SAT solving with higher-order continuations

Posted on October 22, 2022

Finding satisfying assignments to the variables of a boolean formula, called SAT-solving, is a model of a whole class of problems. This means that there are many problems out there whose solutions can be implemented in terms of finding such a satisfying assignment.

I want to share a particularly mind-bending and (I think) elegant way of finding a satisfying assignment to a boolean formula by brute-force search. The idea is to do the search using continuation-passing style (CPS), but with one trick.


At every point of a program’s execution, it is associated to some execution state. This state represents what remains to be done with the result of evaluating the current expression. Normally, this execution state is implicit, and in particular, is implied by the programming language’s order of operations.

For example, suppose we are performing a postorder traversal of a binary tree, collapsing it by summing the integer values contained within each node of the tree.

type tree = Leaf | Node of tree * int * tree
let rec sum = function
  | Leaf -> 0
  | Node (l, x, r) ->
    let a = sum l in
    let b = sum r in
    a + b + x (* writing the order of operations explicitly *)

When we encounter some node of the tree, we had to get there somehow. On one hand, if this node is the root of the tree, then there is nothing to do with the calculated sum; this is the starting position. The calculated sum is simply the result of the program. On the other hand, if we reached this node as a right subtree of a parent node, then what remains to be done with the result of this node is to add it with the parent node’s left subtree sum (a) and integer value (x).

This order of operations is captured concretely by the call stack.

Continuations enter the picture as a representation of the call stack.

Rather than let the language handle the call stack, we will represent it ourselves using functions, in particular using closures.

let rec sum' t k = match t with
  | Leaf -> k 0
  | Node (l, x, r) ->
    sum l @@ fun a ->
    sum r @@ fun b ->
    k (a + b + x)

In this implementation, rather than using let and relying on the language implementation to create stack frames for us, we are passing explicit functions expressing what to do with the result.

The parameter k is a continuation. It contains an accumulation of pending operations to be performed once the result of processing the given tree is ready. We can make a few observations about this code.

Overall, this merely looks like a more cumbersome way of writing code. However, it does buy us some interesting abilities. Since the execution state is explicit, it becomes possible to store it! This enables us to “time travel” back to a previous point of the program’s execution. It is exactly this ability that I exploit in implementing a backtracking SAT-solver.

SAT-solving with continuations

A pedestrian way of implementing a backtracking SAT solver is first to write an evaluator for boolean formulas, and then to write an enumerator of truth assignments. We try each truth assignment with the evaluator until we find one that makes the formula true.

Here is what such an evaluator looks like.

type formula =
  | Conj of formula * formula
  | Disj of formula * formula
  | Neg of formula
  | Var of string
type env = (string * bool) list

let rec eval (r : env) = function
  | Var x -> List.assoc x r
  | Neg e -> not (eval r e)
  | Conj (e1, e2) -> eval r e1 && eval r e2
  | Disj (e1, e2) -> eval r e1 || eval r e2

What remains to implement a solver in this way is to collect all the variable names present in the formula, and enumerate all possible assignments.

Instead, we can do it all in one pass using CPS.

The crucial observation is that when we encounter a variable x, we may or may not have already picked a value for it. If we haven’t picked a value for x yet, then we want to try to evaluate the rest of the tree with x = true and if that doesn’t work, then with x = false.

What if that still doesn’t work? Then we have to go back to the previous variable and try a different value for it.

What if we run out of previous variables to go back to? Then the formula is unsatisfiable; we will have tried every truth assignment and not found one that makes the formula true.

The mind-bending aspect of this solution is that there are two stacks, not just one. We will have an “ordinary” call stack for the evaluator, but also have a secondary stack whose frames are introduced when we pick tentative values for variables. This secondary stack’s frames will contain a copy of the evaluator stack at that point, enabling the solver to jump back to that point and try evaluating the rest of the tree but with an adjusted truth assignment, in particular with x = false.

