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1

I have this following code:

The code is simply stuck when I give input n=6 ,(in jupyter)I don't know what to do please help.

n=int(input("Enter a number to experiment:"))
l=[]
w=[]
nw=[]
c=0
for i in range(2*n):
 if(i<n):
 l.append(0)
 else:
 l.append(1)

import itertools as it
permset=set([i for i in it.permutations(l)])
for x in permset:
 w.append(x)
print (w)
for j in range(len(w)):
 c=1

 for y in range(2*n-1):

 if(w[j][y]==w[j][y+1]):
 c=c
 else:
 c=c+1

nw.append(c)
print(nw)
from collections import Counter
Counter(nw)

jupyter-notebook

share|improve this question

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What is your end goal?
  
  – Nathan Wright
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I want to run this for various n and understand how the frequencies of the 1,2, etc in nw depend on n
  
  – SundarNarasimhan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This code is written to solve a problem in combinatorics...
  
  – SundarNarasimhan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "The phenomenon of just a few inputs producing a large number of outcomes is called a combinatorial explosion and is something to keep in mind when working with combinations(), combinations_with_replacement(), and permutations()" source
  
  – Antonino
  14 hours ago
  

  
  
  
  

add a comment | 

1

I have this following code:

The code is simply stuck when I give input n=6 ,(in jupyter)I don't know what to do please help.

n=int(input("Enter a number to experiment:"))
l=[]
w=[]
nw=[]
c=0
for i in range(2*n):
 if(i<n):
 l.append(0)
 else:
 l.append(1)

import itertools as it
permset=set([i for i in it.permutations(l)])
for x in permset:
 w.append(x)
print (w)
for j in range(len(w)):
 c=1

 for y in range(2*n-1):

 if(w[j][y]==w[j][y+1]):
 c=c
 else:
 c=c+1

nw.append(c)
print(nw)
from collections import Counter
Counter(nw)

jupyter-notebook

share|improve this question

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What is your end goal?
  
  – Nathan Wright
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I want to run this for various n and understand how the frequencies of the 1,2, etc in nw depend on n
  
  – SundarNarasimhan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This code is written to solve a problem in combinatorics...
  
  – SundarNarasimhan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "The phenomenon of just a few inputs producing a large number of outcomes is called a combinatorial explosion and is something to keep in mind when working with combinations(), combinations_with_replacement(), and permutations()" source
  
  – Antonino
  14 hours ago
  

  
  
  
  

add a comment | 

I have this following code:

The code is simply stuck when I give input n=6 ,(in jupyter)I don't know what to do please help.

n=int(input("Enter a number to experiment:"))
l=[]
w=[]
nw=[]
c=0
for i in range(2*n):
 if(i<n):
 l.append(0)
 else:
 l.append(1)

import itertools as it
permset=set([i for i in it.permutations(l)])
for x in permset:
 w.append(x)
print (w)
for j in range(len(w)):
 c=1

 for y in range(2*n-1):

 if(w[j][y]==w[j][y+1]):
 c=c
 else:
 c=c+1

nw.append(c)
print(nw)
from collections import Counter
Counter(nw)

jupyter-notebook

share|improve this question

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

n=int(input("Enter a number to experiment:"))
l=[]
w=[]
nw=[]
c=0
for i in range(2*n):
 if(i<n):
 l.append(0)
 else:
 l.append(1)

import itertools as it
permset=set([i for i in it.permutations(l)])
for x in permset:
 w.append(x)
print (w)
for j in range(len(w)):
 c=1

 for y in range(2*n-1):

 if(w[j][y]==w[j][y+1]):
 c=c
 else:
 c=c+1

nw.append(c)
print(nw)
from collections import Counter
Counter(nw)

share|improve this question

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 14 hours ago

Bogdan Doicin

6041025

edited 14 hours ago

Bogdan Doicin

6041025

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

New contributor

SundarNarasimhan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 15 hours ago

SundarNarasimhanSundarNarasimhan

83

1 Answer
1

active

oldest

votes

0

The problem is the call to itertools.permutations. It is computing every possible ordering of the array built previously, which for N=6 is:

[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

I believe this will return (12!)=479001600 items, which your code is then turning into a set to remove duplicates. It never gets that far of course, but if it did, I believe the 479001600 items would be turned into a set of just 924 items.

Not sure what your goal is but maybe what you need for this part is a way to generate that set of 924 items without building the larger list, which has many many duplicates. As I understand it, that part of the code is building the set of arrays of size (2*n) where exactly half of the elements are 0 and half are 1.

share|improve this answer

answered 14 hours ago

BampferBampfer

1,200517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I see... Thank you I will try to find easier way out
  
  – SundarNarasimhan
  13 hours ago
  
  

  
  
  
  

add a comment | 

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0

The problem is the call to itertools.permutations. It is computing every possible ordering of the array built previously, which for N=6 is:

[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

I believe this will return (12!)=479001600 items, which your code is then turning into a set to remove duplicates. It never gets that far of course, but if it did, I believe the 479001600 items would be turned into a set of just 924 items.

Not sure what your goal is but maybe what you need for this part is a way to generate that set of 924 items without building the larger list, which has many many duplicates. As I understand it, that part of the code is building the set of arrays of size (2*n) where exactly half of the elements are 0 and half are 1.

share|improve this answer

answered 14 hours ago

BampferBampfer

1,200517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I see... Thank you I will try to find easier way out
  
  – SundarNarasimhan
  13 hours ago
  
  

  
  
  
  

add a comment | 

0

The problem is the call to itertools.permutations. It is computing every possible ordering of the array built previously, which for N=6 is:

[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

I believe this will return (12!)=479001600 items, which your code is then turning into a set to remove duplicates. It never gets that far of course, but if it did, I believe the 479001600 items would be turned into a set of just 924 items.

Not sure what your goal is but maybe what you need for this part is a way to generate that set of 924 items without building the larger list, which has many many duplicates. As I understand it, that part of the code is building the set of arrays of size (2*n) where exactly half of the elements are 0 and half are 1.

share|improve this answer

answered 14 hours ago

BampferBampfer

1,200517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes I see... Thank you I will try to find easier way out
  
  – SundarNarasimhan
  13 hours ago
  
  

  
  
  
  

add a comment | 

The problem is the call to itertools.permutations. It is computing every possible ordering of the array built previously, which for N=6 is:

[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

I believe this will return (12!)=479001600 items, which your code is then turning into a set to remove duplicates. It never gets that far of course, but if it did, I believe the 479001600 items would be turned into a set of just 924 items.

Not sure what your goal is but maybe what you need for this part is a way to generate that set of 924 items without building the larger list, which has many many duplicates. As I understand it, that part of the code is building the set of arrays of size (2*n) where exactly half of the elements are 0 and half are 1.

share|improve this answer

answered 14 hours ago

BampferBampfer

1,200517

[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

share|improve this answer

answered 14 hours ago

BampferBampfer

1,200517

answered 14 hours ago

BampferBampfer

1,200517

answered 14 hours ago

BampferBampfer

1,200517