How do I optimize my Python code for balancing a centrifuge with a given input size?
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So, the problem is, given an input N as the size of a centrifuge, I need to find out the number of balanced configurations. The full description is here.
Centrifuge is a piece of equipment that puts multiple test tubes in rotation at very high speeds. It consists of a cylindrical rotor with holes situated evenly along the circumference of the rotor.
Because the device is operating at high speeds, the center of mass of all tubes must coincide with the center of the rotor. We assume all tubes have the same mass.
For the sake of simplicity, we also assume that tubes are point masses, fixed firmly to the rotor. All at the same distance from the center. You may safely assume the
(x,y)coordinates of tubekare $Rsin a, Rcos a$, where $a = 2pifrac{k}{n}$
The problem is: Given a centrifuge with N holes and K tubes, is it possible to balance it?
example 1: given N = 6 possible values for K are 2,3,4,6.
example 2: given N = 12, it is possible to balance 5 tubes: put 3 evenly dispersed (every 4th hole), then find a pair of opposite unoccupied holes and insert remaining 2 tubes there.
example 3: given N = 10, it is impossible to balance 7 tubes: put 5 evenly dispersed, and there's no opposite unoccupied holes left!
This is my code:
import numpy as np
N=int(input())
angle = 2*np.pi/N
z = complex(np.cos(angle), np.sin(angle))
import itertools
combs =
lst = range(1,N)
for i in range(2, N-1):
combs.append(i)
els = [list(x) for x in itertools.combinations(lst, i)]
combs.append(els)
count=1
lis =
while count<=len(combs):
for a in range(len(combs[count])):
s=0
for b in range(len(combs[count][a])):
s+=z**combs[count][a][b]
if abs(s)<1e-10:
lis.append(combs[count][a])
count+=2
lengths =
for i in lis:
if len(i) not in lengths:
lengths.append(len(i))
print(len(lengths)+1)
The code works fine, but slows down after input size of above 20, and is practically unusable after the input grows to 50 or above, due to the for loops. How do I optimize it?
python performance python-3.x programming-challenge
edited 4 mins ago
vnp
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
So, the problem is, given an input N as the size of a centrifuge, I need to find out the number of balanced configurations. The full description is here.
Centrifuge is a piece of equipment that puts multiple test tubes in rotation at very high speeds. It consists of a cylindrical rotor with holes situated evenly along the circumference of the rotor.
Because the device is operating at high speeds, the center of mass of all tubes must coincide with the center of the rotor. We assume all tubes have the same mass.
For the sake of simplicity, we also assume that tubes are point masses, fixed firmly to the rotor. All at the same distance from the center. You may safely assume the
(x,y)coordinates of tubekare $Rsin a, Rcos a$, where $a = 2pifrac{k}{n}$
The problem is: Given a centrifuge with N holes and K tubes, is it possible to balance it?
example 1: given N = 6 possible values for K are 2,3,4,6.
example 2: given N = 12, it is possible to balance 5 tubes: put 3 evenly dispersed (every 4th hole), then find a pair of opposite unoccupied holes and insert remaining 2 tubes there.
example 3: given N = 10, it is impossible to balance 7 tubes: put 5 evenly dispersed, and there's no opposite unoccupied holes left!
This is my code:
import numpy as np
N=int(input())
angle = 2*np.pi/N
z = complex(np.cos(angle), np.sin(angle))
import itertools
combs =
lst = range(1,N)
for i in range(2, N-1):
combs.append(i)
els = [list(x) for x in itertools.combinations(lst, i)]
combs.append(els)
count=1
lis =
while count<=len(combs):
for a in range(len(combs[count])):
s=0
for b in range(len(combs[count][a])):
s+=z**combs[count][a][b]
if abs(s)<1e-10:
lis.append(combs[count][a])
count+=2
lengths =
for i in lis:
if len(i) not in lengths:
lengths.append(len(i))
print(len(lengths)+1)
The code works fine, but slows down after input size of above 20, and is practically unusable after the input grows to 50 or above, due to the for loops. How do I optimize it?
python performance python-3.x programming-challenge
edited 4 mins ago
vnp
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So, the problem is, given an input N as the size of a centrifuge, I need to find out the number of balanced configurations. The full description is here.
