  C RUBY-ON-RAILS MYSQL ASP.NET DEVELOPMENT RUBY .NET LINUX SQL-SERVER REGEX WINDOWS ALGORITHM ECLIPSE VISUAL-STUDIO STRING SVN PERFORMANCE APACHE-FLEX UNIT-TESTING SECURITY LINQ UNIX MATH EMAIL OOP LANGUAGE-AGNOSTIC VB6 MSBUILD # Generate all permutations of fixed length where the elements come from two different sets ## Generate all permutations of fixed length where the elements come from two different sets Tag : python , By : Paul Date : November 26 2020, 03:01 PM

hope this fix your issue If I interpreted your text correctly, the following code should be what you are looking for:
``````import itertools
set1 = [1,2,3]
set2 = [4,5]

for i in itertools.permutations(set1, 2):
for j in itertools.permutations(set2, 2):
print("({},{},{},{})".format(i, j, j, i))
``````
``````(1,4,5,2)
(1,5,4,2)
(1,4,5,3)
(1,5,4,3)
(2,4,5,1)
(2,5,4,1)
(2,4,5,3)
(2,5,4,3)
(3,4,5,1)
(3,5,4,1)
(3,4,5,2)
(3,5,4,2)
``````
``````p1 = itertools.permutations(set1, 2)
p2 = itertools.permutations(set2, 2)

for i in itertools.product(p1, p2):
print(i, i, i, i)
`````` Boards Message : You Must Login Or Sign Up to Add Your Comments .

## Algorithm to generate all unique permutations of fixed-length integer partitions?

Tag : algorithm , By : ZsA
Date : March 29 2020, 07:55 AM
I hope this helps . Okay. First, forget about the permutations and just generate the partitions of length L (as suggested by @Svein Bringsli). Note that for each partition, you may impose an ordering on the elements, such as >. Now just "count," maintaining your ordering. For n = 4, k = 3:
``````(4, 0, 0)
(3, 1, 0)
(2, 2, 0)
(2, 1, 1)
``````
``````def partition_helper(l, i, result):
if i == len(l) - 1:
return
while l[i] - 1 >= l[i + 1] + 1:
l[i]        -= 1
l[i + 1]    += 1
result.append(list(l))
partition_helper(l, i + 1, result)

def partition(n, k):
l = [n] +  * (k - 1)
result = [list(l)]
partition_helper(l, 0, result)
return result
``````

## All binary permutations of a fixed length in C

Tag : c , By : changke
Date : March 29 2020, 07:55 AM
it should still fix some issue The answer goes like this.
In C, the natural way to deal with an array of [0,1] is by treating them as bits. The 2^24 permutations of 24 bits are precisely the values of unsigned int from 0 to 2^24-1. So the question essentially is how to write the code based on that data structure.
``````int all_elements = { ??? };
int all_marked_elements = { ??? };
unsigned combo;
for (combo = 0; combo < 0x1000000; ++combo) {
/* you probably want to take a copy of all_elements here */
for (i = 0; i < 23; ++i) {
unsigned bit = 1 << i;
if (combo & bit) {
int marked_element = all_marked_elements[i];
/* do something I didn't understand, replacing element by its dual */
}
/* now call the operation and do something with the result */
}
}
``````

## Generate all permutations of cyclic shift of length K in a list of length N efficiently

Tag : python , By : OllieDoodle
Date : March 29 2020, 07:55 AM
``````if P in stored_perms:
``````
``````def get_cyclics(p, k):
found = set()      # set of tuples we have seen so far
todo = [tuple(p)]  # list of tuples we still need to explore
n = len(p)
while todo:
x = todo.pop()
for i in range(n - k + 1):
perm = ( x[:i]                    # Prefix
+ x[i+1:i+k] + x[i:i+1]    # Rotated middle
+ x[i+k:]                  # Suffix
)
if perm not in found:
todo.append(perm)
for x in found:
print(x)
``````

## Finding all possible permutations of a fixed length of numbers to reach a given sum

Tag : python , By : socurious
Date : March 29 2020, 07:55 AM
I hope this helps you . Since you've solved the problem of identifying one solution in each equivalence group, my advice is: do not alter that algorithm. Instead, harness itertools.permutations to generate those items:
``````return list(itertools.permutations(numbers))
``````

## fixed-length permutations of a string

Tag : java , By : terrestrial
Date : March 29 2020, 07:55 AM 