Boolean Evaluation

Approach: Basically, we will divide the expression into smaller parts and will try to find the number of ways a smaller expression can result in True and the number of ways a smaller expression can result in False.

With the help of the result of the smaller expression, we will be able to find the number of ways the whole expression can result in True.

Let’s define a recursive function, ‘findWays’, which will receive the following parameter.

(i) ‘exp’ the given expression.

(ii) ‘st’ is an integer variable representing the starting point for the sub-expression.

(iii) ‘end’ is an integer variable representing the ending point for the sub-expression.

(iii) ‘isTrue’ is a bool variable representing whether the sub-expression should evaluate to True or False.

Base Condition :

1. If  ‘st’ >  ‘end’ then do:
   • Return False.
2. If  ‘st’ is equal to  ‘end’ then do:
   • If ‘isTrue’  is equal to True :
     • If current element is true or ‘exp[st]’ == ‘T’
       • Return True
     • Else
       • Return False.
   • If ‘isTrue’  is equal to False :
     • If current element is False or ‘exp[st]’ == ‘F’
       • Return True.
     • Else
       • Return False.

Recursive Calls:

1. Initialize an integer variable ‘ans’=0.
2. Run a loop over ‘exp’ for ‘st’+1 <= ‘k’ <= 'end’-1 and do:
   • We will need to find out the value of the below variables
   • ‘leftTrue’  = The number of ways the expression left to ‘k’ will be true.
   • ‘leftFalse’  = The number of ways the expression left to ‘k’ will be false.
   • ‘rightTrue’  = The number of ways expression right to ‘k’ will be true:
   • ‘rightFalse’  = The number of ways expression right to ‘k’ will be true:
   • We will find these values by making calls to ‘findWays'.
   • To make a call to the left part,  ‘st’ will be ‘st’, and  ‘end’ will be ‘k’-1.
   • To make a call to the right part, ‘st’ will be ‘k’+1, and ‘end’ will be ‘end’.
   • Set the value of ‘isTrue’ = True if we need to find out the number of ways for true else, set False.
3. Now there will be many combinations according to the value of ‘isTrue’ and the value of the current operator(‘exp[k]’).
4. Let’s find out what value will current sub-expression add to ‘ans’ for the given operator ‘|’ and ‘isTrue’ = True.
5. The truth table for OR :
   • T|T=T
   • F|T=T
   • T|F=T
   • F|F=F
6. So ‘ans’+= ‘leftTrue’*‘rightFalse’ +‘leftFalse’ *‘rightTrue’ +‘leftTrue’*‘rightTrue’
7. Similarly, we can find’ for other operators.
8. Return ‘ans’.

Approach: Our last approach was very simple and easy, but its time complexity was of exponential order. We can improve our solution by taking care of the overlapping subproblems. Thus, we will eliminate the need for solving the same subproblems again and again by storing them in a lookup table. This approach is known as Memoization.

We will initialize a 3-D array, say ‘memo’ [n][n][2] with -1, where ‘n' is the size of string ‘exp’.Where ‘memo’[i][j][k] equal to -1 means the current state is not explored. Otherwise, ‘memo’[i][j][0] will store the number of ways expression(i, j)  will be false, and  ‘memo’[i][j][1] will store the number of ways expression(i, j)  will be True.

So there will be two modifications to the earlier approach :

1. In Base Case, first of all, we will check the value of 'memo’ if it is -1, then simply return the value stored in the ‘memo’.
2. In Recursive Calls, before returning the ‘ans’, we will store the value in ‘memo’’.

Approach: As our earlier approach, recursion with memoization surely avoid some subproblems, but we can still improve time complexity using a bottom-up approach.

 

Algorithm:

1. We will initialize a 3-D array, say ‘dp[n][n][1]’ with 0, where ‘n’ is the size of string ‘exp’. ‘dp’[i][j][0] will store the number of ways expression(i, j)  will be false, and ‘dp’[i][j][1] will store the number of ways expression(i, j) will be True.
2. Iterate over the ‘exp’ for 2 <= ‘gap’ < ‘n’ each time increasing ‘gap’ by 2 and do:
   • Iterate over the ‘exp’ for 2 <= ‘j’ < ‘n’ each time increasing 'j' by 2 and do:
     • Iterate over the ‘exp’ for 2 <= ‘k’ < ‘n’ each time increasing ‘k’ by 2 and do:
       • ‘leftTrue’ = ’dp'[j][k][1] (The number of ways expression left to ‘k’ will be true).
       • ‘leftFalse’ = ’dp'[j][k][1] (The number of ways expression left to ‘k’ will be false).
       • ‘rightTrue’ = ’dp'[k+2][j+gap][0] ( The number of ways expression right to ‘k’ will be False).
       • ‘rightFalse’ = ’dp'[k+2][j+gap][1] ( The number of ways expression right to ‘k’ will be False).
       • If 'exp[k+1]’ = ‘|’ then :
         • ‘dp’[j][j+gap][1]+= ‘leftTrue’ * ’rightTrue’+ ‘leftFalse' * ‘rightTrue' + ‘leftTrue’ * ’rightTrue’.
         • ‘dp'[j][j+gap][0]+= ‘leftFalse' * ‘rightFalse’.
         • Similarly, we can solve for other operators too.
3. Return ‘dp’[0][n-1][1].