human
Find length of the largest region in Boolean Matrix | A function to check if a given cell ( row , col ) can be included in DFS.
row number is in range , column number is in range and value is 1 and not yet visited.
A utility function to do DFS for a 2D boolean matrix . It only considers the 8 neighbours as adjacent vertices.
These arrays are used to get row and column numbers of 8 neighbours of a given cell.
Mark this cell as visited.
Recur for all connected neighbours.
increment region length by one.
The main function that returns largest length region of a given boolean 2D matrix.
Make a bool array to mark visited cells . Initially all cells are unvisited.
Initialize result as 0 and travesle through the all cells of given matrix.
If a cell with value 1 is not.
visited yet , then new region found.
maximum region.
Driver Code.
Function call"
How can the above be solved in Python?

What is a code to find length of the largest region in Boolean Matrix and the other steps required?
assistant
#Here is code to find length of the largest region in Boolean Matrix and the other steps required.
 def isSafe ( M , row , col , visited ) :
	global ROW , COL
	return ( ( row >= 0 ) and ( row < ROW ) and ( col >= 0 ) and ( col < COL ) and ( M [ row ] [ col ] and not visited [ row ] [ col ] ) )
def DFS ( M , row , col , visited , count ) :
	rowNbr = [ - 1 , - 1 , - 1 , 0 , 0 , 1 , 1 , 1 ]
	colNbr = [ - 1 , 0 , 1 , - 1 , 1 , - 1 , 0 , 1 ]
	visited [ row ] [ col ] = True
	for k in range ( 8 ) :
		if ( isSafe ( M , row + rowNbr [ k ] , col + colNbr [ k ] , visited ) ) :
			count [ 0 ] += 1
			DFS ( M , row + rowNbr [ k ] , col + colNbr [ k ] , visited , count )
def largestRegion ( M ) :
	global ROW , COL
	visited = [ [ 0 ] * COL for i in range ( ROW ) ]
	result = - 999999999999
	for i in range ( ROW ) :
		for j in range ( COL ) :
			if ( M [ i ] [ j ] and not visited [ i ] [ j ] ) :
				count = [ 1 ]
				DFS ( M , i , j , visited , count )
				result = max ( result , count [ 0 ] )
	return result
ROW = 4
COL = 5
M = [ [ 0 , 0 , 1 , 1 , 0 ] , [ 1 , 0 , 1 , 1 , 0 ] , [ 0 , 1 , 0 , 0 , 0 ] , [ 0 , 0 , 0 , 0 , 1 ] ]
print ( largestRegion ( M ) )