Solving SAT via Positive Supercompilation

Supercompilation is a deep program transformation technique due to V. F. Turchin, a prominent computer scientist, cybernetician, physicist, and Soviet dissident.

He described the concept as follows ¹:

A supercompiler is a program transformer of a certain type. The usual way of thinking about program transformation is in terms of some set of rules which preserve the functional meaning of the program, and a step-by-step application of these rules to the initial program. … The concept of a supercompiler is a product of cybernetic thinking. A program is seen as a machine. To make sense of it, one must observe its operation. So a supercompiler does not transform the program by steps; it controls and observes (SUPERvises) the running of the machine that is represented by the program; let us call this machine M₁. In observing the operation of M₁, the supercompiler COMPILES a program which describes the activities of M₁, but it makes shortcuts and whatever clever tricks it knows in order to produce the same effect as M₁, but faster. The goal of the supercompiler is to make the definition of this program (machine) M₂ self-sufficient. When this is acheived, it outputs M₂ in some intermediate language L^sup and simply throws away the (unchanged) machine M₁.

A supercompiler is interesting not only as a program transformer but also as a very general philosophical concept:

The supercompiler concept comes close to the way humans think and make science. We do not think in terms of rules of formal logic. We create mental and linguistic models of the reality we observe. How do we do that? We observe phenomena, generalize observations, and try to construct a self-sufficient model in terms of these generalizations. This is also what the supercompiler does. … A supercompiler would run M₁, in a general form, with unknown values of variables, and create a graph of states and transitions between possible configurations of the computing system. … To make it finite, the supercompiler performs the operation of generalization on the system configurations in such a manner that it finally comes to a set of generalized configurations, called basic, in terms of which the behavior of the system can be expressed. Thus the new program becomes a self-sufficient model of the old one.

Since the above explanations are hard to comprehend at first sight, I will try to illustrate supercompilation by example.

Let the object language of supercompilation be a simple first-order functional language called SLL, which stands for Simple Lazy Language. It consists of:

• Variables: just regular symbolic identifiers.
  • Examples: a, b, c.
• Constructors: symbolic identifiers that wrap their arguments of arbitrary structure.
  • Examples: Foo(a, b, c), Bar(a, Foo(a), c).
• F-functions: indifferent functions that just transform their arguments.
  • Examples:
    • foo(a, b, c) = Foo(a, b, Bar(c));
    • bar(a, b, c) = foo(Bar(a), b, c);
• G-functions: curious functions that pattern-match on their first argument.
  • Examples: see below.

Suppose that natural numbers are represented as Z, S(Z), S(S(Z)), and so on ². Then, we can define addition as the following g-function (from ³):

add(Z, y) = y;
add(S(x), y) = S(add(x, y));

A supercompiler’s input is a pair of an expression and program, the latter being a set of definitions of f- and g-functions. This pair is usually called a task.

Let the task for a hypothetic supercompiler be the expression add(S(Z), S(S(Z))) together with the definition of add above. In this case, the work of a supercompiler is as simple as sequential reduction of the initial expression to the target expression S(S(S(Z))), according to the rules of add:

However, to think of a supercompiler as of a mere expression evaluator is a grave mistake. Let us consider what happens when it encounters a variable that does not stand for some concrete expression. For example, let the task be add(S(S(Z)), b) with the same definition of add, where b is understood as “any” expression:

It just works!

A supercompiler saw the variable b when trying to reduce S(S(add(Z, b))), and then it just substituted y (from the rule add(Z, y)) for this b. After that, supercompilation has stopped because the actual expression for b is unknown at this point. As a result, we got S(S(b)), which is equivalent to the initial expression.

Now consider what happens if there is a need to pattern-match on an unknown variable. In this case, we cannot just proceed with “direct” computation since there are several possibilities of the form the variable may take. Suppose that the task is add(a, 2) ⁴ with the same function add. What a supercompiler does is that it analyze the expression add(a, 2) according to all the possibilities of a, which are either Z or S(v1), where v1 is some fresh variable identifier. The situation looks like this:

A supercompiler has built an (incomplete) process tree that describes the execution of add(a, 2) in a general sense. In the first branch, a is substituted for Z according to the first rule of add; in the second branch, a is substituted for S(v1) according to the second rule of add. The resulting two nodes are labelled with expressions that resulted in reducing a particular substitution of the parent expression.

