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I have thought of an algorithm to solve 3SAT problem via the below approach :

1) Take all the clauses in the cnf equation who have at least one variable in common. Find all such combinations and put these into a list named “Intersection”. The list “Intersection” now contains the intersecting clauses groups.

2) Next, we convert all the clauses in a particular group of “Intersection” to DNF. Since there would be at least one variable in common I do not think it is going to be in exponential time.

3) Next we convert all the obtained DNFs obtained to a single DNF as all of them are separated with an AND gate in original equation.

4) If there is a single clause in the obtained DNF then the equation is satisfiable or else the equation is unsatisfiable. This is because the non-intersecting clauses(the clauses that do not have any variable in common) are not going to affect the overall equation and finally if we “AND” those with the obtained DNF , it will only multiply and add variables to existing clauses(if any).

My question is:

What is the runtime of this algorithm and does this prove anything related to P=NP as I think this algorithm is quite efficient. I have been let down by others on my previous algorithms so this time please take time to analyze the algorithm as it is my hard work.

vinaych
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    I have found an answer to your question which solves P=NP problem: It is too small to fit in this marg.. err.. comment box.. 42. See ya @Clay Millennium Awards Ceremony.. – toddlermenot Sep 13 '15 at 17:04

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I used exactly this algorithm to make a minesweeper solver.

In practice it works very fast when the variables are spread out (as they are in typical minesweeper levels).

However, for a hard 3-SAT problem, the expansion in steps 2 and 3 to a single DNF ends up taking exponential time because every time we combine clauses the number of potential terms goes up as the product of the length of the clauses.

So, in summary, this is a very useful approach for some SAT problems, but has exponential worst-case performance so has nothing to say about P=NP.

Peter de Rivaz
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