let be a database having the following relational-schemes: R(A,B,D) and S(A,B) with the attributes of same name in the same domain and with the instances r and s respectively.
An instance of r
An instance of s
What is the scheme and what are the tuples of u=r÷s? How to define them in English with r and s?
My attempt
I know that
u=r÷s=
Which leads me to think that it would only be an array of one column A, but I'm not sure enough to know what will be ther result within the array.
Can you help me understand u=r÷s?
An intuitive property of the division operator of the relational algebra is simply that it is the inverse of the cartesian product. For example, if you have two relations R and S, then, if U is a relation defined as the cartesian product of them:
U = R x S
the division is the operator such that:
U ÷ R = S
and:
U ÷ S = R
So, you can think of the result of U ÷ R as: “the projection of U that, multiplied by R, produces U”, and of the operation ÷, as the operation that finds all the “parts” of U that are combined with all the tuples of R.
However, in order to be useful, we want that this operation can be applied to any couple of relations, that is, we want to divide a relation which is not the result of a cartesian product. For this, the formal definition is more complex.
So, supposing that we have two relations R and S with attributes respectively A and B, their division can be defined as:
R ÷ S = πA-B(R) - πA-B((πA-B(R) x S) - R)
that can be read in this way:
πA-B(R) x S: project R over the attributes of R which are not in S, and multiply (cartesian product) this relation with S. This produces a relation with the attributes A of R and with rows all the possible combinations of rows of S and the projection of R;
From the previous result subtract all the tuples originally in R, that is, perform (πA-B(R) x S) - R. In this way we obtain the “extra” tuples, that is the tuples in the cartesian product that were not present in the original relation.
Finally, subtract from the original relation those extra tuples (but, again, perform this operation only on the attributes of R which are not present in S). So, the final operation is: πA-B(R) - πA-B(the result of step 2).
So, coming to your example, the projection of r on D is equal to:
(D)
d1
d2
d3
d4
and the cartesian product with s is:
(A, B, D)
a1 b1 d1
a1 b1 d2
a1 b1 d3
a1 b1 d4
Now we can remove from this set the tuples that were also in the original relation r, i.e. the first two tuples and the last one, so that we obtain the following result:
(A, B, D)
a1 b1 d3
And finally, we can remove the previous tuples (projected on D), from the original relation (again projected on D), that is, we remove:
(D)
d3
from:
(D)
d1
d2
d3
d4
and we obtain the following result, which is the final result of the division:
(D)
d1
d2
d4
Finally, we could double check the result by multiplying it with the original relation s (which is composed only by the tuple (a1, b1)):
(A B D)
a1 b1 d1
a1 b1 d2
a1 b1 d4
And looking at the rows of the original relation r, you can see this fact, that should give you an important insight on the meaning of the division operator:
the only values of the column D in r that are present together with (a1, b1) (the only tuple of s), are d1, d2 and d4.
You can also see another example in Wikipedia, and for a detailed explanation of the division, together with its transformation is SQL, you could look at these slides.
Related
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 4 years ago.
Improve this question
How do I know the cross product of A x B is perpendicular to B.
I'm little confused because there are 3 vectors instead of 2.
A = (0, -2, 5)
B = (2, 2, -5)
C= ( 7, -4, -5)
On R2 plane, (a x b) * b = 0 proves that a x b is perpendicular to b , but how do I find that on R3.
SO, after some of Research I finally figured out how to prove the vectors are perpendicularly to each other on R3.
A= (a1, a2, a3)
B= (b1, b2, b3)
C= (c1, c2, c3)
(AB x AC )* AB = 0
(AB x AC )* AC = 0
I don't think you understand what the cross product does. It gives a vector orthogonal to the two vectors.
The cross product a × b is defined as a vector c that is perpendicular
(orthogonal) to both a and b, with a direction given by the right-hand
rule and a magnitude equal to the area of the parallelogram that the
vectors span.
you can simply show this by using the definition of orthogonality which is from their dot products being zero.
Questions like this come down to precisely what you take to be your definitions.
For instance, one way to define the cross-product A x B is this:
By R^3 we mean three dimensional real space with a fixed orientation.
