Algorithms – Change the structure of 3SAT to 1in3 3SAT

Your question is not clear, but I think you're wondering how to show that 1in3SAT is NP-complete by reducing the well-known NP-complete 3SAT problem.

If this is your question, check the following settings in your notation:
There is an instance of 3SAT defined by a set of $ n $, Boolean variables $ V = {x_1, …, x_n } $ and a Boolean equation of $ m $ clauses $ Phi = C_1 land C_2 land … land C_m $ where everyone $ C_j = (x_ {j_1} l or x_ {j_2} l or x_ {j_3}) $,

To reduce this to 1in3SAT, start with the reduction in the following way.
For each $ C_j $ Create 3 new clauses $ C # _ {j, 1} = ( lnot x_ {j_1} lor alpha_j lor beta_j) $. $ C & # 39; {j, 1} = (x_ {j_2} lor beta_j lor gamma_j) $, and $ C & # 39; {j, 3} = ( lnot x_ {j_3} lor gamma_j lor delta_j) $ from where $ alpha_j, beta_j, gamma_j, delta_j $ are new Boolean random variables.
To let $ Phi & # 39; $ Be the Boolean equation in your 1in3SAT instance, which is now the AND of your new clauses. That means you have now $ 3m $ Clauses and $ n + $ 4 million Variables.

Next, consider what would happen if we had a true assignment for the 3SAT instance, and what would happen if we had a valid assignment for the newly constructed 1in3SAT instance (this second direction will be much simpler).

np complete – reduce subset sum to 3SAT

There are usually no intuitive or insightful reductions between problems in different problem areas.

The proof that 3SAT is NP-complete essentially consists of writing a formula that says "This NP Turing machine accepts this input. "For other problems with logical formulas, you can often translate the formula into a 3SAT instance, and sometimes you can express problems in domains other than 3CNF formulas: For example, you can obtain variable 3-colorability for any combination of" vertex $ v $ has color $ q $"and write a formula that says that each vertex has exactly one color and adjacent vertices have different colors.

However, if you are faced with a problem from a completely different domain, you usually can not do much better than say the subset total is in NP So it's decided by a Turing machine, which I can express as a 3SAT instance. Maybe you can come up with a less generic reduction, but it probably would not teach you anything.

3 sat – How many isomorphic 3SAT formulas?

For a 3SAT formula with $ n $ Variables and $ m $ Clauses, I am interested in counting the number of isomorphic formulas (isomorphic in the sense that they are logically equivalent and have the same number of variables and clauses). I assume that we can reverse the meaning of a variable (swap all) $ x_i $ With $ not x_i $) and permute one of the variables. The first should give $ 2 ^ n $ and the second should give $ n! $,

Are these all isomorphisms, or do I overlook anything?

If that's true, then I assume that the number of unique 3SAT formulas with $ n $ Variables and "different clauses" are the same $ frac {2 ^ m} {2 ^ nn!} $, from where $ m $ the number of 3 clauses is over $ n $ Variables in which all variables are different. I suppose that $ m = 2n (2n-2) (2n-4) $, Does that sound right?

Fulfillability – Is a "stacked", "local" version of 3-SAT NP-hard?

In this previous question I learned that every variable in a string $ C in 3 text {-SAT} $ If only "local" appears, the finding of a satisfactory assignment is no longer NP-hard. Based on this, I ask the question of whether the NP-hardness is restored when there are roughly several "layers" of variables that lead to the formula of interest. (This work of course follows from a study of Bayesian spatial networks.)


question

Look at a diagram of nodes arranged in rows. The number of nodes doubles in each subsequent row. The graph has a node in the first row, two in the second, four in the third, and so on, e.g. $ m $ Series. (There are $ n = 2 ^ m – 1 $ The total number of nodes in the graph.) Each node has the status 0 or 1 and up to three "parent" nodes in the following row (if there is such a row). Every parent is "$ k $-local ": there are less than $ k $ Node between the first and last node for which it is a parent node. ($ k $ is a fixed natural number that does not change $ n $.)

For every node $ v $There is a table $ T_v $ this lists all possible state combinations for the parents of $ v $ and assigns a state of 0 or 1 for each combination $ v $ in a way consistent with $ v $ to be a disjunctive clause of the literals of his parents.

