I have two Functional Dependencies from one relation:
meetid -> pid
meetid, pid -> status
For relation Meetings(meetid,pid,status)
I want to use promotional approach to make multivalued dependencies from this. The problem is that I'm not sure if meetid ->> pid is legal to do in this situation, as the complementation rule will make the other MVD illegal (I think): meetid ->> status.
The other FD will create a dependency that I see as a trivial MVD.
meetid,pid ->> status
Is this relation doomed to be trivial when promoting them to MVDs, or am I missing something in the process?
The two dependencies:
meetid → pid
meetid, pid → status
are not minimal, since they can be simplified (for instance by computing a canonical cover) in:
meetid → pid
meetid → status
and from this we know also that meetid is the only candidate key of the relation. From a FD X → Y one can always derive the MVD X →→ Y, so, you have both the MVD dependencies:
meetid →→ pid
meeditd →→ status
(Note that the second one can also be derived from the first one by complementation).
Note also that none of them is a trivial MVD, since a MVD X →→ Y is trivial (that is, always true) if either Y is the empty set or XY are all the attributes of the relation.
Furthermore, we can note that the schema is in 4NF, since it is in BCNF and each left hand side of non-trivial MVDs is a superkey.
Finally, note that every functional dependency can be “promoted” to a multivalued dependency, so that we have also meetid, pid →→ status (and this is trivial).
Disclosure: I am taking Stanford's online database course. The forum there is dead, and I'm hoping for some help on SO.
Here's the quiz question:
Consider relation R(A,B,C,D,E) with multivalued dependencies:
A -» B, B -» D
and no functional dependencies. Suppose we decompose R into 4th Normal Form. Depending on the order in which we deal with 4NF violations, we can get different final decompositions. Which one of the following relation schemas could be in the final 4NF decomposition?
And here is my thinking:
Since we are given that there are no functional dependencies, the only key is set of attributes (A,B,C,D,E). In other words, both multivalued dependencies in the question are violating, and we must decompose them.
I am following the decomposition algorithm given in lecture:
Compute keys for R [done]
Repeat until all relations are in 4NF
Pick any R' with nontrivial A -» B that violates 4NF
Decompose R' into R_1(A, B) and R_2(A, rest)
Compute functional dependencies and multivalued dependencies for R_1 and R_2
Compute keys for R_1 and R_2
I see two ways to decompose the relations: start with A -» B or B -» D.
Starting with A -» B
R(A,B,C,D,E)
|
+-----------+
| |
R_1(A,B) R_2(A,C,D,E)
Since B and D are no longer in the same relation, we do not have a 4NF violation, and we're done. I'm not sure how to compute the FDs, MVDs, and keys at this point.
Starting with B -» D
R(A,B,C,D,E)
|
+-----------+
| |
R_1(B,D) R_2(B,A,C,E)
|
+----------+
| |
R_3(A,B) R_4(A,C,E)
At this point, (A and B) and (B and D) are decomposed into their own relations, so we have no violations, and we're done.
The answer choices:
At this point, I'm completely stumped. I do not see any of the relations in the answer choices, nor can I come up with an idea that will get me there:
CE
AD
AE
ABD
I don't need the answer outright, but what am I missing?
A correct answer is AD.
How is this obtained?
Consider that, like for functional dependencies, you can have multivalued dependencies implied by other multivalued dependencies. For instance, there is a pseudo-transitivity rule (or multi-valued transitivity rules) that says:
If X →→ Y holds, and Y →→ Z holds, then X →→ Z − Y holds
For this rule, from A →→ B and B →→ D you can derive A →→ D. So, if you decompose the relation in 4NF you could start from this dependency, and get a table with attributes AD. Or, alternatively, in your first decomposition, after finding R_1(A,B) and R_2(A,C,D,E), you should continue to decompose R_2, since it still contain the non-trivial MVD A →→ D, to find R_3(A, D) and R_3(A, C, E).
I am having trouble understanding what my lecturer want me to do from this question. Can anyone help explain to me what he wants me to do?
