Sorry for stupid question. I cannot jog my memory and googling did not help me answer this question.
So basically given a graph G(V,E), I know that O(|V|^2) or O(|E|^2 + |V|^2) is considered to be polynomial complexity, so is O(|E|*|V|) polynomial as well? If not, what kind of complexity is it? I believe it's not pseudo-polynomial either.
Another question is: is O(m*n) considered polynomial as well, given m and n are the sizes of two INDEPENDENT inputs to a problem? I just want to clarify the concept of polynomial time in here and want to know if O(m*n) has a different name for its type of complexity.
it is polynomial O(|V|^3) since the number of edges is bounded O(|V|^2)
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I found this question in my Optimization Algorithm course, the full question is this:
If we can prove all Knapsack problems with capacity limited to 100 can be solved in polynomial time, then all Knapsack problems belong to P. Is this sentence true or false? Justify.
With my book and some research I came out with something like this:
First of all KP is an NP-complete problem. With Dynamic programming it can reach a pseudopolynomial time, but it's not enough.
If, absurdly, we can prove that KP with capacity limited to 100 can be solved in polynomial time then we can assume that KP belongs to P.
What do you think about my answer? I think the absurd is not so right in the last sentence.
Proving that all knapsack problems with a limited capacity can be solved in polynomial time does not prove that all knapsack problems are in P. If a problem is in P, that means that it can be solved in polynomial time. This means that it can be solved in O(n^k) where k is some integer. Big O is an upper bound, meaning that, if an algorithm is O(n), as n approaches infinity, the time it takes to do the algorithms will never be longer than n. By proving that all problems with n<100 can be solved in polynomial time, this makes no guarantee for much larger n. Therefore we cannot say that there is an algorithm that runs in O(n^k) and is therefore in P.
I have a 0-1 second order cone (SOC) problem and I need to know the complexity of solving this problem if branch and cut (B&C) method is used?. The way I addressed this question is as following:
The 0-1 SOC problem can be solved using B&C method which has an exponential worst case complexity, i.e., O(2^n). At each node of B&C method, the relaxed problem is a SOC problem which can be solved using an interior point method which has a polynomial-time complexity. However, I do not have an expression for the complexity of the interior point method yet. Assuming this complexity is O(n). Then, I can claim that the complexity of solving the 0-1 problem using B&C method is O(2^n) times O(n).
Do not think so. You are solving n nodes each with complexity O(n). By my calculations it would come out to be O(2^n*n^2).
I would like to quote from Wikipedia
In mathematics, the minimum k-cut, is a combinatorial optimization
problem that requires finding a set of edges whose removal would
partition the graph to k connected components.
It is said to be the minimum cut if the set of edges is minimal.
For a k = 2, It would mean Finding the set of edges whose removal would Disconnect the graph into 2 connected components.
However, The same article of Wikipedia says that:
For a fixed k, the problem is polynomial time solvable in O(|V|^(k^2))
My question is Does this mean that minimum 2-cut is a problem that belongs to complexity class P?
The min-cut problem is solvable in polynomial time and thus yes it is true that it belongs to complexity class P. Another article related to this particular problem is the Max-flow min-cut theorem.
First of all, the time complexity an algorithm should be evaluated by expressing the number of steps the algorithm requires to finish as a function of the length of the input (see Time complexity). More or less formally, if you vary the length of the input, how would the number of steps required by the algorithm to finish vary?
Second of all, the time complexity of an algorithm is not exactly the same thing as to what complexity class does the problem the algorithm solves belong to. For one problem there can be multiple algorithms to solve it. The primality test problem (i.e. testing if a number is a prime or not) is in P, but some (most) of the algorithms used in practice are actually not polynomial.
Third of all, in the case of most algorithms you'll find on the Internet evaluating the time complexity is not done by definition (i.e. not as a function of the length of the input, at least not expressed directly as such). Lets take the good old naive primality test algorithm (the one in which you take n as input and you check for division by 2,3...n-1). How many steps does this algo take? One way to put it is O(n) steps. This is correct. So is this algorithm polynomial? Well, it is linear in n, so it is polynomial in n. But, if you take a look at what time complexity means, the algorithm is actually exponential. First, what is the length of the input to your problem? Well, if you provide the input n as an array of bits (the usual in practice) then the length of the input is, roughly said, L = log n. Your algorithm thus takes O(n)=O(2^log n)=O(2^L) steps, so exponential in L. So the naive primality test is in the same time linear in n, but exponential in the length of the input L. Both correct. Btw, the AKS primality test algorithm is polynomial in the size of input (thus, the primality test problem is in P).