To let $ (3, k) text {-SATSTACK} $ denote the decision problem of determining whether there is an assignment of states to the nodes of the graph that assigns state 1 to the first node and is consistent with all $ T_v $,

Is there $ k $ for which $ (3, k) text {-SATSTACK} $ is NP-hard?


Remarks

(1) It seems so $ (3, k) text {-SATSTACK} $ is in NP because we can select an assignment for every node in the last row, and then calculate the assignment for the rest of the graph in linear time forward and accept it only when it assigns state 1 to the first node.

(2) It also seems plausible that $ (3, k) text {-SATSTACK} $ is NP-hard for $ k $ big enough, since it seems daunting to do better than a method of continuing from the first line, applying the polynomial time algorithm given above to each line, and taking stock of all the satisfactorily assignments for that line (which may be one) performing exponentially Collection).

(3) However, it seems difficult to find a polynomial time shortening that shows this $ (3, k) text {-SATSTACK} $ is NP-hard, because many of the subsets of $ 3 text {-SAT} $ (and related problems), of course, in $ (3, k) text {-SATSTACK} $ are not NP-complete (see again). So how should we proceed?

Fulfillability – Is a "local" version of 3-SAT NP-hard?

Below is my simplification of part of a larger research project on spatial Bayesian networks:

Suppose a variable is "$ k $-local "in a string $ C in text {3-CNF} $ if it is less than $ k $ Clauses between the first and last clause in which it occurs (where $ k $ is a natural number).

Now consider the subset $ text {3-LSAT} subseteq text {3-SAT} $ defined by the criterion that for each $ C in text {3-LSAT} $, every variable in $ C $ is $ k $-local. For what $ k $ (if available) $ text {3-LSAT} $ NP-hard?


I have considered the following:

(1) Deviations from the detection method $ text {2-SAT} $ is in P by rewriting each disjunction as an implication and examining directed paths on the directed graph of these implications (noted here and detailed on pages 184-185 of Papadimitriou) computational complexity). Unlike in $ text {2-SAT} $there is a branching of directional ways in $ text {3-LSAT} $but perhaps the number of directed paths is limited by the spatial constraints of the variables. So far no success with it.

(2) A polynomial time reduction of $ text {3-SAT} $ (or another known NP-whole problem) $ text {3-LSAT} $, For example, I've tried different schemes for introducing new variables. Merging the clauses that contain the original variable $ x_k $ In general, I have to pull "strings" of extra clauses that contain the new variables, and these affect the spatial constraints of the other variables.

I am certainly not on new territory here. Is there a known NP-hard problem that can be reduced to? $ text {3-LSAT} $ Or do the spatial restrictions prevent the problem from becoming so difficult?

Satisfiability – 3SAT formula with only positive variables

Suppose I try to reduce the mesh generation to 3-SAT

Here's my attempt to create a SAT instance for an algorithm that creates rasters. The grid follows a permute pattern. I try to represent the pattern with a formula.

(~ 1 ~ 9 ~ 8) (~ 2 ~ 1 ~ 9) (~ 3 ~ 2 ~ 1) (~ 4 ~ 3 ~ 2) (~ 5 ~ 4 ~ 3)) (~ 6 ~ 5 ~ 4) ( ~ 7) ∨𝑥6∨𝑦5) ∧ (𝑙8∨𝑥7∨𝑦6) ∧ (𝑙9∨𝑥8∨𝑦7) ∧ (𝑙1∨𝑥9∨𝑦8) (𝑙2∨𝑥1∨𝑦9)

l = [8, 5, 9, 6, 1, 2, 4, 3, 7]

[5, 9, 6, 1, 2, 4, 3, 7, 8]-l1
[9, 6, 1, 2, 4, 3, 7, 8, 5]-l2
[6, 1, 2, 4, 3, 7, 8, 5, 9]-l3
[1, 2, 4, 3, 7, 8, 5, 9, 6]-l4
[2, 4, 3, 7, 8, 5, 9, 6, 1]-l5
[4, 3, 7, 8, 5, 9, 6, 1, 2]-L6
[3, 7, 8, 5, 9, 6, 1, 2, 4]-l7
[7, 8, 5, 9, 6, 1, 2, 4, 3]-l8
[8, 5, 9, 6, 1, 2, 4, 3, 7]-l9

x = [5, 9, 6, 1, 2, 4, 3, 7, 8]