However, the supercompilation is not complete yet: there is still a node labelled as S(add(v1, 2)). A supercompiler decides to decompose it, meaning to move add(v1, 2) out of S(...) in the following way:

After that, if we proceed with supercompiling add(v1, 2), we will eventually arrive at the initial expression add(a, 2). This is because the expressions add(v1, 2) and add(a, 2) are alpha equivalent, meaning that they only differ in the names of variables. A supercompiler should be smart enough to detect this situation of alpha equivalence and, instead of continuing infinite supercompilation, just draw a back arrow from add(v1, 2) to the initial node as depicted below:

Hooray, the supercompilation of add(a, 2) is now complete!

What is left is to perform residualization – the process of converting a process tree to an SLL task, a self-sufficient model of the initial program. Every node in the graph (except for leaf nodes) is transformed into a function whose parameters are the set of free variables from the node expression; each function is given a unique name. For example, if we see the node add(a, 2) pointing to 2 and S(add(v1, 2)), we must generate a new g-function of the following form:

g1(Z) = ...;
g1(S(v1)) = ...;

We can complete the first branch by inserting 2 in place of ...:

g1(Z) = 2;
g1(S(v1)) = ...;

Since the second branch leads to a node that decomposes a constructor, we generate a new f-function and call it inside S:

g1(Z) = 2;
g1(S(v1)) = S(f1(v1));
f1(v1) = ...;

According to the graph, the body of f1 should be add(v1, 2), but since now add stands for g1, the body will be g1(v1):

g1(Z) = 2;
g1(S(v1)) = S(f1(v1));
f1(v1) = g1(v1);

Since f1 is redundant, we can inline it as follows:

g1(Z) = 2;
g1(S(v1)) = S(g1(v1));

This is the final residual program; the residual expression will be g1(a), since a is the only free variable from the initial expression. As you can see, the definition of g1 is simpler than the original function add in the sense that it accepts only one argument; the second argument has been specialized to 2 during supercompilation.

Perhaps a more interesting example would be specializing an interpreter to a concrete program, obtaining a compiled “executable”. In the partial computation literature, this is called the first Futamura projection ⁵.

Consider the following interpreter eq (also from ⁶):

eq(Z, y) = eqZ(y);
eq(S(x), y) = eqS(y, x);
eqZ(Z) = True;
eqZ(S(x)) = False;
eqS(Z, x) = False;
eqS(S(y), x) = eq(x, y);

which compares two Peano numbers for equality, resulting in either True or False.

Now consider the expression eq(S(S(Z)), x) with the above definition of eq; supercompilation will produce the following residual program:

eqZ3(Z) = True;
eqZ3(S(v)) = False;
eqS2(Z) = False;
eqS2(S(v)) = eqZ3(v);
eqS1(Z) = False;
eqS1(S(v)) = eqS2(v);

together with the residual expression eqS1(x). In a sense, we have obtained a compiled version of eq specialized to S(S(Z)), because eqS1 pattern-matches only on an unknown variable x instead of the known “program” S(S(Z)).

However, supercompilation is inherently more powerful than classical partial evaluation. In Turchin’s own words ⁷:

Although supercompilation includes partial evaluation, it does not reduce to it. Supercompilation can lead to a very deep structural transformation of the original program; it can improve the program even if all the actual parameters in the function calls are variable. The supercompilation process aims at the reduction of redundancy in the original program, but this redundancy does not necessarily come from fixed values of variables; it can result from nested loops, repeated variables, and so on.

That being said, a sufficiently smart supercompiler can transform a two-pass list mapping into a single-pass algorithm, essentially performing list fusion:

A similar example can be found in “Rethinking Supercompilation” by Neil Mitchell ⁸ and in ⁹ (section 6, “Examples of supercompilation”).

Although I have deliberately skipped a few important aspects of supercompilation, most notably termination checking (see homeomorphic embedding ^{10 11 12}), the material presented in this section should be enough to understand the rest of the text. An interested reader may go through the references and also look into ^{13 14 15} for some insight into a more general topic of metacomputation. A more thorough introduction to supercompilation can be found in “Supercompilation: Ideas and Methods” by Ilya Klyuchnikov and Dimitur Krustev.

Not let us consider how SAT can be reduced to the problem of dead code elimination.

Consider a language whose expressions are:

• If (x, m, n), where x is a variable, m and n are expressions.
• T and F, which are constants.

