Graph partitioning : an exhaustive search algorithm (no classic algorithms please)

Hello, There are many several definitions of graph partitions: classical says that this in this problem we need to find some number of sets of graphs vertices so that 1) each set isn't empty 2) pairwise intersection of each pair of sets is empty 3) union of all sets contains all vertices of the graph so that some function under this set is minimized (for example number of edges between vertices belonging to different sets or sum of weights of vertices belonging to same / different sets). There's other definition of graph partitioning: Graph partitioning is a partitioning graph's vertices into some number of non-empty sets so that each graph vertex belongs to exactly one set. For given partitioning some function should be minimized. So, I'm software developer with 12+ years of experience and mathematician with a number of publications. Also I'm participant and problem writer of many algorithm competitions (Topcoder, ACM ICPC, etc). Feel free to contact me to discuss any details of the project. Looking forward to hearing from you!

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Answer: Graph partitioning refers to subdividing the vertices into two or more parts such that for the sake of simplicity and ease of calulations and finding homomorphism and isomorphism. Hi there, With over 6 years of experience in the field of algorithm designing and modelling, I am an expert when it comes to programming them in various languages such as MATLAB, C, C++, Python etc. I have completed my university studies in applied mathematics with graph theory being a major part of that. I can provide you insights and implement the best approach for the kind of algorithm you are looking for. I understand that you are hoping to find an original partitioning algorithm that is not derived from any classic algorithms that already exist. I myself have worked on an application of a similar kind of project, where the implementation was that of a combination of Greedy and knapsack algorithm. I can share that details with you too. Please go ahead and discuss further details in the chat. Thank you.

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