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11

A typical SAT solver notices that a satisfying assignment has been found when there are no more variables to assign. So the only time that a SAT solver would save by early notification is the time it would take to assign values to any remaining unassigned variables. This is a one-time cost linear in the number of variables in the formula. It's linear ...

7

You are looking for a maximum weight independent set in an interval graph, which can be solved in linear time (by a deterministic algorithm). By the way, the same is true also for a superclass of interval graphs, namely chordal graphs. One such algorithm is based on dynamic programming, you can see e.g., [1] for details and discussion. [1] Hsiao, Ju Yuan, ...

7

Given a graph $G = (V,E)$, here is a SAT instance which is satisfiable iff the graph is not connected. Pick an arbitrary vertex $v_0 \in V$, and add the following clauses, over the variables $x_v$ for $v \in V$: $x_{v_0}$. For every $(u,v) \in E$, $\lnot x_u \lor x_v$ and $\lnot x_v \lor x_u$. $\bigvee_{v \neq v_0} \lnot x_v$. Here is a SAT instance which ...

6

There is unlikely to be any efficient algorithm. Your first class of constraints are monotone exactly-1 CNF clauses. Your second class of constraints are monotone CNF clauses. The monotone part indicates that negated literals aren't allowed (you can't have $x_1 - x_3 = 1$ or $x_1 - x_4 \ge 1$). Thus, in the special case where you have only type-2 ...

6

What happens next is that we invoke a SMT solver to try to check whether the formula $$x+y < 20 \quad \land \quad x > 10 \quad \land \quad y > 10$$ is satisfiable. The solver tries to find values for $x,y$ that make this formula true. If it finds such a value, then it knows the assume(false) statement is reachable. (I am guessing you meant to ...

5

Given a set of objects in a cell grid, if no 4-set of it forms a rectangle with sides parallel to the sides of the grid, we will call those objects rectangle-free. The more general problem is to determine sequence A072567, whose $n$-th term $M_n$ is the size of maximal set of rectangle-free cells in an $n \times n$ cell grid. Its first 13 entries are 1, ...

4

If you want to model an alldiff() constraint in SAT, there are several options. Here are two different options you can try: One way is to expand $\text{alldiff}(x_1,\dots,x_n)$ into $n(n-1)/2$ inequality constraints: $(x_1 \ne x_2) \land (x_1 \ne x_3) \land \cdots$. Now you can express each inequality constraint $x_i \ne x_j$ on $b$-bit values in turn as ...

4

I don't know, but if you manage, make sure to let us know. The reduction seems easy indeed, with $d$ mapping to the number of colors and each constraint to an edge. However, a polynomial-time algorithm for any $\mathcal{NP}$-complete problem exists if and only if $\mathcal{P} =\mathcal{NP}$; doing what you ask would imply $\mathcal{P} \neq \mathcal{NP}$.

4

Introduce a new variable $Q$, whose domain is $\{0000,0001,0010,\dots,1111\}$. It represents the value of $C1,C2,C3,P$. For instance, if $Q=0001$, that means that $C1=0$, $C2=0$, $C3=1$, $P=0$. Then, you add a constraint that the first bit of $Q$ is equal to $C1$ (that's a binary constraint), a constraint that the second bit of $Q$ is equal to $C2$ (...

4

The theory you are after is universal algebra. See the excellent expository article of Hubie Chen, A rendezvous of logic, complexity, and algebra, which contains a streamlined proof of Schaefer’s dichotomy theorem, which was recently extended to arbitrary alphabets. 3SAT has no polymorphisms. NAE-3SAT has only negation as a polymorphism. All other classes ...

4

The graph below is a positive answer without words. Here is the detailed proof. Definitions Let $X$ be an instance of X3SAT. $X$ is linear if any two clause shares at most one variable. $X$ is monotone if all literals in all clauses are positive. A walk in $X$ is a sequence of clauses where each clause is not disjoint with the clause following it. The ...

3

You may not need a general SMT solver. In this case, you have a bunch of inequalities conjoined together, which you can solve as a linear program. A simpler alternative to an SMT solver is to implement the theory as an add-on to a logic program. Most decent Prolog implementations, for example, have a CLP(R) or CLP(Q) library which can solve equality and ...

3

In complexity theory, CSPs are usually specified as a set of allowed predicates. If the (finite) domain is $D$, a predicate of arity $d$ is an arbitrary subset of $D^d$ of allowed values. In particular, equality (or inequality) is a predicate of arity 2. This is the point of view taken in Schaefer's dichotomy theorem as well as the more recent universal ...

