I am testing some programs involving arithmetics in Clingo 5.0.0 and I don't understand why the below program is unsatisfiable:
#const v = 1.
a(object1).
a(object2).
b(object3).
value(object1,object2,object3) = "1.5".
value(X,Y,Z) > v, a(X), a(Y), b(Z), X!=Y :- go(X,Y,Z).
I expected an answer containing: a(object1) a(object2) b(object3) go(object1,object2,object3).
There is probably something I miss regarding arithmetic with Clingo.
I fear there are quite some misunderstandings about ASP here.
You can not assign values to predicates (value(a,b,c)=1.5). Predicates form atoms, that can be true or false (contained in an answer set or not).
I assume that your last rule shall derive the atom go(X,Y,Z). Rules do work the other way around, what is derived is on the left hand side.
There is no floating point arithmetic possible, you would have to scale your values up to integers.
Your problem might look like this, but this is just groping in the dark:
#const v = 1.
a(object1).
a(object2).
b(object3).
value(object1,object2,object3,2).
go(X,Y,Z) :- value(X,Y,Z,Value), Value > v, a(X), a(Y), b(Z), X!=Y.
The last rule states:
Derive go(object1,object2,object3) if value(object1,object2,object3,2) is true and 2 > 1 and a(object1) is true and a(object2) is true and b(object3) is true and object1 != object2.
I'm going through the question below.
The sequence [0, 1, ..., N] has been jumbled, and the only clue you have for its order is an array representing whether each number is larger or smaller than the last. Given this information, reconstruct an array that is consistent with it.
For example, given [None, +, +, -, +], you could return [1, 2, 3, 0, 4].
I went through the solution on this post but still unable to understand it as to why this solution works. I don't think I would be able to come up with the solution if I had this in front of me during an interview. Can anyone explain the intuition behind it? Thanks in advance!
This answer tries to give a general strategy to find an algorithm to tackle this type of problems. It is not trying to prove why the given solution is correct, but lying out a route towards such a solution.
A tried and tested way to tackle this kind of problem (actually a wide range of problems), is to start with small examples and work your way up. This works for puzzles, but even so for problems encountered in reality.
First, note that the question is formulated deliberately to not point you in the right direction too easily. It makes you think there is some magic involved. How can you reconstruct a list of N numbers given only the list of plusses and minuses?
Well, you can't. For 10 numbers, there are 10! = 3628800 possible permutations. And there are only 2⁹ = 512 possible lists of signs. It's a very huge difference. Most original lists will be completely different after reconstruction.
Here's an overview of how to approach the problem:
Start with very simple examples
Try to work your way up, adding a bit of complexity
If you see something that seems a dead end, try increasing complexity in another way; don't spend too much time with situations where you don't see progress
While exploring alternatives, revisit old dead ends, as you might have gained new insights
Try whether recursion could work:
given a solution for N, can we easily construct a solution for N+1?
or even better: given a solution for N, can we easily construct a solution for 2N?
Given a recursive solution, can it be converted to an iterative solution?
Does the algorithm do some repetitive work that can be postponed to the end?
....
So, let's start simple (writing 0 for the None at the start):
very short lists are easy to guess:
'0++' → 0 1 2 → clearly only one solution
'0--' → 2 1 0 → only one solution
'0-+' → 1 0 2 or 2 0 1 → hey, there is no unique outcome, though the question only asks for one of the possible outcomes
lists with only plusses:
'0++++++' → 0 1 2 3 4 5 6 → only possibility
lists with only minuses:
'0-------'→ 7 6 5 4 3 2 1 0 → only possibility
lists with one minus, the rest plusses:
'0-++++' → 1 0 2 3 4 5 or 5 0 1 2 3 4 or ...
'0+-+++' → 0 2 1 3 4 5 or 5 0 1 2 3 4 or ...
