I read the article on wikipedia but could not understand what exactly are NP problems. Can anyone tell me about them and also what is relation of them with P Problems?
NP problems are problems that given a proposed solution, you can verify the solution in a polynomial time. For example, if you have a list of University courses and need to create a schedule so that courses won't conflict, it would be a really difficult task (complexity-wise). However, given a proposed schedule, you can easily verify its correctness.
Another important example from the field of encryption: given a number which is the result of multiplying two very large prime numbers, it's very difficult to find those primes based only on the result. However, given two numbers, it's very easy to check the solution (multiply them, compare).
I have intentionally chose examples that are in NP and not in P (i.e. problem that are hard to find the solution for) so you can understand the difference. All problems that are easy to solve, are also easy to verify - just solve and compare. That is, P is a subset of NP.
Not really an answer, because Piccolo's link is more useful, but a HP researcher claims having proven P != NP, here is the paper.
www.hpl.hp.com/personal/Vinay_Deolalikar/Papers/pnp12pt.pdf
It was not accepted yet, but I wish him good luck for the 1M$.
Related
I'm having a bit of trouble understanding how one would verify if there is no solution to a given instance of the subset sum problem in polynomial time.
Of course you could easily verify the positive case: simply provide the list of integers which add up to the target sum and check they are all in the original set. (O(N))
How do you verify that the answer "false" is the correct one in polynomial time?
It’s actually not known how to do this - and indeed it’s conjectured that it’s not possible to do so!
The class NP consists of problems where “yes” instances can be verified in polynomial time. The subset sum problem is a canonical example of a problem in NP. Importantly, notice that the definition of NP says nothing about what happens if the answer is “no” - maybe it’ll be easy to show this, or maybe there isn’t an efficient algorithm for doing so.
A counterpart to NP is the class co-NP, which consists of problems where “no” instances can be verified in polynomial time. A canonical example of such a problem is the tautology problem - given a propositional logic formula, is it always true regardless of what values the variables are given? If the formula isn’t a tautology, it’s easy to verify this by having someone tell you how to assign the values to the variables such that the formula is false. But if the formula is always true, it’s unclear how you’d show this efficiently.
Just as the P = NP problem hasn’t been solved, the NP = co-NP problem is also open. We don’t know whether problems where “yes” answers have fast verification are the same as problems where “no” answers have fast verification.
What we do know is that if any NP-complete problem is in co-NP, then NP = co-NP. And since subset sum is NP-complete, there’s no known polynomial time algorithm to verify if the answer to a subset sum instance is “no.”
Both NN and Greedy Search algorithms have a Greed nature, and both have tendency towards the lowest cost/distance (my understanding may be incorrect though). But what makes them different in a way that each one can be classified into a distinct algorithm group is somehow unclear to me.
For instance, if I can solve a particular problem using NN, I can surely solve it with Greedy Search algorithm as well specially if minimization is the case. I came to this conclusion because when I start coding them I come across very similar implementations in code although the general concept behind both might be different. Sometimes I can't even tell if the implementation follows NN or Greedy Search.
I have done my homework well and searched enough on Google, but couldn't find a decent explanation on what distinguishes them from one another. Any such explanation is indeed appreciated.
Hmm, at a very high level they both driven by heuristics in order to evaluate a given solution against an ideal solution. But, whilst a greedy search algo outputs a solution for a given input, the NN trains a model that will generate solutions for given inputs. So at a very very high level, you can think that the NN generates a solution finder, whereas the greedy search is a harcoded solution finder.
In other words, the NN will generate "code" (i.e. the model (aka the weights)) that finds solutions to the problem when provided to the same network topology. The greedy search is you actually writing the code that finds the solution to the problem. This is quite wishy washy though, I'm sure there is a much more concise, academically sound way of expressing what I've just said
All of what I've just said in based on the assumption that by "Greedy search" you meant the algorithms to solve problems such as travelling sales man.
Another way to think of it is:
In greedy search, you write an algorithm that solves a search problem (find me the graph that best describes the relationship, based on provided heuristic(s), between data point A and data point B).
When you write a neural network, you declare a network topology, provide some initially "random" weights and some heuristics to measure output errors and then train the networks weights via a plethora of different methods (back prop, GAN etc). These weights can then be used as a solver for novel problems.