Define a higher order version of the insertion sort algorithm. That is define
functions
insertBy :: Ord b => (a->b) -> a -> [a] -> [a]
inssortBy :: Ord b => (a->b) -> [a] -> [a]
and this bit is where it got me confused:
such that inssort f l sorts the list l such that an element x comes before an elementyif f x < f y.
If you were sorting numbers, then it's clear what x < y means. But what if you were sorting letters? Or customers? Or anything else without a clear (to the computer) ordering?
So you are supposed to create a function f() that defines that ordering for the sorting procedure. That f() will take the letters or customers or whatever and will return an integer for each one that the computer can actually sort on.
At least, that's how the problem is described. I personally would have designed a predicate that accepted two items, x and y and returned a boolean if x < y. But whichever is fine.
The code wants you to rewrite the insertion sort algorithm, but using a function as a parameter - thus a higher order function.
I would like to point out that this code, typo included, seems to stem from a piece of work currently due at a certain university - I found this page while searching for "insertion sort algortihm", as I copy pasted the term out of the document as well, typo included.
Seeking code from the internet is a risky business. Might I recommend the insertion sort algorithm wikipedia entry, or the Haskell code provided in your lecture slides (you are looking for the 'insertion sort algorithm' and for 'higher-order functions), as opposed to the several queries you have placed on Stack Overflow?
I need an algorithm to perform a 2D bisection method for solving a 2x2 non-linear problem. Example: two equations f(x,y)=0 and g(x,y)=0 which I want to solve simultaneously. I am very familiar with the 1D bisection ( as well as other numerical methods ). Assume I already know the solution lies between the bounds x1 < x < x2 and y1 < y < y2.
In a grid the starting bounds are:
^
| C D
y2 -+ o-------o
| | |
| | |
| | |
y1 -+ o-------o
| A B
o--+------+---->
x1 x2
and I know the values f(A), f(B), f(C) and f(D) as well as g(A), g(B), g(C) and g(D). To start the bisection I guess we need to divide the points out along the edges as well as the middle.
^
| C F D
y2 -+ o---o---o
| | |
|G o o M o H
| | |
y1 -+ o---o---o
| A E B
o--+------+---->
x1 x2
Now considering the possibilities of combinations such as checking if f(G)*f(M)<0 AND g(G)*g(M)<0 seems overwhelming. Maybe I am making this a little too complicated, but I think there should be a multidimensional version of the Bisection, just as Newton-Raphson can be easily be multidimed using gradient operators.
Any clues, comments, or links are welcomed.
Sorry, while bisection works in 1-d, it fails in higher dimensions. You simply cannot break a 2-d region into subregions using only information about the function at the corners of the region and a point in the interior. In the words of Mick Jagger, "You can't always get what you want".
I just stumbled upon the answer to this from geometrictools.com and C++ code.
edit: the code is now on github.
I would split the area along a single dimension only, alternating dimensions. The condition you have for existence of zero of a single function would be "you have two points of different sign on the boundary of the region", so I'd just check that fro the two functions. However, I don't think it would work well, since zeros of both functions in a particular region don't guarantee a common zero (this might even exist in a different region that doesn't meet the criterion).
For example, look at this image:
There is no way you can distinguish the squares ABED and EFIH given only f() and g()'s behaviour on their boundary. However, ABED doesn't contain a common zero and EFIH does.
This would be similar to region queries using eg. kD-trees, if you could positively identify that a region doesn't contain zero of eg. f. Still, this can be slow under some circumstances.
If you can assume (per your comment to woodchips) that f(x,y)=0 defines a continuous monotone function y=f2(x), i.e. for each x1<=x<=x2 there is a unique solution for y (you just can't express it analytically due to the messy form of f), and similarly y=g2(x) is a continuous monotone function, then there is a way to find the joint solution.
If you could calculate f2 and g2, then you could use a 1-d bisection method on [x1,x2] to solve f2(x)-g2(x)=0. And you can do that by using 1-d bisection on [y1,y2] again for solving f(x,y)=0 for y for any given fixed x that you need to consider (x1, x2, (x1+x2)/2, etc) - that's where the continuous monotonicity is helpful -and similarly for g. You have to make sure to update x1-x2 and y1-y2 after each step.
