must choose only one of each arc pair, we can ænd an equivalent serial order for schedule S if and only if there is some selection from each arc pair that turns S's polygraph into an acyclic graph. o see why, notice that if there

6 f Figure 19.5: Complete polygraph for Example 19.3 is such an acyclic graph, then any topological sort of the graph gives an order in which no writer may appear between a reader and its source, and every writer appears before its readers. hus, the reader-source connections in the serial order are exactly the same as in S; the two schedules are view-equivalent, and therefore S is view-serializable. Conversely, if S is view-serializable, then there is a view-equivalent serial order S 0. Every arc pair è k! j ; i! k èin S's polygraph must have k either before j or after i in S 0 ; otherwise the writing by k breaks the connection from j to i, which means that S and S 0 are not view-equivalent. Likewise, every arc in the polygraph must be respected by the transaction order of S 0.We conclude that there is a choice of arcs from each arc pair that makes the polygraph into a graph for which the serial order S 0 is consistent with each arc of the graph. hus, this graph is acyclic. Example 19.4 : Consider the polygraph of Fig It is already a graph, and it is acyclic. he only topological order is è 2 ; 1 ; 3 è, which is therefore a view-equivalent serial order for the schedule of Example Now consider the polygraph of Fig We must consider each choice from the one arc pair. If we choose 4! 1, then there is a cycle. However, if we choose 2! 4, the result is an acyclic graph. he sole topological order for this graph is è 1 ; 2 ; 3 ; 4 è. his order yields a view-equivalent serial order and shows that the original schedule is view-serializable Exercises for Section 19.2 Exercise : Draw the polygraph and ænd all view-equivalent serial orders for the following schedules: * aè r 1 èè; r 2 èè; r 3 èè; w 1 èbè; w 2 èbè; w 3 èbè; bè r 1 èè; r 2 èè; r 3 èè; r 4 èè; w 1 èbè; w 2 èbè; w 3 èbè; w 4 èbè;

7 19.2. VIEW SERILIZBILIY 7 cè r 1 èè; r 3 èdè; w 1 èbè; r 2 èbè; w 3 èbè; r 4 èbè; w 2 ècè; r 5 ècè; w 4 èeè; r 5 èeè; w 5 èbè; dè w 1 èè; r 2 èè; w 3 èè; r 4 èè; w 5 èè; r 6 èè;! Exercise : Below are some serial schedules. ell how many schedules are èiè conæict-equivalent and èiiè view-equivalent to these serial schedules. * aè r 1 èè; w 1 èbè; r 2 èè; w 2 èbè; r 3 èè; w 3 èbè; that is, three transactions each read and then write B. bè r 1 èè; w 1 èbè; w 1 ècè; r 2 èè; w 2 èbè; w 2 ècè; that is, two transactions each read and then write B and C.

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