I've recently implemented 2sum and 3sum in leetcode and been wondering if it's possible to find if elements can sum up to a given target without bruteforce.
You're asking if the "subset sum problem" has a non-bruteforce solution. It's not really clear what is and what isn't a bruteforce solution, but NP complete programs (which subset sum is) have no known way to solve them in polynomial time in the worst case, but there are very sophisticated approaches to solving them that work efficiently some of the time.
The wikipedia page has good details about solving the subset sum (either approximately or exactly), and links for further reading.
At its most general, depending on your precise definition of "brute force", this is an open problem in computer science; nobody knows. There are some algorithms that are often fast in practice, but whether there's a fundamentally fast algorithm or not, that's an active area of research
Look up "subset sum problem" and "NP-complete"
So I'm suspecting this problem could be NP-complete or NP-hard at least in certain circumstances, but still, often for NP-complete problems there is a nice solution that runs much faster than naïve brute force.
My problem is, given an NxN matrix A, with non-negative integer entries, and an integer k > 1, how can we determine if A can be written as a product of k NxN matrices whose entries are all either 0 or 1?
Like I said, I think this problem may be
NP-complete even for k = 2 but I maybe wrong,
maybe it's polynomial time for k=2 or even for any fixed k or even for k not fixed.
Also it may help to bound the non-negative entries in the target matrix A, to come up with good running times.
I would just like to find out good algorithms that run asymptotically faster (hopefully much faster) than brute force over all choices of 0/1 matrices to multiply.
Also, I apologize if this question is better suited on CS.stackexchange, if so, please let me know and I'll migrate the question. However we do have an algorithm" tag here, and since I suspect this problem is NP-hard, at least in one of its variants about whether k=2 or k is fixed or unbounded k, it's of more interest to me to just get whatever good algorithms may be available, e.g. that may work in polynomial time for fixed k or run in pseudo-polynomial time for arbitrary k but not polynomial time in general.
This is a problem I am working on
I implemented a brute force solution and a divide and conquer(recursive) solution. They both work with this input(from post 1 to 4)
For the brute force solution, what I did was generate all subsets of {1,2,3,4}, iterate through all the subsets and check only the ones that contained 1 and 4. I know that my brute force solution would run in O(2n) time because there are mathematically 2n subsets and I have to iterate over all of them.
For my recursive solution. what I did was break this problem down so that the only solutions that are generated/will be checked will be ones that contain 1 and 4. For example, from 1 you can go to 2,3,4 and if you go to 2, you can go to 3,4, and so on... till 4 is reached.
After analyzing both algorithms, I realized the only difference is that the divide and conquer wouldn't even check the subsets that didn't have 1 and 4 but the brute force would, say {2,3}. How would this difference affect the time complexity of the recursive algorithm. Would the recursive algorithm also run in O(2n) or something less because it checks less subsets?
Extra step for optimisation in recursive is using memoization. Reasoning goes - once you get and stop at post x it is always same cost till end, no matter where are you coming from.
It turns out that all it is required is array holding at index i lowest price to get to end. This can be populated going right to left in O(n²)time.
There exists a straightforward dynamic programming solution.
Consider each successively longer optimum canoe journey as the function of some shorter optimum canoe journey. I.e., the cheapest way to travel to post 4 is the cheapest way to travel to post 3 plus the best decision for the interval 3 to 4. Then consider the cheapest way to reach post 5 as a function of the previously calculated cheapest way to visit post 4.
Implement this recursion in code and your solution should be O(n^2), similar to the perennial CLRS favorite: the rod-cutting problem: http://www.geeksforgeeks.org/dynamic-programming-set-13-cutting-a-rod/
The divide-and-conquer approach would check singletons at which a rental was initiated, and then "merge" steps would optimize across the new doubleton/4ton/etc. It'll be much better than O(2^n) because you won't be recalculating all possibilities for every subpath, but it's a greedy algorithm--it makes an assumption about the answer before it has perfect information about the answer.
If you assume that the optimum decision occurs during some merge step of the subsets, you will necessarily miss cases where renting a boat at 1 and returning it at n is the best solution, but the runtime is O(n log n). But, if you check the input and optimize for every merge, you're back at O(2^n).
There are a lot of real-world problems that turn out to be NP-hard. If we assume that P ≠ NP, there aren't any polynomial-time algorithms for these problems.