This approach might not be efficient, but should work. Of course, lots of two-variable functions don't intersect the z-plane as continuous monotone functions.
I'm not much experient on optimization, but I built a solution to this problem with a bisection algorithm like the question describes. I think is necessary to fix a bug in my solution because it compute tow times a root in some cases, but i think it's simple and will try it later.
EDIT: I seem the comment of jpalecek, and now I anderstand that some premises I assumed are wrong, but the methods still works on most cases. More especificaly, the zero is garanteed only if the two functions variate the signals at oposite direction, but is need to handle the cases of zero at the vertices. I think is possible to build a justificated and satisfatory heuristic to that, but it is a little complicated and now I consider more promising get the function given by f_abs = abs(f, g) and build a heuristic to find the local minimuns, looking to the gradient direction on the points of the middle of edges.
Introduction
Consider the configuration in the question:
^
| C D
y2 -+ o-------o
| | |
| | |
| | |
y1 -+ o-------o
| A B
o--+------+---->
x1 x2
There are many ways to do that, but I chose to use only the corner points (A, B, C, D) and not middle or center points liky the question sugests. Assume I have tow function f(x,y) and g(x,y) as you describe. In truth it's generaly a function (x,y) -> (f(x,y), g(x,y)).
The steps are the following, and there is a resume (with a Python code) at the end.
Step by step explanation
Calculate the product each scalar function (f and g) by them self at adjacent points. Compute the minimum product for each one for each direction of variation (axis, x and y).
Fx = min(f(C)*f(B), f(D)*f(A))
Fy = min(f(A)*f(B), f(D)*f(C))
Gx = min(g(C)*g(B), g(D)*g(A))
Gy = min(g(A)*g(B), g(D)*g(C))
It looks to the product through tow oposite sides of the rectangle and computes the minimum of them, whats represents the existence of a changing of signal if its negative. It's a bit of redundance but work's well. Alternativaly you can try other configuration like use the points (E, F, G and H show in the question), but I think make sense to use the corner points because it consider better the whole area of the rectangle, but it is only a impression.
Compute the minimum of the tow axis for each function.
F = min(Fx, Fy)
G = min(Gx, Gy)
It of this values represents the existence of a zero for each function, f and g, within the rectangle.
Compute the maximum of them:
max(F, G)
If max(F, G) < 0, then there is a root inside the rectangle. Additionaly, if f(C) = 0 and g(C) = 0, there is a root too and we do the same, but if the root is in other corner we ignore him, because other rectangle will compute it (I want to avoid double computation of roots). The statement bellow resumes:
guaranteed_contain_zeros = max(F, G) < 0 or (f(C) == 0 and g(C) == 0)
In this case we have to proceed breaking the region recursively ultil the rectangles are as small as we want.
Else, may still exist a root inside the rectangle. Because of that, we have to use some criterion to break this regions ultil the we have a minimum granularity. The criterion I used is to assert the largest dimension of the current rectangle is smaller than the smallest dimension of the original rectangle (delta in the code sample bellow).
Resume
This Python code resume:
def balance_points(x_min, x_max, y_min, y_max, delta, eps=2e-32):
width = x_max - x_min
height = y_max - y_min
x_middle = (x_min + x_max)/2
y_middle = (y_min + y_max)/2
Fx = min(f(C)*f(B), f(D)*f(A))
Fy = min(f(A)*f(B), f(D)*f(C))
Gx = min(g(C)*g(B), g(D)*g(A))
Gy = min(g(A)*g(B), g(D)*g(C))
F = min(Fx, Fy)
G = min(Gx, Gy)
largest_dim = max(width, height)
guaranteed_contain_zeros = max(F, G) < 0 or (f(C) == 0 and g(C) == 0)
if guaranteed_contain_zeros and largest_dim <= eps:
return [(x_middle, y_middle)]
elif guaranteed_contain_zeros or largest_dim > delta:
if width >= height:
return balance_points(x_min, x_middle, y_min, y_max, delta) + balance_points(x_middle, x_max, y_min, y_max, delta)
else:
return balance_points(x_min, x_max, y_min, y_middle, delta) + balance_points(x_min, x_max, y_middle, y_max, delta)
else:
return []
Results
I have used a similar code similar in a personal project (GitHub here) and it draw the rectangles of the algorithm and the root (the system have a balance point at the origin):
Rectangles
It works well.
