I have a question regarding Answer Set Programming on how to make an existing fact invalid, when there is already (also) a default statement present in the Knowledge Base.
For example, there are two persons seby and andy, one of them is able to drive at once. The scenario can be that seby can drive as seen in Line 3 but let's say, after his license is cancelled he cannot drive anymore, hence we now have Lines 4 to 7 and meanwhile andy learnt driving, as seen in Line 7. Line 6 shows only one person can drive at a time, besides showing seby and andy are not the same.
1 person(seby).
2 person(andy).
3 drives(seby).
4 drives(seby) :- person(seby), not ab(d(drives(seby))), not -drives(seby).
5 ab(d(drives(seby))).
6 -drives(P) :- drives(P0), person(P), P0 != P.
7 drives(andy).
In the above program, Lines 3 and 7 contradict with Line 6, and the Clingo solver (which I use) obviously outputs UNSATISFIABLE.
Having said all this, please don't say to delete Line 3 and the problem is solved. The intention behind asking this question is to know whether it is possible now to make Line 3 somehow invalid to let Line 4 do its duty.
However, Line 4 can also be written as:
4 drives(P) :- person(P), not ab(d(drives(P))), not -drives(P).
Thanks a lot in advance.
I do not fully understand the problem. Line 3 and line 4 are separate rules, even if line 4's antecedent is false line 3 would still be true. In other words, line 4 seems redundant.
It seems like you want a choice. I assume ab(d(drives(seby))) denotes that seby has lost their license. So, below line four is your constraint on only people with a license driving. Line five is the choice, so by default andy or seby can drive but not both. Notice in the ground program how line five is equivalent to drives(seby) :- not drives(andy). and drives(andy) :- not drives(seby). You can also have seby as being the preferred driver using optimization statements (the choice rule below is like an optimization statement).
person(seby).
person(andy).
ab(d(drives(seby))).
:- person(P), ab(d(drives(P))), drives(P).
1{drives(P) : person(P)}1.
If something is true, it must always be true. Therefore the line:
drives(seby).
will always be true.
However, we can get around this by putting the fact into a choice rule.
0{drives(seby)}1.
This line says that an answer will have 0 to 1 drives(seby).. This means we can have rules that contradict drives(seby). and the answer will still be satisfiable, but we can also have drives(seby). be true.
Both this program:
0{drives(seby)}1.
drives(seby).
And this program:
0{drives(seby)}1.
:- drives(seby).
Are satisfiable.
Most likely you will want drives(seby). to be true if it can be, and false if it can't be. To accomplish this, we need to force Clingo into making drives(seby). true if it can. We can do this by using an optimization.
A naive way to do this is to count how many drives(seby). exist (either 0 or 1) and maximize the count.
We can count the number of drives(seby). with this line:
sebyCount(N) :- N = #count {drives(seby) : drives(seby)}.
N is equal to the number of drives(seby). in the domain drives(seby).. This will either be 0 or 1.
And then we can maximize the value of N with this statement:
#maximize {N#1 : sebyCount(N)}.
This maximizes the value of N with the priority 1 (the lower the number, the lower the priority) in the domain of sebyCount(N).
Related
I have a read/write operation going on in the Fortran code snippet as follows
OPEN(5,FILE='WKDAT.dat', STATUS='OLD')
OPEN(6,FILE='WKLST.dat', STATUS='UNKNOWN')
I know that by default the unit number 5 is used for input from the keyboard and unit number 6 is used to display on the screen. Also I can use *.
But in the above-mentioned Fortran code unit number is 5 and a file name "WKDAT.dat" is given. So this means that the data is being read from "WKDAT.dat" file. Also there is code unit number 6 and a file name "WKLST.dat" is given. So this means that the data is being written to "WKLST.dat" file.
Is my understanding correct?
As per my basic knowledge:
Unit number 5 is only used to take input from keyboard & unit number 6 is only used to print to console so no files should be involved. But in the code snippet it has both unit number 5, 6 as well as file name.
So both are contradicting :(
In this link http://www.oc.nps.edu/~bird/oc3030_online/fortran/io/io.html they have mentioned the following "When I/O is to a file you must ASSOCIATE a UNIT number (which you choose) with the FILENAME. Use any unit number other than 5 and 6. On some computers, some unit numbers are reserved for use by the computer operating system."
Fortran has no magic unit numbers. The Fortran standard says nothing about 5, 6 or any other valid unit number being used for a special purpose. As such you are free to use the open statement to associate any valid unit number with a file. However traditionally for reasons that pre-date me 5 and 6 have been pre-associated with the keyboard and screen, as you say. Now still you can change the association by use of the open statement and that is fine save for the confusion it can cause, so most people I know recommend avoiding this and using unit numbers of 10 and upwards. Also because 5 and 6 are not guaranteed to be associated with the default input and output devices I would recommend against their use, preferring * or, in more modern code, the named constants input_unit, output_unit and error_unit from the iso_fortran_env intrinsic module.
