Division Operation

In relational algebra, several operators are essential due to their unique functionalities. One such operator is the division operator, symbolized by “÷“. This operator is relatively complex but plays a crucial role in solving queries that involve the “all” condition. While many relational algebra operations focus on combining or filtering data, the division operation identifies tuples in one relation that are associated with every tuple in another.

In this chapter, we’ll explore the concept of the division operator in detail, understand its theoretical foundation, and walk through practical examples to grasp its application.

Basics of the Division Operator

The division operation applies to two relations −

• Numerator Relation (R) − Represents the main dataset and contains the superset of possible combinations.
• Denominator Relation (S) − Represents the subset or the set of conditions that must be satisfied.

The result is a relation containing only those tuples from the numerator that are associated with every tuple in the denominator.

Attributes in Relations

Let’s define the attributes involved in the division operator −

• R(Z) is the numerator, where Z = X ∪ Y
• S(X) is the denominator
• T(Y) is the result

Here, T contains the attributes in R that are not in S. For a tuple to appear in T, it must pair with every tuple in S within R.

To better understand the concept, consider a real-world scenario. Suppose we are organizing a workshop. Participants (R) are linked to sessions (X). We want to find those participants who attended all required sessions (S). The division operator helps identify these individuals by evaluating the “all sessions attended” condition.

Step-by-Step Explanation of the Division Operator

The division operation ensures that each result tuple in T(Y) must appear in R(Z) in combination with every tuple in S(X). Any tuple in R that fails to match all tuples in S is excluded from the result.

Mathematical Representation

The division operator can be expressed using a combination of projection, Cartesian product, and difference −

• T1 − Identify all potential result tuples: T1 ← π_Y(R)
• T2 − Find tuples in T1 that do not satisfy the pairing condition with S: T2 ← π_Y((S × T1) − R)
• Final Result − Subtract the unsatisfying tuples from the potential results. T ← T1 − T2

Example: Employees Working on All Projects

Let us take an example to get a clear understanding of how the division operator works.

Query − Find the names of employees who work on all projects that “John Smith” works on.

Relations Involved − Following are the relations involved in this query –

WORKS_ON − Contains tuples of employee IDs (Essn) and project numbers (Pno).

Essn	Pno	Hours
123456789	1	32.5
123456789	2	7.5
666884444	3	40.0
453453453	1	20.0
453453453	2	20.0
333445555	2	10.0
333445555	3	10.0
333445555	10	10.0
333445555	20	10.0
999887777	30	30.0
999887777	10	10.0
987987987	10	35.0
987987987	30	5.0
987654321	30	20.0
987654321	20	15.0
888665555	20	NULL

EMPLOYEE − Contains personal details like Fname, Lname, and Ssn.

Fname	Minit	Lname	Ssn	Bdate	Address	Sex	Salary	Super_ssn	Dno
John	B	Smith	123456789	1965-01-09	731 Fondren, Houston, TX	M	30000	333445555	5
Franklin	T	Wong	333445555	1955-12-08	638 Voss, Houston, TX	M	40000	888665555	5
Alicia	J	Zelaya	999887777	1968-01-19	3321 Castle, Spring, TX	F	25000	987654321	4
Jennifer	S	Wallace	987654321	1941-06-20	291 Berry, Bellaire, TX	F	43000	888665555	4
Ramesh	K	Narayan	666884444	1962-09-15	975 Fire Oak, Humble, TX	M	38000	333445555	5
Joyce	A	English	453453453	1972-07-31	5631 Rice, Houston, TX	F	25000	333445555	5
Ahmad	V	Jabbar	987987987	1969-03-29	980 Dallas, Houston, TX	M	25000	987654321	4
James	E	Borg	888665555	1937-11-10	450 Stone, Houston, TX	M	55000	NULL	1

Now, let’s understand step-by-step how to find the names of employees who work on all projects that “John Smith” works on.

Retrieve John Smith’s Projects

Start by filtering the projects linked to “John Smith.”

SMITH←σFname=′John′ANDLname=′Smith′(EMPLOYEE)

Fname	Minit	Lname	Ssn	Bdate	Address	Sex	Salary	Super_ssn	Dno
John	B	Smith	123456789	1965-01-09	731 Fondren, Houston, TX	M	30000	333445555	5

SMITHPNOS←πPno(WORKSON⋈Essn=SsnSMITH)

Pno
1
2

This gives a relation SMITH_PNOS containing all project numbers “John Smith” is assigned to.

Create All Employee-Project Relationships

Extract a relation showing all employees and their associated projects.

SSNPNOS←πEssn,Pno(WORKSON)

Essn	Pno
123456789	1
123456789	2
666884444	3
453453453	1
453453453	2
333445555	2
333445555	3
333445555	10
333445555	20
999887777	30
999887777	10
987987987	10
987987987	30
987654321	30
987654321	20
888665555	20

Apply Division

Use the division operator to find employees whose project assignments cover all projects in SMITH_PNOS.

SSNS(Ssn)←SSNPNOS÷SMITHPNOS

Ssn
123456789
453453453

Map Employee IDs to Names

Finally, retrieve the names of these employees.

RESULT←πFname,Lname(SSNS⋈EMPLOYEE)

Fname	Lname
John	Smith
Joyce	English

Final output − The final relation contains the names of employees who work on all the projects that “John Smith” works on.

Generalized Example: Products and Suppliers

Scenario − A store sells products (R) supplied by various suppliers (S). We want to identify products available from all suppliers.

Input Relations

• R − Contains ProductID and SupplierID
• S − Contains just SupplierID

Let’s go through its steps.

Extract potential products −

T1←πProductID(R)

Identify mismatched products −

T2←πProductID((S×T1)−R)

Subtract mismatches from potential products −

T←T1−T2

Output − If only certain products are supplied by every supplier, those product IDs will appear in the result relation T.

Key Observations

• Handling Universal Quantification − The division operator is useful for queries that require “for all” conditions. This is like finding students enrolled in all mandatory courses or employees assigned to every project in a list.
• Limitations in SQL − SQL does not directly support the division operator. However, similar results can be achieved using NOT EXISTS or complex joins. Although these approaches may lack the power of relational algebra.
• Practical Relevance − While it is theoretically significant, division is less common in practical database management systems due to its complexity and rare use cases.

Additional Applications

• Matching Preferences − Find customers whose preferences match every feature of a product.
• Cross-Department Participation − Identify employees involved in activities across all departments.