In part I of this tutorial, we introduced the SAT problem and discussed its applications.  The SAT problem operates on Boolean logical formulae and we discussed how to convert these to conjunctive normal form. Here, a set of clauses are logical $\text{AND}$ed together. Each clause logically $\text{OR}$s a set of literals (variables and their complements). For example:

\begin{equation}\label{eq:example_cnf_conditioning}    \phi:= (x_{1} \lor x_{2} \lor x_{3}) \land (\overline{x}_{1} \lor x_{2} \lor x_{3}) \land (x_{1} \lor \overline{x}_{2} \lor x_{3}) \land (x_{1} \lor x_{2} \lor \overline{x}_{3}). \tag{1}\end{equation}

where the notation $\lor$ represents a $\text{OR}$ operation and $\land$ represents a $\text{AND}$ operation.  The satisfiability problem establishes whether there is any way to set the variables $x_{1},x_{2},x_{3}\in$$\{$$\text{true}$,$\text{false}$$\}$ so that the formula $\phi$ evaluates to $\text{true}$.

In this tutorial we focus exclusively on the SAT solver algorithms that are applied to this problem. We’ll start by introducing two ways to manipulate Boolean logic formulae. We’ll then exploit these manipulations to develop algorithms of increasing complexity. We’ll conclude with an introduction to conflict-driven clause learning which underpins most modern SAT solvers.

Operations on Boolean formulae

SAT solvers rely on repeated algebraic manipulation of the formula that we wish to test for satisfiability. Two such manipulations are conditioning and resolution. In this section we will discuss each in turn.


In conditioning, we set a variable $x_{i}$ to a concrete value (i.e., $\text{true}$ or $\text{false}$). When we set $x_{i}$ to $\text{true}$, we can simplify the formula using two rules:

  1. All clauses containing $x_{i}$ can be removed from the formula. These clauses are now satisfied.
  2. Any terms $\overline{x}_{i}$ in the remaining clauses can be removed. These must now evaluate to $\text{false}$ and hence cannot be used to satisfy their clauses. 

For example, consider the formula:

\begin{equation}\label{eq:example_cnf_conditioning_s}    \phi:= (x_{1} \lor x_{2} \lor x_{3}) \land (\overline{x}_{1} \lor x_{2} \lor x_{3}) \land (x_{1} \lor \overline{x}_{2} \lor x_{3}) \land (x_{1} \lor x_{2} \lor \overline{x}_{3}). \tag{2}\end{equation}

When we set $x_{1}=$ $\text{true}$, this becomes 

\begin{equation}\label{eq:example_cnf_conditioning2}    \phi \land x_{1} := (x_{2} \lor x_{3}). \tag{3}\end{equation}

where the first, third and fourth clause have been removed as they are now satisfied (by rule 1) and the term $\overline{x}_{1}$ has been removed from the second clause as this term is now $\text{false}$ (by rule 2). 

Similarly, when we condition by setting a variable to $\text{false}$ all clauses containing $\overline{x}_{i}$ disappear, as do any terms $x_{i}$ in the remaining clauses. Setting $x_{1}$ to $\text{false}$ in equation 2 gives:

\begin{equation}    \phi \land \overline{x}_{1} := (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{3}) \land (x_{2} \lor \overline{x}_{3}). \tag{4}\end{equation}

Note that variable $x_{i}$ must be either $\text{true}$ or $\text{false}$ and so:

\begin{eqnarray}    \phi &=& (\phi \land x_{i}) \lor (\phi \land \overline{x}_{i})\\     &=& (x_{2} \lor x_{3}) \lor ((x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{3}) \land (x_{2} \lor \overline{x}_{3})).\nonumber \tag{5}\end{eqnarray}

Here we apply the conditioning operation twice and the result is to remove the variable $x_{i}$ from the formula $\phi$ to yield two simpler formulae, $(\phi \land x_{i})$ and $(\phi \land \overline{x}_{i})$ which are logically $\text{OR}$ed together. Note though, that the result is not in conjunctive normal form.


The second common operation applied to Boolean formulae is resolution. Consider two clauses $c_{1}$ and $c_{2}$ where $x_{i}\in c_{1}$ and $\overline{x}_{i}\in c_{2}$. When we resolve by $x_{i}$, we replace these two clauses with a single clause $(c_{1}\setminus x_{i})\lor (c_{2}\setminus\overline{x}_{i})$. This clause is known as the resolvent and contains the remaining terms in $c_{1}$ and $c_{2}$ after $x_{i}$ and $\overline{x}_{i}$ are removed.