→ no very obvious pattern seem to emerge
maybe some recursion could help?
given a solution for N, appending one sign more?
appending a plus is easy: just repeat the solution and append the largest plus 1
appending a minus, after some thought: increase all the numbers by 1 and append a zero
→ hey, we have a working solution, but maybe not the most efficient one
the algorithm just appends to an existing list, no need to really write it recursively (although the idea is expressed recursively)
appending a plus can be improved, by storing the largest number in a variable so it doesn't need to be searched at every step; no further improvements seem necessary
appending a minus is more troublesome: the list needs to be traversed with each append
what if instead of appending a zero, we append -1, and do the adding at the end?
this clearly works when there is only one minus
when two minus signs are encountered, the first time append -1, the second time -2
→ hey, this works for any number of minuses encountered, just store its counter in a variable and sum with it at the end of the algorithm
This is in bird's eye view one possible route towards coming up with a solution. Many routes lead to Rome. Introducing negative numbers might seem tricky, but it is a logical conclusion after contemplating the recursive algorithm for a while.
It works because all changes are sequential, either adding one or subtracting one, starting both the increasing and the decreasing sequences from the same place. That guarantees we have a sequential list overall. For example, given the arbitrary
[None, +, -, +, +, -]
turned vertically for convenience, we can see
None 0
+ 1
- -1
+ 2
+ 3
- -2
Now just shift them up by two (to account for -2):
2 3 1 4 5 0
+ - + + -
Let's look at first to a solution which (I think) is easier to understand, formalize and demonstrate for correctness (but I will only explain it and not demonstrate in a formal way):
We name A[0..N] our input array (where A[k] is None if k = 0 and is + or - otherwise) and B[0..N] our output array (where B[k] is in the range [0, N] and all values are unique)
At first we see that our problem (find B such that B[k] > B[k-1] if A[k] == + and B[k] < B[k-1] if A[k] == -) is only a special case of another problem:
Find B such that B[k] == max(B[0..k]) if A[k] == + and B[k] == min(B[0..k]) if A[k] == -.
Which generalize from "A value must larger or smaller than the last" to "A value must be larger or smaller than everyone before it"
So a solution to this problem is a solution to the original one as well.
Now how do we approach this problem?
A greedy solution will be sufficient, indeed is easy to demonstrate that the value associated with the last + will be the biggest number in absolute (which is N), the one associated with the second last + will be the second biggest number in absolute (which is N-1) ecc...
And in the same time the value associated with the last - will be the smallest number in absolute (which is 0), the one associated with the second last - will be the second smallest (which is 1) ecc...
So we can start filling B from right to left remembering how many + we have seen (let's call this value X), how many - we have seen (let's call this value Y) and looking at what is the current symbol, if it is a + in B we put N-X and we increase X by 1 and if it is a - in B we put 0+Y and we increase Y by 1.
In the end we'll need to fill B[0] with the only remaining value which is equal to Y+1 and to N-X-1.
An interesting property of this solution is that if we look to only the values associated with a - they will be all the values from 0 to Y (where in this case Y is the total number of -) sorted in reverse order; if we look to only the values associated with a + they will be all the values from N-X to N (where in this case X is the total number of +) sorted and if we look at B[0] it will always be Y+1 and N-X-1 (which are equal).
So the - will have all the values strictly smaller than B[0] and reverse sorted and the + will have all the values strictly bigger than B[0] and sorted.
This property is the key to understand why the solution proposed here works:
It consider B[0] equals to 0 and than it fills B following the property, this isn't a solution because the values are not in the range [0, N], but it is possible with a simple translation to move the range and arriving to [0, N]
The idea is to produce a permutation of [0,1...N] which will follow the pattern of [+,-...]. There are many permutations which will be applicable, it isn't a single one. For instance, look the the example provided:
[None, +, +, -, +], you could return [1, 2, 3, 0, 4].
But you also could have returned other solutions, just as valid: [2,3,4,0,1], [0,3,4,1,2] are also solutions. The only concern is that you need to have the first number having at least two numbers above it for positions [1],[2], and leave one number in the end which is lower then the one before and after it.
