Boolean Parenthesization Problem | DP-37

Given a boolean expression with following symbols.

Symbols
    'T' ---> true 
    'F' ---> false 

And following operators filled between symbols

Operators
    &   ---> boolean AND
    |   ---> boolean OR
    ^   ---> boolean XOR 

Count the number of ways we can parenthesize the expression so that the value of expression evaluates to true.

Let the input be in form of two arrays one contains the symbols (T and F) in order and other contains operators (&, | and ^}

Examples:

Input: symbol[]    = {T, F, T}
       operator[]  = {^, &}
Output: 2
The given expression is "T ^ F & T", it evaluates true
in two ways "((T ^ F) & T)" and "(T ^ (F & T))"

Input: symbol[]    = {T, F, F}
       operator[]  = {^, |}
Output: 2
The given expression is "T ^ F | F", it evaluates true
in two ways "( (T ^ F) | F )" and "( T ^ (F | F) )". 

Input: symbol[]    = {T, T, F, T}
       operator[]  = {|, &, ^}
Output: 4
The given expression is "T | T & F ^ T", it evaluates true
in 4 ways ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) 
and (T|((T&F)^T)). 


Solution:
Let T(i, j) represents the number of ways to parenthesize the symbols between i and j (both inclusive) such that the subexpression between i and j evaluates to true.
T(i,j)=\sum_{k=i}^{j-1}\begin{Bmatrix} T(i,k)*T(k+1,j) & if&operator&[k]is  '\&'\\  Total(i,k)*Total(k+1,j)-F(i,k)*F(k+1,j) &if&operator&[k]&is'|' \\  T(i,k)*F(k+1,j)+F(i,k)*T(k+1,j) &if&operator&[k]&is '\oplus'  \end{Bmatrix}  Total(i,j)= T(i,j)+F(i,j)
<!–trueeq–>

Let F(i, j) represents the number of ways to parenthesize the symbols between i and j (both inclusive) such that the subexpression between i and j evaluates to false.

F(i,j)=\sum_{k=i}^{j-1} \begin{Bmatrix} Total(i,k)*Total(k+1,j)-T(i,k)*T(k+1,j) & if&operator[k]&is'\&'\\  F(i,k)*F(k+1,j) &if&operator[k] &is'|' \\  T(i,k)*T(k+1,j)+F(i,k)*F(k+1,j) &if&operator[k]&is'\oplus'  \end{Bmatrix}  Total(i,j)=T(i,j)+F(i,j)
<!—falseeq
–>
Base Cases:

T(i, i) = 1 if symbol[i] = 'T' 
T(i, i) = 0 if symbol[i] = 'F' 

F(i, i) = 1 if symbol[i] = 'F' 
F(i, i) = 0 if symbol[i] = 'T'

If we draw recursion tree of above recursive solution, we can observe that it many overlapping subproblems. Like other dynamic programming problems, it can be solved by filling a table in bottom up manner. Following is C++ implementation of dynamic programming solution.

C++

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#include<iostream>
#include<cstring>
using namespace std;
  
// Returns count of all possible parenthesizations that lead to
// result true for a boolean expression with symbols like true
// and false and operators like &, | and ^ filled between symbols
int countParenth(char symb[], char oper[], int n)
{
    int F[n][n], T[n][n];
  
    // Fill diaginal entries first
    // All diagonal entries in T[i][i] are 1 if symbol[i]
    // is T (true).  Similarly, all F[i][i] entries are 1 if
    // symbol[i] is F (False)
    for (int i = 0; i < n; i++)
    {
        F[i][i] = (symb[i] == 'F')? 1: 0;
        T[i][i] = (symb[i] == 'T')? 1: 0;
    }
  
    // Now fill T[i][i+1], T[i][i+2], T[i][i+3]... in order
    // And F[i][i+1], F[i][i+2], F[i][i+3]... in order
    for (int gap=1; gap<n; ++gap)
    {
        for (int i=0, j=gap; j<n; ++i, ++j)
        {
            T[i][j] = F[i][j] = 0;
            for (int g=0; g<gap; g++)
            {
                // Find place of parenthesization using current value
                // of gap
                int k = i + g;
  
