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.

COMING SOON!

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"""
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[0][1]
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
"""
```