How to find the maximum in relational algebra?

This is my database course's homework.
We have this relation.
Trained(A, B) A trained B

How can I figure out who has trained the most people by using relational algebra or it cannot be expressed in relational algebra?
for example,
(John, Kenny)
(John, Kei)
(John, Cohen)
(Willson, John)
(Kenny, Peter)
(Ho, Tina)

In these tables, the answer obviously is John.But is it possible to express in relational algebra?

asked Oct 10 '16 at 0:40

John

add a comment |

This is my database course's homework.
We have this relation.
Trained(A, B) A trained B

In these tables, the answer obviously is John.But is it possible to express in relational algebra?

asked Oct 10 '16 at 0:40

John

add a comment |

This is my database course's homework.
We have this relation.
Trained(A, B) A trained B

In these tables, the answer obviously is John.But is it possible to express in relational algebra?

asked Oct 10 '16 at 0:40

John

This is my database course's homework.
We have this relation.
Trained(A, B) A trained B

In these tables, the answer obviously is John.But is it possible to express in relational algebra?

relation-algebra

asked Oct 10 '16 at 0:40

John

asked Oct 10 '16 at 0:40

John

asked Oct 10 '16 at 0:40

John

asked Oct 10 '16 at 0:40

John

asked Oct 10 '16 at 0:40

John

add a comment |

2 Answers
2

active

oldest

votes

You use GROUP and HAVING. I don't know how your class does notation, but in mine we do something like this:

SELECT A FROM Trained GROUP BY A HAVING COUNT(A)

I'm not sure if this is correct, but it should be similar.

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

add a comment |

Yes, it is possible to express it in relational algebra (RA).

The following line groups column A by COUNT(A). And plots COUNT(A) as 'frequencyOf_A'.

γ A; COUNT(A)→frequencyOf_A (Trained)

Resulting in this:



Trained.A   frequencyOf_A

John        3

Willson     1

Kenny       1

Ho          1

I used this RA engine. And uploaded your relation via github's gist. Thanks. (On the side "relation algebra" is mathematics, while "relational algebra" is considered computer science I think.)

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You use GROUP and HAVING. I don't know how your class does notation, but in mine we do something like this:

SELECT A FROM Trained GROUP BY A HAVING COUNT(A)

I'm not sure if this is correct, but it should be similar.

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

add a comment |

You use GROUP and HAVING. I don't know how your class does notation, but in mine we do something like this:

SELECT A FROM Trained GROUP BY A HAVING COUNT(A)

I'm not sure if this is correct, but it should be similar.

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

add a comment |

You use GROUP and HAVING. I don't know how your class does notation, but in mine we do something like this:

SELECT A FROM Trained GROUP BY A HAVING COUNT(A)

I'm not sure if this is correct, but it should be similar.

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

You use GROUP and HAVING. I don't know how your class does notation, but in mine we do something like this:

SELECT A FROM Trained GROUP BY A HAVING COUNT(A)

I'm not sure if this is correct, but it should be similar.

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

edited Oct 23 '16 at 17:01

answered Oct 23 '16 at 16:56

user1766555

3092618

answered Oct 23 '16 at 16:56

user1766555

3092618

answered Oct 23 '16 at 16:56

user1766555

3092618

add a comment |

Yes, it is possible to express it in relational algebra (RA).

The following line groups column A by COUNT(A). And plots COUNT(A) as 'frequencyOf_A'.

γ A; COUNT(A)→frequencyOf_A (Trained)

Resulting in this:



Trained.A   frequencyOf_A

John        3

Willson     1

Kenny       1

Ho          1

I used this RA engine. And uploaded your relation via github's gist. Thanks. (On the side "relation algebra" is mathematics, while "relational algebra" is considered computer science I think.)

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

add a comment |

Yes, it is possible to express it in relational algebra (RA).

The following line groups column A by COUNT(A). And plots COUNT(A) as 'frequencyOf_A'.

γ A; COUNT(A)→frequencyOf_A (Trained)

Resulting in this:



Trained.A   frequencyOf_A

John        3

Willson     1

Kenny       1

Ho          1

I used this RA engine. And uploaded your relation via github's gist. Thanks. (On the side "relation algebra" is mathematics, while "relational algebra" is considered computer science I think.)

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

add a comment |

Yes, it is possible to express it in relational algebra (RA).

The following line groups column A by COUNT(A). And plots COUNT(A) as 'frequencyOf_A'.

γ A; COUNT(A)→frequencyOf_A (Trained)

Resulting in this:



Trained.A   frequencyOf_A

John        3

Willson     1

Kenny       1

Ho          1

I used this RA engine. And uploaded your relation via github's gist. Thanks. (On the side "relation algebra" is mathematics, while "relational algebra" is considered computer science I think.)

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

Yes, it is possible to express it in relational algebra (RA).

The following line groups column A by COUNT(A). And plots COUNT(A) as 'frequencyOf_A'.

γ A; COUNT(A)→frequencyOf_A (Trained)

Resulting in this:



Trained.A   frequencyOf_A

John        3

Willson     1

Kenny       1

Ho          1

I used this RA engine. And uploaded your relation via github's gist. Thanks. (On the side "relation algebra" is mathematics, while "relational algebra" is considered computer science I think.)

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

edited Oct 18 '17 at 0:14

answered Oct 17 '17 at 23:44

Petr Kosvanec

answered Oct 17 '17 at 23:44

Petr Kosvanec

answered Oct 17 '17 at 23:44

Petr Kosvanec

add a comment |

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