For what it's worth, I don't think an NN would be a good approach to generate a solver for travelling sales man problem. You would be far better off just using a common graph search algorithm..
Currently learning about the A* search algorithm and using it to find the quickest solution to the N-Puzzle. For some random seed of the initial starting state, the puzzle may be unsolvable which would result in extremely long wait times until the algorithm has search the entire search-space and determined there is not solution to the give start state.
I was wondering if there is a method of precalculating whether the A* algorithm will fail to avoid such a scenario. I've read a bit about how it is possible but can't find a direct answer as to a method in which to do it.
Any guidance or options are appreciated.
I think A* does not offer you a mechanism to know whether or not a problem is solvable. Specifically for N-Puzzle, I think this could help you to check if it can be solved or not:
http://www.geeksforgeeks.org/check-instance-8-puzzle-solvable/
It seems that if you are in a state where you have an odd amount inversion, you know for sure the problem for that permutation is infeasible.
For the N-puzzle specifically, there are only two possible parities, so you just need to check which parity the current puzzle is.
There is an in-depth explanation on how to do this on the math stackexchange
For general A* problems, no, there is no way to pre-compute if the graph is solvable.
Let's start here:
It is said that all NP problems can be reduced to SAT(boolean satisfiability problem). To be more accurate to Circuit SAT, because all decision problems like NP should end up with answer Yes or No.
But now, if I have a random NP problem, how to build a boolean circuit to test, how to group my input, what kind of gates(AND, NOT, OR etc..) should connect those inputs. So basically, my question how to design boolean Circuit which gives an answer TRUE or FALSE.
Last thing, what that answer means. I understand TRUE as this NP problem can be solved in polynomial time and FALSE cannot, am I correct?
It is huge mess in my mind, don't be really outrageous if I made logical mistakes explaining my question :) I hope you understood it.
Excitingly waiting for answers.
I understand the confusion but your understanding is not quite how it works.
NP-hardness is a qualification of decision problems, that is, a problem with answer is yes or no. If we want to show that a decision problem is NP-hard, we do so by showing that it is as least as hard as a problem of which we know it is NP-hard already, for instance SAT.
How can we show that problem A is at least as hard as problem B? Well, we can phrase that as
if we can solve A, we can also solve B
So, given an instance of problem B, we convert it to an instance of problem A, use our solution to A to solve it, and convert it back to a solution to B. Assuming that both conversations are easy, we know that A cannot be easier than B, since a solution to A is also a solution to B.
Your understanding thus had it backwards. In order to show that some problem is NP-hard, we to show that it is at least as hard as SAT, that is, given an arbitrary instance of SAT, convert it to an instance of your problem, and then solve that problem. If the answer is "yes", then the original SAT problem was satisfiable, otherwise it wasn't.
Now, as I wrote in a comment, there is no standard way to do the conversion. You somehow need to manipulate your problem, such that it "looks like SAT", in order to make the conversion. For some problems that's easier then others, but I'd claim it's the hardest part of the NP-hardness proof.
What people typically do instead is that they look for another problem, which is known to be NP-hard already, but looks a bit more like their own problem. That way, the reduction becomes a bit easier. But still it requires a lot of work and creativity. I recommend you look at some existing proofs to see how others do this.
Ok, so here's the problem:
I need to find any number of intem groups from 50-100 item set that add up to 1000, 2000, ..., 10000.
Input: list of integers
Integer can be on one list only.
Any ideas on algorithm?
Googling for "Knapsack problem" should get you quite a few hits (though they're not likely to be very encouraging -- this is quite a well known NP-complete problem).
Edit: if you want to get technical, what you're describing seems to really be the subset sum problem -- which is a special case of the knapsack problem. Of course, that's assuming I'm understanding your description correctly, which I'll admit may be open to some question.
You might find Algorithm 3.94 in The Handbook of Applied Cryptography helpful.
I'm not 100% on what you are asking, but I've used backtracking searches for something like this before. This is a brute force algorithm that is the slowest possible solution, but it will work. The wiki article on Backtracking Search may help you. Basically, you can use a recursive algorithm to examine every possible combination.
This is the knapsack problem. Are there any constraints on the integers you can choose from? Are they divisible? Are they all less than some given value? There may be ways to solve the problem in polynomial time given such constraints - Google will provide you with answers.