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(Database CSc460 Systems) Assignment 5 Solutions Fall 2007:14 Question 1 Express the following queries in relational algebra, domain relational calculus, and tuple relational calculus, using the relations handed out in homework 4 solution. Attempt to use the different relational operators that we have covered, including the derived operators, such as natural join, as much as possible. a. List the ID#s of all artifacts which are displayed in Room #10. Relational Algebra Artifact ID ( Room Number = 10 (Displayed In)) Tuple Relational Calculus {t.Artif act ID|Displayed In(t) t.Room N umber = 10} Domain Relational Calculus Rename the attributes for the relations used as follows. Displayed In(A, V, R) A = Artifact ID, V = Visitor Name, R = Room Number Is Owned By(V, A, P) P = Purchase Price Artifact(A, T, I, W, H, L) T = Title, I = Image, W = Width, H = Height, L = Length Style(S, A) S = Style Institution(C, N, V, E, G) C = City, N = Institution Name, E = Endowment, G = G or M Owner(V) {A| V, R(Displayed In(A, V, R) R = 10)} 1 CSc460 (Database Systems) Assignment 5 Solutions Fall 2007:14.2 b. List the name of all owners with artifacts in the expressionist style. Relational Algebra Visitor Name ( Style ="Expressionism" (Is Owned By TRC Style)) {t.V isitor N ame| s(Is Owned By(t) Style(s) s.Style = "Expressionism" t.Artif act ID = s.Artif act ID)} DRC {V | A, P (Is Owned By(V, A, P ) Style("Expressionism", A))} c. List all artifacts who are owned by those who display them. Relational Algebra Artifact ID (Displayed In TRC Is Owned By) {t.Artif act ID| s(Is Owned By(t) Displayed In(s) t.V isitor N ame = s.V isitor N ame t.Artif act ID = s.Artif act ID)} DRC {A| V, R(Displayed In(A, V, R) P (Is Owned By(V, A, P )))} d. List the room number(s) containing artifacts owned by the Guggenheim Museum in Bilbao and artifacts owned by the San Francisco MoMA along with the styles of these artifacts. Relational Algebra This query is so difficult because we have to make sure that the Guggenheim artifact is in the same room as the MoMA artifact, which can be done with an intersection or an equijoin (we use the latter), but then we have to get the style from each of these, which could be a different style. We use a self-join and a operator to differentiate the two Artifact IDs after the selfjoin. 2 CSc460 (Database Systems) Assignment 5 Solutions Fall 2007:14.3 1.d cont. Q1 = Institution Is Owned By Displayed In Q2 = Room Number, Artifact ID ( Institution Name ="Guggenheim"City ="Bilbao" (Q1 )) Q3 = Room Number, Artifact ID ( Institution Name ="MoMA"City ="San Francisco" (Q1 )) Q4 = Q2 Room Number = M.Room Number (M (Q3 )) Room Number, Style ( Room Number, Artifact ID (Q4 ) Style) Room Number, Style ( M.Room Number, Artifact ID (Q4 ) Style) TRC For such complex tuple relational calculus expressions, the names of the variables and how you break it up across lines really make a difference. As with the RA, we need two sets of variables for Institution, Displayed In, and Owned By. The G are the Guggenheim ones; the M are the MoMA ones. The Room Numbers should be identical, but the Artifact IDs won't be. The relational union is done using logical "or" in the TRC. Note that the parentheses around the "or" can't be omitted. Once you have the algebra for this query, the TRC isn't that hard: just a step-by-step translation. {DI G.Room N umber, S.Style | I G I M DI M OB G OB M ( Displayed In(DI G) Displayed In(DI M ) Style(S) Institution(I Institution(I G) M ) Is Owned By(OB G) Is Owned By(OB M ) I G.Institution N ame = "Guggenheim" I G.City = "Bilbao" I M.Institution N ame = "M oM A" I M.City = "San F rancisco" I G.V isitor N ame = OB G.V isitor N ame I M.V isitor N ame = OB M.V isitor N ame DI G.V isitor N ame = I G.V isitor N ame DI