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                    Analysis of Algorithms - Homework 2

                              Gigel and Mafia

                         Buf Sorina-Anamaria 321CA

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    The purpose of this homework is to manoeuvre an investigation on Mafia
families, having as ultimate goal to abolish the relations between the clan
members through multiple arrests. In order to do so, the current problem will be
reduced to the SAT problem, along the different stages of the mission, thus
streamlining the investigation process by communicating with the Antic Oracle.
Generally, transforming the current problem to SAT will mean creating a boolean
formula, that will later be evaluated by the Oracle, determining if there exists
an interpretation that satisfies the given conjunctive normal form, similar to
other famous graph problems (e.g. reducing graph k-coloring to SAT).

    ▶ Task 1 - SPY PLANTING

        The first task of Gigel consists in planting a number of spies in the
    Mafia families, so each family would get infiltrated, but two related clans
    could not have the same spy.
        In order to explain the logic behind the solution of this problem, in
    the demonstration that is to follow, we will consider the next variables:

        ◌ N - number of the Mafia families;
        ◌ M - number of the connections between the families (undirected
        relations);
        ◌ K - the maximum number of spies that may be planted.

        The reduction to SAT problem and consequently the boolean formula that
    the current problem is transformed to was implemented in the function:

        ◉ formulateOracleQuestion()

        The conjunctive normal form will be represented by a number of clauses
    and variables, component of the clauses formulated. The clausal normal form
    is a conjunction of one or more clauses, while a clause is a disjunction of
    literals, here represented by the variables that are to be declared. We will
    consider the following:

        ◌ number of variables = N * K, attributed as following:

        variable: 1 → family no.1 may be infiltrated by spy no.1
        variable: 2 → family no.2 may be infiltrated by spy no.1
        variable: 3 → family no.3 may be infiltrated by spy no.1
        .
        .
        .
        variable: N → family no.N may be infiltrated by spy no.1
        variable: N + 1 → family no.1 may be infiltrated by spy no.2
        . . .

        variable: N * (K - 1) + 1 → family no.1 may be infiltrated by spy no.K
        variable: N * (K - 1) + 2 → family no.2 may be infiltrated by spy no.K
        variable: N * (K - 1) + 3 → family no.3 may be infiltrated by spy no.K
        .
        .
        .
        variable: N * (K - 1) + N → family no.N may be infiltrated by spy no.K

        ◌ number of clauses = N + M * K + N * K * (K - 1) / 2, thus explained:

            The first series of clauses is represented by the first "for loop"
            present in the current function, that displays the conditions that
            must be respected regarding the spies distribution amongst the
            families: each Mafia family must have at least one spy. Thus, there
            will be "N" clauses, one clause related to each family, formed of a
            number of "K" literals, represented by the corresponding variables
            that illustrated the fact that in each family may be planted one of
            the spies:

                clause no.1 → 1 ∨ (N + 1) ∨ . . . ∨ (N * (K - 1) + 1)
                clause no.2 → 2 ∨ (N + 2) ∨ . . . ∨ (N * (K - 1) + 2)
                . . .

                clause no.N → N ∨ (N + N) ∨ . . . ∨ (N * (K - 1) + N)

                ● Time complexity of constructing the first clauses is
                equivalent to parsing the first "for" of families and the second
                one (used for variable construction): O(N * K).

            The second series of clauses is represented by the second "for loop"
            block, illustrating the strictness over spy distribution between
            connected families: two related families can not have the same spy.
            Thus, there will be "M * K" clauses; each pair of families must have
            distinct spy, therefore the following clauses:
                . . .
                clause no.i → ¬family no.k with spy no.1 ∨ ¬family no.l with spy
                            no.1
                clause no.i+1 → ¬family no.k with spy no.2 ∨ ¬ family no.l with
                            spy no.2
                . . .
                 ● Time complexity of constructing the second clauses is
                 equivalent to parsing the first "for" of relations and the
                 second one of spies: O(M * K).