Observe that two linearly independent vectors A and B in R^3 span a plane, so every vector perpendicular to them lies on the (unique) line through the origin perpendicular to this plane.
Observe that for any positive magnitude, there are precisely two vectors along this line with that magnitude.
Observe that if we consider the ordered basis {A, B, C} of R^3, where C is one of the two vectors from the previous step, then one choice matches the orientation of R^3 and the other does not.
Define A x B as the vector C from the previous step for which {A, B, C} matches the orientation of R^3.
For instance, this is how the cross product is defined in the Wikipedia article:
"The cross product a × b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span."
If this is your definition, then there is literally nothing to prove, because the definition already has the word "perpendicular" in it.
Another definition might go like this:
By R^3 we mean three-dimensional real space with a fixed orientation.
For an ordered basis { e1, e2, e3 } of R^3 with the same orientation as R^3, we can write any two vectors A and B as A = a1 e1 + a2 e2 + a3 e3 and B as B = b1 e1 + b2 e2 + b3 e3.
Observe that, regardless of the choice of { e1, e2, e3 } we make in step 2, the vector C := (a2 b3 - b2 a3) e1 - (a1 b3 - b3 a1) e2 + (a1 b2 - b1 a2) e3 is always the same.
Take the vector C from the previous step as the definition of A x B.
This isn't a great definition, because step 3 is both a lot of work and complete black magic, but it's one you'll commonly see. If this is your definition, the best way to prove that A x B is perpendicular to A and B would be to show that the other definition gives you the same vector as this one, and then the perpendicularity comes for free.
A more direct way would be to show that vectors with a dot product of zero are perpendicular, and then to calculate the dot product by doing a bunch of algebra. This is, again, a fairly popular way to do it, but it's essentially worthless because it doesn't offer any insight into what's going on.
I am trying to understand what will be the result of performing a natural join
between two relations R and S, where they have no common attributes.
By following the below definition, I thought the answer might be an empty set:
Natural Join definition.
My line of thought was because the condition in the 'Select' symbol is not met, the projection of all of the attributes won't take place.
When I asked my lecturer about this, he said that the output will be the same as doing a cartezian product between R and S.
I can't seem to understand why, would appreciate any help )
Natural join combines a cross product and a selection into one
operation. It performs a selection forcing equality on those
attributes that appear in both relation schemes. Duplicates are
removed as in all relation operations.
There are two special cases:
• If the two relations have no attributes in common, then their
natural join is simply their cross product.
• If the two relations have more than one attribute in common,
then the natural join selects only the rows where all pairs of
matching attributes match.
Notation: r s
Let r and s be relation instances on schema R and S
respectively.
The result is a relation on schema R ∪ S which is
obtained by considering each pair of tuples tr from r and ts from s.
If tr and ts have the same value on each of the attributes in R ∩ S, a
tuple t is added to the result, where
– t has the same value as tr on r
– t has the same value as ts on s
Example:
R = (A, B, C, D)
S = (E, B, D)
Result schema = (A, B, C, D, E)
r s is defined as:
πr.A, r.B, r.C, r.D, s.E (σr.B = s.B r.D = s.D (r x s))
The definition of the natural join you linked is:
It can be broken as:
1.First take the cartezian product.
2.Then select only those row so that attributes of the same name have the same value
3.Now apply projection so that all attributes have distinct names.
If the two tables have no attributes with same name, we will jump to step 3 and therefore the result will indeed be cartezian product.
I have a function that uses 4 parameters, called tile . It is designed to work the following way :
tile(?E, ?S, ?W, ?N, ?ID)
I would like a getter function that given an ID, it returns the first 4 parameters: E, S, W and N.
I have tried something like:
coordonates(tile(E,S,W,N,L), (E,S,W,N)).
But it does not return the actual values, only true.
If I type tile(E, S, W, N, #1) in the terminal I get the desired result but I do not know what exactly is returned (a list maybe?).
Let's suppose our facts describing tile looks as follows:
tile(p1,p2,p3,p4,id1).
tile(q1,q2,q3,q4,id2).
tile(r1,r2,r3,r4,id3).
In this we have a finite number of facts. That can be checked by the most general query for tile:
?- tile(E,S,W,N,I).
