How to Use Recursion to Generate Arrays in Python

I am trying to generate all possible arrays with values from 1 - 9 using recursion in python. My code is below:

totalArr = 



def recursion(arr, n):

    for i in range(9):

        if (arr[i] == 0):

            arr[i] = n

            if (n < 8):

                recursion(arr, n + 1)

            else:

                print(arr)

                totalArr.append(arr)



recursion([0, 0, 0, 0, 0, 0, 0, 0, 0], 0)



print(len(totalArr))

When I run this code, all I get is the single array below:

[1, 2, 3, 4, 5, 6, 7, 8, 9]

I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run.

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

2

arr[i] == n does not assign n to arr[i]. You need =, not equality operator.

– Austin
Nov 25 '18 at 3:38

2

What output are you expecting?

– 0liveradam8
Nov 25 '18 at 3:50

Welcome to SO! "I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run." Permutation != recursion. The definition of "all possible arrays with values 1-9" is all permutations of an input array with those values, and you can do this recursively, which is what you appear to be attempting. What is the problem with your code (infinite loop due to == typo aside)?

– ggorlen
Nov 25 '18 at 3:57

"however for my use case of these arrays, I believe that recursion is better in the long run"... why do you believe that?

– Mateen Ulhaq
Nov 25 '18 at 4:43

Yeah, sorry about that typo. The problem with my code is that all the arrays it creates are the same [1, 2, 3, 4, 5, 6, 7, 8, 9]

– Luke Schultz
Nov 25 '18 at 4:43

|
show 3 more comments

I am trying to generate all possible arrays with values from 1 - 9 using recursion in python. My code is below:

totalArr = 



def recursion(arr, n):

    for i in range(9):

        if (arr[i] == 0):

            arr[i] = n

            if (n < 8):

                recursion(arr, n + 1)

            else:

                print(arr)

                totalArr.append(arr)



recursion([0, 0, 0, 0, 0, 0, 0, 0, 0], 0)



print(len(totalArr))

When I run this code, all I get is the single array below:

[1, 2, 3, 4, 5, 6, 7, 8, 9]

I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run.

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

2

arr[i] == n does not assign n to arr[i]. You need =, not equality operator.

– Austin
Nov 25 '18 at 3:38

2

What output are you expecting?

– 0liveradam8
Nov 25 '18 at 3:50

Welcome to SO! "I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run." Permutation != recursion. The definition of "all possible arrays with values 1-9" is all permutations of an input array with those values, and you can do this recursively, which is what you appear to be attempting. What is the problem with your code (infinite loop due to == typo aside)?

– ggorlen
Nov 25 '18 at 3:57

"however for my use case of these arrays, I believe that recursion is better in the long run"... why do you believe that?

– Mateen Ulhaq
Nov 25 '18 at 4:43

Yeah, sorry about that typo. The problem with my code is that all the arrays it creates are the same [1, 2, 3, 4, 5, 6, 7, 8, 9]

– Luke Schultz
Nov 25 '18 at 4:43

|
show 3 more comments

I am trying to generate all possible arrays with values from 1 - 9 using recursion in python. My code is below:

totalArr = 



def recursion(arr, n):

    for i in range(9):

        if (arr[i] == 0):

            arr[i] = n

            if (n < 8):

                recursion(arr, n + 1)

            else:

                print(arr)

                totalArr.append(arr)



recursion([0, 0, 0, 0, 0, 0, 0, 0, 0], 0)



print(len(totalArr))

When I run this code, all I get is the single array below:

[1, 2, 3, 4, 5, 6, 7, 8, 9]

I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run.

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

I am trying to generate all possible arrays with values from 1 - 9 using recursion in python. My code is below:

totalArr = 



def recursion(arr, n):

    for i in range(9):

        if (arr[i] == 0):

            arr[i] = n

            if (n < 8):

                recursion(arr, n + 1)

            else:

                print(arr)

                totalArr.append(arr)



recursion([0, 0, 0, 0, 0, 0, 0, 0, 0], 0)



print(len(totalArr))

When I run this code, all I get is the single array below:

[1, 2, 3, 4, 5, 6, 7, 8, 9]

I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run.

python arrays python-3.x list recursion

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

edited Nov 25 '18 at 4:40

asked Nov 25 '18 at 3:36

Luke Schultz

464

asked Nov 25 '18 at 3:36

Luke Schultz

464

asked Nov 25 '18 at 3:36

Luke Schultz

464

2

arr[i] == n does not assign n to arr[i]. You need =, not equality operator.

– Austin
Nov 25 '18 at 3:38

2

What output are you expecting?

– 0liveradam8
Nov 25 '18 at 3:50

Welcome to SO! "I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run." Permutation != recursion. The definition of "all possible arrays with values 1-9" is all permutations of an input array with those values, and you can do this recursively, which is what you appear to be attempting. What is the problem with your code (infinite loop due to == typo aside)?

