I also blog on Medium medium.com/@seetharam.anand. I write on data science, AI, machine learning and computer science in general. Please FOLLOW me on Medium for more articles.I also provide FREE courses on Udemy on these topics www.udemy.com/user/anand-seetharam/. Check it out!
for extra credit points, couldve explained why the first part of the 3 cnf sat problem's statement was irrelevant in the example you provided. When you have an OR between x1 and its complement, it doesnt matter what the rest of the statement says in your example.
check this out: let's say all the litterals are independent, meaning if you have say X in one clause there will not be ¬X in the same or another clause. In that case the solutions are trivial. and the number of those solutions is (2^3 - 1)^n were n is the number of clauses. Now let's suppose we have dependent litterals. the trick is to set an nxor gate (nxor = not(or and nand) or simply nor or and) between all x and their respective ¬x, if the return is 1 it means x = ¬x and that solution is removed from the pool of all possible solutions. Try it out and tell me how'd u do!
I REALLY HOPE YOU ARE ACTIVE TO ANSWER THIS QUESTION if we have a 3-CNF problem with only three terms instead of 4..(x1,x2, and x3) would I be able to assign x1 = 1, x2 = 1, x3 = 0 and check?
When I assigned those values just for the first clause, I got a 0. already showing it isn't satisfiable with my assignment. Does that mean it cannot be satisfied? Or am I supposed to keep checking?
I have a serious question: let's say all the litterals are independent, meaning if you have say X in one clause there will not be ¬X in another clause. in that case the solutions are trivial. and the number of those solutions is (2^3 - 1)^n were n is the number of clauses. now let's suppose we have dependent litterals like in your example. the trick is to set an nxor gate between all x and their respective ¬x, if the return is 1 it means x = ¬x and that solution is removed from the pool of all possible solutions. what you're left with in the end are true solutions. I believe if done correctly this can solve 3sat in linear time. I can already picture the custom made circuit for it. With that said I believe this algorithm is too simple, so if it works somebody would've done it by now, I guess my question is: what's u algorithm?
I also blog on Medium medium.com/@seetharam.anand. I write on data science, AI, machine learning and computer science in general. Please FOLLOW me on Medium for more articles.I also provide FREE courses on Udemy on these topics www.udemy.com/user/anand-seetharam/. Check it out!
Dude the narration as well as the content is top notch ... Helped me out a lot .. Keep making vids you'll be big in no time!
Glad to hear that you liked the video.
Me too 🙋♀🙋♀🙋♀🙏🙏👍
Nyc explanation sir
for extra credit points, couldve explained why the first part of the 3 cnf sat problem's statement was irrelevant in the example you provided. When you have an OR between x1 and its complement, it doesnt matter what the rest of the statement says in your example.
Nice explanation
Thank you... very good explanation.
Sir how I show that unrestricted formulas can be converted into
restricted forms, and solution for both is the same
give me an approximate satisfiability problem algorithm and thank you
check this out: let's say all the litterals are independent, meaning if you have say X in one clause there will not be ¬X in the same or another clause.
In that case the solutions are trivial. and the number of those solutions is (2^3 - 1)^n were n is the number of clauses.
Now let's suppose we have dependent litterals. the trick is to set an nxor gate (nxor = not(or and nand) or simply nor or and) between all x and their respective ¬x, if the return is 1 it means x = ¬x and that solution is removed from the pool of all possible solutions. Try it out and tell me how'd u do!
@@mohamedb737 homie she asked any year ago
Top notch!
I REALLY HOPE YOU ARE ACTIVE TO ANSWER THIS QUESTION
if we have a 3-CNF problem with only three terms instead of 4..(x1,x2, and x3) would I be able to assign x1 = 1, x2 = 1, x3 = 0 and check?
When I assigned those values just for the first clause, I got a 0. already showing it isn't satisfiable with my assignment. Does that mean it cannot be satisfied? Or am I supposed to keep checking?
I have a serious question: let's say all the litterals are independent, meaning if you have say X in one clause there will not be ¬X in another clause. in that case the solutions are trivial. and the number of those solutions is (2^3 - 1)^n were n is the number of clauses. now let's suppose we have dependent litterals like in your example. the trick is to set an nxor gate between all x and their respective ¬x, if the return is 1 it means x = ¬x and that solution is removed from the pool of all possible solutions. what you're left with in the end are true solutions. I believe if done correctly this can solve 3sat in linear time. I can already picture the custom made circuit for it. With that said I believe this algorithm is too simple, so if it works somebody would've done it by now, I guess my question is: what's u algorithm?
very useful
Or conjunction
Conjunction and