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In order to be certain that we can do this even if planar 3-SAT isn't NP-hard, we can build a gadget which allows paths to cross.We can now do arbitrary 3-SAT.this can either be covered with one domino like.With that in mind, we can then observe that in this case the value of the input x is equal to the value of b c and the input y is equal to the value of a c (note that this would be logical.So, I'll start with a single tile which can be in two possible places.(If you don't want to count all the dominoes you can note that I used every letter except for Y and.) If the maximal tiling had contained either 22 or 23 dominoes, then we would know that one of the clauses was not satisfied.I'll also use sets of three capital letters to represent dominoes and lower case letters to represent "signals" which are spaces which may or may not be filled depending on the state of the system.
If we were to allow both ends of something boite cadeau de noel a imprimer to be true, this would result in incorrect solutions, but we do not do this in this scheme.
So we just need to split c and then use a logical and gadget to connect connect the values of c with a and b respectively and we will then have successfully completed our cross over.Essentially, my old splitter looked like this and could be packed with 6 when x does not protrude (rather than the 5 I had intended) like this: y* *z aaac dbbb * * C D * C*D * EEE *x* FFF So I revised.Gggd* *cggg *CGD* * * G eeef eeef geee.We can interconnect these gadgets using paths.But it's really quite complex (if anyone sees a simpler way, please let me know) so first I'm going to do an example of solving planar 3-SAT with this system.



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