This is best illustrated with an example. Consider the formula:

\begin{equation}\label{eq:example_cnf_resolution}    \phi:= (x_{1}\lor x_{2} \lor \overline{x}_{3}) \land (\overline{x}_{2} \lor x_{4}) \land (x_{2} \lor x_{4}\lor x_{5}). \tag{6}\end{equation}

We note that $x_{2}$ is in the first clause and $\overline{x}_{2}$ is in the second clause and so we can resolve with respect to $x_{2}$ by combining the remaining terms from the first and second clause:

\begin{equation}    \phi:= (x_{1}\lor \overline{x}_{3} \lor x_{4}) \land (x_{2} \lor x_{4}\lor x_{5}). \tag{7}\end{equation}

Note that the third clause is unaffected by this operation.

The underlying logic is as follows. If $x_{2}$ is $\text{false}$, then for the first clause to be satisfied we must have $x_{1}\lor \overline{x}_{3}$. However, if $x_{2}$ is $\text{true}$, then for the second clause to be satisfied, we must have $x_{4}$. Since either $x_{2}$ or $\overline{x}_{2}$ must be the case, it follows that we must have $x_{1}\lor \overline{x}_{3} \lor x_{4}$.

Unit resolution

An important special case is unit resolution. Here, at least one of the clauses that we are resolving with respect to is a unit clause (i.e., only contains a single literal). For example,

\begin{equation}    \phi:= (x_{1}\lor \overline{x}_{3} \lor \overline{x}_{4}) \land x_{4} \tag{8}\end{equation}

Resolution between these two clauses works as normal. However, we can go further. Since we know that $x_{4}$ must be $\text{true}$ from the second clause, the effect of resolution here is the same as conditioning. We can remove all clauses containing $x_{4}$ and remove all terms $\overline{x}_{4}$ from the remaining clauses. So unit resolution can be seen as either a special case of resolution or as a conditioning operation depending how you look at it.

Unit propagation

A unit resolution operation may create more unit clauses. In this case, we can repeatedly apply unit resolution to the expression and at each stage we eliminate one of the variables from consideration. This procedure is known as unit propagation.

SAT solving algorithms based on resolution

We now present a series of learning algorithms that use conditioning and resolution to solve the satisfiability problem. In this section, we will use resolution to solve the 2-SAT problem and show why this can be solved in polynomial time. Then we’ll introduce the directional resolution algorithm which uses resolution to solve 3-SAT problems and above, but we’ll see that this becomes more computationally complex. In the next section, we’ll move to algorithms that primarily exploit the conditioning algorithm to solve SAT problems.

Solving 2-SAT by unit propagation

To solve a 2-SAT problem we first condition on an arbitrarily chosen variable. This sets off a unit propagation process (a chain of unit resolutions) in which variables are removed one-by-one. This continues until either the formula is satisfied or we are left with a contradiction $x_{i}\land \overline{x}_{i}$. 

Worked example: This process is easiest to understand using a concrete example. Consider the following 2-SAT problem in four variables:

\begin{equation}\phi:= (x_{1}\lor \overline{x}_{2}) \land (\overline{x}_{1}\lor \overline{x}_{3}) \land (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{4}) \land (x_{3}\lor \overline{x}_{4}). \tag{9}\end{equation}

We start with a single step of conditioning on an arbitrarily chosen variable. Here we’ll choose $x_{1}$ and apply the formula $\phi = (\phi \land x_{1}) \lor (\phi \land \overline{x}_{1})$. We could work directly with this cumbersome expression, but in practice we set $x_{1}$ to $\text{true}$ and test for satisfiability. If this is not satisfiable, then we set $x_{1}$ to $\text{false}$ and try again and if neither are satisfiable, then the expression is not satisfiable as a whole.

Let’s work through this process explicitly. Setting $x_{1}$ to $\text{true}$ gives:

\begin{equation}    \phi \land x_{1} =(x_{1}\lor \overline{x}_{2}) \land (\overline{x}_{1}\lor \overline{x}_{3}) \land (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{4}) \land (x_{3}\lor \overline{x}_{4}) \land x_{1}. \tag{10}\end{equation}

We now perform unit resolution with respect to $x_{1}$ which means removing any clauses that contain $x_{1}$ and removing $\overline{x}_{1}$ from the rest of the formula to get:

\begin{equation}\phi \land x_{1} = \overline{x}_{3} \land (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{4}) \land (x_{3} \lor \overline{x}_{4}). \tag{11}\end{equation}

Notice that we are left with another unit clause $\overline{x}_{3}$ so we know $x_{3}$ must be $\text{false}$ and we can perform unit resolution again to yield:

\begin{equation}\phi \land x_{1}\land \overline{x}_{3} = x_{2} \land (\overline{x}_{2} \lor x_{4})\land \overline{x}_{4}. \tag{12}\end{equation}

