be a spanning subgraph of G (in other words, it includes all vertices from G).If G only those edges (x,y) which satisfy the following condition: , then it is an equality subgraph.
be a spanning subgraph of G (in other words, it includes all vertices from G).If G only those edges (x,y) which satisfy the following condition: , then it is an equality subgraph.Then, after the contest, you find out in the editorial that this problem can be simply reduced to a classical one.Tags: Alcohol Poisoning ThesisEssay On Compare And Contrast Two PeopleLegal Writing HelpResearch Paper Psychology ClassOutline For A Narrative EssayThesis Statement For Animal Adoption
Discuss this article in the forums Are you familiar with the following situation?
You open the Div I Medium and don’t know how to approach it, while a lot of people in your room submitted it in less than 10 minutes.
In order to avoid this, on each step we can just modify the matching from the previous step, which only takes O(n2) operations.
It’s easy to see that no more than n2 iterations will occur, because every time at least one edge becomes 0-weight. This is simply a function (for each vertex we assign some number called a label).
See picture below for explanation: At last, here's the function that implements Hungarian algorithm: To see all this in practice let's complete the example started on step 0.| Build alternating tree V| Augmenting V path found| Build V alternating tree| Update V labels(Δ=1)| Build alternating V tree| Update V labels(Δ=1)→| Build V alternating tree| Augmenting V path found| Build V alternating tree| Update labels V (Δ=2)| Build V alternating tree| Update labels V (Δ=1)| Build V alternating tree| Augmenting V path found Optimalmatching found Finally, let's talk about the complexity of this algorithm.
On each iteration we increment matching so we have n iterations.If we can’t find perfect matching on the current step, then the Hungarian algorithm changes weights of the available edges in such a way that the new 0-weight edges appear and these changes do not influence the optimal solution.To clarify, let’s look at the step-by-step overview: Step 0)A. Apply the same procedure for the vertices in the right part (jobs). Otherwise find the minimum vertex cover (for the subgraph with 0-weight edges only), the best way to do this is to use Köning’s graph theorem.For each vertex from left part (workers) find the minimal outgoing edge and subtract its weight from all weights connected with this vertex. Actually, this step is not necessary, but it decreases the number of main cycle iterations. Find the maximum matching using only 0-weight edges (for this purpose you can use max-flow algorithm, augmenting path algorithm, etc.). Step 2) Let and adjust the weights using the following rule: Step 3) Repeat Step 1 until solved.But there is a nuance here; finding the maximum matching in step 1 on each iteration will cause the algorithm to become O(n5).Let’s call this labeling feasible if it satisfies the following condition: .In other words, the sum of the labels of the vertices on both sides of a given edge are greater than or equal to the weight of that edge.The main idea of the method is the following: consider we’ve found the perfect matching using only edges of weight 0 (hereinafter called “0-weight edges”).Obviously, these edges will be the solution of the assignment problem.We’ll handle the assignment problem with the Hungarian algorithm (or Kuhn-Munkres algorithm).I’ll illustrate two different implementations of this algorithm, both graph theoretic, one easy and fast to implement with O(n4) complexity, and the other one with O(n3) complexity, but harder to implement.