davis_putnam_logemann_loveland.py

#!/usr/bin/env python3

"""
Davis-Putnam-Logemann-Loveland (DPLL) algorithm is a complete, backtracking-based
search algorithm for deciding the satisfiability of propositional logic formulae in
conjunctive normal form, i.e, for solving the Conjunctive Normal Form SATisfiability
(CNF-SAT) problem.

For more information about the algorithm: https://en.wikipedia.org/wiki/DPLL_algorithm
"""

from __future__ import annotations

import random
from collections.abc import Iterable


class Clause:
    """
    | A clause represented in Conjunctive Normal Form.
    | A clause is a set of literals, either complemented or otherwise.

    For example:
        * {A1, A2, A3'} is the clause (A1 v A2 v A3')
        * {A5', A2', A1} is the clause (A5' v A2' v A1)

    Create model

    >>> clause = Clause(["A1", "A2'", "A3"])
    >>> clause.evaluate({"A1": True})
    True
    """

    def __init__(self, literals: list[str]) -> None:
        """
        Represent the literals and an assignment in a clause."
        """
        # Assign all literals to None initially
        self.literals: dict[str, bool | None] = dict.fromkeys(literals)

    def __str__(self) -> str:
        """
        To print a clause as in Conjunctive Normal Form.

        >>> str(Clause(["A1", "A2'", "A3"]))
        "{A1 , A2' , A3}"
        """
        return "{" + " , ".join(self.literals) + "}"

    def __len__(self) -> int:
        """
        To print a clause as in Conjunctive Normal Form.

        >>> len(Clause([]))
        0
        >>> len(Clause(["A1", "A2'", "A3"]))
        3
        """
        return len(self.literals)

    def assign(self, model: dict[str, bool | None]) -> None:
        """
        Assign values to literals of the clause as given by model.
        """
        for literal in self.literals:
            symbol = literal[:2]
            if symbol in model:
                value = model[symbol]
            else:
                continue
            # Complement assignment if literal is in complemented form
            if value is not None and literal.endswith("'"):
                value = not value
            self.literals[literal] = value

    def evaluate(self, model: dict[str, bool | None]) -> bool | None:
        """
        Evaluates the clause with the assignments in model.

        This has the following steps:
          1. Return ``True`` if both a literal and its complement exist in the clause.
          2. Return ``True`` if a single literal has the assignment ``True``.
          3. Return ``None`` (unable to complete evaluation)
             if a literal has no assignment.
          4. Compute disjunction of all values assigned in clause.
        """
        for literal in self.literals:
            symbol = literal.rstrip("'") if literal.endswith("'") else literal + "'"
            if symbol in self.literals:
                return True

        self.assign(model)
        for value in self.literals.values():
            if value in (True, None):
                return value
        return any(self.literals.values())


class Formula:
    """
    | A formula represented in Conjunctive Normal Form.
    | A formula is a set of clauses.
    | For example,
    |   {{A1, A2, A3'}, {A5', A2', A1}} is ((A1 v A2 v A3') and (A5' v A2' v A1))
    """

    def __init__(self, clauses: Iterable[Clause]) -> None:
        """
        Represent the number of clauses and the clauses themselves.
        """
        self.clauses = list(clauses)

    def __str__(self) -> str:
        """
        To print a formula as in Conjunctive Normal Form.

        >>> str(Formula([Clause(["A1", "A2'", "A3"]), Clause(["A5'", "A2'", "A1"])]))
        "{{A1 , A2' , A3} , {A5' , A2' , A1}}"
        """
        return "{" + " , ".join(str(clause) for clause in self.clauses) + "}"