This time, we have two unit clauses. We can perform unit resolution with respect to either. We’ll choose $x_{2}$ so we now now that $x_{2}$ is $\text{true}$ and we get:

\begin{equation}\phi \land x_{1}\land\overline{x}_{3}\land x_{2} = x_{4} \land \overline{x}_{4} =\text{false}. \tag{13}\end{equation}

Clearly this is a contradiction, and so we conclude that the formula is not satisfiable if we set $x_{1}$ to $\text{true}$

We now repeat this process with $x_{1}$ = $\text{false}$, which gives

\begin{eqnarray}    \phi \land \overline{x}_{1} &=&(x_{1}\lor \overline{x}_{2}) \land (\overline{x}_{1}\lor \overline{x}_{3}) \land (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{4})\land (\overline{x}_{4} \lor x_{3}) \land \overline{x}_{1} \nonumber \\    &=& \overline{x}_{2} \land (x_{2} \lor x_{3}) \land (\overline{x}_{2} \lor x_{4}) \land (\overline{x}_{4} \lor x_{3}). \tag{14}\end{eqnarray}

Once more, this leaves a unit clause $\overline{x}_{2}$, so we set $x_{2}$ to $\text{false}$ and perform unit resolution again to get

\begin{equation}\phi \land \overline{x}_{1}\land \overline{x}_{2} = x_{3} \land (\overline{x}_{4} \lor x_{3}) \tag{15}\end{equation}

which gives the unit clause $x_{3}$ and so we set $x_{3}$ to $\text{true}$. Now something different happens. The entire of the right hand side disappears. Since there are no clauses left to be satisfied, the formula is satisfiable:

\begin{equation}\phi \land \overline{x}_{1}\land \overline{x}_{2} \land x_{3}= \text{true} \tag{16}\end{equation}

Note that the formula is satisfiable regardless of the value of $x_{4}$ (it is on neither side of the equation) so we have found two satisfiable solutions $\{\overline{x}_{1},\overline{x}_{2},x_{3},x_{4}\}$ and $\{\overline{x}_{1},\overline{x}_{2},x_{3},\overline{x}_{4}\}$.

Complexity: If there are $V$ variables, there are at most $V$ rounds of unit resolution for each of the two values of the initial conditioned variable. Each unit resolution procedure is linear in the number of clauses $C$ so the algorithm has total complexity $\mathcal{O}[CV]$.

It’s possible to reach a case where the chain of unit propagation stops and we have to condition on one of the remaining variables to start it again. However, this only occurs when subsets of the variables have no interaction with one another and so it does not add to the complexity.

Directional resolution

Now consider what happens if we apply the unit resolution approach above to a 3-SAT problem. When we condition on the first variable $x_{i}$, we remove clauses that contain $x_{i}$ and remove $\overline{x}_{i}$ from the rest of the clauses. Unfortunately, this doesn’t create another unit clause (at best it just changes a subset of the 3-clauses to 2-clauses), and so it’s not clear how to proceed.

Directional resolution is a method that uses resolution to tackle $3$-SAT and above. The idea is to choose an ordering of the variables and then perform all possible resolution operations with each variable in turn before moving on. We continue until we find a contradiction or reach the end. In the latter case, we work back in the reverse order to find the values that satisfy the expression.

Worked example: Again, this is best understood via a worked example. Consider the formula:

\begin{eqnarray}&\phi:= &(x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}) \land (\overline{x}_{1}\lor x_{3}\lor \overline{x}_{5}) \land \\&&\hspace{1cm}(x_{2} \lor x_{3}\lor \overline{x}_{4}) \land (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}) \land (\overline{x}_{3} \lor x_{4}\lor x_{5}).\nonumber \tag{17}\end{eqnarray}

We sort the clauses into bins. Those containing $x_{1}$ or $\overline{x}_{1}$ are put in the bin 1 and any remaining clauses containing $x_{2}$ or $\overline{x}_{2}$ are put in bin 2 and so on:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor x_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: (x_{2} \lor x_{3}\lor \overline{x}_{4}), (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}) \nonumber \\&& x_{3}: (\overline{x}_{3} \lor x_{4}\lor x_{5}) \nonumber \\&& x_{4}: \nonumber \\&& x_{5}: \tag{18}\end{eqnarray}