I twice emphasize “evaluate the rest of tree” above because that is precisely the current continuation of the evaluator at the point that we encounter the variable.

Let’s see how the code looks.

let rec solve (r : env) (fail : unit -> 'r) (phi : formula)
    (assign : env -> bool -> (unit -> 'r) -> 'r) : 'r =
  match phi with
  | Var x -> begin match List.assoc_opt x r with
    | Some b -> (* we already have a value for `x` *)
      assign r b fail
    | None -> (* we don't have a value for `x` *)
      assign ((x, true) :: r) true @@ fun () ->
      assign ((x, false) :: r) false fail
  | Conj (e1, e2) -> 
    solve r fail e1 @@ fun r b1 fail ->
    if b1 then (* short-circuiting *)
      solve r fail e2 @@ fun r b2 fail -> assign r (b1 && b2) fail
      assign r false fail
  | Disj (e1, e2) ->
    solve r fail e1 @@ fun r b1 fail ->
    if b1 then
      assign r true fail
      solve r fail e2 @@ fun r b2 fail ->
      assign r (b1 || b2) fail

When I first wrote this code, I was frankly surprised that it worked, because I didn’t really understand why it worked. Since then I’ve thought a lot about it, and I will try to offer an intuitive explanation.

Let’s understand the fail continuation as a stack of variables and execution states to return to. To see this, consider that the only place where we construct a value for this continuation is in the case Var, and the free variables of that continuation are assign, x, r, and fail.

Given that assign is the main continuation, we should understand its inputs as what are supposed to be the outputs of solve.

There is a little something missing from solve above. How exactly are we supposed to call it? In other words, what initial continuations should we provide, to recover a function formula -> env option that decides whether a formula is satisfiable, giving in that case a satisfying assignment?

We need to pick an initial frame for the fail stack. This frame will be reached only if we try all the values of all the variables – we return None here as the formula is unsatisfiable.

We need to pick an initial frame for the assign stack. This frame will be reached when we finish evaluating the formula, and we will be given the assignment that resulted in the given bool. Also, we will receive the failure continuation, which we can call to try different values of the variables until we get a bool we like. Of course, the initial env that we pass will be empty, as we haven’t seen any variables yet.

let solve' (phi : formula) : env option =
  solve [] (fun () -> None) phi @@ fun r result fail ->
  if result then Some r else fail ()

Unexpected benefits

In contrast with the so-called pedestrian solver that enumerates assignments and tries them, the implementation of solve above has an interesting performance benefit: it caches partial results. Shout-out to Max Kopinsky (github) for noticing this. To be clear, it isn’t obvious that this is a net benefit, since we are using extra memory to hold an evaluator stack for each variable, but it is interesting nonetheless!

To see this caching, suppose we have a fairly large formula with eight variables. Let’s order the variables according to when they first appear in a left-to-right traversal. This is exactly the order in which variables are added to the environment by solve. Under this order, by the time we encounter the eighth variable, we have already evaluated some chunk of the formula that didn’t depend on that variable. When we finish evaluating the formula and see that result = false, we call fail, jumping back to the eighth variable, and we reevaluate with x8 = false only the part of the formula that depended on that variable!


This article explains execution states, continuations as representations thereof, and shows an interesting use of them to implement a backtracking search by storing continuations in the closures of other continuations. This layering of continuations gives rise to a fascinating kind of control flow, enabling backtracking to certain leaves of a tree to make different choices at those points.

It is not immediately obvious how one can recover a nice, “direct” implementation of solve above, without using continuations. There is, however, a general way of eliminating higher-order functions called defunctionalization. In short, whereas CPS gives us control of stack frames, defunctionalization forces us moreover to manually construct and destruct closures. Defunctionalizing solve gives us a program using explicit stacks in the form of lists of frames, represented as explicit data structures. This representation makes it plainly obvious that the failure continuation stores a copy of the evaluator continuation.

Looking at the defunctionalization of solve in more details would be the topic of a future post!