E = p1,
S = p2,
W = p3,
N = p4,
I = id1 ; % <---- user input ; to continue
E = q1,
S = q2,
W = q3,
N = q4,
I = id2 ; % <---- user input ; to continue
E = r1,
S = r2,
W = r3,
N = r4,
I = id3. % <---- toplevel outputs . -- we're done
So in theory, we could define coordonates as follows:
coordonates(id1, t(p1, p2, p3, p4)).
coordonates(id2, t(q1, q2, q3, q4)).
coordonates(id3, t(r1, r2, r3, r4)).
which could be queried for id2 as follows:
?- coordonates(id2,X).
X = t(q1, q2, q3, q4).
I used the functor t to group the solution, to make clear that it is not the predicate tile we defined earlier. There's also a lot of repetition in this definition which is already a hint, that we can do better. What we are looking for is a rule which tells us how, given we have a answer for tile, we can describe coordonates. In logical terms, this is written as an implication of the form: goal1 ∧ ... ∧ goalN → head. which means "Suppose I know that goal1 to goalN is true, then I also know that head is true." In Prolog, this is written backwards:
head :-
goal1,
% ...
goalN.
Let's go back to our task: we know something about a tile and we want to describe how the projection looks like. This means, our code looks as follows:
coordonates( ... ) :-
% ...
tile(E,S,W,N,I).
The body tile(E,S,W,N,I) is the most general form we can write (see our query above) and can be read as "suppose I have any tile at coordinates E S W N with id I". Now we only need to fill in, how coordonates should look like. We know it has two arguments, because it relates the id with the four other elements. Lets give them names, say Id and Coords:
coordonates(Id, Coords) :-
% ...
tile(E,S,W,N,I).
Now we only need to find out how to relate E,S,E,N and I with Id and Coords. One is easy: Id is just I. The other one is also not too hard, we just need to group the coordinates into one term. We can pick an arbitrary one, but already above decided to take t, so we will stick with it:
coordonates(Id, Coords) :-
Id = I,
Coords = t(E,S,W,N),
tile(E,S,W,N,I).
This already works as we expect:
?- coordonates(X,Y).
X = id1,
Y = t(p1, p2, p3, p4) ;
X = id2,
Y = t(q1, q2, q3, q4) ;
X = id3,
Y = t(r1, r2, r3, r4).
Now we can make one observation: if two terms are equal, we can use one instead of the other. So instead of writing Id = I, we can just reuse Id. The same goes for Coords and t(E,S,W,N):
coordonates(I, t(E,S,W,N)) :-
tile(E,S,W,N,I).
It couldn't be much shorter :-)
You have to declare 'E, S, W and N' so that prolog can unify those parameters with the input when you make the query. Something like (In the most basic case):
tile(['cordE1','cordS1','cordW1','cordN1'],1).
tile(['cordE2','cordS2','cordW2','cordN2'],2).
tile(['cordE3','cordS3','cordW3','cordN3'],3).
Query:
?- tile(C,2).
C = [cordE2, cordS2, cordW2, cordN2].
?- tile(C,1).
C = [cordE1, cordS1, cordW1, cordN1].
?- tile(C,3).
C = [cordE3, cordS3, cordW3, cordN3].
I'm not sure how best to describe this without using sets:
Assume we have two distinct, finite sets A & B, and a set P which contains a subset of all the different pairs of A & B (it's a predicate join, basically).
for example:
P = { (a1, b1), (a1, b3), (a2, b1), (a2, b3), (a1, b2) }
I want to find a set C which contains contains the fewest (or close to) number of pairs (as,bs) of subsets of A & B, eg:
C = { ( {a1, a2}, {b1, b3} ), ( {a1}, {b2} ) }
such that for each (as,bs) in C, for each combination of a in as and b in bs, (a,b) is in P and each element of P appears once and only once in this 'expansion' of C.
i'm not exactly sure how to describe this, but it seems somewhat analagous to rectangle covering in computational geometry. maybe someone has seen something like this before?
Given a DAG, in which each node belongs to a category, how can this graph be transformed into a table with a column for each category? The transformation doesn't have to be reversible, but should preserve useful information about the structure of the graph; and should be a 'natural' transformation, in the sense that a person looking at the graph and the table should not be surprised by any of the rows. It should also be compact, i.e. have few rows.