We work through these bins in turn. For each bin we perform all possible resolutions and move the resulting generated clauses into subsequent bins. So for bin 1 we resolve the clauses $(x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4})$ and $(\overline{x}_{1}\lor x_{3}\lor \overline{x}_{5})$ with respect to $x_{1}$ to get the new clause $\color{BurntOrange} (\overline{x}_{2}\lor \overline{x}_{4}\lor x_{3}\lor \overline{x}_{5})$. We add this to bin 2 as it contains a $x_{2}$ term:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor x_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: (x_{2} \lor x_{3}\lor \overline{x}_{4}), (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}), \color{BurntOrange}(\overline{x}_{2}\lor \overline{x}_{4}\lor x_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{3}: (\overline{x}_{3} \lor x_{4}\lor x_{5}) \nonumber \\&& x_{4}: \nonumber \\&& x_{5}: \tag{19}\end{eqnarray}

We then consider bin 2 and resolve the clauses with respect to $x_{2}$ in all possible ways. In bin 2 there is one clause containing $x_{2}$ and we can resolve it against the two clauses containing $\overline{x}_{2}$. This creates two new clauses that we simplify and add to bin 3 since they contain terms in $x_{3}$:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: (x_{2} \lor x_{3}\lor \overline{x}_{4}), (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}), (\overline{x}_{2}\lor \overline{x}_{4}\lor x_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{3}: (\overline{x}_{3} \lor x_{4}\lor x_{5}), \color{BurntOrange} (x_{3}\lor \overline{x}_{4} \lor x_{5})\color{Black}, \color{BurntOrange}(x_{3} \lor \overline{x}_{4}\lor \overline{x}_{5}) \nonumber \\&& x_{4}: \nonumber \\&& x_{5}: \tag{20}\end{eqnarray}

Now we consider bin 3. Again, there are three clauses here and combining them with resolution creates two new clauses. Resolving the first and second clause with respect to $x_{3}$ creates $(x_{4}\lor \overline{x}_{4} \lor x_{5})$ which evaluates to $\text{true}$ since either $x_{4}$or $\overline{x}_{4}$ must always be $\text{true}$. Similarly, combining the first and third clause creates the clause $x_{4}\lor x_{5}\lor \overline{x}_{4}\lor \overline{x}_{5}$ which evaluates to $\text{true}$ and so we are done. At this point, we can say that the formula is $\text{SAT}$ as we have not created any contradictions of the form $x_{i}\land\overline{x}_{i}$ during this resolution process

Finding the certificate: To find an example that satisfies the expression, we work backwards, setting the bin value to $\text{true}$ or $\text{false}$ in such a way that it satisfies the clause. There are no clauses in bin 5 and so we are free to choose either value. We’ll set $x_{5}$ to be $\text{true}$. Similarly, there are no clauses in bin 4 and so we will arbitrarily set $x_{4}$ to $\text{true}$ as well. After these changes we have:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: (x_{2} \lor x_{3}\lor \overline{x}_{4}), (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}), (\overline{x}_{2}\lor \overline{x}_{4}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{3}: (\overline{x}_{3} \lor x_{4}\lor x_{5}), (x_{3}\lor \overline{x}_{4} \lor x_{5}) \nonumber \\&& x_{4}: \text{true} \nonumber \\&& x_{5}: \text{true} \tag{21}\end{eqnarray}

Now we consider the third bin. We substitute in the values for $x_{4}$ and $x_{5}$ and see that both clauses evaluate to $\text{true}$, regardless of the value of $x_{3}$, so again, we can choose any value that we want. We’ll set $x_{3}$ to $\text{false}$ to give:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: (x_{2} \lor x_{3}\lor \overline{x}_{4}), (\overline{x}_{2} \lor \overline{x}_{4}\lor x_{5}), (\overline{x}_{2}\lor \overline{x}_{4}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{3}: \text{false}\nonumber \\&& x_{4}: \text{true} \nonumber \\&& x_{5}: \text{true} \tag{22}\end{eqnarray}

Progressing to the second bin, we observe that the second and third clause are already satisfied by the previous assignments, but the first clause is not since $x_{3}$ is $\text{false}$ and $x_{4}$ is $\text{true}$. Consequently, we must satisfy this clause by setting $x_{2}$ to $\text{true}$:

\begin{eqnarray}&& x_{1}: (x_{1}\lor \overline{x}_{2} \lor \overline{x}_{4}), (\overline{x}_{1}\lor \overline{x}_{3}\lor \overline{x}_{5}) \nonumber \\&& x_{2}: \text{true} \nonumber \\&& x_{3}: \text{false} \nonumber \\&& x_{4}: \text{true} \nonumber \\&& x_{5}: \text{true} \tag{23}\end{eqnarray}

Finally, we consider the first bin. We note that the second clause is satisfied because $x_{3}$ is $\text{false}$ but the first clause is not and so to satisfy it, we must set $x_{1}$ to $\text{true}$ and now we have a satisfying example.

