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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

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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[0], j[0], j[1], i[1]))
(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[0][0], i[1][0], i[1][1], i[0][1])

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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] + [0] * (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[160] = { ??? };
int all_marked_elements[24] = { ??? };
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
This might help you , The check
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:
        found.add(perm)
        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
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