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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 :
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def __init__ ( self , task_performed , total ) :
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self . total_tasks = total # total no of tasks (N)
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# DP table will have a dimension of (2^M)*N
# initially all values are set to -1
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self . dp = [
[ - 1 for i in range ( total + 1 ) ] for j in range ( 2 * * len ( task_performed ) )
]
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self . task = defaultdict ( list ) # stores the list of persons for each task
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# finalmask is used to check if all persons are included by setting all bits to 1
self . finalmask = ( 1 << len ( task_performed ) ) - 1
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def CountWaysUtil ( self , mask , taskno ) :
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# if mask == self.finalmask all persons are distributed tasks, return 1
if mask == self . finalmask :
return 1
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# if not everyone gets the task and no more tasks are available, return 0
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if taskno > self . total_tasks :
return 0
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# if case already considered
if self . dp [ mask ] [ taskno ] != - 1 :
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return self . dp [ mask ] [ taskno ]
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# Number of ways when we don't this task in the arrangement
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total_ways_util = self . CountWaysUtil ( mask , taskno + 1 )
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# 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 ] :
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# if p is already given a task
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if mask & ( 1 << p ) :
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continue
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# assign this task to p and change the mask value. And recursively assign tasks with the new mask value.
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total_ways_util + = self . CountWaysUtil ( mask | ( 1 << p ) , taskno + 1 )
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# save the value.
self . dp [ mask ] [ taskno ] = total_ways_util
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return self . dp [ mask ] [ taskno ]
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def countNoOfWays ( self , task_performed ) :
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# 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 )
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# call the function to fill the DP table, final answer is stored in dp[0][1]
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return self . CountWaysUtil ( 0 , 1 )
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if __name__ == " __main__ " :
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total_tasks = 5 # total no of tasks (the value of N)
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# 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
)
)
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"""
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
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"""
