Solutions for Exercise no. 4

1. Tuple Relational Calculus

{t | $f Î Employees $d Î Departments(t[FName] = f[FName] Ù
        d[MGRCPR] = 654 Ù d[DNUMBER] = f[DNO])}

2:
{t |  $e Î Employees $p Î Projects $a Î Allocations (t[CPR] = e[CPR] Ù
        e[BDATE] < 1961-01-01 Ùp[PNAME] = 'InteractiveTV' Ù
        p[PNUMBER] = a[PNO] Ù a[ECPR] = e[CPR])}

3:
{t | $e Î Employees $p ÎProjects $aÎ Allocations
     (t[FNAME] = e[FNAME] Ù t[MINIT] = e[MINIT] Ù t[LNAME] = e[LNAME] Ù
      e[DNO] = 10 Ù p[PNAME] = 'PalmTop' Ù a[HOURS] 15 Ù
      p[PNUMBER] = a[PNO] Ù a[ECPR] = e[CPR])}

2. Domain Relational Calculus

2:
Anonymous variables are in the following query implicitly existentially quantified, e.g., á _, _ñ Î p stands for $v₁, v₂ áv₁, v₂ñ Î p.

{áCPR, BDATEñ | $DNO (á_, _, _, CPR, BDATE, _, _, _, DNOñ Î Employees Ù
    $MGRCPR, BDATE1 á_, _, _, MGRCPR, BDATE1, _, _, _, _ñ Î Employees Ù
    á_, DNO, MGRCPR, _ñ Î Departments Ù BDATE < BDATE1)}

3. Relational Algebra

2:
    R₁ ¬((Employees n _{CPR = ECPR} Allocations) n _{PNO = PNUMBER} Projects))
    p_CPR (s_{BDATE < 1961-01-01 Ù PNAME = 'InteractiveTV'}(R₁)
 

3:
    R₁¬Employees n _{CPR = ECPR} Allocations
    R₂ ¬R₁ n _{PNO = PNUMBER} Projects
    R₃ ¬s_{DNO = 10 Ù PNAME = 'PalmTop' Ù HOURS 15} (R₂)
    p_{FNAME, MINIT, LNAME} (R₃)

4:
    R₁  ¬Locations n D_{DNBR = DNUM Ù DLOCATION = PLOCATION} Projects
    R ₂ ¬p_{DNBR, DLOCATION} (R₁)
    R₃  ¬ Locations \ R₂
    p_DNBR (R₃)

5:
    R₁ ¬(Employees n _{DNO = DNUMBER} Departments) n E_{Employees.CPR = Managers.CPR} r_Managers(Employees)
    R₂ ¬s_{Employees.BDATE < Managers.BDATE} (R₁)
    p_{Employees.CPR, Employees.BDATE} (R₂)

Result sets

1: Empty

2:

FNAME

Kristian

Charlotte

Niels

3:

CPR

111

654

4:

DNBR

12

5:

CPR	BDATE
123	1955-12-10