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Slide 8.13: Assignment and alias Slide 8.15: Native relational operations Home         Print version

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Set Theoretic Operation (Cont.)

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Product
Assume that R and S are two tables with Head(R)=A1…An and Head(S)=B1…Bm. The product of the tables R and S is a table T whose heading is Head(T)=R.A1…R.An S.B1…S.Bm. We say t is a row in T if and only if there are two rows u in R and v in S such that t is the concatenation of u with v, u∥v.

t is a row in T if and only if there are two rows u in R and v in S such that t(R.A1) = u(Ai) for 1≤i≤n and t(S.Bj) = v(Bj) for 1≤i≤m. The product of R and S is denoted by R×S. The attribute name of the form W.A is referred as qualified attribute names. If an attribute name appears in only one table, we can refer to it using its unqualified name.

R×S R.A R.B R.C S.A S.B S.C a1 b1 c1 a1 b1 c1 a1 b1 c1 a1 b1 c2 a1 b1 c1 a1 b2 c3 a1 b1 c1 a3 b2 c3 a1 b2 c3 a1 b1 c1 a1 b2 c3 a1 b1 c2 a1 b2 c3 a1 b2 c3 a1 b2 c3 a3 b2 c3 a2 b1 c2 a1 b1 c1 a2 b1 c2 a1 b1 c2 a2 b1 c2 a1 b2 c3 a2 b1 c2 a3 b2 c3

• Since R×R would have the heading R.A1…R.An R.A1…R.An, use the alias to solve the problem, that is, S:=R, then S×R.

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Slide 8.13: Assignment and alias Slide 8.15: Native relational operations Home         Print version

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I feel bad for the homeless guy,           but I feel really bad for the homeless guy’s dog,           because he must be thinking           “Man, this is the longest walk ever.”