3

I think you could use a "spatial join". I haven't played the game, but I assume $d_{max}$ is rather small, i.e. there are in the order of 10 or so $n$ and $m$ in the neighborhood of each $m$. I further assume that $N$ and $M$ are large, say 1,000,000 or more. Put all $m$ as 2D points in a spatial index For each $m_1$ in the index, perform a spatial range ...

3

Good question. Your instinctive reaction is quite valid: if you stick with the variables $A,B,C$, there is no way to introduce some set of binary constraints that will be equivalent to $A+B=C$ (where each constraint is only allowed to mention two of $A,B,C$). The trick is to introduce new variables, or change the set of variables. If you do that, it can ...

3

Yes, these two heuristics does sound like inconsistent. Most Constrained Variable (MCV) (also called MRV for Minimum Remaining Values) tries to reduce the size of the next branch to search while Least Constraining Values tries to enlarge the size of the next branch to branch. However, if you take a close a look, they both serve the same goal, which is, ...

3

It means that you have a partial solution, and you extend that partial solution by assigning at least one more variable to some value, thus producing either a new partial solution or a solution where every variable is assigned a value. Put differently, if $x$ extends an assignment $a$, it holds that $x$ does precisely the same assignments as $a$ and assigns ...

3

Your problem is solvable in polynomial time. Your problem is equivalent to ILP with difference constraints, i.e., every linear inequality has the form $x_j - x_i \ge c_{i,j}$. (Why? Set $x_i = v_1 + \dots + v_i$; then $v_{i+1} + \dots + v_j = x_j - x_i$.) Such a system of inequalities can be solved with shortest-path algorithms. In particular, if we ...

3

Here's a strategy for solving Kakuro with a SAT solver. Make a nine variables for each cell, each variable indicating whether that cell contains $1$, $2$, etc. Add a exactly-one-out-nine constraint for the variables of each cell. You can do this naively in 37 clauses by adding a clause that at least one of the nine must be true ($x_1 \vee x_2 \vee \dots \... 2 The variable$T$is not a timer but the temperature. It starts very high, making it more likely that transitions are taken, and then slowly cools. It's called annealing due to the metallurgical technique of the same name. Simulated annealing handles one problematic aspect of the hill climbing algorithm, namely that it can get stuck at a local optimum which ... 2 see this link: https://people.cs.pitt.edu/~wiebe/courses/CS2710/lectures/constraintSat.example.txt It first picks variable "O" and then tests "O" with all of it's legal values "i" to see the number of reductions on "O"'s neighbors "N". It adds all of them. and picks an "i" that causes less reductions: sums = {0:0,1:0,2:0,3:0,4:0,5:0,6:0,7:0,8:0,9:0} ... 2 As you observe, restricting the domain of a variable has exactly the same effect as applying a unary constraint to it. One situation where you might prefer to use unary constraints rather than restricted domains is when you want to control very tightly the relations that are allowed to be used in constraints. For example, if you want to investigate the ... 2 I suggest you use integer linear programming. Suppose you have$n$recipes. Introduce zero-or-one variables$x_1,x_2,\dots,x_n$, with the idea that$x_i=1$represents including the$i$th recipe and$x_i=0$represents omitting it. Now you can obtain a bunch of linear equations, one per macronutrient requirement. For instance, in your example, we get ... 2 Let's say that there's a [$k$-ary] relation$R$and [$m$-ary] function$f$such that$m>k$. Is$f$a polymorphism of$R$? Maybe, maybe not: it depends on the function and the relation. A given$k$-ary relation may have polymorphisms of arity arity less than, equal to, and/or greater than$k$. In fact, every relation has polymorphisms of all arities.... 2 I think I found the answer here! It is sufficient to use the big-M method by introducing the following constraints: B >= C + 1 - M*(1-A); C >= B + 1 - M*A 2 One example of a tiling problem that was successfully attacked by reducing it to a SAT instance was rectangular grid coloring. In "Extremely Complex 4-Colored Rectangle-Free Grids: Solution of Open Multiple-Valued Problems" the authors describe how they tackled 17x17, 17x18, and 18x18 grid colorings using a SAT solver plus some previous knowledge of grid ... 2 The conjecture is true, and the proof strategy listed there can be made to work. Let$s = (s_1,\dots,s_n)$be a satisfying assignment for the variables that makes all of the constraints hold. We'll write$E(s)$for the result of evaluating the expression$E$with the variables$S_i$set to$s_i$. In the following, define$U$by the procedure listed in the ... 2 I'll list two approaches, either of which should work. Projected gradient descent One general approach would be to use projected gradient descent, which provides a way to accommodate these kinds of constraints. For specific kinds of constraints there may be a better way. Let's recall how we train neural networks. We have an objective function$\Psi\$ ...

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