So in summary you've got the right idea, and I'm not surprised you're confused.
Nothing in the standard says units 5 and 6 have any special meaning although in practice standard input and standard output are often pre-connected to 5 and 6.
Module iso_fortran_env from Fortran 2008 contains constants
INPUT_UNIT
OUTPUT_UNIT
ERROR_UNIT
with the unit numbers where standard input, standard output and standard error are connected. These are allowed to be different than 5 and 6.
Opening a file in unit that is in use causes the unit to be associated with the new file.
For example the Cray Fortran manual says:
Unit numbers 100, 101, and 102 are permanently associated with the
standard input, standard output, and standard error files,
respectively.
That means if you open some other file as unit 5 or 6 standard input and standard output still have some other unit where they are pre-connected and they will not be closed.
Is there a betther way to do this condition:
if ( x > 3 || y > 3 || z > 3 ) {
...
}
I was thinking of some bitwise operation but could'nt find anything.
I searched over Google but it is hard to find something related to that kind of basic question.
Thanks!
Edit
I was thinkg in general programming. Would it be different according to a specific language? Such as C/C++, Java...
What you have is good. Assuming C/C++ or Java:
Intent is clear
Short circuit optimisation means the expression will be true as soon as any one part is true.
Looking at that 2nd point- if any of x, y or z is more likely to be >3, then put them to the left of the expression so they are evaluated first which means the others may not need to be evaluated at all.
For argument's sake, if you must have a bitwise check x|y|z > 3 works, but it normally won't be reduced, so it's (probably*) always 2 bitwise ops and a compare, where the other way could be as fast as 1 compare.
(* This is where the language lawyers arrive an add comments why this edit is wrong and the bitwise version can be optimised;-)
There was a comment here (now deleted) along the lines of "new programmer shouldn't worry about this level of optimisation" - and it was 100% correct. Write easy to follow, working code and THEN try to squeeze performance out of it AFTER you know it is "too slow".
I was trying to solve Google Code Jam problems and there is one of them that I don't understand. Here is the question (World Finals 2013 - problem C): https://code.google.com/codejam/contest/2437491/dashboard#s=p2&a=2
And here follows the problem analysis: https://code.google.com/codejam/contest/2437491/dashboard#s=a&a=2
I don't understand why we can use binary search. In order to use binary search the elements have to be sorted. In order words: for a given element e, we can't have any element less than e at its right side. But that is not the case in this problem. Let me give you an example:
Suppose we do what the analysis tells us to do: we start with a left bound angle of 90° and a right bound angle of 0°. Our first search will be at angle of 45°. Suppose we find that, for this angle, X < N. In this case, the analysis tells us to make our left bound 45°. At this point, we can have discarded a viable solution (at, let's say, 75°) and at the same time there can be no more solutions between 0° and 45°, leading us to say that there's no solution (wrongly).
I don't think Google's solution is wrong =P. But I can't figure out why we can use a binary search in this case. Anyone knows?
I don't understand why we can use binary search. In order to use
binary search the elements have to be sorted. In order words: for a
given element e, we can't have any element less than e at its right
side. But that is not the case in this problem.
A binary search works in this case because:
the values vary by at most 1
we only need to find one solution, not all of them
the first and last value straddle the desired value (X .. N .. 2N-X)
I don't quite follow your counter-example, but here's an example of a binary search on a sequence with the above constraints. Looking for 3:
1 2 1 1 2 3 2 3 4 5 4 4 3 3 4 5 4 4
[ ]
[ ]
[ ]
[ ]
*
I have read the problem and in the meantime thought about the solution. When I read the solution I have seen that they have mostly done the same as I would have, however, I did not thought about some minor optimizations they were using, as I was still digesting the task.
Solution:
Step1: They choose a median so that each of the line splits the set into half, therefore there will be two provinces having x mines, while the other two provinces will have N - x mines, respectively, because the two lines each split the set into half and
2 * x + 2 * (2 * N - x) = 2 * x + 4 * N - 2 * x = 4 * N.
If x = N, then we were lucky and accidentally found a solution.
Step2: They are taking advantage of the "fact" that no three lines are collinear. I believe they are wrong, as the task did not tell us this is the case and they have taken advantage of this "fact", because they assumed that the task is solvable, however, in the task they were clearly asking us to tell them if the task is impossible with the current input. I believe this part is smelly. However, the task is not necessarily solvable, not to mention the fact that there might be a solution even for the case when three mines are collinear.
Thus, somewhere in between X had to be exactly equal to N!
Not true either, as they have stated in the task that
You should output IMPOSSIBLE instead if there is no good placement of
borders.
Step 3: They are still using the "fact" described as un-true in the previous step.
