Question

I have a logical expression that I would like to evaluate.
The expression can be nested and consists of T (True) or F (False) and parenthesis.
The parenthesis “(” means “logical OR”.
Two terms TF beside each others (or any other two combinations beside each others), should be ANDED (Logical AND).

For example, the expression:

((TFT)T) = true

I need an algorithm for solving this problem. I thought of converting the expression first to disjunctive or conjunctive normal form and then I can easily evaluate the expression. However, I couldn’t find an algorithm that normalizes the expression. Any suggestions? Thank you.

The problem statement can be found here:
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=2&category=378&page=show_problem&problem=2967

Edit: I misunderstood part of the problem. In the given logical expression, the AND/OR operators alternate with every parenthesis “(“. If we are to represent the expression by a tree, then the AND/OR operators depend on the the sub-tree’s depth-level. However, it’s initially given that the trees at the deepest level are AND-trees. My task is to evaluate the given expression possibly by constructing the tree.
Thanks for the answers below which clarified the correct requirement of the problem.

Answer 1

I solved this problem using a different technique than the ones mentioned. And I got it Accepted by the online system judge.

After figuring out the operator at the first level of the tree (Thanks to @Asiri Rathnayake for his idea), I recursively construct the expression tree. During the construction, I scan the string. If the character is ‘(‘, then I create a node with the current operator value and add it to the tree. Then, I alternate the operator and go for a deeper recursion level. If the character is ‘T’, then I create a node with value “True”, add it to the tree and continue scanning. If the character is ‘F’, then I create a node with the value “False”, add it to the tree and continue scanning. Finally, if the character is ‘)’, then I return to one level up of the recursion.

At the end, I will have the expression tree completed. Now, all I need to do is a simple evaluation for the tree using basic recursive function.

Below is my C++ code:

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>

using namespace std;

struct Node {

    char value;
    vector<Node*> children;
};


void ConstructTree (int &index, string X, Node *&node, int op)
{

    for(; index<X.size(); index++)
    {
        if(X[index]=='T')
        {
            Node *C= new Node;
            C->value='T';
            node->children.push_back(C);
        }


        else if(X[index]=='F')
        {
            Node* C= new Node;
            C->value='F';
            node->children.push_back(C);
        }


        else if(X[index]=='(')
        {
            if(op==0)
            {
                Node* C= new Node;
                C->value='O';
                node->children.push_back(C);
            }


            else
            {
                Node* C= new Node;
                C->value='A';
                node->children.push_back(C);
            }

            index++;
            ConstructTree(index,X,node->children[node->children.size()-1],1-op);
        }

        else
            return;
    }



}

bool evaluateTree(Node* node)
{
    if(node->value=='T')
        return true;
    else if(node->value=='F')
        return false;
    else if(node->value=='O')
    {
        for(int i=0; i<node->children.size(); i++)
            if(evaluateTree(node->children[i])==true)
                return true;

        return false;
    }

    else if(node->value=='A')
    {

        for(int i=0; i<node->children.size(); i++)
            if(evaluateTree(node->children[i])==false)
                return false;

        return true;
    }
}


int main()
{
    string X;
    int testCase=1;

    while(cin>>X)
    {
        if(X=="()")
            break;


        int index=0;

        int op=-1;

        int P=0;

        int max=0;
        for(int i=0; i<X.size(); i++)
        {
            if(X[i]=='(')
                P++;
            if(X[i]==')')
                P--;

            if(P>max)
                max=P;
        }


        if(max%2==0)
            op=0; //OR
        else
            op=1; //AND


        Node* root = new Node;

        if(op==0)
        root->value='O';
        else
        root->value='A';

        index++;
        ConstructTree(index,X,root,1-op);

        if(evaluateTree(root))
            cout<<testCase<<". true"<<endl;
        else
            cout<<testCase<<". false"<<endl;

        testCase++;
    }
}