In situation, there are two positive numbers a,b,c,d and it is satisfied with a+b>c+d.(this is fact)
When we only know the values ab, cd (we don't know a,b,c,d respectively),
how can we prove this fact a+b>c+d?
for example, a=2,b=8,c=3,d=6 => a+b(10)>c+d(9).
but we only know ab(16), cd(18). how can we know the fact a+b>c+d?
thank you
Related
I'm attempting to find the amount of inversions within a list. Inversions would be defined as any pair a,b from a list, where ai is the index of a and bi is the index of b that satisfies a > b and ai < bi. Essentially a comes before b but yet is larger than b.
The first thing I did was write a predicate to find out what the index is.
indexOf(Index, Element, List) :-
nth1(Index, List, Element).
Then I wrote a predicate to determine if any set of two numbers is an inversion
isInversion(A, B, List) :-
A \= B, indexOf(AI, A, List), indexOf(BI, B, List), A > B, AI < BI.
At this point I have a lot of questions, especially as I'm very unfamiliar with logic programming languages. My first question is, indexOf won't actually give me the index will it? I'm confused how that would actually work as it seems like it'd essentially have to try every number, which I'm not explicitly telling it to do.
If somehow indexOf will automatically determine the index and store it in AI/BI like I'm expecting, then I believe my isInversion predicate will evaluate correctly, if I'm wrong please let me know.
My main concern is how to actually determine the amount of inversions. In something like python I would do
count = 0
for a in puzzle
for b in puzzle
if a is b continue
if isInversion(a, b, puzzle)
count = count + 1
That would give me my amount of inversions. But how can I do this in prolog? For loops don't seem very stylistic so I don't want to use that.
Something to note, I have searched for other questions. It's a little tough since I obviously don't know exactly what I'm trying to look for. However I just wanted to make it clear that I felt things such as Prolog do predicate for all pairs in List? didn't help me answer the question.
You should remove the constraint A\=B as it will fail with unbound variables.
Then use aggregate_all/3 to count all the inversions (you don't actually need the values A/B of the inversion):
isInversion(A, B, List):-
indexOf(AI, A, List),
indexOf(BI, B, List),
A > B,
AI < BI.
countInversions(List, N):-
aggregate_all(count, isInversion(_, _, List), N).
Sample run:
?- countInversions([4,3,2,1,9], N).
N = 6.
You may see which inversions exists using findall/3 on isInversion:
?- findall(A-B, isInversion(A,B,[4,3,2,1,9]), LInversions).
LInversions = [4-3, 4-2, 4-1, 3-2, 3-1, 2-1].
The goal is to select shapes that don't touch each other using constraints (clpfd). Calling start(Pairs,4) would return Pairs = [1,3,5,7].
One problem I noticed is that if I print Final before labeling, it prints [1,3,5,7]. Which means labeling isn't doing anything.
What could I change/add to this code in order to fix that and also remove possible backtracking?
:-use_module(library(clpfd)).
:-use_module(library(lists)).
% init initialises Pairs and Max
% Pairs - The elements inside the Nth list in Pairs,
% represent the index of the shapes that shape N can touch
init([[3,5,6,7],[4,5,7],[1,4,5,7],[2,3,7],[1,2,3,7],[1],[1,2,3,4,5]],7).
start(Final, N):-
init(Pairs, Max),
length(Final, N),
domain(Final, 1, Max),
ascending(Final),
all_different(Final),
rules(Pairs,Final),
labeling([],Final).
rules(_,[]).
rules(Pairs,[H|T]):-
nth1(H,Pairs,PairH),
secondrule(PairH,T),
rules(Pairs,T).
secondrule(_, []).
secondrule(PairH, [H|T]):-
element(_,PairH,H),
secondrule(PairH, T).
ascending([_|[]]).
ascending([H|[T1|T2]]):-
H #< T1,
ascending([T1|T2]).
This is an Independent Set problem, which is an NP-hard problem. Therefore, it is unlikely that anybody will ever find a way to do it without search (backtracking) for general instances.
Regarding your code, labeling/2 does nothing, because your rules/2 is in fact a search procedure that returns the solution it it can find it. all_different/1 is useless too, because it is implied by ascending/1.
Presumably, your goal is a program that sets up constraints (without any search) and then searches for a solution with labeling/2. For that, you need to rethink your constraint model. Read up a bit on independent sets.
I have a question regarding Knights and Knaves and logical proposition. If I want to solve the puzzle and I assume I have two kinds of citizens: Knights, who always tell the truth, and knaves, who always tell lies. On the basis of utterances from some citizens, I must decide what kind they are.
There are three kinds of citizens: a, b and c, who are talking about themselves:
a says: ”All of us are knaves.”
b says: ”Exactly one of us is a knight.”
To solve the puzzle I should determine: What kinds of citizens are a, b and c? I should solve the puzzle by modelling the two utterances above using propositional logic, and I assume that I can use p to describe a knight and ¬p to describe a knave. How would I go about doing that? Any hint for someone who hasn't done any noticeable discrete mathematics in college?
