# Examples of DIVISION RELATIONAL ALGEBRA and SQL. r s is used when we wish to express queries with all :

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1 Examples of DIVISION RELATIONAL ALGEBRA and SQL r s is used when we wish to express queries with all : Ex. Which persons have a loyal customer's card at ALL the clothing boutiques in town X? Which persons have a bank account at ALL the banks in the country? Which students are registered on ALL the courses given by Soini? Which students are registered on ALL the courses that are taught in period 1? Which boys are registered on those courses that are taken by ALL the girls? Which girls are registered on ALL the courses taken by student nr ? In all these queries the description after ALL defines a set with a number of elements. The result is composed of those data units (persons, students etc.) who satisfy these requirements. The logic these sentences express is that of implication: for which persons is it true that IF there is a clothing boutique in town X THEN the person has a loyal customer's card there, IF Soini gives a course THEN the student is registered on it or IF is registered on a course THEN the girl is registered on the same course. We shall study the last example in detail here. On the one hand we have a list (a relation) with girls and the courses they are registered on: this will be called the relation r. On the other hand we have a list (a relation) of ALL the courses taken by This will be the relation s. Now we wish to know which of the girls take ALL these courses. Suppose that takes the following three courses: G555, and. To be a part of the answer the girl in question must be registered on ALL of these three. (In addition to that she may be registered on other courses not taken by 40101, these will not affect the result.) In relational algebra the query will be expressed as follows (the example database kursdatabas is found on the home page of this course. matr = student number, namn = name, kurskod = course code, kön = sex, kursanmälan = course registration): Π matr, namn, kurskod ( kön = 'K' (student) x kursanmälan) Π kurskod ( matr = (kursanmälan)) The result will be a relation with the attributes namn and matr. The attribute kurskod that we are dividing by will disappear in the division. NOTE! Which persons have passed ALL the courses they have registered on? has the same surface form, but a different logic: the set of courses that is tested will vary from one person to another, depending on which courses the current person has registered on. There is no common set of courses that should be tested for each person. The formulation in SQL will often be easier than that using division; to express the example query it is enough to create

2 the set of persons registered on courses and then subtract those who have NULL as the value of some course mark (vitsord, one of the attributes in kursanmalan in the sql code of the example database). Which female students take ALL the courses that is taking? Π matr, namn, kurskod ( kön = 'K' (student) x kursanmälan) Π kurskod ( matr = matr namn up kön kurskod matr kurskod ================================ =========== Brita DT K G Brita DT K Ann Helen DT K Ann Helen DT K Ann Helen DT K Siru DT K G Siru DT K Siru DT K Sara IS K Sara IS K Sara IS K (kursanmälan)) Π matr, namn, kurskod ( kön = 'K' (student) x kursanmälan) Π kurskod ( matr = (kursanmälan)) matr namn up kön kurskod matr kurskod ================================ =========== Brita DT K G Brita DT K Ann Helen DT K Ann Helen DT K Ann Helen DT K Siru DT K G Siru DT K Siru DT K Sara IS K Sara IS K Sara IS K We can see that only 40128, Siru has all the required courses.

3 r s is defined as Π R S (r) Π R S ((Π R S (r) s) Π R S,S (r)) r < Π matr, namn, kurskod ( kön = 'K' (student) x kursanmälan) s < Π kurskod ( matr = (kursanmälan)) Π R S (r) is here Π matr, namn (r) (No duplicate elements in sets!) Π R S,S (r) is here Π matr, namn, kurskod (r), in this case = r. The projection will guarantee that the attributes come in the right order for the subtraction. Π R S (r) s is here Π matr, namn (r) x {G555,, }, i. e. all the possible combinations of a girl registered on some course and the courses taken by From this set we shall subtract those tuples that represent real elements (actual course registrations). From what is left (the combinations representing false, unactualized information, we shall project the student number (matr) and the name (namn), the attributes represented by (Π R S ). Finally these will be subtracted from Π matr, namn (r). The answer is what is left. Π R S (r) Π R S (r) s (tänkbara) Π R S,S (r) (verkliga) ============== ===================== ===================== Brita Brita G Brita Ann Helen Brita Brita Siru Brita Ann Helen Sara Ann Helen G Ann Helen Ann Helen Ann Helen SVAR: Ann Helen Siru G555 ===== Siru G Siru Siru Siru Siru Siru Sara Sara G Sara Sara Sara Sara The false The tuples representing (Tuples that are not marked with candidates true information have green represent real have been stricken been stricken through. information about other courses through. than the relevant ones.)