This language is enough to encode any SAT problem. We can encode each clause in the formula as follows (in pseudocode):

OR(x, ...) = If (x, T, ...);
OR(NOT x, ...) = If (x, ..., T);

Then OR(x, OR(y, OR(NOT z, F))) would correspond to “x OR y OR NOT z”:

Now consider the encoding of a conjunction of clauses ¹⁶:

AND(If(x, m, n), ...) = If(x, AND(m, ...), AND(n, ...));
AND(T, ...) = ...;
AND(F, ...) = F;

Then AND(OR(x, F), AND(OR(NOT y, F), T)) would correspond to “x AND NOT y”:

Is the formula satisfiable? Yes, because we can assign x to T and y to F.

Now consider the formula AND(OR(x, F), AND(OR(NOT x, F), T)), which is equivalent to “x AND NOT x”:

Is the formula satisfiable? To figure out, let us remove dead paths from the formula:

We have replaced the lower if x node with its first child F because x was already assigned T in this path. Since there are no T leafs in the resulting tree, it is correct to say that the formula is unsatisfiable: with any value of x we will arrive at F.

One more example is AND(OR(x, F), AND(OR(x, OR(y, F)), AND(OR(NOT x, F), T))), which is equivalent to “x AND (y OR z) AND NOT x”:

After removing dead paths:

The general observation is that, after encoding a CNF formula as an if-tree and removing dead paths from it, if there is at least one T leaf, the initial formula is satisfiable; otherwise, the formula is unsatisfiable because there is no path from the root that will take us to T. Think about it for a moment.

Having this technique at our disposal, we can proceed to the key section.

Positive supercompilation is a particular model of supercompilation that propagates only positive information when doing case analysis. For example, when supercompiling add(a, 2), a positive supercompiler will substitute (i.e., propagate) a=Z and a=S(v1) to the respective children nodes. On the other hand, negative information means that some variable is not equal to a particular value (or values); a typical example is the “default” case in a pattern-matching construction. A perfect supercompiler is a supercompiler that is able to propagate both positive and negative information.

In this post, we only deal with positive supercompilation. Consider the schematic representation of the CNF formula “x AND NOT x” again:

Imagine that if, T, and F are SLL constructors, with if holding three arguments: two branches and a variable, which is T in the first branch and F in the second. If we analyze the uppermost if x, we will get the following “process tree” ¹⁷:

Supercompilation acted as a dead code eliminator! This is because x=T was propagated to the first branch of the uppermost if x, resulting in the elimination of the branch T of the innermost if x. The second uppermost branch remains unchanged.

What if we somehow launch analysis of the whole if-tree? If so, we would essentially eliminate all dead paths from the corresponding expression, which is equivalent to solving the SAT problem! As I have mentioned in the previous section, if there is at least one T leaf in the final tree, the formula is satisfiable because this T is reachable from the root. Otherwise (if there is no T), the formula is unsatisfiable ¹⁸.

Here is a complete SAT solver written in SLL:

or(Var(x), rest) = If(x, T, rest);
or(Not(x), rest) = If(x, rest, T);

and(If(x, m, n), rest) = If(x, and(m, rest), and(n, rest));
and(T, rest) = rest;
and(F, rest) = F;

solve(If(x, m, n)) = analyze(x, solve(m), solve(n));
solve(T) = T;
solve(F) = F;

analyze(T, m, n) = m;
analyze(F, m, n) = n;

Without blank lines, it is 10 lines of code! This is probably the simplest (and the slowest!) SAT solver in the history of humanity.

The functions or and and are the encodings of the logical OR and AND operations, respectively. The function solve is the main entry point of the solver, and analyze is a helper function that selects one of the two branches of an if-expression according to the value of the first argument. Since solve traverses the whole expression and invokes analyze for each If (x, m, n), all If-nodes in the expression are case-analyzed when solve(formula) is being supercompiled.