A and C are Knaves. B is a Knight.
If A is a Knight, "All of us are knaves" is true. So, A would also be a Knave. This is a contradiction. Hence, A is a Knave.
If B is a Knave, then "Exactly one of us is a knight." is false. Meaning that 2 or more are Knights. But neither A nor B is a Knight. How can possibly be 2 or more Knights (since C is the only one with a possibility of being a Knight). This is also a contradiction. So, B is a Knight.
We just showed that B is a Knight. So, he himself is the only Knight he is talking about. So, C is a Knave.
Now, I don't think you can model this argument in the Propositional Logic. For one, notice the universal and existential quantifiers ("All" and "Exactly One") in the statements "All of us are knaves" and "Exactly one of us is a knight.". For another one, notice that A and B are talking about themselves. Modeling situations like this is one of the hardest problems in the history (not kidding!). Look at the following links for more info:
https://en.wikipedia.org/wiki/Liar_paradox
https://en.wikipedia.org/wiki/Self-reference
you can create a Truth table,
by first look at it i can say A must be knave, and B is a knight. because if A is a knight he can't say he's a knave(lie), also he can't be right about that all are knaves (can't say the truth) so B is a knight(if B Knave he can't say the truth that makes A a liar and he must be one) and then C is a Knave.
I'm currently stuck on a part of an application I am working on. I don't want to copy all the code and paste it here but let me go directly straight to the point with a simple example:
Suppose I have a string "abcdefg", I am trying to find an algorithm that would get all the possible grouping without exchanging the characters, for example:
abcdeg
a, b, c, d, e, f, g
ab, c, d, e, f, g
..
..
abc, def, g
..
ab, cd, efg
..
and so on...
I think the example is pretty much the point. Can anyone provide me a pseudo-code? I understand Java, C, and C++ as well, so maybe a code snippet on those language is better, but if not pseudo-code is fine and I'll try to implement it. Thanks in advance.
It's surprisingly simple. Lop off the first letter and associate a 0 or a 1 with the remaining letters. A 1 means place a comma just before the letter. A 0 means don't.
E.g. 001100 corresponds to abc,d,efg.
The notation I'm using is a simple map of a number increasing from zero expressed in binary.
So three things, (i) count integers, (ii) convert to binary, (iii) use that binary as the comma positioning rule.
The stopping condition is obvious.
Can anyone explain the algorithm for 2-satisfiability problem or provide me the links for the same? I could not find good links to understand it.
If you have n variables and m clauses:
Create a graph with 2n vertices: intuitively, each vertex resembles a true or not true literal for each variable. For each clause (a v b), where a and b are literals, create an edge from !a to b and from !b to a. These edges mean that if a is not true, then b must be true and vica-versa.
Once this digraph is created, go through the graph and see if there is a cycle that contains both a and !a for some variable a. If there is, then the 2SAT is not satisfiable (because a implies !a and vica-versa). Otherwise, it is satisfiable, and this can even give you a satisfying assumption (pick some literal a so that a doesn't imply !a, force all implications from there, repeat). You can do this part with any of your standard graph algorithms, ala Breadth-First Search , Floyd-Warshall, or any algorithm like these, depending on how sensitive you are to the time complexity of your algorithm.
You can solve it with greedy approach.
Or using Graph theory, here is link which explains the solution using graph theory.
http://www.cs.tau.ac.il/~safra/Complexity/2SAT.ppt
Here is the Wikipedia page on the subject, which describes a polynomial time algorithm. (The brute force algorithm of just trying all the different truth assignments is exponential time.) Maybe a bit of further explanation will help.
The expression "if P then Q" is only false when P is true and Q is false. So the expression has the same truth table values as "Q or not P". It is also equivalent to its contrapositive, "if not Q then not P", and that in turn is equivalent to "not P or Q" (the same as the other one).
So the algorithm involves replacing every expression of the form "A or B", with the two expressions, "if not A then B" and "if not B then A". (Putting it another way, A and B can't both be false.)
Next, construct a graph representing these implications. Create nodes for each "A" and "not A", and add links for each of the implications obtained above.
The last step is to make sure that none of the variables is equivalent to its own negation. That is, for each variable A (or not A), follow the links to discover all the nodes that can be reached from it, taking care to detect loops.
If one of the variables, A, can reach "not A", and "not A" can also reach A, then the original expression is not satisfiable. (It is a paradox.) If none of the variables do this, then it is satisfiable.
(It's okay if A implies "not A", but not the other way around. That just means that A must be negated to satisfy the expression.)
2 satisfiabilty:
if x & !x is strongly connected
then from !x we can reach to x
from x we can reach to !x
so in our operation,
in case of x,
we have 2 options only,
1.taking x (x) that leads to !x
2.rejecting x (!x) that leads to x
and both the choices are leading to the paradox of taking and rejecting a choice at the same time
so the satisfiability is impossible :D