Improvements
In some cases the algorithm compute tow times the same zero. I thinh it can have tow reasons:
I the case the functions gives exatly zero at neighbour rectangles (because of an numerical truncation). In this case the remedy is to incrise eps (increase the rectangles). I chose eps=2e-32, because 32 bits is a half of the precision (on 64 bits archtecture), then is problable that the function don't gives a zero... but it was more like a guess, I don't now if is the better. But, if we decrease much the eps, it extrapolates the recursion limit of Python interpreter.
The case in witch the f(A), f(B), etc, are near to zero and the product is truncated to zero. I think it can be reduced if we use the product of the signals of f and g in place of the product of the functions.
I think is possible improve the criterion to discard a rectangle. It can be made considering how much the functions are variating in the region of the rectangle and how distante the function is of zero. Perhaps a simple relation between the average and variance of the function values on the corners. In another way (and more complicated) we can use a stack to store the values on each recursion instance and garantee that this values are convergent to stop recursion.
This is a similar problem to finding critical points in vector fields (see http://alglobus.net/NASAwork/topology/Papers/alsVugraphs93.ps).
If you have the values of f(x,y) and g(x,y) at the vertexes of your quadrilateral and you are in a discrete problem (such that you don't have an analytical expression for f(x,y) and g(x,y) nor the values at other locations inside the quadrilateral), then you can use bilinear interpolation to get two equations (for f and g). For the 2D case the analytical solution will be a quadratic equation which, according to the solution (1 root, 2 real roots, 2 imaginary roots) you may have 1 solution, 2 solutions, no solutions, solutions inside or outside your quadrilateral.
If instead you have analytic functions of f(x,y) and g(x,y) and want to use them, this is not useful. Instead you could divide your quadrilateral recursively, however as it was already pointed out by jpalecek (2nd post), you would need a way to stop your divisions by figuring out a test that would assure you would have no zeros inside a quadrilateral.
Let f_1(x,y), f_2(x,y) be two functions which are continuous and monotonic with respect to x and y. The problem is to solve the system f_1(x,y) = 0, f_2(x,y) = 0.
The alternating-direction algorithm is illustrated below. Here, the lines depict sets {f_1 = 0} and {f_2 = 0}. It is easy to see that the direction of movement of the algorithm (right-down or left-up) depends on the order of solving the equations f_i(x,y) = 0 (e.g., solve f_1(x,y) = 0 w.r.t. x then solve f_2(x,y) = 0 w.r.t. y OR first solve f_1(x,y) = 0 w.r.t. y and then solve f_2(x,y) = 0 w.r.t. x).
Given the initial guess, we don't know where the root is. So, in order to find all roots of the system, we have to move in both directions.
I was reading this SO question and I got intrigued by what are Boolean Networks, I've looked it up in Wikipedia but the explanation is too vague IMO. Can anyone explain me what Boolean Networks are? And if possible with some examples too?
Boolean networks represent a class of networks where the nodes have states and the edges represent transitions between states. In the simplest case, these states are either 1 or 0 – i.e. boolean.
Transitions may be simple activations or inactivations. For example, consider nodes a and b with an edge from a to b.
f
a ------> b
Here, f is a transition function. In the case of activation, f may be defined as:
f(x) = x
i.e. b's value is 1 if and only if a's value is 1. Conversely, an inactivation (or repression) might look like this:
f(x) = NOT x
More complex networks use more involved boolean functions. E.g. consider:
a b
\ /
\ /
\ /
v
c
Here, we've got edges from a to c and from b to c. c might be defined in terms of a and b as follows.
f(a, b) = a AND NOT b
Thus, c is activated only if a is active and b is inactive, at the same time.
Such networks can be used to model all kinds of relations. One that I know of is in systems biology where they are used to model (huge) interaction networks of chemicals in living cells. These networks effectively model how certain aspects of the cells work and they can be used to find deficiencies, points of attack for drugs and similarities between unrelated components that point to functional equivalence. This is fundamentally important in understanding how life works.