4 How to translate this in MySQL? Problem: There is neither division ( ) nor set difference ( ) i MySQL, so you are obliged to find other ways around this problem. There are (at least) three ways to express these queries: 1) Not exists ( ) combined with not in ( ): ( There may not be a course that takes that is not among the courses taken by the girl in question ) select distinct matr, namn from student as R where kon = 'K' and not exists (select kurskod where matr = and kurskod not in (select kurskod as R2 where R.matr = R2.matr)); matr namn Siru 1 row in set (0.00 sec) The outmost selection will decide which columns (matr, namn) we shall see in the result. The outmost select from where will create a table including all the female students and their course registrations. These students will now be tested one by one: for the female student to be part of the result, the not exists clause for this student must be true (the list after the not exists must remain empty). What does this list consist of? First we put there all the courses that the student is registered on (3 items: G555, and ). These are represented by kurskod. Then these courses are tested one by one against the courses the current female student is registered on: a set of her courses is created in the innermost select. These are represented by kurskod. This set is thus composed of all the courses that the female student is registered on. As soon as we find a kurskod among 40101's courses that is not among the course codes of the current girl, this kurskod will part of the select list in the middle, which accordingly will no longer be empty. This in turn leads to the not exists becoming false, and that means that the female student in question does not satisfy the requirements (she does not have all the courses required) and she will not be part of the result. On the other hand, if all the course codes in the select in the middle are found in the set of the course codes taken by the current female student, this select will stay empty and the not exists will accordingly be true. Now this student will be part of the result.

5 The example query: 40101's courses: G555 Brita's (40112) courses: G555 Ann Helen's (40113) courses: G555 Siru's (40128) courses: G555 G555 Sara's (40240) courses: G555 Shows which course(s) are lacking in the girls' selections. These courses will be part of the selection in the middle and make not exists false. If all the courses taken by can be found among the courses chosen by the girl, she will be part of the result.

6 2) Two times not exists: ( There may not be a course that takes that is not taken by the current girl ) select distinct matr, namn from student as R where kon = 'K' and not exists (select kurskod as S where matr = and not exists (select kurskod as R2 where R.matr = R2.matr and S.kurskod = R2.kurskod )); matr namn Siru 1 row in set (0.01 sec) This solution is very similar to the last one, but now the test with set membership is replaced by another not exists. The outer not exists will choose the courses taken by The inner not exists will control these courses one by one; will we find the same course for the girl in question? If there is a kurskod (among the courses taken by the girl) equal to the kurskod (of 40101's courses) that we are testing at the moment, this kurskod will be the result of the innermost select. In this case this selection will not be empty, and the inner not exists will thus be false. Now the kurskod we just tested will not be a part of the selection in the middle. Each kurskod for will be tested like this. If we can find them all among the courses taken by the girl, none of them will be selected in the middle and the outer not exists will thus be true. This means that the girl in question had all the specified courses and that she will accordingly be part of the result. On the other hand, if there are one or several course codes that will not find their counterpart in the innermost selection, this selection will be empty for them, making the inner not exists true. These course codes will be a part of the select in the middle, making the outer not exists false, the girl in question (who did not have these courses) will not be a part of the result. See the figure on next page!

7 40101's courses Brita's (40112) courses G555 Can't be found chosen by the middle selection Ann Helen's (40113) courses G555 Can't be found chosen by the middle selection Siru's (40128) courses G555 G555 Sara's (40240) courses G555 Can't be found chosen by the middle selection Next page!

8 3) Count and compare The idea behind this method is to count how many different values there are in the divisor (i. e. all the values that one should have to be a part of the result). When this is applied to our example we first count how many courses takes. Then we count how many of these courses our female students take. If we get the same result, the girl in question has all the courses necessary and will appear in the result. mysql> select matr, namn natural join student where kon = 'K' and kurskod in (select kurskod where matr = 40101) group by matr having count(kurskod) = (select count(*) from (select kurskod where matr = 40101) as tab); matr namn Siru 1 row in set (0.00 sec) Here is the result of kursanmalan natural join student where kon = 'K' and kurskod in (select kurskod where matr = 40101): matr kurskod vitsord namn up kon NULL Brita DT K NULL Brita DT K NULL Ann Helen DT K NULL Ann Helen DT K NULL Siru DT K NULL Siru DT K G555 NULL Siru DT K NULL Sara IS K NULL Sara IS K 9 rows in set (0.00 sec) Next page!

9 These tuples are now grouped according to the student number: matr kurskod vitsord namn up kon NULL Brita DT K NULL Brita DT K NULL Ann Helen DT K NULL Ann Helen DT K NULL Siru DT K NULL Siru DT K G555 NULL Siru DT K NULL Sara IS K NULL Sara IS K having count(kurskod) counts the number of course codes/group, i. e. the number of courses each of these girls has registered on (note that in this selection we are only concerned with those courses that takes; it is quite possible that the girls are also registered on other courses in addition to these). having count(kurskod) = (select count(*) from (select kurskod where matr = 40101) as tab); compares the number of courses/group (the number of the relevant courses taken by the girl in question) with the number of courses taken by (3 courses). If we get the same number, we know that we have the same courses, and then the girl will appear in the result. This happens in the selection in the beginning of the query: mysql> select matr, namn natural join student where kon = 'K' and kurskod in (select kurskod where matr = 40101) group by matr having count(kurskod) = (select count(*) from (select kurskod where matr = 40101) as tab); OBS! Here we can only have one registration/course. In other cases we might need to use count distinct to eliminate duplicates. OBS! It would be a natural solution to use a temporary table to represent the courses taken by

10 Unfortunately, the same temporary table may not be opened twice in MySQL 5.1, so that solution will only give you error messages.

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