In computer science, a mask or bitmask is data that is used for bitwise operations, particularly in a bit field.can be put either on, off or inverted from on to off (or vice versa) in a individual bitwise operation.
""" This is a Python implementation for questions involving task assignments between people. Here Bitmasking and DP are used for solving this. Question :- We have N tasks and M people. Each person in M can do only certain of these tasks. Also a person can do only one task and a task is performed only by one person. Find the total no of ways in which the tasks can be distributed. """ from collections import defaultdict class AssignmentUsingBitmask: def __init__(self, task_performed, total): self.total_tasks = total # total no of tasks (N) # DP table will have a dimension of (2^M)*N # initially all values are set to -1 self.dp = [ [-1 for i in range(total + 1)] for j in range(2 ** len(task_performed)) ] self.task = defaultdict(list) # stores the list of persons for each task # final_mask is used to check if all persons are included by setting all bits # to 1 self.final_mask = (1 << len(task_performed)) - 1 def CountWaysUtil(self, mask, task_no): # if mask == self.finalmask all persons are distributed tasks, return 1 if mask == self.final_mask: return 1 # if not everyone gets the task and no more tasks are available, return 0 if task_no > self.total_tasks: return 0 # if case already considered if self.dp[mask][task_no] != -1: return self.dp[mask][task_no] # Number of ways when we don't this task in the arrangement total_ways_util = self.CountWaysUtil(mask, task_no + 1) # now assign the tasks one by one to all possible persons and recursively # assign for the remaining tasks. if task_no in self.task: for p in self.task[task_no]: # if p is already given a task if mask & (1 << p): continue # assign this task to p and change the mask value. And recursively # assign tasks with the new mask value. total_ways_util += self.CountWaysUtil(mask | (1 << p), task_no + 1) # save the value. self.dp[mask][task_no] = total_ways_util return self.dp[mask][task_no] def countNoOfWays(self, task_performed): # Store the list of persons for each task for i in range(len(task_performed)): for j in task_performed[i]: self.task[j].append(i) # call the function to fill the DP table, final answer is stored in dp return self.CountWaysUtil(0, 1) if __name__ == "__main__": total_tasks = 5 # total no of tasks (the value of N) # the list of tasks that can be done by M persons. task_performed = [[1, 3, 4], [1, 2, 5], [3, 4]] print( AssignmentUsingBitmask(task_performed, total_tasks).countNoOfWays( task_performed ) ) """ For the particular example the tasks can be distributed as (1,2,3), (1,2,4), (1,5,3), (1,5,4), (3,1,4), (3,2,4), (3,5,4), (4,1,3), (4,2,3), (4,5,3) total 10 """