To test the solver, I will use my own positive supercompiler written in OCaml. The definitions of or, and, solve, and analyze will be:

let g_rules =
  [
    ( "or",
      [
        ( ("Var", [ "x" ]),
          [ "rest" ],
          ctr_call ("If", [ Var "x"; ctr_call ("T", []); Var "rest" ]) );
        ( ("Not", [ "x" ]),
          [ "rest" ],
          ctr_call ("If", [ Var "x"; Var "rest"; ctr_call ("T", []) ]) );
      ] );
    ( "and",
      [
        ( ("If", [ "x"; "m"; "n" ]),
          [ "rest" ],
          ctr_call
            ( "If",
              [
                Var "x";
                g_call ("and", Var "m", [ Var "rest" ]);
                g_call ("and", Var "n", [ Var "rest" ]);
              ] ) );
        (("T", []), [ "rest" ], Var "rest");
        (("F", []), [ "rest" ], ctr_call ("F", []));
      ] );
    ( "solve",
      [
        ( ("If", [ "x"; "m"; "n" ]),
          [],
          g_call
            ( "analyze",
              Var "x",
              [ g_call ("solve", Var "m", []); g_call ("solve", Var "n", []) ]
            ) );
        (("T", []), [], ctr_call ("T", []));
        (("F", []), [], ctr_call ("F", []));
      ] );
    ( "analyze",
      [ (("T", []), [ "m"; "n" ], Var "m"); (("F", []), [ "m"; "n" ], Var "n") ]
    );
  ]

And some helper functions in OCaml to conveniently construct CNF formulas:

let rec clause = function
  | `Var x :: rest -> g_call ("or", ctr_call ("Var", [ Var x ]), [ clause rest ])
  | `Not x :: rest -> g_call ("or", ctr_call ("Not", [ Var x ]), [ clause rest ])
  | [] -> ctr_call ("F", [])

let rec formula = function
  | hd :: rest -> g_call ("and", hd, [ formula rest ])
  | [] -> ctr_call ("T", [])

The function check calls optimize, which is supercompilation + residualization, on a given formula, and compares the resudual task with expected:

let check ~expected formula =
  check ~equal:[%derive.eq: string * string list]
    ~show:[%derive.show: string * string list] ~expected
    ~actual:(optimize ~g_rules (g_call ("solve", formula, [])))

Testing “x AND y”:

let () =
  check
    ~expected:
      ( "g0(x, y)",
        [
          "g0(F(), y) = F()";
          "g0(T(), y) = g1(y)";
          "g1(F()) = F()";
          "g1(T()) = T()";
        ] )
    (formula [ clause [ `Var "x" ]; clause [ `Var "y" ] ]);

The residual function g0 acts as the AND operation. The formula is satisfiable.

Testing “(x OR y) AND NOT z”:

  check
    ~expected:
      ( "g0(x, y, z)",
        [
          "g0(F(), y, z) = g1(y, z)";
          "g0(T(), y, z) = g3(z)";
          "g1(F(), z) = F()";
          "g1(T(), z) = g2(z)";
          "g2(F()) = T()";
          "g2(T()) = F()";
          "g3(F()) = T()";
          "g3(T()) = F()";
        ] )
    (formula [ clause [ `Var "x"; `Var "y" ]; clause [ `Not "z" ] ]);

The formula is TRUE when either “x” or “y” are TRUE (or both), whilst “z” must always be FALSE. The formula is satisfiable.

Testing “(x OR NOT z) AND (y OR z OR NOT x)”:

  check
    ~expected:
      ( "g0(x, y, z)",
        [
          "g0(F(), y, z) = g1(z, y)";
          "g0(T(), y, z) = g3(y, z)";
          "g1(F(), y) = g2(y)";
          "g1(T(), y) = F()";
          "g2(F()) = T()";
          "g2(T()) = T()";
          "g3(F(), z) = g4(z)";
          "g3(T(), z) = T()";
          "g4(F()) = F()";
          "g4(T()) = T()";
        ] )
    (formula
       [
         clause [ `Var "x"; `Not "z" ]; clause [ `Var "y"; `Var "z"; `Not "x" ];
       ]);

Ditto.

Testing “(a OR b) AND (NOT b OR NOT d) AND (c OR d) AND (NOT d OR NOT e) AND (e OR NOT f) AND (f OR NOT g) AND (f OR g) AND (g OR NOT p) AND (h OR NOT i) AND (NOT h OR NOT n) AND (i OR g) AND (i OR NOT g) AND (NOT g OR NOT k) AND (g OR l) AND (k OR l) AND (m OR n) AND (n OR NOT o) AND (o OR p)”:

