Discuss the differences between the five Join operations: Theta join, Equijoin, Natural join, Outer join, and Semijoin. Give examples to illustrate your answer.
What will be an ideal response?
Tuple relational calculus
A (well-formed) formula is made out of one or more atoms, where an atom has one of the
following forms:
? R(S i ), where S i is a tuple variable and R is a relation.
? S i .a 1 ??S j .a 2 , where Si and Sj are tuple variables, a 1 is an attribute of the relation over
which S i ranges, a 2 is an attribute of the relation over which S j ranges, and ? is one of the
comparison operators (?, ??, ?, ??, ?, ?); the attributes a 1 and a 2 must have domains
whose members can be compared by ??
? S i .a 1 ??c, where Si is a tuple variable, a 1 is an attribute of the relation over which S i
ranges, c is a constant from the domain of attribute a 1 , and ? is one of the comparison
operators?
We recursively build up formulae from atoms using the following rules:
? an atom is a formula;
? if F1 and F2 are formulae, so are their conjunction F1 ? F2, their disjunction F1 ? F2, and
the negation ?F1;
? if F is a formula with free variable X, then (?X)(F) and (?X)(F) are also formulae.
Domain relational calculus
A (well-formed) formula is made out of one or more atoms, where an atom has one of the
following forms:
? R(d 1 , d 2 ,…, d n ), where R is a relation of degree n and each d i is a domain variable.
? d i ??d j , where di and dj are domain variables and ? is one of the comparison operators (?,
??, ?, ??, ?, ?); the domains di and dj must have members that can be compared by ??
? d i ??c, where di is a domain variable, c is a constant from the domain of di, and ? is one of
the comparison operators?
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What does the following algorithm do? What is its time complexity?
``` public static int doSomething( int array[], int n ) { int k = 0; // find the largest value in the range array[0] to array[n-1] for ( int i = 1; i < n; i++ ) if ( array[i] > array[k] ) k = i; // postcondition: array[k] is the largest value in array // Swap the largest value with the value in position 0 int temp = array[0]; array[0] = array[k]; array[k] = temp; k = 1; // find the largest value in the range array[1] to array[n-1] for ( int i = 2; i < n; i++ ) if ( array[i] > array[k] ) k = i; // postcondition: array[k] is the SECOND largest value in array return array[k]; } ```