M.V isitor N ame = I M.V isitor N ame OB G.Artif act ID = DI G.Artif act ID OB M.Artif act ID = DI M.Artif act ID (S.Artif act ID = DI G.Artif act ID S.Artif act ID = DI M.Artif act ID) DI G.Room N umber = DI M.Room N umber)} DRC It turns out that the DRC is much shorter for this query, for a couple of reasons. First, the strings as constants works well, and there are four of them. Secondly, the equijoins are just reused variables. And thirdly, we use unprimed variables for the MoMA and primed for the Guggenheim. {R, S| V, E, G(Institution("San F rancisco", "M oM A", V, E, G) V , E , G (Institution("Bilbao", "Guggenheim", V , E , G ) P, P (Is Owned By(V, A, P ) Is Owned By(V , A , P ) Displayed In(A, V, R) Displayed In(A , V , R) (Style(S, A) Style(S, A ))))} 3 CSc460 (Database Systems) Assignment 5 Solutions Fall 2007:14.4 1.e. List the owners who do not display any artifacts. Relational Algebra This is a great opportunity for using the semi-join. Visitor Name (Owner) - Visitor Name (Is Owned By TRC Artifact ID (Displayed In)) {t.V isitor N ame|Owner(t) s, o(Displayed In(s) Is Owned By(o) s.Artif act ID = o.Artif atct ID t.V isitor N ame = s.V isitor N ame)} DRC {V |Owner(V ) A, R, V , P (Displayed In(A, V, R) Is Owned By(V , A, P ))} f. List the owners who display every single style. Relational Algebra The observant reader will note that the following expression doesn't involve the Owner relation. We could natural join this with Owner and then project out Visitor Name. However, in this particular case, that is not necessary, since Visitor Name is just a foreign key to Owner. Visitor Name, Style (Displayed In TRC Style) Style (Style) This is a direct translation with the following slight modification: "List the owners for which for every style, that owner displays an artifact of that style." Here we need the Owner relation because we need a tuple variable on the outside of the forall. We could have used another d variable with Displayed In(d ), though it wouldn't have matched the English as well. Note that s is needed in the "then" part because s just stands for style, not for a particular artifact. {o.V isitor N ame| Owner(o) s(Style(s) d s (Displayed In(d) Style(s ) s.Style = s .Style d.Artif act ID = s .Artif act ID d.V isitor N ame = o.V isitor N ame))} DRC Here again the DRC is shorter, in this case, just one line. Note that the A in the "if" part is different from the A in the "then" part. The first is an arbitrary artifact; we use that to get a style. The second A is a particular artifact displayed by the owner. {V |Owner(V ) S(( A(Style(S, A))) A R(Displayed In(A, V, R) Style(S, A)) )} 4 CSc460 (Database Systems) Assignment 5 Solutions Fall 2007:14.5 Question 2 Give a query in English on the UA CATS schema that cannot be expressed using the five basic relational algebra operators, and justify briefly. There are many possible answers. Here's one: "List the artifacts in ascending order of artifact id". This cannot be be expressed using the five basic relational algebra operators since all the schemas and instances of relations are sets, and, by definition, there is no notion of order in sets. 5
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Arizona >> ISTA >> 6 (Fall, 2009)
ACUTE TOXICITY OF ETHANOLIC EXTRACT OF Derris elliptica ROOTS TO Oreochromis niloticus FINGERLINGS M. O. Akinbulumo, O. A. Fagbenro and E. A. Fasakin Department of Fisheries, Federal University of Technology, PMB 704 Akure, Nigeria Abstract Replica...
Arizona >> ISTA >> 6 (Fall, 2009)
32 El-Seify M.A., Desouky,A.Y. & Bazh,Eman Parasitology Department, Kafr ElSheikh, Tanta University, Egypt. Introduction 1. Fish is considered as a source of protein. 2. Fish production is greatly affected with parasitic infection. 3....
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