For example given a graph of nodes a1,b1,b2,c1 with edges a1->b1, a1->b2, b1->c1, b2->c1 (i.e. a diamond-shaped graph) I would expect to see the following table:
a b c
--------
a1 b1 c1
a1 b2 c1
I've thought about this problem quite a bit, but I'm having trouble coming up with an algorithm that gives intuitive results on certain graphs. Consider the graph a1,b1,c1 with edges a1->c1, b1->c1. I'd like the algorithm to produce this table:
a b c
--------
a1 b1 c1
But maybe it should produce this instead:
a b c
--------
a1 c1
a1 b1
I'm looking for creative ideas and insights into the problem. Feel free to vary to simplify or constrain the problem if you think it will help.
Brainstorm away!
Edit:
The transformation should always produce the same set of rows, although the order of rows does not matter.
The table should behave nicely when sorting and filtering using, e.g., Excel. This means that mutliple nodes cannot be packed into a single cell of the table - only one node per cell.
What you need is a variation of topological sorting. This is an algorithm that "sorts" graph vertexes as if a---->b edge meant a > b. Since the graph is a DAG, there is no cycles in it and this > relation is transitive, so at least one sorting order exists.
For your diamond-shaped graph two topological orders exist:
a1 b1 b2 c1
a1 b2 b1 c1
b1 and b2 items are not connected, even indirectly, therefore, they may be placed in any order.
After you sorted the graph, you know an approximation of order. My proposal is to fill the table in a straightforward way (1 vertex per line) and then "compact" the table. Perform sorting and pick the sequence you got as output. Fill the table from top to bottom, assigning a vertex to relevant column:
a b c
--------
a1
b2
b1
c1
Now compact the table by walking from top to bottom (and then make similar pass from bottom to top). On each iteration, you take a closer look to a "current" row (marked as =>) and to the "next" row.
If in a column nodes in current and next node differ, do nothing for this column:
from ----> to
X b c X b c
-------- --------
=> X1 . . X1 . .
X2 . . => X2 . .
If in a column X in the next row there is no vertex (table cell is empty) and in the current row there is vertex X1, then you sometimes should fill this empty cell with a vertex in the current row. But not always: you want your table to be logical, don't you? So copy the vertex if and only if there's no edge b--->X1, c--->X1, etc, for all vertexes in current row.
from ---> to
X b c X b c
-------- --------
=> X1 b c X1 b c
b1 c1 => X1 b1 c1
(Edit:) After first (forward) and second (backward) passes, you'll have such tables:
first second
a b c a b c
-------- --------
a1 a1 b2 c1
a1 b2 a1 b2 c1
a1 b1 a1 b1 c1
a1 b1 c1 a1 b1 c1
Then, just remove equal rows and you're done:
a b c
--------
a1 b2 c1
a1 b1 c1
And you should get a nice table. O(n^2).
How about compacting all reachable nodes from one node together in one cell ? For example, your first DAG should look like:
a b c
---------------
a1 [b1,b2]
b1 c1
b2 c1
It sounds like a train system map with stations within zones (a,b,c).
You could be generating a table of all possible routes in one direction. In which case "a1, b1, c1" would seem to imply a1->b1 so don't format it like that if you have only a1->c1, b1->c1
You could decide to produce a table by listing the longest routes starting in zone a,
using each edge only once, ending with the short leftover routes. Or allow edges to be reused only if they connect unused edges or extend a route.
In other words, do a depth first search, trying not to reuse edges (reject any path that doesn't include unused edges, and optionally trim used edges at the endpoints).
Here's what I ended up doing:
Find all paths emanating from a node without in-edges. (Could be expensive for some graphs, but works for mine)
Traverse each path to collect a row of values
Compact the rows
Compacting the rows is dones as follows.
For each pair of columns x,y
Construct a map of every value of x to it's possible values of y
Create another map For entries that only have one distinct value of y, mapping the value of x to its single value of y.
Fill in the blanks using these maps. When filling in a value, check for related blanks that can be filled.
This gives a very compact output and seems to meet all my requirements.