            The third series of clauses is represented by the third loop block,
            responsible for creating the clauses regarding the distinction
            distribution of infiltrates over a family: a family must not have
            more than a spy. Thus, there will be "N * K * (K - 1) / 2" clauses,
            as each pair of non-repetitive, distinct spies can not coexist in a
            clan:

                clause no.1 → ¬1 ∨ ¬(N + 1)
                clause no.2 → ¬1 ∨ ¬(N * 2 + 1)
                . . .
                clause no.K-1 → ¬1 ∨ ¬(N * (K - 1) + 1)
                clause no.K → ¬(N + 1) ∨ ¬(N * 2 + 1)
                . . .
                ● Time complexity of constructing the third clauses is
                equivalent to parsing the first "for" of relations, the second
                one of spies and the third one of remaining spies: O(M *
                (1 + 2 + ... + (K - 1))) = O(N * K * (K - 1) / 2).

    ⇒ Final complexity of SAT reduction: O(N * K + M * K + N * K * (K - 1) / 2).

        The rest of the functions used to resolve the task are:

        ◉ readProblemData() → responsible for populating the main attributes of
    the problem and for storing the relations between the family in a O(M)
    complexity;
        ◉ decipherOracleAnswer() → extracts the answer of the oracle consisting
    in a list of size "N * K" of variables, positive (the ones marked as TRUE)
    and negative (the ones marked as FALSE), and converts the list into the
    problem's answer in a O(N * K) time complexity;
        ◉ writeAnswer() → writes the index of the spy assigned to each family
    (O(N)).

    ▶ Task 2 - EXTENDED FAMILIES INVESTIGATIONS

        The second task of Gigel consists in determining the most influential
    groups of families, represented by the extended ones, in which each two
    families must be related.
        In order to explain the logic behind the solution of this problem, in
    the demonstration that is to follow, we will consider the next variables:

        ◌ N - number of the Mafia families;
        ◌ M - number of the connections between the families (undirected
        relations);
        ◌ K - size of the extended searched family;
        ◌ X - number of non-edges = N * (N - 1) / 2 - M.

        The reduction to SAT problem and consequently the boolean formula that
    the current problem is transformed to was implemented in the function:

        ◉ formulateOracleQuestion()

        The conjunctive normal form will be represented by a number of clauses
    and variables, component of the clauses formulated, similar to the previous
    task. This algorithm can be seen as the generalized clique problem, where we
    want to find a clique of size "K" in the graph of family relations.

        ◌ number of variables = N * K, attributed as following:

        variable: 1 → family no.1 may be node no.1 in the clique
        variable: 2 → family no.2 may be node no.1 in the clique
        variable: 3 → family no.3 may be node no.1 in the clique
        .
        .
        .
        variable: N → family no.N may be node no.1 in the clique
        variable: N + 1 → family no.1 may be node no.2 in the clique
        . . .

        variable: N * (K - 1) + 1 → family no.1 may be node no.K in the clique
        variable: N * (K - 1) + 2 → family no.2 may be node no.K in the clique
        variable: N * (K - 1) + 3 → family no.3 may be node no.K in the clique
        .
        .
        .
        variable: N * (K - 1) + N → family no.N may be node no.K in the clique

        ◌ number of clauses = K + K * N * (N - 1) / 2 + X * K * (K - 1) +
        N * K * (K - 1) / 2, thus explained:

            The first series of clauses is represented by the first "for loop"
            present in the current function, that displays the conditions that
            must be respected regarding the family distribution amongst the
            clique: each node of the clique must be represented by a family.
            There will be "K" clauses, one clause related to each node, formed
            of a number of "N" literals, represented by the corresponding
            variables:

                clause no.1 → 1 ∨ 2 ∨ . . . ∨ N
                clause no.2 → (N + 1) ∨ (N + 2) ∨ . . . ∨ (N + N)
                . . .

                clause no.K → (N * (K - 1) + 1) ∨ . . . ∨ (N * (K - 1) + N)

                ● Time complexity of constructing the first clauses is
                equivalent to parsing the first "for" of nodes and the second
                one (used for variable construction): O(K * N).

            The second series of clauses is represented by the second loop
            block, responsible for creating the clauses regarding the
            distinction distribution of families over a node: a node must
            not be represented by more than a family. Thus, there will be
            "K * N * (N - 1) / 2" clauses, as each pair of non-repetitive,
            distinct families can not coexist as the same clique node:

                 clause no.1 → ¬1 ∨ ¬2
                 clause no.2 → ¬1 ∨ ¬3
                 . . .
                 clause no.N-1 → ¬1 ∨ ¬N
                 clause no.N → ¬2 ∨ ¬3
                 . . .
                 ● Time complexity of constructing the second clauses is
                 equivalent to parsing the first "for" of nodes, the second
                 one of families and the third one of remaining families: O(K *
                 (1 + 2 + ... + (N - 1))) = O(K * N * (N - 1) / 2).

