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https://github.com/Lorenzinco/colombiniSAT

A complete 3-SAT solver with polynomial time euristic.
https://github.com/Lorenzinco/colombiniSAT

ai sat sat-solver

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A complete 3-SAT solver with polynomial time euristic.

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# Colombini-SAT

[![MacOS](https://github.com/Lorenzinco23/colombiniSAT/actions/workflows/macos.yml/badge.svg)](https://github.com/Lorenzinco23/colombiniSAT/actions/workflows/macos.yml)

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A simple 3-SAT solver written in Rust.



#### Links

[GitHub](https://github.com/Lorenzinco23/colombiniSAT "GitHub Repository page of the project.")



## How to use

Create an istance of the Solver struct passing it the path to the DMACS file you wish to solve.



```rust

let solver = Solver::new("path/to/file.cnf");

```



Then call the solve method on the solver istance.



```rust

let solution = solver.solve();

```



The solve method returns an Option enum, which can be either Some or None.

If the solve method returns Some, it means that the formula is satisfiable and the solution is contained in the Option.

If the solve method returns None, it means that the formula is unsatisfiable.



```rust

match solution{

        Some(solution) => println!("SAT: ({:?})", solution),

        None => println!("UNSAT")

    }

```



## Compiling

To compile the project you need to have Rust installed on your machine.

You can download Rust from [here](https://www.rust-lang.org/tools/install "Rust download page").



Once you have Rust installed, you can run the following command to see if everything is working:



```bash

cargo test

```



If everything is working correctly you can now run the following command to compile and run the project:



```bash

cargo run --release

```



# How it works

My euristics is a lookahead algorithm that tries to find implications between literals. First I need to address 

the reason why I chose to solve 3-SAT instances instead of $k$-SAT instances with $k \geq 3$.



As you'll see in the following sections, my algorithm takes advantage of the fact that in 3-SAT instances,

every clause contains exactly 3 literals. This allows me to find implications between literals in a very efficient way.

If I were to solve $k$-SAT instances with $k \geq 4$, I would have to find implications between $k$ literals, which (in my case) is a much harder problem.



You may be wondering now, why is it easier for you to find implications between 3 literals than between $k$ literals?

Tag along and you'll find out.



First we need to define some terms:



### Definition of $\phi'$:

let $n$ be the number of literals in $\phi$



$\forall i \in [1, n]$ let $\phi'_i \subseteq \phi$ be the subformula containing only the clauses from $\phi$ that contain the literal $x_i$.



#### Example:



$\phi\ = (x_1 \vee x_2 \vee x_3) \wedge (x_1 \vee x_3 \vee x_4) \wedge (x_1 \vee x_3 \vee x_5) \wedge (x_2 \vee x_3 \vee x_4)$



$\phi'_1 = (x_1 \vee x_2 \vee x_3) \wedge (x_1 \vee x_3 \vee x_4) \wedge (x_1 \vee x_3 \vee x_5)$



$\phi'_2 = (x_1 \vee x_2 \vee x_3) \wedge (x_2 \vee x_3 \vee x_4)$



$\phi'_3 = (x_1 \vee x_2 \vee x_3) \wedge (x_1 \vee x_3 \vee x_4) \wedge (x_2 \vee x_3 \vee x_4)$



$\vdots$



In each $\phi'_i$ performing the lookahead algorithm on $x_i$ leaves us with a 2-SAT instance by definition, which is trivial to solve.



By enumerating all the possible 2-SAT satisfing assignments of $\phi'_i$, we can find the implications between $x_i$ and the other literals in $\phi'_i$, which holds for the whole current branch of $\phi$.



Same goes for the fact that while performing the lookahead algorithm on $x_i$, if we find that no assignments of the remaining literals satisfies $\phi'_i$ for any assignment of $x_i$, then the problem is UNSAT and we can kill the branch much quicker.



These are just a handfull of glimpses about the implications that can be made while looking at $\phi'_i$, the following sections will explore more in detail the proprierties  with some examples, as hopefully you'll understand better.



## If the literal $x_i$ satisfies its $\phi'_i$ with only one assignment $A$ either $true\ or\ false$, then that literal is implied to be $A$ in the current branch of $\phi$.



#### Example:

$\phi'_1 = (x_1 \vee \neg x_5) \wedge (\neg x_1 \vee \neg x_3) \wedge (x_1 \vee x_3 \vee x_5) \wedge (x_1 \vee \neg x_3)$ 



$x_1$ is satisfied with only one assignment $true$ in $\phi'_1$, so $x_1$ is implied to be $true$ in the current branch of $\phi$.



## While solving some $\phi'_i$, if some other literal $x_j$ takes the same assignment in all satisfing assignments of $\phi'$, then $x_j$ is also implied in the current branch of $\phi$.



#### Example:

$\phi'_2 = (\neg x_2 \vee x_3) \wedge (x_2 \vee x_3) \wedge (x_2 \vee x_3 \vee x_4)$



$x_2$ appears with both assignments in each satisfing assignment in $\phi'_2$, so we can say nothing about $x_2$,

however $x_3$ must be true to satisfy $\phi'_2$, so $x_3$ is implied to be $true$ in the current branch of $\phi$.



## While solving some $\phi'_i$, let $x_i$ the implication found , if some other literal $x_j$ takes the same assignment in all satisfing assignments of Phi', then $x_i\implies x_j$ in the current branch of $\phi$.



#### Example:

$\phi'_3 = (x_3 \vee x_5) \wedge (x_1 \vee \neg x_3) \wedge (x_2 \vee x_3 \vee x_4)$



In all satisfing assignments where $x_3 = true$, $x_1 = true$ too: that means $x_3\implies x_1$ in the current branch of $\phi$.