  check
    ~expected:
      ( "g0(a, b, c, d, e, f, g, h, i, k, l, m, n, o, p)",
        [
          "g0(F(), b, c, d, e, f, g, h, i, k, l, m, n, o, p) = g1(b, c, d, e, \
           f, g, h, i, k, l, m, n, o, p)";
          "g0(T(), b, c, d, e, f, g, h, i, k, l, m, n, o, p) = g30(b, c, d, e, \
           f, g, h, i, k, l, m, n, o, p)";
          "g1(F(), c, d, e, f, g, h, i, k, l, m, n, o, p) = F()";
          "g1(T(), c, d, e, f, g, h, i, k, l, m, n, o, p) = g2(d, c, e, f, g, \
           h, i, k, l, m, n, o, p)";
          "g10(F(), h, i, k, l, m, n, o) = g11(h, i, k, l, m, n, o)";
          "g10(T(), h, i, k, l, m, n, o) = F()";
          "g11(F(), i, k, l, m, n, o) = g12(i)";
          "g11(T(), i, k, l, m, n, o) = g13(n, i, k, l, m, o)";
          "g12(F()) = F()";
          "g12(T()) = F()";
          "g13(F(), i, k, l, m, o) = g14(i, k, l, m, o)";
          "g13(T(), i, k, l, m, o) = F()";
          "g14(F(), k, l, m, o) = F()";
          "g14(T(), k, l, m, o) = g15(l, k, m, o)";
          "g15(F(), k, m, o) = F()";
          "g15(T(), k, m, o) = g16(k, m, o)";
          "g16(F(), m, o) = g17(m, o)";
          "g16(T(), m, o) = g19(m, o)";
          "g17(F(), o) = F()";
          "g17(T(), o) = g18(o)";
          "g18(F()) = F()";
          "g18(T()) = F()";
          "g19(F(), o) = F()";
          "g19(T(), o) = g20(o)";
          "g2(F(), c, e, f, g, h, i, k, l, m, n, o, p) = g3(c, e, f, g, h, i, \
           k, l, m, n, o, p)";
          "g2(T(), c, e, f, g, h, i, k, l, m, n, o, p) = F()";
          "g20(F()) = F()";
          "g20(T()) = F()";
          "g21(F(), i, k, l, m, n, o, p) = g22(i)";
          "g21(T(), i, k, l, m, n, o, p) = g23(n, i, k, l, m, o, p)";
          "g22(F()) = F()";
          "g22(T()) = F()";
          "g23(F(), i, k, l, m, o, p) = g24(i, k, l, m, o, p)";
          "g23(T(), i, k, l, m, o, p) = F()";
          "g24(F(), k, l, m, o, p) = F()";
          "g24(T(), k, l, m, o, p) = g25(k, l, m, o, p)";
          "g25(F(), l, m, o, p) = g26(l, m, o, p)";
          "g25(T(), l, m, o, p) = F()";
          "g26(F(), m, o, p) = F()";
          "g26(T(), m, o, p) = g27(m, o, p)";
          "g27(F(), o, p) = F()";
          "g27(T(), o, p) = g28(o, p)";
          "g28(F(), p) = g29(p)";
          "g28(T(), p) = F()";
          "g29(F()) = F()";
          "g29(T()) = T()";
          "g3(F(), e, f, g, h, i, k, l, m, n, o, p) = F()";
          "g3(T(), e, f, g, h, i, k, l, m, n, o, p) = g4(e, f, g, h, i, k, l, \
           m, n, o, p)";
          "g30(F(), c, d, e, f, g, h, i, k, l, m, n, o, p) = g31(c, d, e, f, \
           g, h, i, k, l, m, n, o, p)";
          "g30(T(), c, d, e, f, g, h, i, k, l, m, n, o, p) = g66(d, c, e, f, \
           g, h, i, k, l, m, n, o, p)";
          "g31(F(), d, e, f, g, h, i, k, l, m, n, o, p) = g32(d, e, f, g)";
          "g31(T(), d, e, f, g, h, i, k, l, m, n, o, p) = g36(d, e, f, g, h, \
           i, k, l, m, n, o, p)";
          "g32(F(), e, f, g) = F()";
          "g32(T(), e, f, g) = g33(e, f, g)";
          "g33(F(), f, g) = g34(f, g)";
          "g33(T(), f, g) = F()";
          "g34(F(), g) = g35(g)";
          "g34(T(), g) = F()";
          "g35(F()) = F()";
          "g35(T()) = F()";
          "g36(F(), e, f, g, h, i, k, l, m, n, o, p) = g37(e, f, g, h, i, k, \
