# SQL puzzles and answers: Finding equal sets

## A SQL puzzle in the December 1993 issue of *Database Programming & Design* magazine was to use a supplier and parts table to find pairs of suppliers who provide exactly the same parts. This is the
same thing as finding two equal sets. How many ways can you find to do this problem?

*This is an excerpt from*Joe Celko's SQL Puzzles and Answers, Second Edition

*by Joe Celko. Printed with permission from Morgan Kaufmann, a division of Elsevier. Copyright 2007. For more information about this book, and other similar titles, please visit www.mkp.com.*

Set theory has two symbols for subsets. One is a "horseshoe" on its side, which means that set A is contained within set B and is sometimes called a proper subset. The other is the same symbol with a horizontal bar under it, which means "contained in or equal to," which is sometimes called just a subset or containment operator.

Standard SQL has never had an operator to compare tables against each other. Several college textbooks on relational databases mention a CONTAINS predicate that does not exist in standard SQL. Two such offenders are *An Introduction to Data Base Systems* by Bipin C. Desai (West Publishing, 1990, ISBN 0-314-66771-7) and *Fundamentals of Database Systems* by Elmasri and Navthe (Benjamin Cummings, 1989, ISBN 0-8053-0145-3). This predicate used to exist in the original System R, IBM's first experimental SQL system, but it was dropped from later SQL implementations because of the expense of running it.

The IN() predicate is a test for membership, not for subsets. For those of you who remember your high school set theory, membership is shown with a stylized epsilon with the containing set of the right side, thus ∈. Membership is for one element, while a subset is itself a set, not just an element.

Chris Date's puzzle in the December 1993 issue of *Database Programming & Design* magazine ("A Matter of Integrity, Part II" According to Date, December 1993) was to use a supplier and parts table to find pairs of suppliers who provide *exactly* the same parts. This is the same thing as finding two equal sets. Given his famous table:

CREATE TABLE SupParts (sno CHAR(2) NOT NULL, pno CHAR(2) NOT NULL, PRIMARY KEY (sno, pno));

How many ways can you find to do this problem?

**Answer #1**

One approach would be to do a FULL OUTER JOIN on each pair of suppliers. Any parts that are not common to both would show up, but would have generated NULLs in one of the columns derived from the supplier who was not in the INNER JOIN portion. This tells you which pairs are not matched, not who is. The final step is to remove these nonmatching pairs from all possible pairs.

SELECT SP1.sno, SP2.sno FROM SupParts AS SP1 INNER JOIN SupParts AS SP2 ON SP1.pno = SP2.pno AND SP1.sno < SP2.sno EXCEPT SELECT DISTINCT SP1.sno, SP2.sno FROM SupParts AS SP1 FULL OUTER JOIN SupParts AS SP2 ON SP1.pno = SP2.pno AND SP1.sno < SP2.sno) WHERE SP1.sno IS NULL OR SP2.sno IS NULL;

This is probably going to run very slowly. The EXCEPT operator is the SQL equivalent of set difference.

**Answer #2**

The usual way of proving that two sets are equal to each other is to show that set A contains set B, and set B contains set A. What you would usually do in standard SQL would be to show that there exists no element in set A that is not in set B, and therefore A is a subset of B. So the first attempt is usually something like this:

SELECT DISTINCT SP1.sno, SP2.sno FROM SupParts AS SP1, SupParts AS SP2 WHERE SP1.sno < SP2.sno AND SP1.pno IN (SELECT SP22.pno FROM SupParts AS SP22 WHERE SP22.sno = SP2.sno) AND SP2.pno IN (SELECT SP11.pno FROM SupParts AS SP11 WHERE SP11.sno = SP1.sno));

Oops, this does not work because if a pair of suppliers has one item in common, they will be returned.

**Answer #3**

You can use the NOT EXISTS predicate to imply the traditional test mentioned in Answer #2.

SELECT DISTINCT SP1.sno, SP2.sno FROM SupParts AS SP1, SupParts AS SP2 WHERE SP1.sno < SP2.sno AND NOT EXISTS (SELECT SP3.pno -- part in SP1 but not in SP2 FROM SupParts AS SP3 WHERE SP1.sno = SP3.sno AND SP3.pno NOT IN (SELECT pno FROM SupParts AS SP4 WHERE SP2.sno = SP4.sno)) AND NOT EXISTS (SELECT SP5.pno -- part in SP2 but not in SP1 FROM SupParts AS SP5 WHERE SP2.sno = SP5.sno AND SP5.pno NOT IN (SELECT pno FROM SupParts AS SP4 WHERE SP1.sno = SP4.sno));

**Answer #4**

Instead of using subsets, I thought I would look for another way to do set equality. First, I join one supplier to another on their common parts, eliminating the situation where supplier 1 is the same as supplier 2, so that I have the intersection of the two sets. If the intersection has the same number of pairs as each of the two sets has elements, then the two sets are equal.

