【正文】 s. No operation can distinguish among them, so they all survive any operation, or none do. Thus, we can never produce a result with exactly two tuples with two 0 ponents. Solutions for Section Exercise (a)SIGMA_{speed1000 AND price1500} (PC) = EMPTYSETExercise (d)This plex expression is best seen as a sequence of steps in which we define temporary relations R1 through R4 that stand for nodes of expression trees. Here is the sequence: R1(maker, model, speed) := PROJ_{maker,model。s in different ponents without using a product. However, then we get at least four identical tuples (or none, if we have eliminated the 039。 it does not preserve the multipicity of tuples in R. The next two expressions work for bags. R JOIN DELTA(PI_{R INTERSECT S}(S)) In this expression, each projection of a tuple from S onto the attributes that are also in R appears exactly once in the second argument of the join, so it preserves multiplicity of tuples in R, except for those that do not join with S, which disappear. The DELTA operator removes duplicates, as described in Section . R [R PI_R(R JOIN S)] Here, the strategy is to find the dangling tuples of R and remove them. Solutions for Section Exercise As a bag, the value is {700, 1500, 866, 866, 1000, 1300, 1400, 700, 1200, 750, 1100, 350, 733}. The order is unimportant, of course. The average is 959. As a set, the value is {700, 1500, 866, 1000, 1300, 1400, 1200, 750, 1100, 350, 733}, and the average is 967. H3Exercise (a) As sets, an element x is in the leftside expression (R UNION S) UNION T if and only if it is in at least one of R, S, and T. Likewise, it is in the rightside expression R UNION (S UNION T) under exactly the same conditions. Thus, the two expressions have exactly the same members, and the sets are equal. As bags, an element x is in the leftside expression as many times as the sum of the number of times it is in R, S, and T. The same holds for the right side. Thus, as bags the expressions also have the same value. Exercise (h)As sets, element x is in the left side R UNION (S INTERSECT T) if and only if x is either in R or in both S and T. Element x is in the right side (R UNION S) INTERSECT (R UNION T) if and only if it is in both R UNION S and R UNION T. If x is in R, then it is in both unions. If x is in both S and T, then it is in both union. However, if x is neither in R nor in both of S and T, then it cannot be in both unions. For example, suppose x is not in R and not in S. Then x is not in R UNION S. Thus, the statement of when x is in the right side is exactly the same as when it is in the left side: x is either in R or in both of S and T. Now, consider the expression for bags. Element x is in the left side the sum of the number of times it is in R plus the smaller of the number of times x is in S and the number of times x is in T. Likewise, the number of times x is in the right side is the smaller of The sum of the number of times x is in R and in S. The sum of the number of times x is in R and in T. A moment39。39。39。s name. Likewise, there is an arc from each account node to each attribute of that account, ., an arc labeled balance to the value of the balance. To represent ownership of accounts by customers, we place an arc labeled owns from each customer node to the node of each account that customer holds (possibly jointly). Also, we place an arc labeled ownedBy from each account node to the customer node for each owner of that account. Exercise In the semistructured model, nodes represent data elements, ., entities rather than entity sets. In the E/R model, nodes of all types represent schema elements, and the data is not represented at all. Solutions for Section Exercise (a)STARSMOVIES STAR starId = cf starredIn = sw, esb, rj NAMECarrie Fisher/NAME ADDRESSSTREET123 Maple St./STREET CITYHollywood/CITY/ADDRESS ADDRESSSTREET5 Locust Ln./STREET CITYMalibu/CITY/ADDRESS /STAR STAR starId = mh starredIn = sw, esb, rj NAMEMark Hamill/NAME ADDRESSSTREET456 Oak Rd.STREET CITYBrentwood/CITY/ADDRESS /STAR STAR starId = hf starredIn = sw, esb, rj, wit NAMEHarrison Ford/NAME ADDRESSSTREETwhatever/STREET CITYwhatever/CITY/ADDRESS /STAR MOVIE movieId = sw starsOf = cf, mh TITLEStar Wars/TITLE YEAR1977/YEAR /MOVIE MOVIE movieId = esb starsOf = cf, mh TITLEEmpire Strikes Back/TITLE YEAR1980/YEAR /MOVIE MOVIE movieId = rj starsOf = cf, mh TITLEReturn of the Jedi/TITLE YEAR1983/YEAR /MOVIE MOVIE movieID = wit starsOf = hf TITLEWitness/TITLE YEAR1985/YEAR /MOVIE/STARSMOVIESExercise !DOCTYPE Bank [ !ELEMENT BANK (CUSTOMER* ACCOUNT*) !ELEMENT CUSTOMER (NAME, ADDRESS, PHONE, SSNO) !ATTLIST CUSTOMER custId ID owns IDREFS !ELEMENT NAME (PCDATA) !ELEMENT ADDRESS (PCDATA) !ELEMENT PHONE (PCDATA) !ELEMENT SSNO (PCDATA) !ELEMENT ACCOUNT (NUMBER, TYPE, BALANCE) !ATTLIST ACCOUNT acctId ID ownedBy IDREFS !ELEMENT NUMBER (PCDATA) !ELEMENT TYPE (PCDATA) !ELEMENT BALANCE (PCDATA) ]Database Systems: The Complete Book Solutions for Chapter 5Solutions for Section Exercise (a)PI_model( SIGMA_{speed = 1000} ) (PC) Exercise (f)The trick is to thetajoin PC with itself on the condition that the hard disk sizes are equal. That gives us tuples that have two PC model numbers with the same value of hd. However, these two PC39。 }Alternatively, PlayerHand can be defined directly within the declaration of attribute theDeal. Exercise (h)Since keys f