So the question isn't finding the one and only pattern which is scrambled, but to produce any permutation which will work with these rules.
This algorithm answers two questions for the next member of the list: get a number who’s both higher/lower from previous - and get a number who hasn’t been used yet. It takes a starting point number and essentially create two lists: an ascending list for the ‘+’ and a descending list for the ‘-‘. This way we guarantee that the next member is higher/lower than the previous one (because it’s in fact higher/lower than all previous members, a stricter condition than the one required) and for the same reason we know this number wasn’t used before.
So the intuition of the referenced algorithm is to start with a referenced number and work your way through. Let's assume we start from 0. The first place we put 0+1, which is 1. we keep 0 as our lowest, 1 as the highest.
l[0] h[1] list[1]
the next symbol is '+' so we take the highest number and raise it by one to 2, and update both the list with a new member and the highest number.
l[0] h[2] list [1,2]
The next symbol is '+' again, and so:
l[0] h[3] list [1,2,3]
The next symbol is '-' and so we have to put in our 0. Note that if the next symbol will be - we will have to stop, since we have no lower to produce.
l[0] h[3] list [1,2,3,0]
Luckily for us, we've chosen well and the last symbol is '+', so we can put our 4 and call is a day.
l[0] h[4] list [1,2,3,0,4]
This is not necessarily the smartest solution, as it can never know if the original number will solve the sequence, and always progresses by 1. That means that for some patterns [+,-...] it will not be able to find a solution. But for the pattern provided it works well with 0 as the initial starting point. If we chose the number 1 is would also work and produce [2,3,4,0,1], but for 2 and above it will fail. It will never produce the solution [0,3,4,1,2].
I hope this helps understanding the approach.
This is not an explanation for the question put forward by OP.
Just want to share a possible approach.
Given: N = 7
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Go from 0 to N
[1] fill all '-' starting from right going left.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 2 1 0
[2] fill all the vacant places i.e [X & +] starting from left going right.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 3 4 5 6 7
Final:
Pattern: X + - + - + - + //X = None
Answer: 3 4 2 5 1 6 0 7
My answer definitely is too late for your problem but if you need a simple proof, you probably would like to read it:
+min_last or min_so_far is a decreasing value starting from 0.
+max_last or max_so_far is an increasing value starting from 0.
In the input, each value is either "+" or "-" and for each increase the value of max_so_far or decrease the value of min_so_far by one respectively, excluding the first one which is None. So, abs(min_so_far, max_so_far) is exactly equal to N, right? But because you need the range [0, n] but max_so_far and min_so_far now are equal to the number of "+"s and "-"s with the intersection part with the range [0, n] being [0, max_so_far], what you need to do is to pad it the value equal to min_so_far for the final solution (because min_so_far <= 0 so you need to take each value of the current answer to subtract by min_so_far or add by abs(min_so_far)).
In OWL ontology, let's have:
P Domain A
P Range B
A subClassOf P max 1 Thing
Asking a DL query
(1) P max 1 Thing
will return A; OK
Asking
(2) P exactly 1 Thing or P exactly 0 Thing
will return A as well.
However; asking
(3) P exactly 1 Thing
will return Nothing. And asking
(4) P exactly 0 Thing
will return Nothing as well.
I thought that the union of (3) + (4) results is equivalent to the result of (2). Unfortunately, it's not! Why?
Because OWL semantics is not extensional. The "or" is not a set union. Based on your axioms, there is just no named class that is a subclass of (3) or (4).
In particular, when you ask DL Queries about classes, you ask queries about axioms that are entailed by your theory/ontology. They must be true in all possible interpretations of your theory. This includes (at least) one where all A's stand in P to exactly one other thing, one where all A's stand in P to exactly zero other things, and one in which there are no instances of A. DLQuerys will only return things that are true in all interpretations, and in some interpretations, the instances of A do not satisfy either (3) or (4).