SELECT SP1.sno, SP2.sno FROM SupParts AS SP1 INNER JOIN SupParts AS SP2 ON SP1.pno = SP2.pno AND SP1.sno < SP2.sno GROUP BY SP1.sno, SP2.sno HAVING (SELECT COUNT(*) -- one to one mapping EXISTS FROM SupParts AS SP3 WHERE SP3.sno = SP1.sno) = (SELECT COUNT(*) FROM SupParts AS SP4 WHERE SP4.sno = SP2.sno);

If there is an index on the supplier number in the SupParts table, it can provide the counts directly as well as help with the join operation.

**Answer #5**

This is the same as Answer #4, but the GROUP BY has been replaced with a SELECT DISTINCT clause:

SELECT DISTINCT SP1.sno, SP2.sno FROM (SupParts AS SP1 INNER JOIN SupParts AS SP2 ON SP1.pno = SP2.pno AND SP1.sno < SP2.sno) WHERE (SELECT COUNT(*) FROM SupParts AS SP3 WHERE SP3.sno = SP1.sno) = (SELECT COUNT(*) FROM SupParts AS SP4 WHERE SP4.sno = SP2.sno);

**Answer #6**

This is a version of Answer #3, from Francisco Moreno, which has the NOT EXISTS predicate replaced by set difference. He was using Oracle, and its EXCEPT operator (called MINUS in their SQL dialect) is pretty fast.

SELECT DISTINCT SP1.sno, SP2.sno FROM SupParts AS SP1, SupParts AS SP2 WHERE SP1.sno < SP2.sno AND NOT EXISTS (SELECT SP3.pno -- part in SP1 but not in SP2 FROM SupParts AS SP3 WHERE SP1.sno = SP3.sno EXCEPT SELECT SP4.pno FROM SupParts AS SP4 WHERE SP2.sno = SP4.sno AND NOT EXISTS (SELECT SP5.pno -- part in SP2 but notin SP1 FROM SupParts AS SP5 WHERE SP2.sno = SP5.sno EXCEPT SELECT SP6.pno FROM SupParts AS SP6 WHERE SP1.sno = SP6.sno);

**Answer #7**

Alexander Kuznetsov once more has a submission that improves the old "counting matches in a join" approach:

SELECT A.sno, B.sno AS sno1 FROM (SELECT sno, COUNT(*), MIN(pno), MAX(pno) FROM SubParts GROUP BY sno) AS A(cnt, min_pno, max_pno) INNER JOIN (SELECT sno, COUNT(*), MIN(pno), MAX(pno) FROM SubParts GROUP BY sno) AS B(cnt, min_pno, max_pno) -- four conditions filter out most permutations ON A.cnt = B.cnt AND A.min_pno = B.min_pno AND A.max_pno = B.max_pno AND A.sno < B.sno -- Expensive inner select below does not have to execute for every pair WHERE A.cnt = (SELECT COUNT(*) FROM SubParts AS A1, SubParts AS B1 WHERE A1.pno = B1.pno AND A1.sno = A.sno AND B1.sno = B.sno); sn sn ======= ab bb aq pq

The clever part of this query is that most optimizers can quickly find the MIN() and MAX() values on a column because they are stored in the statistics table.

**Answer #8**

Let's look at notation and some of the usual tests for equality:

((A ⊆ B) = (B ⊆ A)) ⇒ (A = B) ((A ∪ B) = (B ∩ A)) ⇒ (A = B)

The first equation is really the basis for the comparisons that use joins. The second equation is done at the set level rather than the subset level, and it implies this answer:

SELECT DISTINCT 'not equal' FROM (SELECT * FROM A) INTERSECT SELECT * FROM B) EXCEPT (SELECT * FROM A) UNION SELECT * FROM B);

The idea is to return an empty set if tables A and B are equal. You have to be careful about using the ALL clauses on the set operators if you have duplicates. The good news is that these operators work with rows and not at the column level, so this template will generalize to any pairs of union-compatible tables. You do not have to know the column names.

**About the author**

Joe Celko is a noted consultant, lecturer, teacher and one of the most-read SQL authors in the world. He is well-known for his 10 years of service on the ANSI SQL standards committee, his dependable help on assorted SQL newsgroups, his column in *Intelligent Enterprise* (which won several Reader's Choice Awards) and the war stories he tells to provide real-world insights into SQL programming. His best-selling books include *Joe Celko's SQL for Smarties: Advanced SQL Programming*, *Joe Celko's SQL Puzzles and Answers*, *Joe Celko's Trees and Hierarchies in SQL for Smarties* and *Joe Celko's SQL Style*, all published by Morgan Kaufmann.