【正文】
where is the antisemijoinExercise The form of a constraint as E1 = E2 can be expressed as the other two constraints. Using the “equating an expression to the empty set” method, we can simply say:E1 – E2 = 248。Japan and Gt. Britain violate the constraint.Exercise This plex expression is best seen as a sequence of steps in which we define temporary relations R1 through R5 that stand for nodes of expression trees. Here is the sequence: R1(ship,battle,result,class) := πship,battle,result,class(Outes (ship = name) Ships) R2(ship,battle,result,numGuns) := πship,battle,result,numGuns(R1 Classes) R3(ship,battle) := πship,battle(σnumGuns 9 AND result = sunk (R2)) R4(ship2,battle2) := ρR4(ship2,battle2)(πship,battle(σnumGuns 9(R2))) R5(ship2) := πship2(R3 (battle = battle2) R4)The constraint is R5 = 248。No violations to the constraint.Exercise This plex expression is best seen as a sequence of steps in which we define temporary relations R1 through R5 that stand for nodes of expression trees. Here is the sequence: R1(class,name) := πclass,name(Classes Ships) R2(class2,name2) := ρR2(class2,name2)(R1) R3(class3,name3) := ρR3(class3,name3)(R1) R4(class,name,class2,name2) := R1 (class = class2 AND name name2) R2 R5(class,name,class2,name2,class3,name3) := R4 (class=class3 AND name name3 AND name2 name3) R3The constraint is R5 = 248。Models 2002,2006,2008 violate the constraint.Exercise πclass(σbore 16(Classes)) = 248。Model 2004 violates the constraint.Exercise πmaker(σtype = laptop(Product)) ∩ πmaker(σtype = pc(Product)) = 248。Exercise R1 := σspeed ≥ (PC)R2 := πmodel(R1)model100510061013 Exercise R1 := σhd ≥ 100 (Laptop)R2 := Product (R1)R3 := πmaker (R2)makerEABFG Exercise R1 := σmaker=B (Product PC)R2 := σmaker=B (Product Laptop)R3 := σmaker=B (Product Printer)R4 := πmodel,price (R1)R5 := πmodel,price (R2)R6: = πmodel,price (R3)R7 := R4 R5 R6modelprice100464910056301006104920071429Exercise R1 := σcolor = true AND type = laser (Printer)R2 := πmodel (R1) model30033007 Exercise R1 := σtype=laptop (Product)R2 := σtype=PC(Product)R3 := πmaker(R1)R4 := πmaker(R2)R5 := R3 – R4makerFG Exercise R1 := ρPC1(PC) R2 := ρPC2(PC) R3 := R1 ( = AND ) R2 R4 := πhd(R3)hd25080160 Exercise R1 := ρPC1(PC) R2 := ρPC2(PC) R3 := R1 ( = AND = AND ) R2 R4 := ,(R3)10041012 Exercise R1 := πmodel(σspeed ≥ (PC)) πmodel(σspeed ≥ (Laptop)) R2 := πmaker,model(R1 Product) R3 := ρR3(maker2,model2)(R2)R4 := R2 (maker = maker2 AND model model2) R3R5 := πmaker(R4)makerBEExercise R1 := πmodel,speed(PC) R2 := πmodel,speed(Laptop) R3 := R1 R2 R4 := ρR4(model2,speed2)(R3)R5 := πmodel,speed (R3 (speed speed2 ) R4)R6 := R3 – R5R7 := πmaker(R6 Product)makerBExercise R1 := πmaker,speed(Product PC) R2 := ρR2(maker2,speed2)(R1) R3 := ρR3(maker3,speed3)(R1) R4 := R1 (maker = maker2 AND speed speed2) R2R5 := R4 (maker3 = maker AND speed3 speed2 AND speed3 speed) R3R6 :=