           l, m, n, o, p)";
          "g36(T(), e, f, g, h, i, k, l, m, n, o, p) = g63(e, f, g)";
          "g37(F(), f, g, h, i, k, l, m, n, o, p) = g38(f, g)";
          "g37(T(), f, g, h, i, k, l, m, n, o, p) = g40(f, g, h, i, k, l, m, \
           n, o, p)";
          "g38(F(), g) = g39(g)";
          "g38(T(), g) = F()";
          "g39(F()) = F()";
          "g39(T()) = F()";
          "g4(F(), f, g, h, i, k, l, m, n, o, p) = g5(f, g)";
          "g4(T(), f, g, h, i, k, l, m, n, o, p) = g7(f, g, h, i, k, l, m, n, \
           o, p)";
          "g40(F(), g, h, i, k, l, m, n, o, p) = g41(g)";
          "g40(T(), g, h, i, k, l, m, n, o, p) = g42(g, h, i, k, l, m, n, o, p)";
          "g41(F()) = F()";
          "g41(T()) = F()";
          "g42(F(), h, i, k, l, m, n, o, p) = g43(p, h, i, k, l, m, n, o)";
          "g42(T(), h, i, k, l, m, n, o, p) = g54(h, i, k, l, m, n, o, p)";
          "g43(F(), h, i, k, l, m, n, o) = g44(h, i, k, l, m, n, o)";
          "g43(T(), h, i, k, l, m, n, o) = F()";
          "g44(F(), i, k, l, m, n, o) = g45(i)";
          "g44(T(), i, k, l, m, n, o) = g46(n, i, k, l, m, o)";
          "g45(F()) = F()";
          "g45(T()) = F()";
          "g46(F(), i, k, l, m, o) = g47(i, k, l, m, o)";
          "g46(T(), i, k, l, m, o) = F()";
          "g47(F(), k, l, m, o) = F()";
          "g47(T(), k, l, m, o) = g48(l, k, m, o)";
          "g48(F(), k, m, o) = F()";
          "g48(T(), k, m, o) = g49(k, m, o)";
          "g49(F(), m, o) = g50(m, o)";
          "g49(T(), m, o) = g52(m, o)";
          "g5(F(), g) = g6(g)";
          "g5(T(), g) = F()";
          "g50(F(), o) = F()";
          "g50(T(), o) = g51(o)";
          "g51(F()) = F()";
          "g51(T()) = F()";
          "g52(F(), o) = F()";
          "g52(T(), o) = g53(o)";
          "g53(F()) = F()";
          "g53(T()) = F()";
          "g54(F(), i, k, l, m, n, o, p) = g55(i)";
          "g54(T(), i, k, l, m, n, o, p) = g56(n, i, k, l, m, o, p)";
          "g55(F()) = F()";
          "g55(T()) = F()";
          "g56(F(), i, k, l, m, o, p) = g57(i, k, l, m, o, p)";
          "g56(T(), i, k, l, m, o, p) = F()";
          "g57(F(), k, l, m, o, p) = F()";
          "g57(T(), k, l, m, o, p) = g58(k, l, m, o, p)";
          "g58(F(), l, m, o, p) = g59(l, m, o, p)";
          "g58(T(), l, m, o, p) = F()";
          "g59(F(), m, o, p) = F()";
          "g59(T(), m, o, p) = g60(m, o, p)";
          "g6(F()) = F()";
          "g6(T()) = F()";
          "g60(F(), o, p) = F()";
          "g60(T(), o, p) = g61(o, p)";
          "g61(F(), p) = g62(p)";
          "g61(T(), p) = F()";
          "g62(F()) = F()";
          "g62(T()) = T()";
          "g63(F(), f, g) = g64(f, g)";
          "g63(T(), f, g) = F()";
          "g64(F(), g) = g65(g)";
          "g64(T(), g) = F()";
          "g65(F()) = F()";
          "g65(T()) = F()";
          "g66(F(), c, e, f, g, h, i, k, l, m, n, o, p) = g67(c, e, f, g, h, \
           i, k, l, m, n, o, p)";
          "g66(T(), c, e, f, g, h, i, k, l, m, n, o, p) = F()";
          "g67(F(), e, f, g, h, i, k, l, m, n, o, p) = F()";
          "g67(T(), e, f, g, h, i, k, l, m, n, o, p) = g68(e, f, g, h, i, k, \
           l, m, n, o, p)";
          "g68(F(), f, g, h, i, k, l, m, n, o, p) = g69(f, g)";