Centrifuge is a piece of equipment that puts multiple test tubes in rotation at very high speeds. It consists of a cylindrical rotor with holes situated evenly along the circumference of the rotor.
Because the device is operating at high speeds, the center of mass of all tubes must coincide with the center of the rotor. We assume all tubes have the same mass.
For the sake of simplicity, we also assume that tubes are point masses, fixed firmly to the rotor. All at the same distance from the center. You may safely assume the
(x,y)coordinates of tubekare $Rsin a, Rcos a$, where $a = 2pifrac{k}{n}$
The problem is: Given a centrifuge with N holes and K tubes, is it possible to balance it?
example 1: given N = 6 possible values for K are 2,3,4,6.
example 2: given N = 12, it is possible to balance 5 tubes: put 3 evenly dispersed (every 4th hole), then find a pair of opposite unoccupied holes and insert remaining 2 tubes there.
example 3: given N = 10, it is impossible to balance 7 tubes: put 5 evenly dispersed, and there's no opposite unoccupied holes left!
This is my code:
import numpy as np
N=int(input())
angle = 2*np.pi/N
z = complex(np.cos(angle), np.sin(angle))
import itertools
combs =
lst = range(1,N)
for i in range(2, N-1):
combs.append(i)
els = [list(x) for x in itertools.combinations(lst, i)]
combs.append(els)
count=1
lis =
while count<=len(combs):
for a in range(len(combs[count])):
s=0
for b in range(len(combs[count][a])):
s+=z**combs[count][a][b]
if abs(s)<1e-10:
lis.append(combs[count][a])
count+=2
lengths =
for i in lis:
if len(i) not in lengths:
lengths.append(len(i))
print(len(lengths)+1)
The code works fine, but slows down after input size of above 20, and is practically unusable after the input grows to 50 or above, due to the for loops. How do I optimize it?
python performance python-3.x programming-challenge
edited 4 mins ago
vnp
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
So, the problem is, given an input N as the size of a centrifuge, I need to find out the number of balanced configurations. The full description is here.
Centrifuge is a piece of equipment that puts multiple test tubes in rotation at very high speeds. It consists of a cylindrical rotor with holes situated evenly along the circumference of the rotor.
Because the device is operating at high speeds, the center of mass of all tubes must coincide with the center of the rotor. We assume all tubes have the same mass.
For the sake of simplicity, we also assume that tubes are point masses, fixed firmly to the rotor. All at the same distance from the center. You may safely assume the
(x,y)coordinates of tubekare $Rsin a, Rcos a$, where $a = 2pifrac{k}{n}$
The problem is: Given a centrifuge with N holes and K tubes, is it possible to balance it?
example 1: given N = 6 possible values for K are 2,3,4,6.
example 2: given N = 12, it is possible to balance 5 tubes: put 3 evenly dispersed (every 4th hole), then find a pair of opposite unoccupied holes and insert remaining 2 tubes there.
example 3: given N = 10, it is impossible to balance 7 tubes: put 5 evenly dispersed, and there's no opposite unoccupied holes left!
This is my code:
import numpy as np
N=int(input())
angle = 2*np.pi/N
z = complex(np.cos(angle), np.sin(angle))
import itertools
combs =
lst = range(1,N)
for i in range(2, N-1):
combs.append(i)
els = [list(x) for x in itertools.combinations(lst, i)]
combs.append(els)
count=1
lis =
while count<=len(combs):
for a in range(len(combs[count])):
s=0
for b in range(len(combs[count][a])):
s+=z**combs[count][a][b]
if abs(s)<1e-10:
lis.append(combs[count][a])
count+=2
lengths =
for i in lis:
if len(i) not in lengths:
lengths.append(len(i))
print(len(lengths)+1)
The code works fine, but slows down after input size of above 20, and is practically unusable after the input grows to 50 or above, due to the for loops. How do I optimize it?
python performance python-3.x programming-challenge
python performance python-3.x programming-challenge
edited 4 mins ago
vnp
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
vnp
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
vnp
38.4k13096
edited 4 mins ago
vnp
38.4k13096
edited 4 mins ago
vnp
38.4k13096
38.4k13096
asked 18 mins ago
Kristada673
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 mins ago
Kristada673
1011
asked 18 mins ago
Kristada673
1011
1011
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kristada673 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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