If you have to solve one of these problems, is there any hope that you'll be able to do so efficiently? Or are you just out of luck?
If a problem is NP-hard, under the assumption that P ≠ NP there is no algorithm that is
deterministic,
exactly correct on all inputs all the time, and
efficient on all possible inputs.
If you absolutely need all of the above guarantees, then you're pretty much out of luck. However, if you're willing to settle for a solution to the problem that relaxes some of these constraints, then there very well still might be hope! Here are a few options to consider.
Option One: Approximation Algorithms
If a problem is NP-hard and P ≠ NP, it means that there's is no algorithm that will always efficiently produce the exactly correct answer on all inputs. But what if you don't need the exact answer? What if you just need answers that are close to correct? In some cases, you may be able to combat NP-hardness by using an approximation algorithm.
For example, a canonical example of an NP-hard problem is the traveling salesman problem. In this problem, you're given as input a complete graph representing a transportation network. Each edge in the graph has an associated weight. The goal is to find a cycle that goes through every node in the graph exactly once and which has minimum total weight. In the case where the edge weights satisfy the triangle inequality (that is, the best route from point A to point B is always to follow the direct link from A to B), then you can get back a cycle whose cost is at most 3/2 optimal by using the Christofides algorithm.
As another example, the 0/1 knapsack problem is known to be NP-hard. In this problem, you're given a bag and a collection of objects with different weights and values. The goal is to pack the maximum value of objects into the bag without exceeding the bag's weight limit. Even though computing an exact answer requires exponential time in the worst case, it's possible to approximate the correct answer to an arbitrary degree of precision in polynomial time. (The algorithm that does this is called a fully polynomial-time approximation scheme or FPTAS).
Unfortunately, we do have some theoretical limits on the approximability of certain NP-hard problems. The Christofides algorithm mentioned earlier gives a 3/2 approximation to TSP where the edges obey the triangle inequality, but interestingly enough it's possible to show that if P ≠ NP, there is no polynomial-time approximation algorithm for TSP that can get within any constant factor of optimal. Usually, you need to do some research to learn more about which problems can be well-approximated and which ones can't, since many NP-hard problems can be approximated well and many can't. There doesn't seem to be a unified theme.
Option Two: Heuristics
In many NP-hard problems, standard approaches like greedy algortihms won't always produce the right answer, but often do reasonably well on "reasonable" inputs. In many cases, it's reasonable to attack NP-hard problems with heuristics. The exact definition of a heuristic varies from context to context, but typically a heuristic is either an approach to a problem that "often" gives back good answers at the cost of sometimes giving back wrong answers, or is a useful rule of thumb that helps speed up searches even if it might not always guide the search the right way.
As an example of the first type of heuristic, let's look at the graph-coloring problem. This NP-hard problem asks, given a graph, to find the minimum number of colors necessary to paint the nodes in the graph such that no edge's endpoints are the same color. This turns out to be a particularly tough problem to solve with many other approaches (the best known approximation algorithms have terrible bounds, and it's not suspected to have a parameterized efficient algorithm). However, there are many heuristics for graph coloring that do quite well in practice. Many greedy coloring heuristics exist for assigning colors to nodes in a reasonable order, and these heuristics often do quite well in practice. Unfortunately, sometimes these heuristics give terrible answers back, but provided that the graph isn't pathologically constructed the heuristics often work just fine.
As an example of the second type of heuristic, it's helpful to look at SAT solvers. SAT, the Boolean satisfiability problem, was the first problem proven to be NP-hard. The problem asks, given a propositional formula (often written in conjunctive normal form), to determine whether there is a way to assign values to the variables such that the overall formula evaluates to true. Modern SAT solvers are getting quite good at solving SAT in many cases by using heuristics to guide their search over possible variable assignments. One famous SAT-solving algorithm, DPLL, essentially tries all possible assignments to see if the formula is satisfiable, using heuristics to speed up the search. For example, if it finds that a variable is either always true or always false, DPLL will try assigning that variable its forced value before trying other variables. DPLL also finds unit clauses (clauses with just one literal) and sets those variables' values before trying other variables. The net effect of these heuristics is that DPLL ends up being very fast in practice, even though it's known to have exponential worst-case behavior.
