DPLL algorithm (#3866)

* DPLL algorithm

* Corrections complete

* Formatting

* Codespell hook

* Corrections part 2

* Corrections v2

* Corrections v3

* Update and rename dpll.py to davis–putnam–logemann–loveland.py

Co-authored-by: Christian Clauss <cclauss@me.com>

other/davis–putnam–logemann–loveland.py

@@ -0,0 +1,357 @@
+#!/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
+"""
+
+import random
+from typing import Dict, List
+
+
+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[int]) -> None:
+        """
+        Represent the literals and an assignment in a clause."
+        """
+        # Assign all literals to None initially
+        self.literals = {literal: None for literal in 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:
+        """
+        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
+            if value is not None:
+                # Complement assignment if literal is in complemented form
+                if literal.endswith("'"):
+                    value = not value
+            self.literals[literal] = value
+
+    def evaluate(self, model: Dict[str, bool]) -> bool:
+        """
+        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: List[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()
+    no_of_clauses = random.randint(1, 10)
+    while len(clauses) < no_of_clauses:
+        clauses.add(generate_clause())
+    return Formula(set(clauses))
+
+
+def generate_parameters(formula: Formula) -> (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]
+) -> (List[str], Dict[str, bool]):
+    """
+    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()
+    literals = []
+
+    for clause in clauses:
+        if clause.evaluate(model) is True:
+            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]
+) -> (List[str], Dict[str, bool]):
+    """
+    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(list(clause.literals.keys())[0])
+        else:
+            Fcount, Ncount = 0, 0
+            for literal, value in clause.literals.items():
+                if value is False:
+                    Fcount += 1
+                elif value is None:
+                    sym = literal
+                    Ncount += 1
+            if Fcount == len(clause) - 1 and Ncount == 1:
+                unit_symbols.append(sym)
+    assignment = dict()
+    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]
+) -> (bool, Dict[str, bool]):
+    """
+    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 = [i for i in 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 = [i for i in 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.")