Complexity: The directional resolution procedure works, but is not especially efficient. For large problems, the number of clauses can expand very quickly: if there were $C$ clauses and half contain $x_{1}$ and the other half $\overline{x}_{1}$ then we could create $C^{2}/4$ new clauses in the first step. For a $K$-SAT problem, each of these clauses are larger than the original ones with size $2(K-1)$.

It is possible to improve the efficiency. Any time we generate a unit clause, we can perform unit propagation which may eliminate many variables. Also in our example we organized the bins by the variable index, but this was an arbitrary choice. This order can have a big effect on the total computational cost and so careful selection can improve efficiency. However, even with these improvements, this approach is not considered viable for large problems.

SAT solving algorithms based on conditioning

In this section, we will develop algorithms that are fundamentally centered around the conditioning operation (although they also have unit resolution embedded). We’ll describe both the DPLL algorithm and clause learning algorithms which underpin most modern SAT solvers. To understand these methods, we first need to examine the connection between conditioning and tree search.

SAT as binary search

We’ll use the running example of the following Boolean formula with $C=7$ clauses and $V=4$ variables:

\begin{eqnarray}\label{eq:SAT_working_example}\phi&:=&(x_{1} \lor x_{2}) \land (x_{1} \lor \overline{x}_{2} \lor \overline{x}_{3} \lor x_{4}) \land (x_{1} \lor \overline{x}_{3} \lor \overline{x}_{4}) \land \\&&\hspace{0.5cm} (\overline{x}_{1} \lor x_{2} \lor \overline{x}_{3}) \land (\overline{x}_{1} \lor x_{2} \lor \overline{x}_{4}) \land (\overline{x}_{1} \lor x_{3} \lor x_{4}) \land (\overline{x}_{2} \lor x_{3}) \nonumber, \tag{24}\end{eqnarray}

Consider conditioning on variable $x_{1}$ so that we have:

\begin{equation}    \phi = (\phi \land \overline{x}_{1}) \lor (\phi \land x_{1}). \tag{25}\end{equation}

This equation makes the obvious statement that in any satisfying solution $x_{1}$ is either $\text{true}$ or $\text{false}$. We could first investigate the case where $x_{1}$ is $\text{false}$. If we establish this is $\text{SAT}$ then we are done, and if not we consider the case where $x_{1}$ is $\text{true}$. Taking this one step further, we could condition each of these two cases on $x_{2}$ to get:

\begin{equation}    \phi = ((\phi \land \overline{x}_{1}) \land \overline{x}_{2}) \lor ((\phi \land \overline{x}_{1}) \land \overline{x}_{2}) \lor ((\phi \land x_{1})\land \overline{x}_{2} ) \lor ((\phi \land x_{1})\land x_{2} ). \tag{26}\end{equation}

and now we could consider each of the four combinations $\{\overline{x}_{1}\overline{x}_{2}\},\{\overline{x}_{1}x_{2}\},\{x_{1}\overline{x}_{2}\}$ and $\{x_{1}x_{2}\}$ in turn, terminating when we find a solution that is $\text{SAT}$.

One way to visualise this process is as searching through a binary tree (figure 1). At each node of the tree we branch on one of the variables. When we reach a leaf, we have known values for each variable and we can just check if the solution is $\text{SAT}$.

Image - T10_1.png

Figure 1. SAT as binary search. At each node of the search tree we condition on a variable, splitting into a left sub-tree in which this variable is set to $\text{false}$ and a right sub-tree in which it is set to $\text{true}$. There is one level in the tree for each variable so that at each leaf all the variables are set and we can test if the formula is satisfied. For the example in equation 24, the formula evaluates to $\text{false}$ for the first 14 leaves and the individual clauses that are violated are indicated in grey at each leaf. The last two leaves are both satisfying solutions. In practice, we would stop searching when we found the first satisfying solution and so we would only need to test 15 leaves in this example.

This example was deliberately constructed to be pathological in that the first 14 combinations (or equivalently leaves of the tree) all make the formula evaluate to $\text{false}$. These are signified in the plot by red crosses. We number the clauses:

\begin{eqnarray}&& 1:(x_{1} \lor x_{2}) \nonumber\\&& 2:(x_{1} \lor \overline{x}_{2} \lor \overline{x}_{3} \lor x_{4}) \nonumber\\&& 3:(x_{1} \lor \overline{x}_{3} \lor \overline{x}_{4}) \nonumber\\&& 4:(\overline{x}_{1} \lor x_{2} \lor \overline{x}_{3}) \nonumber\\&& 5:(\overline{x}_{1} \lor x_{2} \lor \overline{x}_{4}) \nonumber\\&& 6:(\overline{x}_{1} \lor x_{3} \lor x_{4}) \nonumber\\&& 7:(\overline{x}_{2} \lor x_{3}) \tag{27}\end{eqnarray}

and for each leaf of the tree in figure 1, the clauses that were contradicted are indicated in grey. In this case, both of the last two combinations (leaves) satisfy the formula, and once we find the first one ($x_{1}, x_{2}, x_{3},\overline{x}_{4})$ we can return SAT.

