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TheAlgorithms-Python/dynamic_programming/bitmask.py at master · windNight/TheAlgorithms-Python · GitHub
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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
"""
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