[9, 6, 1, 2, 4, 3, 7, 8, 5]x1
[6, 1, 2, 4, 3, 7, 8, 5, 9]-x2
[1, 2, 4, 3, 7, 8, 5, 9, 6]-x3
[2, 4, 3, 7, 8, 5, 9, 6, 1]X4
[4, 3, 7, 8, 5, 9, 6, 1, 2]-x5
[3, 7, 8, 5, 9, 6, 1, 2, 4]-X6
[7, 8, 5, 9, 6, 1, 2, 4, 3]-X7
[8, 5, 9, 6, 1, 2, 4, 3, 7]X8
[5, 9, 6, 1, 2, 4, 3, 7, 8]X9

y = [9, 6, 1, 2, 4, 3, 7, 8, 5]

[6, 1, 2, 4, 3, 7, 8, 5, 9]-y1
[1, 2, 4, 3, 7, 8, 5, 9, 6]y2
[2, 4, 3, 7, 8, 5, 9, 6, 1]-y3
[4, 3, 7, 8, 5, 9, 6, 1, 2]-Y4
[3, 7, 8, 5, 9, 6, 1, 2, 4]Y5
[7, 8, 5, 9, 6, 1, 2, 4, 3]-Y6
[8, 5, 9, 6, 1, 2, 4, 3, 7]-y7
[5, 9, 6, 1, 2, 4, 3, 7, 8]-y8
[9, 6, 1, 2, 4, 3, 7, 8, 5]-y9

(l1 ∨ x9 ∨ y8) ∧ (l2 ∨ x1 ∨ y9)

l1 = [5, 9, 6, 1, 2, 4, 3, 7, 8]

x9 = [5, 9, 6, 1, 2, 4, 3, 7, 8]

y8 = [5, 9, 6, 1, 2, 4, 3, 7, 8]

The formula should always generate this grid.

    [5, 9, 6, 1, 2, 4, 3, 7, 8]-l1
    [9, 6, 1, 2, 4, 3, 7, 8, 5]-l2
    [6, 1, 2, 4, 3, 7, 8, 5, 9]-l3
    [1, 2, 4, 3, 7, 8, 5, 9, 6]-l4
    [2, 4, 3, 7, 8, 5, 9, 6, 1]-l5
    [4, 3, 7, 8, 5, 9, 6, 1, 2]-L6
    [3, 7, 8, 5, 9, 6, 1, 2, 4]-l7
    [7, 8, 5, 9, 6, 1, 2, 4, 3]-l8
    [8, 5, 9, 6, 1, 2, 4, 3, 7]-l9

Is this a correct 3-SAT reduction from the hypothetical problem?

Would not I have to extend it?

np complete – Reduces 3SAT to a set splitting problem

I want to be able to reduce the 3SAT problem to a problem of the set-splitting problem. Basically given $ n $ Articles and $ m $ subsets $ S_1, S_2, …, S_m $ Of these elements, I would like a yes or no answer based on the fact that it is possible to color the elements red and blue so that each sentence is set $ S_i $ contains at least one red object and at least one blue object. I know this problem exists in NP and it's pretty easy to prove (at least I think so). In fact, I would like to prove that the problem is harsh by the reduction of 3SAT to NP. I want to be able to transform an instance of 3SAT into an instance of the aforementioned set split so that a yes instance of 3SAT is mapped to a yes instance of set layout, and vice versa. I have a basic intuition about how to do this: a 3SAT formula $ f $ on $ n $ Variables that I will create $ 2n + 1 $ Items. One to designate everyone $ x_i $ One for each $ overline {x_i} $ and an extra article.
I am not quite sure how to go beyond this basic intuition. This could be a gap in my understanding of the problem. Is this the right way to achieve this? If so, how can I use this intuition to actually prove that this is hard by reducing 3SAT to NP?

Many Thanks!