I have some data that looks like this, and identifies pairs that are related:
From_ID To_ID
A C
B C
D E
E D (note this is the same pair as above, in a different order)
E F
A F
G H
Using the logic of 'if x is paired with y, and y is paired with z, then x is paired with z', how can I run an SQL query to return all members of a group?
So for the table above I would like a set of results that identifies or returns two groups: 'A, B, C, D, E, F' and 'G, H', not fussy about how this is done.
It feels like some kind of iterative query but I really have no idea where to start with this so any pointers would be appreciated.
edit: could be run in SQL Developer or HiveQL.
Related
I've some variables, Lets say a, b, c, d. All belongs to a fixed interval [0, e]
Now i've some relations between them like
a > b
a > c
b > d
Something like this; I want to make a function which print all the possible cases for this.
Example:
a b c d
a c b d
a b d c
a c b d
In essence, what you have is a directed acyclic graph.
A relatively simple approach is to store, for each variable, a set of the variables that must precede them. (In your example, this storage would map b to {a}, c to {a}, and d to {b}.) You can then write a recursive function that generates all valid tails consisting of a subset of these variables (in your case, for example, the subset {c,d} produces two valid tails: [c,d] and [d,c]). This recursive function examines each variable in the subset and determines whether its prerequisites are already met. (For example, since b maps to {a}, any subset including both a and b cannot produce a tail that begins with b.) If so, then it can recursively call itself on the subset excluding that variable.
There are some optimizations you can then perform, if desired. For example, you can use dynamic programming to avoid repeatedly re-computing the set of valid tails for the same subset.
Let's say we have 6 polygons ( A,B,C,D,E, F)in Spatial database.
A touches B,
B touches C and D,
E and F are not connected to other polygons.
A - B
/ \
C D E F
Having Polygon A, I need to query all polygons connected to it , and do it "recursivelly" Cannot describe it better. So the query by A should return A, B, C, D.
Of course, it is possible to do programmatically, first using SDO_RELATE query B by A, and then query C and D by B. But is it possible to do the task with a single query?
I'm developing a rudimentary word finder app with sql and ruby where I have an array of letters to find available words. It's easier to make the query by narrowing down what alphabetic letters aren't in the array. For ex.
alphabet= %w{a b c d e f g h i j k l m n o p q r s t u v w x y z}
available_letters = %w{p k z l p m t l n g g r u a r t n d z w a l m n e}
I can then subtract from alphabet, letters to exclude from my search and end up with an sql query like the one below.
select * from words
where word not like '%b%' and word not like '%c%' and word not like '%f%'.....
This gives me all the available words with a combination of all available letters. It does not narrow them down by the number of times that letter occurs. So if I only have one "E", I would like the query to narrow down words that only contain one e. I'm not sure if this can be done with an sql query or whether I will need to use a procedure. Anyone know a good way of solving this?
You will probably want to look into faster ways of implementing this, but to answer your question, you can exclude words with more than one "e" with not like '%e%e%'.
Hi all I have an exam coming up and am not getting much help from the lecturer on two questions on the practice exam. She has provided the answer but has not responded to my questions about the answer, I'm hoping someone here would be able to explain why the answer is the way it is.
Consider the following two tables R and S with their instances:
R S
A B C D E
a x y x y
a z w z w
b x k
b m j
c x y
f g h
a) πA(R[natural join]B=D S)
the answer being (a,b,c), why isn't it (a,a,b,c)? does a projection make it distinct?
b) π A(R[natural join] B<>D S)
the answer being (a,b,c,f), why is a an answer? b=d both times when values are x and z, so why is this being printed out?
a)In Relation Algebra, the projection operator provides duplicate elimination. In SQL this is not the default operation, but it is for relational algebra. Here is my source. At the moment, I can't recall why it does duplicate elimination, but this was my professor for databases and he is very knowledgeable. (I think it's because Relation Algebra uses set-logic and sets do not have duplicates.)
b)The joining of 2 tables creates a CROSS PRODUCT between the 2 tables. You have 6 rows and 2 rows. So the cross product is 6x2 = 12 rows. For row 1 of table R, you have a x y. This will be paired with x y AND z w resulting in [a x y x y] and [a x y z w]. The second pairing is valid for this relational algebra statement. Columns B and D do not match x != z.
a) πA(R[natural join]B=D S)
the answer being (a,b,c), why isn't it (a,a,b,c)? does a projection make it distinct?
In relational algebra, duplicate tuples are not permitted; that a main difference between sql (where distinct is needed) and relational algebra
b) π A(R[natural join] B<>D S)
the answer being (a,b,c,f), why is a an answer? b=d both times when values are x and z, so why is this being printed out?
Natural join operation returns the set of all combinations of tuples in R and S, so in this case returns also tuples (a x y z w) and (a z w x y); thus a has to be in the resulting projection.