Option Three: Pseudopolynomial-Time Algorithms
If P ≠ NP, then no NP-hard problem can be solved in polynomial time. However, in some cases, the definition of "polynomial time" doesn't necessarily match the standard intuition of polynomial time. Formally speaking, polynomial time means polynomial in the number of bits necessary to specify the input, which doesn't always sync up with what we consider the input to be.
As an example, consider the set partition problem. In this problem, you're given a set of numbers and need to determine whether there's a way to split the set into two smaller sets, each of which has the same sum. The naive solution to this problem runs in time O(2n) and works by just brute-force testing all subsets. With dynamic programming, though, it's possible to solve this problem in time O(nN), where n is the number of elements in the set and N is the maximum value in the set. Technically speaking, the runtime O(nN) is not polynomial time because the numeric value N is written out in only log2 N bits, but assuming that the numeric value of N isn't too large, this is a perfectly reasonable runtime.
This algorithm is called a pseudopolynomial-time algorithm because the runtime O(nN) "looks" like a polynomial, but technically speaking is exponential in the size of the input. Many NP-hard problems, especially ones involving numeric values, admit pseudopolynomial-time algorithms and are therefore easy to solve assuming that the numeric values aren't too large.
For more information on pseudopolynomial time, check out this earlier Stack Overflow question about pseudopolynomial time.
Option Four: Randomized Algorithms
If a problem is NP-hard and P ≠ NP, then there is no deterministic algorithm that can solve that problem in worst-case polynomial time. But what happens if we allow for algorithms that introduce randomness? If we're willing to settle for an algorithm that gives a good answer on expectation, then we can often get relatively good answers to NP-hard problems in not much time.
As an example, consider the maximum cut problem. In this problem, you're given an undirected graph and want to find a way to split the nodes in the graph into two nonempty groups A and B with the maximum number of edges running between the groups. This has some interesting applications in computational physics (unfortunately, I don't understand them at all, but you can peruse this paper for some details about this). This problem is known to be NP-hard, but there's a simple randomized approximation algorithm for it. If you just toss each node into one of the two groups completely at random, you end up with a cut that, on expectation, is within 50% of the optimal solution.
Returning to SAT, many modern SAT solvers use some degree of randomness to guide the search for a satisfying assignment. The WalkSAT and GSAT algorithms, for example, work by picking a random clause that isn't currently satisfied and trying to satisfy it by flipping some variable's truth value. This often guides the search toward a satisfying assignment, causing these algorithms to work well in practice.
It turns out there's a lot of open theoretical problems about the ability to solve NP-hard problems using randomized algorithms. If you're curious, check out the complexity class BPP and the open problem of its relation to NP.
Option Five: Parameterized Algorithms
Some NP-hard problems take in multiple different inputs. For example, the long path problem takes as input a graph and a length k, then asks whether there's a simple path of length k in the graph. The subset sum problem takes in as input a set of numbers and a target number k, then asks whether there's a subset of the numbers that dds up to exactly k.
Interestingly, in the case of the long path problem, there's an algorithm (the color-coding algorithm) whose runtime is O((n3 log n) · bk), where n is the number of nodes, k is the length of the requested path, and b is some constant. This runtime is exponential in k, but is only polynomial in n, the number of nodes. This means that if k is fixed and known in advance, the runtime of the algorithm as a function of the number of nodes is only O(n3 log n), which is quite a nice polynomial. Similarly, in the case of the subset sum problem, there's a dynamic programming algorithm whose runtime is O(nW), where n is the number of elements of the set and W is the maximum weight of those elements. If W is fixed in advance as some constant, then this algorithm will run in time O(n), meaning that it will be possible to exactly solve subset sum in linear time.
Both of these algorithms are examples of parameterized algorithms, algorithms for solving NP-hard problems that split the hardness of the problem into two pieces - a "hard" piece that depends on some input parameter to the problem, and an "easy" piece that scales gracefully with the size of the input. These algorithms can be useful for finding exact solutions to NP-hard problems when the parameter in question is small. The color-coding algorithm mentioned above, for example, has proven quite useful in practice in computational biology.
However, some problems are conjectured to not have any nice parameterized algorithms. Graph coloring, for example, is suspected to not have any efficient parameterized algorithms. In the cases where parameterized algorithms exist, they're often quite efficient, but you can't rely on them for all problems.
For more information on parameterized algorithms, check out this earlier Stack Overflow question.