          "g68(T(), f, g, h, i, k, l, m, n, o, p) = g71(f, g, h, i, k, l, m, \
           n, o, p)";
          "g69(F(), g) = g70(g)";
          "g69(T(), g) = F()";
          "g7(F(), g, h, i, k, l, m, n, o, p) = g8(g)";
          "g7(T(), g, h, i, k, l, m, n, o, p) = g9(g, h, i, k, l, m, n, o, p)";
          "g70(F()) = F()";
          "g70(T()) = F()";
          "g71(F(), g, h, i, k, l, m, n, o, p) = g72(g)";
          "g71(T(), g, h, i, k, l, m, n, o, p) = g73(g, h, i, k, l, m, n, o, p)";
          "g72(F()) = F()";
          "g72(T()) = F()";
          "g73(F(), h, i, k, l, m, n, o, p) = g74(p, h, i, k, l, m, n, o)";
          "g73(T(), h, i, k, l, m, n, o, p) = g85(h, i, k, l, m, n, o, p)";
          "g74(F(), h, i, k, l, m, n, o) = g75(h, i, k, l, m, n, o)";
          "g74(T(), h, i, k, l, m, n, o) = F()";
          "g75(F(), i, k, l, m, n, o) = g76(i)";
          "g75(T(), i, k, l, m, n, o) = g77(n, i, k, l, m, o)";
          "g76(F()) = F()";
          "g76(T()) = F()";
          "g77(F(), i, k, l, m, o) = g78(i, k, l, m, o)";
          "g77(T(), i, k, l, m, o) = F()";
          "g78(F(), k, l, m, o) = F()";
          "g78(T(), k, l, m, o) = g79(l, k, m, o)";
          "g79(F(), k, m, o) = F()";
          "g79(T(), k, m, o) = g80(k, m, o)";
          "g8(F()) = F()";
          "g8(T()) = F()";
          "g80(F(), m, o) = g81(m, o)";
          "g80(T(), m, o) = g83(m, o)";
          "g81(F(), o) = F()";
          "g81(T(), o) = g82(o)";
          "g82(F()) = F()";
          "g82(T()) = F()";
          "g83(F(), o) = F()";
          "g83(T(), o) = g84(o)";
          "g84(F()) = F()";
          "g84(T()) = F()";
          "g85(F(), i, k, l, m, n, o, p) = g86(i)";
          "g85(T(), i, k, l, m, n, o, p) = g87(n, i, k, l, m, o, p)";
          "g86(F()) = F()";
          "g86(T()) = F()";
          "g87(F(), i, k, l, m, o, p) = g88(i, k, l, m, o, p)";
          "g87(T(), i, k, l, m, o, p) = F()";
          "g88(F(), k, l, m, o, p) = F()";
          "g88(T(), k, l, m, o, p) = g89(k, l, m, o, p)";
          "g89(F(), l, m, o, p) = g90(l, m, o, p)";
          "g89(T(), l, m, o, p) = F()";
          "g9(F(), h, i, k, l, m, n, o, p) = g10(p, h, i, k, l, m, n, o)";
          "g9(T(), h, i, k, l, m, n, o, p) = g21(h, i, k, l, m, n, o, p)";
          "g90(F(), m, o, p) = F()";
          "g90(T(), m, o, p) = g91(m, o, p)";
          "g91(F(), o, p) = F()";
          "g91(T(), o, p) = g92(o, p)";
          "g92(F(), p) = g93(p)";
          "g92(T(), p) = F()";
          "g93(F()) = F()";
          "g93(T()) = T()";
        ] )
    (formula
       [
         clause [ `Var "a"; `Var "b" ];
         clause [ `Not "b"; `Not "d" ];
         clause [ `Var "c"; `Var "d" ];
         clause [ `Not "d"; `Not "e" ];
         clause [ `Var "e"; `Not "f" ];
         clause [ `Var "f"; `Not "g" ];
         clause [ `Var "f"; `Var "g" ];
         clause [ `Var "g"; `Not "p" ];
         clause [ `Var "h"; `Not "i" ];
         clause [ `Not "h"; `Not "n" ];
         clause [ `Var "i"; `Var "g" ];
         clause [ `Var "i"; `Not "g" ];
         clause [ `Not "g"; `Not "k" ];
         clause [ `Var "g"; `Var "l" ];
         clause [ `Var "k"; `Var "l" ];
         clause [ `Var "m"; `Var "n" ];
         clause [ `Var "n"; `Not "o" ];
         clause [ `Var "o"; `Var "p" ];
       ]);