                // Store Total[i][k] and Total[k+1][j]
                int tik = T[i][k] + F[i][k];
                int tkj = T[k+1][j] + F[k+1][j];
  
                // Follow the recursive formulas according to the current
                // operator
                if (oper[k] == '&')
                {
                    T[i][j] += T[i][k]*T[k+1][j];
                    F[i][j] += (tik*tkj - T[i][k]*T[k+1][j]);
                }
                if (oper[k] == '|')
                {
                    F[i][j] += F[i][k]*F[k+1][j];
                    T[i][j] += (tik*tkj - F[i][k]*F[k+1][j]);
                }
                if (oper[k] == '^')
                {
                    T[i][j] += F[i][k]*T[k+1][j] + T[i][k]*F[k+1][j];
                    F[i][j] += T[i][k]*T[k+1][j] + F[i][k]*F[k+1][j];
                }
            }
        }
    }
    return T[0][n-1];
}
  
// Driver program to test above function
int main()
{
    char symbols[] = "TTFT";
    char operators[] = "|&^";
    int n = strlen(symbols);
  
    // There are 4 ways
    // ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and (T|((T&F)^T))
    cout << countParenth(symbols, operators, n);
    return 0;
}

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Python3

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# Returns count of all possible 
# parenthesizations that lead to 
# result true for a boolean 
# expression with symbols like  
# true and false and operators 
# like &, | and ^ filled between symbols 
def countParenth(symb, oper, n):
    F = [[0 for i in range(n + 1)] 
            for i in range(n + 1)]
    T = [[0 for i in range(n + 1)] 
            for i in range(n + 1)]
              
    # Fill diaginal entries first 
    # All diagonal entries in 
    # T[i][i] are 1 if symbol[i] 
    # is T (true). Similarly, all
    # F[i][i] entries are 1 if 
    # symbol[i] is F (False) 
    for i in range(n):
        if symb[i] == 'F':
            F[i][i] = 1
        else:
            F[i][i] = 0
  
        if symb[i] == 'T':
            T[i][i] = 1
        else:
            T[i][i] = 0
              
    # Now fill T[i][i+1], T[i][i+2], 
    # T[i][i+3]... in order And 
    # F[i][i+1], F[i][i+2], 
    # F[i][i+3]... in order
    for gap in range(1, n):
        i = 0
        for j in range(gap, n): 
            T[i][j] = F[i][j] = 0
            for g in range(gap):
                  
                # Find place of parenthesization 
                # using current value of gap
                k = i + g
                  
                # Store Total[i][k] and Total[k+1][j] 
                tik = T[i][k] + F[i][k]; 
                tkj = T[k + 1][j] + F[k + 1][j];
                  
                # Follow the recursive formulas 
                # according to the current operator
                if oper[k] == '&':
                    T[i][j] += T[i][k] * T[k + 1][j]
                    F[i][j] += (tik * tkj - T[i][k] * 
                                            T[k + 1][j])
                if oper[k] == '|':
                    F[i][j] += F[i][k] * F[k + 1][j]
                    T[i][j] += (tik * tkj - F[i][k] * 
                                            F[k + 1][j])
                if oper[k]=='^':
                    T[i][j] += (F[i][k] * T[k + 1][j] + 
                                T[i][k] * F[k + 1][j]) 
                    F[i][j] += (T[i][k] * T[k + 1][j] + 
                                F[i][k] * F[k + 1][j])
            i += 1
    return T[0][n - 1
      
# Driver Code
symbols = "TTFT"
operators = "|&^"
n = len(symbols)
  
# There are 4 ways 
# ((T|T)&(F^T)), (T|(T&(F^T))), 
# (((T|T)&F)^T) and (T|((T&F)^T)) 
print(countParenth(symbols, operators, n))
  
# This code is contributed by 
# sahil shelangia

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Output:

4


Time Complexity:
O(n3)
Auxiliary Space: O(n2)

References:
http://people.cs.clemson.edu/~bcdean/dp_practice/dp_9.swf

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Improved By : sahilshelangia