Option Six: Fast Exponential-Time Algorithms
Exponential-time algorithms don't scale well - their runtimes approach the lifetime of the universe for inputs as small as 100 or 200 elements.
What if you need to solve an NP-hard problem, but you know the input is reasonably small - say, perhaps its size is somewhere between 50 and 70. Standard exponential-time algorithms are probably not going to be fast enough to solve these problems. What if you really do need an exact solution to the problem and the other approaches here won't cut it?
In some cases, there are "optimized" exponential-time algorithms for NP-hard problems. These are algorithms whose runtime is exponential, but not as bad an exponential as the naive solution. For example, a simple exponential-time algorithm for the 3-coloring problem (given a graph, determine if you can color the nodes one of three colors each so that no edge's endpoints are the same color) might work checking each possible way of coloring the nodes in the graph, testing if any of them are 3-colorings. There are 3n possible ways to do this, so in the worst case the runtime of this algorithm will be O(3n · poly(n)) for some small polynomial poly(n). However, using more clever tricks and techniques, it's possible to develop an algorithm for 3-colorability that runs in time O(1.3289n). This is still an exponential-time algorithm, but it's a much faster exponential-time algorithm. For example, 319 is about 109, so if a computer can do one billion operations per second, it can use our initial brute-force algorithm to (roughly speaking) solve 3-colorability in graphs with up to 19 nodes in one second. Using the O((1.3289n)-time exponential algorithm, we could solve instances of up to about 73 nodes in about a second. That's a huge improvement - we've grown the size we can handle in one second by more than a factor of three!
As another famous example, consider the traveling salesman problem. There's an obvious O(n! · poly(n))-time solution to TSP that works by enumerating all permutations of the nodes and testing the paths resulting from those permutations. However, by using a dynamic programming algorithm similar to that used by the color-coding algorithm, it's possible to improve the runtime to "only" O(n2 2n). Given that 13! is about one billion, the naive solution would let you solve TSP for 13-node graphs in roughly a second. For comparison, the DP solution lets you solve TSP on 28-node graphs in about one second.
These fast exponential-time algorithms are often useful for boosting the size of the inputs that can be exactly solved in practice. Of course, they still run in exponential time, so these approaches are typically not useful for solving very large problem instances.
Option Seven: Solve an Easy Special Case
Many problems that are NP-hard in general have restricted special cases that are known to be solvable efficiently. For example, while in general it’s NP-hard to determine whether a graph has a k-coloring, in the specific case of k = 2 this is equivalent to checking whether a graph is bipartite, which can be checked in linear time using a modified depth-first search. Boolean satisfiability is, generally speaking, NP-hard, but it can be solved in polynomial time if you have an input formula with at most two literals per clause, or where the formula is formed from clauses using XOR rather than inclusive-OR, etc. Finding the largest independent set in a graph is generally speaking NP-hard, but if the graph is bipartite this can be done efficiently due to König’s theorem.
As a result, if you find yourself needing to solve what might initially appear to be an NP-hard problem, first check whether the inputs you actually need to solve that problem on have some additional restricted structure. If so, you might be able to find an algorithm that applies to your special case and runs much faster than a solver for the problem in its full generality.
Conclusion
If you need to solve an NP-hard problem, don't despair! There are lots of great options available that might make your intractable problem a lot more approachable. No one of the above techniques works in all cases, but by using some combination of these approaches, it's usually possible to make progress even when confronted with NP-hardness.
I'm given a task to write an algorithm to compute the maximum two dimensional subset, of a matrix of integers. - However I'm not interested in help for such an algorithm, I'm more interested in knowing the complexity for the best worse-case that can possibly solve this.
Our current algorithm is like O(n^3).
I've been considering, something alike divide and conquer, by splitting the matrix into a number of sub-matrices, simply by adding up the elements within the matrices; and thereby limiting the number of matrices one have to consider in order to find an approximate solution.
Worst case (exhaustive search) is definitely no worse than O(n^3). There are several descriptions of this on the web.
Best case can be far better: O(1). If all of the elements are non-negative, then the answer is the matrix itself. If the elements are non-positive, the answer is the element that has its value closest to zero.
Likewise if there are entire rows/columns on the edges of your matrix that are nothing but non-positive integers, you can chop these off in your search.
I've figured that there isn't a better way to do it. - At least not known to man yet.
And I'm going to stick with the solution I got, mainly because its simple.