Note, we have not yet obviously made the algorithm more efficient. We might still have to search all $2^{V}$ combinations of variables to establish satisfiability or lack thereof. However, viewing SAT solving as tree search is the foundation that supports more efficient algorithms.

Efficient binary search

We can immediately improve the efficiency of the binary search method by some simple bookkeeping. As we pass through the tree we keep track of which clauses are satisfied and which are not. As soon as we find one that is not satisfied, we do not need to explore further and we can backtrack. Similarly, if we find a situation where all of the clauses are already satisfied before we reach a leaf then we can return $\text{SAT}$ without exploring further. This means that the variables below this point can take any value.

In our worked example, when we pass down the first branch and set $x_{1}$ to $\text{false}$ and $x_{2}$ to $\text{false}$ we have already contradicted clause 1 which was $(x_{1} \lor x_{2})$, and so there is no reason to proceed further. Continuing in this way we only need to search a subset of the full tree (figure 2). We find the first satisfying solution when $x_{1},x_{2},x_{3}=$$\text{true}$,$\text{true}$,$\text{true}$ and need not continue to the leaf. As we saw from the full tree in figure 1, the setting of $x_{4}$ is immaterial.

Image - T10_2.png

Figure 2. Efficient binary search. With some simple bookkeeping, tree search can be made much more efficient. If we track the status of each clause then we can backtrack as soon as one of the clauses is violated. Again, the index of the violated clause is shown in grey. Similarly, when we have satisfied all of the clauses we can return $\text{SAT}$ even though we have not yet reached a leaf. The variables below this can be set to any value.


We can also consider the tree search from an algebraic point of view. Each time we make a decision at a node in the tree, we are conditioning on a given variable. So when we set $x_{1}$ to $\text{false}$, the resulting formula is 

\begin{eqnarray}\label{eq:sat_tree_cond}\phi\land \overline{x}_{1} := x_{2} \land (\overline{x}_{2} \lor \overline{x}_{3} \lor x_{4}) \land (\overline{x}_{3} \lor \overline{x}_{4}) \land (\overline{x}_{2} \lor x_{3}), \tag{28}\end{eqnarray}

where we have used the usual recipe of removing all clauses containing $\overline{x}_{1}$ and removing the term $x_{1}$ from the remaining clauses.

The Davis–Putnam–Logemann-Loveland (DPLL) algorithm takes tree search one step further, by embedding unit propagation into the search algorithm (figure 3). For example, when we condition on $\overline{x}_{1}$ and yield the new expression in equation 28, we generate the unit clause $x_{2}$. We can perform unit resolution using $x_{2}$ to get:

\begin{eqnarray}\phi\land \overline{x}_{1}\land x_{2} := (\overline{x}_{3} \lor x_{4}) \land (\overline{x}_{3} \lor \overline{x}_{4}) \land x_{3}, \tag{29}\end{eqnarray}

which creates another unit clause $x_{3}$. Applying unit resolution again we yield the contradiction $x_{4}\land \overline{x}_{4}$ and need proceed no further.


​Figure 3. DPLL algorithm. By performing unit propagation where possible, we can eliminate many variables very efficiently. In this case, once we condition on $\overline{x}_{1}$, this creates a unit clause in $x_{2}$, which starts a chain of unit resolution operations that establishes a contradiction. Similar effects happen at other points in the tree.

To summarize, the DPLL algorithm consists of tree search, where we perform unit propagation whenever unit clauses are produced. Since unit resolution can be done in linear time, this is much more efficient than the tree search that it replaces.

Note that in our worked example, the unit propagation process always generated a contradiction or a $\text{SAT}$ solution. However, this is not necessarily the case in a larger problem. After unit resolution there will usually be non-unit clauses left containing the remaining variables have neither been conditioned on, nor eliminated using unit resolution. At this point, we condition on the next available variable and continue down the tree, performing unit resolution when we can (figure 4).

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Figure 4. DPLL algorithm in practice. In a real problem the DPLL algorithm will alternate between conditioning on variables and performing unit resolution. The effect of this is that we condition on different variables in different paths of the tree.