Find the number of functional dependencies in a relation having n attributes?
On first thought , I figure out that left hand side can have N+1 possibilities(null can also be there.) and similarly N+1 possibility for right hand side.
Hence total no of FD should be
(n+1)*(n+1) - 1
But ans given is 2^(n+1) .
On analyzing the answer , We can see they are not including trivital ones like ABC -> A etc .
So what should be the correct ans ???
2^n possible attribute sets on the LHS, and again 2^n possible attribute sets on the RHS. Both counts include the empty set.
Number of possible distinct pairs between those is 2^n * 2^n.
While technically correct, this answer also implies that FDs such as {AB} -> {} are also considered. How many of those are there ? For each cardinality l of the LHS, there are 2^l possible subsets, each of those giving a trivial FD if it appears on the RHS. So the number of trivial FDs is 2^0 + 2^1 + 2^2 + ... + 2^n = 2^(n+1) - 1. Leaving in total 2^(2*n) + 1 - 2^(n+1).
But now we have excluded only {AB} -> {A} and the like, and not {AB} -> {AC}. If we want the RHS to mention only attributes that are not mentioned on the LHS, then for each cardinality l of the LHS, there are 2^(n-l)-1 possible subsets on the RHS (that extra minus one is needed because the empty set must be excluded). Summing up to 2^0 - 1 + 2^1 - 1 + 2^2 - 1 + ... + 2^n - 1 = 2^(n+1) - 1 - (n + 1).
Still different from the answer given. And at any rate, the question was formulated hopelessly poorly. The question DID NOT STATE that trivial FDs were to be excluded. The question DID NOT STATE that "partially trivial" FDs were to be excluded.
BTW let's take the answers to the test. Choose a relation of degree 1, {A}. There are four possible FDs :
{} -> {} trivial, RHS subset of LHS
{} -> {A}
{A} -> {} trivial, RHS subset of LHS
{A} -> {A} trivial, RHS subset of LHS.
The correct answer, if trivial FDs are to be excluded, is "one". Your textbook says it's "four".
I think 2^n *2 is the answer because it includes all basic minimal FD’s and the rest can be derived from axioms or are trivial. A -> BC will have A -> B, A -> C, etc. ABC -> A, etc. are trivial.
Although it’s not clear from question whether to include all trivial cases. In that case the answer will be 2^n * 2^n. Because we can have 2^n choices for both LHS and RHS .
Pretty simple, really. I want to negate an integer which is represented in 2's complement, and to do so, I need to first flip all the bits in the byte. I know this is simple with XOR--just use XOR with a bitmask 11111111. But what about without XOR? (i.e. just AND and OR). Oh, and in this crappy assembly language I'm using, NOT doesn't exist. So no dice there, either.
You can't build a NOT gate out of AND and OR gates.
As I was asked to explain, here it is nicely formatted. Let's say you have any number of AND and OR gates. Your inputs are A, 0 and 1. You have six possibilities as you can make three pairs out of three signals (pick one that's left out) and two gates. Now:
Operation Result
A AND A A
A AND 1 A
A AND 0 0
A OR A A
A OR 1 1
A OR 0 A
So after you fed any of your signals into the first gate, your new set of signals is still just A, 0 and 1. Therefore any combination of these gates and signals will only get you A, 0 and 1. If your final output is A, then this means that for both values of A it won't equal !A, if your final output is 0 then A = 0 is such a value that your final value is not !A same for 1.
Edit: that monotony comment is also correct! Let me repeat here: if you change any of the inputs of AND / OR from 0 to 1 then the output won't decrease. Therefore if you claim to build a NOT gate then I will change your input from 0 to 1 , your output also can't decrease but it should -- that's a contradiction.
Does (foo & ~bar) | (~foo & bar) do the trick?
Edit: Oh, NOT doesn't exist. Didn't see that part!