            The third series of clauses is represented by the third loop
            block, responsible for creating the clauses regarding the
            representation over a node, regarding the families: a family must
            not represent more than a clique node. Thus, there will be
            "N * K * (K - 1) / 2" clauses, as each pair of non-repetitive,
            distinct nodes can not coexist as having the same family:

                 clause no.1 → ¬1 ∨ ¬(N + 1)
                 clause no.2 → ¬1 ∨ ¬(N * 2 + 1)
                 . . .
                 clause no.K-1 → ¬1 ∨ ¬(N * (K - 1) + 1)
                 clause no.K → ¬(N + 1) ∨ ¬(N * 2 + 1)
                 . . .
                 ● Time complexity of constructing the third clauses is
                 equivalent to parsing the first "for" of families, the second
                 one of nodes and the third one of remaining nodes: O(N *
                 (1 + 2 + ... + (K - 1))) = O(N * K * (K - 1) / 2).

            The last series of clauses is defined in the final "for" segment,
            aiming to create clauses that illustrate the fact that two families
            that are unconnected can not coexist in the searched clique.
            Therefore, we write "X * K (K - 1)" clauses, as two families with no
            direct relation between them can not represent different nodes in
            the clique:
                 . . .
                 clause no.i → ¬family no.k as node no.1 ∨ ¬family no.l as node
                             no.2
                 clause no.i+1 → ¬family no.k as node no.1 ∨ ¬family no.l as
                                 node spy no.3
                 . . .
                 ● Time complexity of constructing the final clauses is
                 equivalent to parsing the first two "for" of relations and the
                 second ones of nodes: O(N^2 * K^2).

    ⇒ Final complexity of SAT reduction: O(K * N + K * N * (N - 1) / 2
    + N * K * (K - 1) / 2 + N^2 * K^2).

        The rest of the functions used to resolve the task are:

        ◉ readProblemData() → responsible for populating the main attributes of
    the problem and for storing the relations between the family in a
    O(N + M + 1) complexity;
        ◉ decipherOracleAnswer() → extracts the answer of the oracle consisting
    in a list of size "N * K" of variables, positive (the ones marked as TRUE)
    and negative (the ones marked as FALSE), and converts the list into the
    problem's answer in a O(N * K) time complexity;
        ◉ writeAnswer() → writes the index of the family assigned to each node
    of the clique (O(K)).

    ▶ Task 3 - MAFIA ARRESTS

        Final task of Gigel consists in arresting the minimum number of families
    in order to disconnect all the clans and, consequently, destroy the Mafia
    grid.
        In order to explain the logic behind the solution of this problem, in
    the demonstration that is to follow, we will consider the next variables:

        ◌ N - number of the Mafia families;
        ◌ M - number of the connections between the families (undirected
        relations);
        ◌ K - size of the extended searched family;
        ◌ X - number of non-edges = N * (N - 1) / 2 - M.

        As a way to resolve this problem, we will use the previously implemented
    solution from task 2, requesting multiple answers from the Oracle. The main
    idea of the programme is to try finding the maximal clique in the
    complementary graph of relations between Mafia families, as the minimum
    number of arrests that need to be undertaken will be consisted of the
    families that are not found in the answer requested by the Antic Oracle.
        The utility functions used for resolving this task are:

        ◉ readProblemData() → responsible for populating the main attributes of
    the problem and for storing the relations between the families in the list
    of adjacency lists, while doing the same for the complementary graph, in a
    O(N + M + 1 + N^2) time complexity;
        ◉ reduceToTask2() → creates the input file of the Task 2 with the
    attributes of the current problem transferred to its complement;
        ◉ extractAnswerFromTask2() → extracts the answer given by the call of
    Task2, consisting in the maximal clique of size "K" present in the
    complementary graph;
        ◉ writeAnswer() → writes the minimal arrests needed to be performed.