def generate_clause() -> Clause:
    """
    | Randomly generate a clause.
    | All literals have the name Ax, where x is an integer from ``1`` to ``5``.
    """
    literals = []
    no_of_literals = random.randint(1, 5)
    base_var = "A"
    i = 0
    while i < no_of_literals:
        var_no = random.randint(1, 5)
        var_name = base_var + str(var_no)
        var_complement = random.randint(0, 1)
        if var_complement == 1:
            var_name += "'"
        if var_name in literals:
            i -= 1
        else:
            literals.append(var_name)
        i += 1
    return Clause(literals)


def generate_formula() -> Formula:
    """
    Randomly generate a formula.
    """
    clauses: set[Clause] = set()
    no_of_clauses = random.randint(1, 10)
    while len(clauses) < no_of_clauses:
        clauses.add(generate_clause())
    return Formula(clauses)


def generate_parameters(formula: Formula) -> tuple[list[Clause], list[str]]:
    """
    | Return the clauses and symbols from a formula.
    | A symbol is the uncomplemented form of a literal.

    For example,
      * Symbol of A3 is A3.
      * Symbol of A5' is A5.

    >>> formula = Formula([Clause(["A1", "A2'", "A3"]), Clause(["A5'", "A2'", "A1"])])
    >>> clauses, symbols = generate_parameters(formula)
    >>> clauses_list = [str(i) for i in clauses]
    >>> clauses_list
    ["{A1 , A2' , A3}", "{A5' , A2' , A1}"]
    >>> symbols
    ['A1', 'A2', 'A3', 'A5']
    """
    clauses = formula.clauses
    symbols_set = []
    for clause in formula.clauses:
        for literal in clause.literals:
            symbol = literal[:2]
            if symbol not in symbols_set:
                symbols_set.append(symbol)
    return clauses, symbols_set


def find_pure_symbols(
    clauses: list[Clause], symbols: list[str], model: dict[str, bool | None]
) -> tuple[list[str], dict[str, bool | None]]:
    """
    | Return pure symbols and their values to satisfy clause.
    | Pure symbols are symbols in a formula that exist only in one form,
    | either complemented or otherwise.
    | For example,
    |   {{A4 , A3 , A5' , A1 , A3'} , {A4} , {A3}} has pure symbols A4, A5' and A1.

    This has the following steps:
      1. Ignore clauses that have already evaluated to be ``True``.
      2. Find symbols that occur only in one form in the rest of the clauses.
      3. Assign value ``True`` or ``False`` depending on whether the symbols occurs
         in normal or complemented form respectively.

    >>> formula = Formula([Clause(["A1", "A2'", "A3"]), Clause(["A5'", "A2'", "A1"])])
    >>> clauses, symbols = generate_parameters(formula)
    >>> pure_symbols, values = find_pure_symbols(clauses, symbols, {})
    >>> pure_symbols
    ['A1', 'A2', 'A3', 'A5']
    >>> values
    {'A1': True, 'A2': False, 'A3': True, 'A5': False}
    """
    pure_symbols = []
    assignment: dict[str, bool | None] = {}
    literals = []

    for clause in clauses:
        if clause.evaluate(model):
            continue
        for literal in clause.literals:
            literals.append(literal)

    for s in symbols:
        sym = s + "'"
        if (s in literals and sym not in literals) or (
            s not in literals and sym in literals
        ):
            pure_symbols.append(s)
    for p in pure_symbols:
        assignment[p] = None
    for s in pure_symbols:
        sym = s + "'"
        if s in literals:
            assignment[s] = True
        elif sym in literals:
            assignment[s] = False
    return pure_symbols, assignment


def find_unit_clauses(
    clauses: list[Clause],
    model: dict[str, bool | None],  # noqa: ARG001
) -> tuple[list[str], dict[str, bool | None]]:
    """
    Returns the unit symbols and their values to satisfy clause.

    Unit symbols are symbols in a formula that are:
      - Either the only symbol in a clause
      - Or all other literals in that clause have been assigned ``False``

    This has the following steps:
      1. Find symbols that are the only occurrences in a clause.
      2. Find symbols in a clause where all other literals are assigned ``False``.
      3. Assign ``True`` or ``False`` depending on whether the symbols occurs in
         normal or complemented form respectively.