So let us close the book and think ourselves. Their solution is not bad, but they assume something which is not necessarily true. I believe them that all their inputs contained mines corresponding to their assumption, but this is not necessarily the case, as the task did not clearly state this and I can easily create a solvable input having three collinear mines.
Their idea for median choice is correct, so we must follow this procedure, the problem gets more complicated if we do not do this step. Now, we could search for a solution by modifying the angle until we find a solution or reach the border of the period (this was my idea initially). However, we know which provinces have too much mines and which provinces do not have enough mines. Also, we know that the period is pi/2 or, in other terms 90 degrees, because if we move alpha by pi/2 into either positive (counter-clockwise) or negative (clockwise) direction, then we have the same problem, but each child gets a different province, which is irrelevant from our point of view, they will still be rivals, I guess, but this does not concern us.
Now, we try and see what happens if we rotate the lines by pi/4. We will see that some mines might have changed borders. We have either not reached a solution yet, or have gone too far and poor provinces became rich and rich provinces became poor. In either case we know in which half the solution should be, so we rotate back/forward by pi/8. Then, with the same logic, by pi/16, until we have found a solution or there is no solution.
Back to the question, we cannot arrive into the situation described by you, because if there was a valid solution at 75 degrees, then we would see that we have not rotated the lines enough by rotating only 45 degrees, because then based on the number of mines which have changed borders we would be able to determine the right angle-interval. Remember, that we have two rich provinces and two poor provinces. Each rich provinces have two poor bordering provinces and vice-versa. So, the poor provinces should gain mines and the rich provinces should lose mines. If, when rotating by 45 degrees we see that the poor provinces did not get enough mines, then we will choose to rotate more until we see they have gained enough mines. If they have gained too many mines, then we change direction.
i am currenty studying BDD, but i have a different doubt, can you tell me if the following is right or not:
1) Feature = it means "the problem" isn't it?
2) Scenario = the way (the beahaviour) to resolve the feature
I find very difficult to find the "given when then" sentences.
In this problem for example:
As a student
I would like / i want to calculate the rectangle perimeters if i have 2 number
Or the circle area if i have one
So i don't make mistake with the computation
I wrote down the scenario, is that correct?
Given 1 number
Or 2 number
When i have 1 positive number
Or 2 positive number
Then calculate the area
Or the Perimeters
About the terminology:
1) feature is not a "problem". It would rather be a solution.
In software programming, a feature is a something that your program does to solve a problem.
A feature could be the ability to compute the area of a rectangle.
2) a scenario is a description of the usage of your feature. Like an example.
Like a test case, but usually in a more human-readable form.
3) a story (in Agile terminology, in which BDD stands) is a way to describe the a need/problem.
Your problem ("as a student...") is presented as a story.
This story will lead to a new feature in your soft.
This new feature will be tested by scenarios.
About your scenarios.
Yours are not correct.
There is no way to know that if you have 1 nb you should compute an area.
You should have several scenarios, like
Given I send the number 2
When I launch the computation
Then I get the result 12,56
Given I send the number 2 and 3
When I launch the computation
Then I get the result 10
Given I send the number -4
When I launch the computation
Then I get the result error
Given I send the number 1 3 7
When I launch the computation
Then I get the result error
I wish to use Senseval-2 Coarse Sense Dataset but there is description available for the same (about the format of the dataset).
It is supposed to have the decision data i.e. whether two senses should be merged or not. Is the middle value a confidence measure? Also, they used a prerelease of Wordnet 1.7. Can I use Wordnet 1.7 for the same?
A sample from the file looks like :
material%5:00:00:physical:00 3 material%5:00:00:worldly:00
material%3:00:03:: 3 material%5:00:00:worldly:00
material%3:00:04:: 2 material%3:00:01::
material%3:00:02::
post%5:00:00:succeeding(a):00
present%3:00:01::
present%3:00:02::
present%3:01:00::
stone%3:01:00::
stone%5:00:00:chromatic:00
air%1:15:00:: 4 air%1:27:00::
air%1:19:00:: 4 air%1:27:00::
air%1:27:01:: 4 air%1:27:00::
air%1:04:00::
air%1:10:02::
air%1:07:00::
air%1:10:01::
appeal%1:04:00:: 3 appeal%1:10:00::
appeal%1:10:02:: 3 appeal%1:10:00::
Through inspection, the middle number actually describes how many senses are in the same merged sense. For example:
matrial%5:00:00:physical:00 3 material%5:00:00:worldly:00
material%3:00:03:: 3 material%5:00:00:worldly:00
basically says that there are 3 senses which is considered the same as material%5:00:00:worldly:00, which are the two senses provided in the two lines, and the sense itself.
You can see also that there are no number for senses that do not get merged, such as air%1:04:00, and for the sense material%3:00:04:: 2 material$2:00:01:: you can see that there are two senses. So you can do the merging by mapping the senses in the first position into the sense in the second position.