        The main method of the programme, responsible for controlling the flow
    of multiple requests to the Oracle if defined as it follows:

        ◉ solve()

        In order to find the maximal clique in the complementary graph, by
    multiply requesting help from the Oracle, we need to set up the starting
    value of "K", considered as maximum, that is going to decrease in the
    formulated problem until a valid solution is discovered.
        Taking into account that, in a complete graph, the maximal amount of
    edges is "n * (n - 1) / 2" (where "n" is the number of nodes), we can set
    the superior limit of "K" by considering that all the edges given in the
    problem are part of the same complete subgraph:

                            ⇒ K * (K - 1) = 2 * X
                              ⇒ K^2 - K = 2 * X
                         ⇒ (K - 0.5)^2 = 2 * X + 0.25
                        ⇒ K - 0.5 = sqrt(2 * X + 0.25)
                  ⇒ K = 0.5 + sqrt(2 * X + 0.25) (final formula)

        Considering the complexity previously calculated for resolving task 2,
    we can agree that the complexity of this method will consist in the one
    already set at the second stage of the mission, multiplied by the difference
    between the optimized "K" discovered on the basis of the number of non-edges
    and the actual size of a maximal clique that is to be found.

    ▶ Task 4 - MAFIA ARRESTS (BONUS)

        In this special part of the mission, we are trying to reduce the
    questions addressed to the Oracle to a single one by transforming the input
    format to the Weighted Partial Max-SAT Input format. Therefore, the main
    functions will be identical to the Task 2, transferring the logic of the
    problem to this task, but applied to the complementary graph of relations,
    in the same way as we managed Task 3. Therefore, we can say that this task
    will consist in a combination between the second and the third one,
    modified in such a way as we get the correct answer through one Oracle call.
        In Task 2, we considered four types of clauses, now applied on the
    complementary problem (we consider the notations from previous tasks):

            ◌ "K" clauses representing the fact that a node of the clique must
        be represented by one family;
            ◌ "K * N * (N - 1) / 2" clauses corresponding to the fact that a
        node of a clique can not be represented by more than a family;
            ◌ "N * K * (K - 1) / 2" clauses corresponding to the fact that a
        family can not represent more than a node in the clique;
            ◌ "M * K * (K - 1)" clauses representing the fact that for every
        two families that form a relation, they can not both be present in the
        complementary clique.

        In the Weighted Partial Max-SAT Input format, it must be defined the
    weight value of hard clauses (the obligatory ones), and the weight of soft
    clauses (the optional ones), that will precede every clause in the input
    oracle file.
        In order to find the maximal possible clique, we will consider all the
    clauses as being hard, except the first set, that defined the existence of
    the respective nodes. Also, as the minimal size of a clique is considered as
    two, we will define two of the first clauses as obligatory to be fulfilled.
        I considered the soft clauses, with their respective weights, thus:

            clause no.3 → 1: (N * 2 + 1) ∨ (N * 2 + 2) ∨ . . . ∨ (N * 2 + N)
            clause no.4 → 2 : (N * 3 + 1) ∨ (N * 3 + 2) ∨ . . . ∨ (N * 3 + N)
            clause no.5 → 4 : (N * 4 + 1) ∨ (N * 4 + 2) ∨ . . . ∨ (N * 4 + N)
            . . .

            clause no.K → 2^(K-3): (N * (K - 1) + 1) ∨ . . . ∨ (N * (K - 1) + N)

        ⇒ hard clauses weight = sum of all the weights of soft clauses + 1
        ⇒ hard clauses weight = 1 + 2 + 2^2 + ... + 2^(K - 3) + 1
        ⇒ hard clauses weight = 2^(K - 2) - 1 + 1
        ⇒ hard clauses weight = 2^(K - 2)

        Therefore, we created the respective clauses of this problem. A final
    aspect to be reminded is the chosen value of "K", as I tried to optimize the
    previous value set for the size of the possible maximal clique:

        ⇒ before: K = 0.5 + sqrt(2 * X + 0.25)

         A method to reduce the size of "K" was trying to find the maximal
    degree of a family in a complete graph by verifying the corresponding number
    of relations of their neighbours, which should be equal or greater than the
    initial family's.

        For more details related to effective implementation, verify the
    comments in the code.

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