FilesExpand file tree

Python-3.9rc2

/

bitmask.py

Copy path

Blame

More file actions

Blame

More file actions

Latest commit

 

History

History

History

93 lines (68 loc) · 3.04 KB

Python-3.9rc2

/

bitmask.py

Copy path

Top

File metadata and controls

• Code
• Blame

93 lines (68 loc) · 3.04 KB

Raw

Copy raw file

Download raw file

Open symbols panel

Edit and raw actions

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

"""

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

"""