– ggorlen
Nov 25 '18 at 3:57

"however for my use case of these arrays, I believe that recursion is better in the long run"... why do you believe that?

– Mateen Ulhaq
Nov 25 '18 at 4:43

Yeah, sorry about that typo. The problem with my code is that all the arrays it creates are the same [1, 2, 3, 4, 5, 6, 7, 8, 9]

– Luke Schultz
Nov 25 '18 at 4:43

|
show 3 more comments

2

arr[i] == n does not assign n to arr[i]. You need =, not equality operator.

– Austin
Nov 25 '18 at 3:38

2

What output are you expecting?

– 0liveradam8
Nov 25 '18 at 3:50

Welcome to SO! "I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run." Permutation != recursion. The definition of "all possible arrays with values 1-9" is all permutations of an input array with those values, and you can do this recursively, which is what you appear to be attempting. What is the problem with your code (infinite loop due to == typo aside)?

– ggorlen
Nov 25 '18 at 3:57

"however for my use case of these arrays, I believe that recursion is better in the long run"... why do you believe that?

– Mateen Ulhaq
Nov 25 '18 at 4:43

Yeah, sorry about that typo. The problem with my code is that all the arrays it creates are the same [1, 2, 3, 4, 5, 6, 7, 8, 9]

– Luke Schultz
Nov 25 '18 at 4:43

arr[i] == n does not assign n to arr[i]. You need =, not equality operator.

– Austin
Nov 25 '18 at 3:38

What output are you expecting?

– 0liveradam8
Nov 25 '18 at 3:50

Welcome to SO! "I am aware that I can use permutations to the arrays, however for my use case of these arrays, I believe that recursion is better in the long run." Permutation != recursion. The definition of "all possible arrays with values 1-9" is all permutations of an input array with those values, and you can do this recursively, which is what you appear to be attempting. What is the problem with your code (infinite loop due to == typo aside)?

– ggorlen
Nov 25 '18 at 3:57

"however for my use case of these arrays, I believe that recursion is better in the long run"... why do you believe that?

– Mateen Ulhaq
Nov 25 '18 at 4:43

Yeah, sorry about that typo. The problem with my code is that all the arrays it creates are the same [1, 2, 3, 4, 5, 6, 7, 8, 9]

– Luke Schultz
Nov 25 '18 at 4:43

|
show 3 more comments

2 Answers
2

active

oldest

votes

# Python program to print all permutations with

# duplicates allowed



# Function to print permutations of my_array

# This function takes three parameters:

# 1. my_array

# 2. Starting index of the my_array

# 3. Ending index of the my_array.

def permute(a, l, r):

    if l==r:

        print a

    else:

        for i in xrange(l,r+1):

            a[l], a[i] = a[i], a[l]

            permute(a, l+1, r)

            a[l], a[i] = a[i], a[l] # backtrack



# Driver program to test the above function

my_array = [0,1,2,3,4,5,6,7,8,9]

n = len(my_array)

a = list(my_array)

permute(a, 0, n-1)

We are doing permutation by using recursion with backtracking :
More info on https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

answered Nov 25 '18 at 5:29

NiksVij

676

add a comment |

totalArr=

def permute(lst,n):

    ''' O(n!), optimal'''

    if n==1:totalArr.append(lst.copy())

    else:

        for i in range(n):

            lst[i],lst[n-1] = lst[n-1],lst[i]

            permute(lst,n-1)

            lst[i],lst[n-1] = lst[n-1],lst[i]

lst = [i for i in range(1,9)] # any other lst with unique elements is ok

permute(lst,len(lst))

print(totalArr)

This method can generate all permutations using divide-and-conquer algorithm

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

# Python program to print all permutations with

# duplicates allowed



# Function to print permutations of my_array

# This function takes three parameters:

# 1. my_array

# 2. Starting index of the my_array

# 3. Ending index of the my_array.

def permute(a, l, r):

    if l==r:

        print a

    else:

        for i in xrange(l,r+1):

            a[l], a[i] = a[i], a[l]

            permute(a, l+1, r)

            a[l], a[i] = a[i], a[l] # backtrack



# Driver program to test the above function

my_array = [0,1,2,3,4,5,6,7,8,9]

n = len(my_array)

a = list(my_array)

permute(a, 0, n-1)

We are doing permutation by using recursion with backtracking :
More info on https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

answered Nov 25 '18 at 5:29

NiksVij

676

add a comment |

# Python program to print all permutations with

# duplicates allowed



# Function to print permutations of my_array

# This function takes three parameters:

# 1. my_array

# 2. Starting index of the my_array

# 3. Ending index of the my_array.

def permute(a, l, r):

    if l==r:

        print a

    else:

        for i in xrange(l,r+1):

            a[l], a[i] = a[i], a[l]

            permute(a, l+1, r)

            a[l], a[i] = a[i], a[l] # backtrack



# Driver program to test the above function

my_array = [0,1,2,3,4,5,6,7,8,9]

n = len(my_array)

a = list(my_array)

permute(a, 0, n-1)

