human
What is a code to boolean Parenthesization Problem and the other steps required given the following:
Boolean Parenthesization Problem | DP | 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.
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 ).
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.
Find place of parenthesization using current value of gap.
Store Total [ i ] [ k ] and Total [ k + 1 ] [ j ].
Follow the recursive formulas according to the current operator.
Driver Code.
There are 4 ways ( ( T T ) & ( F ^ T ) ) , ( T | ( T & ( F ^ T ) ) ) , ( ( ( T T ) & F ) ^ T ) and ( T | ( ( T & F ) ^ T ) )"
Solution in Python:
assistant
#Here is code to boolean Parenthesization Problem and the other steps required.
 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 ) ]
	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
	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 ) :
				k = i + g
				tik = T [ i ] [ k ] + F [ i ] [ k ]
				tkj = T [ k + 1 ] [ j ] + F [ k + 1 ] [ j ]
				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 ]
symbols = " TTFT "
operators = " | & ^ "
n = len ( symbols )
print ( countParenth ( symbols , operators , n ) )