Fourth of all, what is P in the first place? Well, it is a class of problems that contains all decision problems that can be solved in polynomial time. What is a decision problem? A problem that can be answered with yes or no. Check these two Wikipedia pages for more details: P (complexity) and decision problems.
Coming back to your question, the answer is no (but pretty close to yes :p). The minimum 2-cut problem is in P if formulated as a decision problem (your formulation requires an answer that is not just a yes-or-no). In the same time the algorithm that solves the problem in O(|V|^4) steps is a polynomial algorithm in the size of the input. Why? Well, the input to the problem is the graph (i.e. vertices, edges and weights), to keep it simple lets assume we use an adjacency/weights matrix (i.e. the length of the input is at least quadratic in |V|). So solving the problem in O(|V|^4) steps means polynomial in the size of the input. The algorithm that accomplishes this is a proof that the minimum 2-cut problem (if formulated as decision problem) is in P.
A class related to P is FP and your problem (as you formulated it) belongs to this class.
Is an approximation algorithm the same as a Polynomial Time Approximation Algorithm (PTAS)? E.g. It can be shown that A(I) <= 2 * OPT(I) for vertex cover. Does it mean that Vertex Cover has a 2-polynomial time approximation algorithm or a PTAS?
Thanks!
Note: The text in Italics is the edit I made after I posted my question.
No, this isn't necessarily the case. A PTAS is an algorithm where given any ε > 0, you can approximate the answer to a factor of (1 + ε) in polynomial time. In other words, you can get arbitrarily good approximations.
Some problems are known (for example, MAX-3SAT) that have approximation algorithms for specific factors (for example, 5/8), but where it's known that unless P = NP there is a hard limit to how well the problem can be approximated in polynomial time. For example, the PCP theorem says that MAX-3SAT doesn't have a polynomial-time 7/8 approximation unless P = NP. It's therefore possible that MAX-3SAT has a PTAS, but only if P = NP.
Hope this helps!
Vertex cover having 2-approximation algorithm is not same as having a PTAS algorithm. Sometimes, there are problems where much better approximation is possible. These problems then admit PTAS.
Such algorithms take an instance of the problem as input, with another input parameter epsilon>0. And it gives an output whose value is at most (1+epsilon).OPT for minimisation problem; and (1/(1+epsilon)).OPT for maximisation problem.
Run time of a PTAS algorithm is polynomial in n (size of problem instance). Sometimes, the runtime is also polynomial in epsilon, then its called to admit FPTAS(fully PTAS).
Example:
Dynamic programming algorithm for KNAPSACK with integer-profits gives optimal solution.
While, KNAPSACK problem with real-valued profits do not admit polynomial-time algorithm. But it admits a FPTAS, where real-value profits are converted into integer profits; and DP algorithm is used to calculate the solution with "rounded" profits.
Another example, Max Independent Set does not admit a PTAS or FPTAS. Because, in this case, we can set a value for epsilon, which will always give optimal solution for any graph using that PTAS algorithm; which is not possible until P=NP.
i have NP hard problem. Let imagine I have found some polynomial algorithm that find ONLY one of many existing solutions of that problem, but at least one solution (if present in the probem). Is that algorithm considered as solution of NP=P question (if that algorithm transformed to mathematical proof)?
Thanks for answers
NP is a class of decision problems. Your algorithm should answer "yes" or "no" correctly to all possible instances (questions).
For example, the problem: "given graph G and number k, does G contain a clique of size >= k" is NP-hard. If you have a polynomial time algorithm that answers "yes" or "no" correctly each time, then it is a valid proof of P=NP. The algorithm doesn't need to explicitly show the clique - only answer if it exists for all possible G and k.
If you find a NP-hard problem and you can detect some cases that you can solve in polynomial time (leaving others for exponential time), then only if the fraction of cases remaining is on the order of log(N)/N will you change the order of the entire problem, and even then only if you can restrict your exponential case to examining only log(N) not all N possibilities.
Also, if you find a NP-hard problem where you think you can solve every case in polynomial time, you have probably made a mistake, either in posing a NP-hard problem correctly, or in finding the more troublesome examples. Try a larger test set before believing yourself!