Conflict Driven Clause learning

The DPLL algorithm makes SAT solving by tree search much more efficient, but there can still be considerable wasted computation. Consider the case where we have set $x_{1}$ to $\text{false}$ and then set $x_{2}$ to $\text{false}$ (figure 5). However, imagine that there are clauses that mean that when $x_{2}$ is $\text{false}$, there is no way to set the variables $x_{3}$ and $x_{4}$ in a valid way. For example, the following combination of clauses will achieve this:

\begin{equation}    (x_{2} \lor x_{3}\lor x_{4}) \land (x_{2} \lor x_{3}\lor \overline{x}_{4}) \land (x_{2} \lor \overline{x}_{3}\lor x_{4}) \land (x_{2} \lor \overline{x}_{3}\lor \overline{x}_{4}). \tag{30}\end{equation}

As we work through the sub-tree in the blue region in figure 5, we duly establish that there is no possible solution.

As we search through the tree, we will eventually come to another place where we set $x_{2}$ to $\text{false}$ and now we must work through exactly the same calculations again to establish that there is no valid solution (yellow region in figure 5). In a large problem this may happen many times.

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Figure 5. Motivation for conflict-driven clause learning. Consider the case where setting $x_{2}$ to $\text{false}$ inevitably results in a conflict where there is no way to set $x_{3}$ and $x_{4}$. Without taking action, we will have to repeat the computations to find this conflict in every sub-tree where $x_{2}$ is $\text{false}$ (blue and yellow rectangles). When conflict-driven clause learning algorithms find a conflict in a sub-tree, they add a new clause to the original expression that prevents redundant exploration of sub-trees.

Conflict-driven clause learning aims to reduce this redundancy. When a conflict occurs, the cause is found and we add a new clause to the original statement that prevents exploration of redundant sub-trees. For example, in this simple case, we could add the clause $(x_{2})$ which would prevent exploration of trees where $x_{2}$ is $\text{false}$.

Unfortunately, the causes of a conflict are usually more complex than a single variable. To find the combinations of variables that are ultimately responsible for the conflict, we build a structure called an implication graph as we search through the tree.

Figure 6a provides a concrete example of a SAT problem where there are 11 clauses and 10 variables. Figure 6b illustrates the situation where we are mid-way through the DPLL search in which we have interleaved processes of conditioning (blue shaded areas) and unit resolution (yellow shaded areas). We have just established a conflict at clause 11 (at the blue arrow) which cannot be satisfied when we set $x_{5}$ to $\text{true}$.

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Figure 6. Implication graphs. a) A SAT problem with 11 clauses in 10 variables. b) We are mid-way though the DPLL process (at the blue arrow) and have just found a conflict at clause 11. c) The implication graph representing the current state. Each vertex corresponds to a variable (blue if conditioned, yellow if inferred by unit resolution). The incoming edges to yellow vertices correspond to the variables that caused the vertex variable to be inferred and are labelled with the relevant clause. So, for example, $x_{6}$ is set to $\text{false}$ by clause $c_{2}$ because $x_{1}$ is $\text{false}$ and $x_{2}$ is $\text{true}$. The conflict results when setting variable $x_{5}$ implies both $x_{7}$ and it’s complement $\overline{x}_{7}$.

Figure 6c is the implication graph associated with this point in the search, which contains all of the variables that we have established so far. The literals $\overline{x}_{1}, x_{2},x_{3}, x_{5}$ that we conditioned on are depicted with blue vertices and the literals $x_{4}, \overline{x}_{6},x_{10},x_{9}, x_{7}$ and $\overline{x}_{7}$ that resulted from unit propagation are shown as yellow vertices. Each edge depicts a contribution to the unit resolution process. For example, the edge between $\overline{x}_{1}$ and $x_{4}$ represents the fact that when $x_{1}$ is set to $\text{false}$, we must set $x_{4}$ to $\text{true}$. This is due to clause 1 and the edge is accordingly labelled with $c_{1}$. Similarly, we can see that $x_{10}$ has become $\text{true}$ by clause $c_{3}$ because previously in the search process $x_{1}$ and $x_{6}$ were both set to $\text{false}$. 

You can see from this implication graph exactly how the conflict happened. When we condition on $x_{5}$, clause $c_{6}$ implied that $x_{7}$ must be $\text{true}$ given that $x_{2}$ and $x_{5}$ were both $\text{true}$, but clause $c_{11}$ implied that $x_{7}$ must be $\text{false}$ given that $x_{5}$ and $x_{10}$ were both $\text{true}$. So, one interpretation is that the conflict is inevitable given states $x_{2},x_{5}$ and $x_{10}$ that were inputs to these contradictory clauses.

