1

Relational Algebra

April 18, 2023

2

Exercise 1 Consider the following relational database scheme:

Treatment (disease, medication) Doctor(name, disease-of-specialization) Treated (doctor_name, patient_name, date, procedure,

diagnostic) Patients and doctors are uniquely identified by their name A patient may suffer from several diseases, and may take

several medications for each disease. It is possible to a doctor to be a patient and vice-versa. The domains of the attributes "disease" and "disease-of-

specialization" are the same, namely, the set of all diseases. In "treated" relation, the procedure could be consultation, or

intervention (surgery etc.) and the diagnostic could be the name of a disease or the type of intervention.

Write the following queries in relational algebra:

3

Exercise 1 (a) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

Give the name of doctors who don't suffer from any disease.

4

Solution 1 (a) Give the name of doctors who don't suffer

from any disease.DoctorsWhoSuffered(name (name=patient_name

(Doctor Treated)))

name (Doctor) - DoctorsWhoSuffered

5

Exercise 1 (b) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

List patients who suffer from more than one disease.

6

Solution 1 (b) List patients who suffer from more than one

disease.Patient1(name1, diagnostic1 (patient_name, diagnostic

( Treated))Patient2(name2, diagnostic2 (Patient1)

name1 (name1=name2 AND diagnostic1 <> diagnostic2

(Patient1 Patient2))

7

Exercise 1 (c) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

Give the names of doctors who are also patients suffering from a disease in their own specialization.

8

Solution 1 (c) Give the names of doctors who are also

patients suffering from a disease in their own specialization.

name (name=patient_name AND disease-of-specialty = diagnostic (Doctor Treated))

9

Exercise 1 (d) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

Find diseases for which there is only one medication.

10

Solution 1 (d) Find diseases for which there is only one

medication.DiseaseWithMoreThanOneTreatment (t1.disease

(t1.disease=t2.disease AND t1.medication <> t2.medication (t1(Treatment) t2(Treatment))

disease (Treatment) - DiseaseWithMoreThanOneTreatment

11

Exercise 1 (e) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

Find the names of patient who had an operation done by a doctor with HIV

12

Solution 1 (e) Find the names of patient who had an

operation done by a doctor with HIV.DoctorsWithHIV(doc_name) (name (name=patient_name AND

diagnostic = ‘HIV’ (Doctor Treated))

patient_name (doc_name=doctor_name AND

procedure=‘intervention’ (DoctorsWithHIV Treated)

13

Exercise 1 (f) Treatment (disease, medication) Doctor (name, disease-of-specialization) Treated (doctor_name, patient_name, date,

procedure, diagnostic)

List the patients who consulted 2 or more doctors and was given exactly the same diagnosis. List patient’s name, doctor’s name, the date, and the diagnosis.

14

Solution 1 (f) List the patients who consulted 2 or

more doctors and was given exactly the same diagnosis. List patient’s name, doctor’s name, the date, and the diagnosis.

t1.patient_name, t1.doctor_name, t1.date, t1.disgnostic (t1.patient_name=t2.patient_name AND t1.doctor_name<>t2.doctor_name AND

t1.diagnostic=t2.diagnostic ( t1(Treated) t2(Treated))

15

Exercise 2 Consider the following relation scheme about hockey teams:

Team (tname, year, coach, salary) Player (pname, position) Winner (tname, year) Played (pname, tname, year, salary)

Assume that players and coaches appointments with teams are on a calendar year basis.

The team relation indicates teams, their coaches on yearly basis and their salaries.

The player relation indicates players and the position each player plays.

The winner relation indicates the Stanley Cup winners. The played relation indicates players and the team they have

played for on a yearly basis. Write the following queries in Relational Algerbra:

16

Exercise 2 (a) Team (tname, year, coach, salary) Player (pname, position) Winner (tname, year) Played (pname, tname, year, salary)

List all the players who played for all the teams in the league, with a salary higher than 2M$.

17

Teams:

t1 t2t3

Played:

p1, t1, 3p1, t1, 1p1, t2, 4p1, t3, 5p2, t1, 1p2, t2, 4p2, t3, 4p3, t1, 4

Step 1: get all the players with salary more than 2M

R1:pname,tname(salary> 2M (Played))

p1, t1p1, t2p1, t3p2, t2p2, t3p3, t1

Players:

p1p2p3

Sample instances Solution 2 (a) –

1

18

Step 2:[all possible combination]

R2: pname,tname (Player x Team)

p1, t1p1, t2 p1, t3p2, t1 p2, t2 p2, t3 p3, t1 p3, t2 p3, t3

Teams:

t1 t2t3

Played:

p1, t1, 3Mp1, t1, 1Mp1, t2, 4Mp1, t3, 5Mp2, t1, 1Mp2, t2, 4Mp2, t3, 4Mp3, t1, 4M

Players:

p1p2p3

Sample instances Solution 2 (a) –

2

19

Step 3: get players that have not played for all the teams

R3: R2-R1p2, t1 p3, t2 p3, t3 Step 4: R4: pname(R3):

P2p3

Played:

p1, t1, 3p1, t1, 1p1, t2, 4p1, t3, 5p2, t1, 3p2, t2, 4p2, t3, 4p3, t1, 4

Players:

p1p2p3

Sample instances Solution 2 (a) –

3Teams:

t1 t2t3

Step 5: get players that have played for all the teams

pname(Player)-R4

p1

20

Solution 2 (a) -4 List all the players who played for all the

teams in the league, with a salary higher than 2M$.

Final answer in two lines!

1) R4(pname(pname,tname (Player x Team)-pname,tname (salary > 2M (Played)))

2) pname(Player)-R4

21

Exercise 2 (b) Team (tname, year, coach, salary) Player (pname, position) Winner (tname, year) Played (pname, tname, year, salary)

List coaches who have also been players, and have coached only teams they once played for.

22

Solution 2 (b) List coaches who have also been players,

and have coached only teams they once played for.

R1(Tname,name)(tname,pname (Played))

R2(Tname, name)(tname,coach (Team))

Answer= name (R2) - name (R2-R1)

{a1,a2} – {a1}={a2}

R2

T1 a1

T2 a1

T3 a1

T1 a2

T4 a2

R1

T1 a1

T2 a1

T4 a1

T1 a2

T3 a2

T4 a2

Answer would be a2.

23

Exercise 2 (c) Team (tname, year, coach, salary) Player (pname, position) Winner (tname, year) Played (pname, tname, year, salary)

List all the players who won the Stanley Cup during two consecutive years with two different teams.

24

Solution 2 (c) List all the players who won the Stanley Cup

during two consecutive years with two different teams.

Winner1(pname1, year1, tname1) (Played.pname, Played.year,

Winner.tname (Winner tname, year Played))

Winner2(pname2, year2, tname2) (Winner1)

pname1(pname1=pname2 AND tname1<>tname2

AND (year1=year2+1 OR year1=year2-1) (Winner1 Winner2))

25

Exercise 2 (d) Team (tname, year, coach, salary) Player (pname, position) Winner (tname, year) Played (pname, tname, year, salary)

List all coaches who earn more than the highest player's salary in the team.

26

Solution 2 (d) List all coaches who earn more than the

highest player's salary in the team.NotHighestPaidPlayers(pname, tname, salary)

(p1.pname, p1.tname, p1.salary

(p1.salary<p2.salary AND p1.tname=p2.tname

(p1(Played) p2(Played)))

HighestPaidPlayer (pname, tname, salary(Played) – NotHighestPaidPlayers)

coach( Team.tname=HighestPaidPlayer.tname AND Team.salary>HighestPaidPlayer.salary

(Team HighestPaidPlayer))