Fix sphinx/build_docs warnings for other (TheAlgorithms#12482)

* Fix sphinx/build_docs warnings for other

* Fix

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Fix

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>

Author: MaximSmolskiy

other/bankers_algorithm.py

@@ -10,9 +10,10 @@
 predetermined maximum possible amounts of all resources, and then makes a "s-state"
 check to test for possible deadlock conditions for all other pending activities,
 before deciding whether allocation should be allowed to continue.
-[Source] Wikipedia
-[Credit] Rosetta Code C implementation helped very much.
- (https://rosettacode.org/wiki/Banker%27s_algorithm)
+
+| [Source] Wikipedia
+| [Credit] Rosetta Code C implementation helped very much.
+| (https://rosettacode.org/wiki/Banker%27s_algorithm)
 """
 
 from __future__ import annotations
@@ -75,7 +76,7 @@ def __available_resources(self) -> list[int]:
     def __need(self) -> list[list[int]]:
         """
         Implement safety checker that calculates the needs by ensuring that
-        max_claim[i][j] - alloc_table[i][j] <= avail[j]
+        ``max_claim[i][j] - alloc_table[i][j] <= avail[j]``
         """
         return [
             list(np.array(self.__maximum_claim_table[i]) - np.array(allocated_resource))
@@ -86,7 +87,9 @@ def __need_index_manager(self) -> dict[int, list[int]]:
         """
         This function builds an index control dictionary to track original ids/indices
         of processes when altered during execution of method "main"
-            Return: {0: [a: int, b: int], 1: [c: int, d: int]}
+
+            :Return: {0: [a: int, b: int], 1: [c: int, d: int]}
+
         >>> index_control = BankersAlgorithm(
         ...     test_claim_vector, test_allocated_res_table, test_maximum_claim_table
         ... )._BankersAlgorithm__need_index_manager()
@@ -100,7 +103,8 @@ def __need_index_manager(self) -> dict[int, list[int]]:
     def main(self, **kwargs) -> None:
         """
         Utilize various methods in this class to simulate the Banker's algorithm
-        Return: None
+            :Return: None
+
         >>> BankersAlgorithm(test_claim_vector, test_allocated_res_table,
         ...    test_maximum_claim_table).main(describe=True)
                  Allocated Resource Table

other/davis_putnam_logemann_loveland.py

@@ -17,13 +17,15 @@
 
 class Clause:
     """
-    A clause represented in Conjunctive Normal Form.
-    A clause is a set of literals, either complemented or otherwise.
+    | 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)
+        * {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
@@ -39,6 +41,7 @@ def __init__(self, literals: list[str]) -> None:
     def __str__(self) -> str:
         """
         To print a clause as in Conjunctive Normal Form.
+
         >>> str(Clause(["A1", "A2'", "A3"]))
         "{A1 , A2' , A3}"
         """
@@ -47,6 +50,7 @@ def __str__(self) -> str:
     def __len__(self) -> int:
         """
         To print a clause as in Conjunctive Normal Form.
+
         >>> len(Clause([]))
         0
         >>> len(Clause(["A1", "A2'", "A3"]))
@@ -72,11 +76,13 @@ def assign(self, model: dict[str, bool | None]) -> None:
     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.
+          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 + "'"
@@ -92,10 +98,10 @@ def evaluate(self, model: dict[str, bool | None]) -> bool | None:
 
 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))
+    | 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:
@@ -107,16 +113,17 @@ def __init__(self, clauses: Iterable[Clause]) -> None:
     def __str__(self) -> str:
         """
         To print a formula as in Conjunctive Normal Form.
-        str(Formula([Clause(["A1", "A2'", "A3"]), Clause(["A5'", "A2'", "A1"])]))
+
+        >>> 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.
+    | 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)
@@ -149,11 +156,12 @@ def generate_formula() -> Formula:
 
 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.
+    | 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.
+      * 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)
@@ -177,21 +185,20 @@ 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.
+    | 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.
+      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']
@@ -231,20 +238,21 @@ def find_unit_clauses(
 ) -> 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
+      - 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.
+      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']
@@ -278,16 +286,16 @@ 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
+    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.
+      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

other/scoring_algorithm.py

@@ -1,25 +1,26 @@
 """
-developed by: markmelnic
-original repo: https://github.com/markmelnic/Scoring-Algorithm
+| developed by: markmelnic
+| original repo: https://github.com/markmelnic/Scoring-Algorithm
 
 Analyse data using a range based percentual proximity algorithm
 and calculate the linear maximum likelihood estimation.
 The basic principle is that all values supplied will be broken
-down to a range from 0 to 1 and each column's score will be added
+down to a range from ``0`` to ``1`` and each column's score will be added
 up to get the total score.
 
-==========
 Example for data of vehicles
-price|mileage|registration_year
-20k  |60k    |2012
-22k  |50k    |2011
-23k  |90k    |2015
-16k  |210k   |2010
+::
+
+    price|mileage|registration_year
+    20k  |60k    |2012
+    22k  |50k    |2011
+    23k  |90k    |2015
+    16k  |210k   |2010
 
 We want the vehicle with the lowest price,
 lowest mileage but newest registration year.
 Thus the weights for each column are as follows:
-[0, 0, 1]
+``[0, 0, 1]``
 """
 
 
@@ -97,10 +98,11 @@ def procentual_proximity(
     source_data: list[list[float]], weights: list[int]
 ) -> list[list[float]]:
     """
-    weights - int list
-    possible values - 0 / 1
-    0 if lower values have higher weight in the data set
-    1 if higher values have higher weight in the data set
+    | `weights` - ``int`` list
+    | possible values - ``0`` / ``1``
+
+        * ``0`` if lower values have higher weight in the data set
+        * ``1`` if higher values have higher weight in the data set
 
     >>> procentual_proximity([[20, 60, 2012],[23, 90, 2015],[22, 50, 2011]], [0, 0, 1])
     [[20, 60, 2012, 2.0], [23, 90, 2015, 1.0], [22, 50, 2011, 1.3333333333333335]]