We are doing permutation by using recursion with backtracking :
More info on https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

answered Nov 25 '18 at 5:29

NiksVij

676

add a comment |

# Python program to print all permutations with

# duplicates allowed



# Function to print permutations of my_array

# This function takes three parameters:

# 1. my_array

# 2. Starting index of the my_array

# 3. Ending index of the my_array.

def permute(a, l, r):

    if l==r:

        print a

    else:

        for i in xrange(l,r+1):

            a[l], a[i] = a[i], a[l]

            permute(a, l+1, r)

            a[l], a[i] = a[i], a[l] # backtrack



# Driver program to test the above function

my_array = [0,1,2,3,4,5,6,7,8,9]

n = len(my_array)

a = list(my_array)

permute(a, 0, n-1)

We are doing permutation by using recursion with backtracking :
More info on https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

answered Nov 25 '18 at 5:29

NiksVij

676

# Python program to print all permutations with

# duplicates allowed



# Function to print permutations of my_array

# This function takes three parameters:

# 1. my_array

# 2. Starting index of the my_array

# 3. Ending index of the my_array.

def permute(a, l, r):

    if l==r:

        print a

    else:

        for i in xrange(l,r+1):

            a[l], a[i] = a[i], a[l]

            permute(a, l+1, r)

            a[l], a[i] = a[i], a[l] # backtrack



# Driver program to test the above function

my_array = [0,1,2,3,4,5,6,7,8,9]

n = len(my_array)

a = list(my_array)

permute(a, 0, n-1)

We are doing permutation by using recursion with backtracking :
More info on https://www.geeksforgeeks.org/write-a-c-program-to-print-all-permutations-of-a-given-string/

answered Nov 25 '18 at 5:29

NiksVij

676

answered Nov 25 '18 at 5:29

NiksVij

676

answered Nov 25 '18 at 5:29

NiksVij

676

answered Nov 25 '18 at 5:29

NiksVij

676

add a comment |

totalArr=

def permute(lst,n):

    ''' O(n!), optimal'''

    if n==1:totalArr.append(lst.copy())

    else:

        for i in range(n):

            lst[i],lst[n-1] = lst[n-1],lst[i]

            permute(lst,n-1)

            lst[i],lst[n-1] = lst[n-1],lst[i]

lst = [i for i in range(1,9)] # any other lst with unique elements is ok

permute(lst,len(lst))

print(totalArr)

This method can generate all permutations using divide-and-conquer algorithm

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

add a comment |

totalArr=

def permute(lst,n):

    ''' O(n!), optimal'''

    if n==1:totalArr.append(lst.copy())

    else:

        for i in range(n):

            lst[i],lst[n-1] = lst[n-1],lst[i]

            permute(lst,n-1)

            lst[i],lst[n-1] = lst[n-1],lst[i]

lst = [i for i in range(1,9)] # any other lst with unique elements is ok

permute(lst,len(lst))

print(totalArr)

This method can generate all permutations using divide-and-conquer algorithm

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

add a comment |

totalArr=

def permute(lst,n):

    ''' O(n!), optimal'''

    if n==1:totalArr.append(lst.copy())

    else:

        for i in range(n):

            lst[i],lst[n-1] = lst[n-1],lst[i]

            permute(lst,n-1)

            lst[i],lst[n-1] = lst[n-1],lst[i]

lst = [i for i in range(1,9)] # any other lst with unique elements is ok

permute(lst,len(lst))

print(totalArr)

This method can generate all permutations using divide-and-conquer algorithm

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

totalArr=

def permute(lst,n):

    ''' O(n!), optimal'''

    if n==1:totalArr.append(lst.copy())

    else:

        for i in range(n):

            lst[i],lst[n-1] = lst[n-1],lst[i]

            permute(lst,n-1)

            lst[i],lst[n-1] = lst[n-1],lst[i]

lst = [i for i in range(1,9)] # any other lst with unique elements is ok

permute(lst,len(lst))

print(totalArr)

This method can generate all permutations using divide-and-conquer algorithm

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

edited Nov 25 '18 at 12:31

answered Nov 25 '18 at 4:25

mbinary

answered Nov 25 '18 at 4:25

mbinary

answered Nov 25 '18 at 4:25

mbinary

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

add a comment |

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

When I run your code, all it generates is arrays filled with zeros [0, 0, 0, 0, 0, 0, 0, 0, 0]

– Luke Schultz
Nov 25 '18 at 4:42

@LukeSchultz, there's a line missing in this code where it actually calls permute(). My guess is something like permute([1, 2, 3, 4, 5, 6, 7, 8, 9], 9) before the print() statement.

– cdlane
Nov 25 '18 at 6:37

Yes. You should firstly apply function permute in a lst with its length, and then get the result in totalArr

– mbinary
Nov 25 '18 at 12:29

add a comment |

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