However, this is not the only interpretation. For example, if $x_{10}$ is one of the proximal causes, then this was only set to $\text{true}$ because we previously set $x_{1}$ and $x_{6}$ to $\text{false}$. So maybe we should attribute the contradiction not to variables $x_{2},x_{5},x_{10}$ but to variables $x_{2},x_{5},\overline{x}_{6},\overline{x}_{1}$. We can use the implication graph to find alternative explanations. Any cut of the graph that separates the conditioning (blue) variables from the conflict defines an explanation (figure 7). The explanatory variables are the source vertices of the edges that were cut.

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​Figure 7. Cutting the implication graph to find the causes of a conflict. Any cut that separates the conditioned nodes (in blue) from the conflict can be interpreted as a possible cause. a) In this case, the cause of the conflict is attributed to variables $x_{2}, x_{5}, x_{10}$, since the edges from these variables are cut. b) In this second case, the cause is attributed to variables $\overline{x}_{1}, x_{2}, x_{5}$.

Having established a cause, we must now derive a new clause that prevents the SAT solver from exploring similar dead-ends in the future. If the case was attributed to $\overline{x}_{1}, x_{2}, x_{5}$, then we would add the clause $(x_{1}\lor \overline{x}_{2} \lor \overline{x}_{5})$ to prevent this combination happening. We continue exploring the tree, by jumping back up the tree structure to a sensible point and resuming with this new constraint.

Machine leaning powered CDCL

The previous discussion outlined the main ideas of conflict-driven clause learning algorithms, but there are many additional choices to be made in a modern system. For example, we must decide the order of variables to condition on. In our examples, we have done this in numerical order, but this choice was arbitrary and there is no particular reason to evaluate them in the same order as we go down different branches in the tree. Much work is devoted to developing heuristics to making this choice. For example, we might prioritize variables that are in short clauses, with the goal of triggering unit propagation earlier. Alternatively, we might prioritize variables that are in lots of clauses as this will simplify the expression a great deal. 

There are also many other decisions to make. In CDCL, we must choose which of many potential explanations for a conflict is superior and decide exactly where we should jump back to in the tree. Some solvers periodically restart the solution process to avoid wasting the available computation time fruitlessly searching a single branch, and we must decide when exactly to perform these restarts. One approach to making these decisions is to use machine learning to guide the choices.

For example, Liang et al., (2016) developed uses a reward function to choose the order in which variables in a CDCL solver are considered. A reward function $r[i]$ is defined for each variable $x_{i}$:

\begin{equation}r[i] \propto \frac{1}{numConflicts-lastConflict[i])} \tag{31}\end{equation}

Here $numConflicts$ keeps track of the total number of conflicts the solver has encountered so far whereas $lastConflict[i]$ keeps track of the last time variable $x_{i}$ was involved in a conflict. From this we can see that a variable which was recently involved in a conflict would get a high reward. This reward is then incorporated into a score function:

\begin{equation}Q[i] \longleftarrow (1-\alpha)Q[i] + \alpha r[i] \tag{32}\end{equation}

At any iteration where a new variable must be selected for conditioning, the variable with the highest score is picked, provided that it is not currently already conditioned on. This is known as the conflict history branching heuristic.

In the above formulation, the terms appearing in the reward will always be known and can hence be computed exactly. However, Liang et al., (2016) also dealt with a case where the reward function is defined such that the terms appearing in it have associated uncertainty. This new reward definition improves the branching heuristic and the authors show how tools from a Multi-Armed Bandit framework in RL can be used to estimate the uncertainty and further improve performance.

In further work Liang et al., 2017 the same authors show how gradient based methods can be used to optimize another branching heuristic, one based on how many learnt clauses can be obtained from each decision. Other examples of machine learning in the SAT community include Nejati et al., (2017) where reinforcement learning is used to decide when to restart the solver.


In this blog, we introduced resolution and conditioning operations. We then developed a series of algorithms based on these operations. We started with using resolution (via unit propagation) to efficiently solve 2-SAT and then investigated the directional resolution for 3-SAT and above. We re-framed the SAT solving problem as tree search where conditioning is used at each branch in the tree. This led to the DPLL and CDCL algorithms.

For further information about SAT solving including the DPLL and CDCL algorithms, consult the Handbook of Satisfiability. Most chapters are available on the internet if you search for their titles individually. A second useful resource is Donald Knuth’s Facsicle 6.

In part III of this article, we’ll investigate a completely different approach to SAT solving that relies on belief propagation in factor graphs. Finally, we’ll show how the machinery of SAT solving can be extended to continuous variables by introducing satisfiability module theory (SMT) solvers.