    >>> clause1 = Clause(["A4", "A3", "A5'", "A1", "A3'"])
    >>> clause2 = Clause(["A4"])
    >>> clause3 = Clause(["A3"])
    >>> clauses, symbols = generate_parameters(Formula([clause1, clause2, clause3]))
    >>> unit_clauses, values = find_unit_clauses(clauses, {})
    >>> unit_clauses
    ['A4', 'A3']
    >>> values
    {'A4': True, 'A3': True}
    """
    unit_symbols = []
    for clause in clauses:
        if len(clause) == 1:
            unit_symbols.append(next(iter(clause.literals.keys())))
        else:
            f_count, n_count = 0, 0
            for literal, value in clause.literals.items():
                if value is False:
                    f_count += 1
                elif value is None:
                    sym = literal
                    n_count += 1
            if f_count == len(clause) - 1 and n_count == 1:
                unit_symbols.append(sym)
    assignment: dict[str, bool | None] = {}
    for i in unit_symbols:
        symbol = i[:2]
        assignment[symbol] = len(i) == 2
    unit_symbols = [i[:2] for i in unit_symbols]

    return unit_symbols, assignment


def dpll_algorithm(
    clauses: list[Clause], symbols: list[str], model: dict[str, bool | None]
) -> tuple[bool | None, dict[str, bool | None] | None]:
    """
    Returns the model if the formula is satisfiable, else ``None``

    This has the following steps:
      1. If every clause in clauses is ``True``, return ``True``.
      2. If some clause in clauses is ``False``, return ``False``.
      3. Find pure symbols.
      4. Find unit symbols.

    >>> formula = Formula([Clause(["A4", "A3", "A5'", "A1", "A3'"]), Clause(["A4"])])
    >>> clauses, symbols = generate_parameters(formula)
    >>> soln, model = dpll_algorithm(clauses, symbols, {})
    >>> soln
    True
    >>> model
    {'A4': True}
    """
    check_clause_all_true = True
    for clause in clauses:
        clause_check = clause.evaluate(model)
        if clause_check is False:
            return False, None
        elif clause_check is None:
            check_clause_all_true = False
            continue

    if check_clause_all_true:
        return True, model

    try:
        pure_symbols, assignment = find_pure_symbols(clauses, symbols, model)
    except RecursionError:
        print("raises a RecursionError and is")
        return None, {}
    p = None
    if len(pure_symbols) > 0:
        p, value = pure_symbols[0], assignment[pure_symbols[0]]

    if p:
        tmp_model = model
        tmp_model[p] = value
        tmp_symbols = list(symbols)
        if p in tmp_symbols:
            tmp_symbols.remove(p)
        return dpll_algorithm(clauses, tmp_symbols, tmp_model)

    unit_symbols, assignment = find_unit_clauses(clauses, model)
    p = None
    if len(unit_symbols) > 0:
        p, value = unit_symbols[0], assignment[unit_symbols[0]]
    if p:
        tmp_model = model
        tmp_model[p] = value
        tmp_symbols = list(symbols)
        if p in tmp_symbols:
            tmp_symbols.remove(p)
        return dpll_algorithm(clauses, tmp_symbols, tmp_model)
    p = symbols[0]
    rest = symbols[1:]
    tmp1, tmp2 = model, model
    tmp1[p], tmp2[p] = True, False

    return dpll_algorithm(clauses, rest, tmp1) or dpll_algorithm(clauses, rest, tmp2)


if __name__ == "__main__":
    import doctest

    doctest.testmod()

    formula = generate_formula()
    print(f"The formula {formula} is", end=" ")

    clauses, symbols = generate_parameters(formula)
    solution, model = dpll_algorithm(clauses, symbols, {})

    if solution:
        print(f"satisfiable with the assignment {model}.")
    else:
        print("not satisfiable.")