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bitmask.py

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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 __future__ import print_function

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

#finalmask is used to check if all persons are included by setting all bits to 1

self.finalmask = (1<<len(task_performed)) - 1

def CountWaysUtil(self,mask,taskno):

# if mask == self.finalmask all persons are distributed tasks, return 1

if mask == self.finalmask:

return 1

#if not everyone gets the task and no more tasks are available, return 0

if taskno > self.total_tasks:

return 0

#if case already considered

if self.dp[mask][taskno]!=-1:

return self.dp[mask][taskno]

# Number of ways when we dont this task in the arrangement

total_ways_util = self.CountWaysUtil(mask,taskno+1)

# now assign the tasks one by one to all possible persons and recursively assign for the remaining tasks.

if taskno in self.task:

for p in self.task[taskno]:

# 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),taskno+1)

# save the value.

self.dp[mask][taskno] = total_ways_util

return self.dp[mask][taskno]

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

"""