[natural join] B=D
This is not a natural join because "natural join" is a join that joins relations exclusively over attributes of the same name. The construct you describe might in some places be labeled/termed an "equijoin" or so, but it is certainly noy a "natural join".
[natural join] B<>D
This is not a natural join because "natural join" is a join that joins together tuples of the argument relations if and only if the attribute values are equal.
You are being hopelessly mistaught and miseducated. Reference material : "an introduction to database systems", C.J.Date. It won't do you any good for your exams, but if you seek a later career in database technology it might be worthwhile to remember this.
But to answer your actual questions (in line with preceding answers) :
a) The attribute value 'a' cannot appear twice in the result of a projection, because a projection produces a relation, and a relation is defined to be a set, and sets cannot contain duplicates.
b) The [non-] natural join contains both the tuples (axyzw) and (azwxy). "First" tuple from R with "second" tuple from S, and other way round. The projection includes the result (a).
I have a table with multiple indexes, several of which duplicate the same columns:
Index 1 columns: X, B, C, D
Index 2 columns: Y, B, C, D
Index 3 columns: Z, B, C, D
I'm not very knowledgeable on indexing in practice, so I'm wondering if somebody can explain why X, Y and Z were paired with these same columns. B is an effective date. C is a semi-unique key ID for this table for a specific effective date B. D is a sequence that identifies the priority of this record for the identifier C.
Why not just create 6 indexes, one for each X, Y, Z, B, C, D?
I want to add an index to another column T, but in some contexts I'll only be querying on T alone while in others I will also be specifying the B, C and D columns... so should I create just one index like above or should I create one for T and one for (T, B, C, D)?
I've not had as much luck as expected when googling for comprehensive coverage of indexing. Any resources where I can get a through explanation and lots of examples of B-tree indexing?
The rule with indexing is that an index can be used to filter on any list of columns that constitute a prefix of the columns used for that index.
In other words, we can use Index 1 when we filter on X and B, or X, B and C, or just X, or all four.
However, we cannot use the index to filter "in the middle". This is because indexes work not entirely unlike concatenating the values of those columns for each row, and sorting the result. If we know what the thing we're looking for begins with, we can figure out where in the index to look - just like when doing binary search.
That's why a single index is no good: if we need to filter on B, C, D, and one of X, Y and Z, we need three indexes; X, Y is no good as an index for just filtering on Y, because the prefix of the values we're looking for - the X - is not known.
As Daniel mentioned, a covering index is a possible explanation for repeating B, C, and D: even if D is never filtered on, it may be the case that we need exactly the columns which you see in your indexes, and we can then just read the columns from the index instead of just using the index to locate the row.
One reason for having B, C and D in those indexes might be to have a covering index for frequently used queries. You will have a covering index when the index itself contains all the required data fields for a particular query.
A covering index can dramatically speed up data retrieval, since only the index pages, not the data pages, will be used to retrieve the data.
Below is an example query where index 1 would be a covering index:
SELECT B, C, D FROM table WHERE X = '10'
You should create it in (T, B, C, D).
Let's say you have two fields with an index in a table: A and B. When you create a separate index on each one of the columns, and have a query such as:
SELECT * FROM table WHERE A = 10 AND B = 20
What happens is either:
1) The DB creates two intermediate result-sets, one with rows where A = 10, and another one with rows where B = 20. It then has to merge these two result-sets into one (and also check for duplicate rows).
2) The DB creates one result-set with rows where A = 10. It then has to go manually through all of the rows in this intermediate result-set and check in each one where B = 10.
However when you know that index B depends on index A, and your query uses A before B, you can create one index for both of the columns: (A, B)
What this means that now the DB will first find all rows where A = 10, but because B is part of the same index, it can use the same index information to filter the result-set into rows where B is also 20. It doesn't have to make two intermediate result-sets + merge them, or only use one of the indexes and do manual scan for the other.
There might be other ways that the DB deals with these situations as well, it largely depends on an implementation.
The indexes in the form (X, B, C, D) can be used to optimize queries like:
... WHERE X rel sthg (possibly ORDER BY B, C, D)
... WHERE X = sthg AND B rel sthg (possibly ORDER BY C, D)
... WHERE X = sthf AND B = sthg AND C rel sthg (possibly ORDER BY D)
etc. where rel are arbitrary relation operators (<, >, =, <=, >=) and sthg are values or expressions. Especially the second two, and the sorting variants wouldn't be optimized by the "single column indexes variant".
OTOH, it cannot optimize a query
... WHERE B = sthg
because it starts in the middle of the index; here, the single column index would work.
For a resource where you can get a through explanation and lots of examples regarding indexes on Oracle (and any other Oracle-related issue), you should visit and bookmark askTom.