Ditto.

Testing “x AND NOT x”:

  check
    ~expected:("g0(x)", [ "g0(F()) = F()"; "g0(T()) = F()" ])
    (formula [ clause [ `Var "x" ]; clause [ `Not "x" ] ]);

Since there is no path in g0 that leads to T, the formula is unsatisfiable.

Testing “x AND (y OR z) AND NOT x”:

  check
    ~expected:
      ( "g0(x, y, z)",
        [
          "g0(F(), y, z) = F()";
          "g0(T(), y, z) = g1(y, z)";
          "g1(F(), z) = g2(z)";
          "g1(T(), z) = F()";
          "g2(F()) = F()";
          "g2(T()) = F()";
        ] )
    (formula
       [
         clause [ `Var "x" ]; clause [ `Var "y"; `Var "z" ]; clause [ `Not "x" ];
       ]);

Ditto.

Testing “(x OR x OR x) AND (NOT x OR NOT x OR NOT x)”:

  check
    ~expected:("g0(x)", [ "g0(F()) = F()"; "g0(T()) = F()" ])
    (formula
       [
         clause [ `Var "x"; `Var "x"; `Var "x" ];
         clause [ `Not "x"; `Not "x"; `Not "x" ];
       ]);

Ditto.

Testing “(x OR y OR z) AND (x OR y OR NOT z) AND (x OR NOT y OR z) AND (x OR NOT y OR NOT z) AND (NOT x OR y OR z) AND (NOT x OR y OR NOT z) AND (NOT x OR NOT y OR z) AND (NOT x OR NOT y OR NOT z)”:

  check
    ~expected:
      ( "g0(x, y, z)",
        [
          "g0(F(), y, z) = g1(y, z)";
          "g0(T(), y, z) = g4(y, z)";
          "g1(F(), z) = g2(z)";
          "g1(T(), z) = g3(z)";
          "g2(F()) = F()";
          "g2(T()) = F()";
          "g3(F()) = F()";
          "g3(T()) = F()";
          "g4(F(), z) = g5(z)";
          "g4(T(), z) = g6(z)";
          "g5(F()) = F()";
          "g5(T()) = F()";
          "g6(F()) = F()";
          "g6(T()) = F()";
        ] )
    (formula
       [
         clause [ `Var "x"; `Var "y"; `Var "z" ];
         clause [ `Var "x"; `Var "y"; `Not "z" ];
         clause [ `Var "x"; `Not "y"; `Var "z" ];
         clause [ `Var "x"; `Not "y"; `Not "z" ];
         clause [ `Not "x"; `Var "y"; `Var "z" ];
         clause [ `Not "x"; `Var "y"; `Not "z" ];
         clause [ `Not "x"; `Not "y"; `Var "z" ];
         clause [ `Not "x"; `Not "y"; `Not "z" ];
       ])

Ditto.

In this essay, I have demonstrated how supercompilation can solve any SAT problem with only 10 lines of code in SLL.

There are several theoretical and practical consequences of this discovery:

• Theoretically, this means that a positive supercompiler can solve SAT “automatically” – possibly mixed with a number of other optimizations and code transformations. In other words, it is possible that supercompilation can optimize your code based on the knowledge of a solution to a particular SAT problem.
  • A direct consequence of this is that supercompiling a particular task can have exponential time and space complexity. This is both good and bad: the good part is that supercompilation is very smart, and the bad part is that it can be too smart to serve as a predictable code optimizer.
  • Since it is generally believed that there is no polynomial-time algorithm to solve SAT, positive supercompilation is also likely not to have it.
  • Since SAT is an NP-complete problem, meaning that it is at least as hard as any other problem in NP, supercompilation is also at least as hard as any other problem in NP. In other words, supercompilation has enough expressive power to solve any NP problem during the process.
• Practically, using naive supercompilation as an optimization pass would not be a good idea. However, since supercompilation is a general method rather than a particular algorithm, it should always be possible to restrict the expressive power of a particular supercompiler to make it more predictable – albeit less smart. See the technique of speculative supercompilation from ¹⁹.
  • The SLL solver demonstrated in the previous section can serve as a good stress test for a practical supercompiler.

Although the idea of supercompilation dates back to the 1970’s, developing an industrial supercompiler is still an open problem.

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