Program to find number of elements in all permutation which are following given conditions in Python

PythonServer Side ProgrammingProgramming

Suppose we have a set A where all elements from 1 to n are present. And P(A) represents all permutations of elements present in A. We have to find number of elements in P(A) which satisfies the given conditions

  • For all i in range [1, n], A[i] is not same as i
  • There exists a set of k indices {i1, i2, ... ik} such that A[ij] = ij+1 for all j < k and A[ik] = i1 (cyclic).

So, if the input is like n = 3 k = 2, then the output will be 0 because -

Consider the Array's are 1 indexed. As N = 3 and K = 2, we can find 2 sets of A that satisfy the first property a[i] ≠ i, they are [3,1,2] and [2,3,1]. Now as K = 2, we can have 6 such elements.

[1,2], [1,3],[2,3], [2,1], [3,1], [3,2]. Now if we consider the first element of

P(A) -> [3,1,2]

  • [1,2], A[1] ≠ 2
  • [1,3], A[1] = 3 but A[3] ≠ 1
  • [2,3], A[2] ≠ 3
  • [2,1], A[2] = 1 but A[1] ≠ 2
  • [3,1], A[3] = 1 but A[1] ≠ 3
  • [3,2], A[3] ≠ 2

P(A) -> [2,3,1]

  • [1,2], A[1] = 2 but A[2] ≠ 1
  • [1,3], A[1] ≠ 3
  • [2,3], A[2] = 3 but A[3] ≠ 3
  • [2,1], A[2] ≠ 1
  • [3,1], A[3] = but A[1] ≠ 3
  • [3,2], A[3] ≠ 2

As none of the elements of a satisfy the properties above, hence 0.

To solve this, we will follow these steps −

  • ps := all permutations of arrays with elements in range [1, n]
  • c := 0
  • for each p in ps, do
    • for each index i and value a in p, do
      • if a is same as i, then
        • come out from the loop
    • otherwise,
      • for j in range 0 to n - 1, do
        • current := p[j]
        • cycle_length := 1
        • while current is not same as j, do
          • current := p[current]
          • cycle_length := cycle_length + 1
        • if cycle_length is same as k, then
          • c := c + 1
          • come out from the loop
  • return c

Example

Let us see the following implementation to get better understanding −

import itertools

def solve(n, k):
   ps = itertools.permutations(range(n), n)
   c = 0
   for p in ps:
      for i, a in enumerate(p):
         if a == i:
            break
      else:
         for j in range(n):
            current = p[j]
            cycle_length = 1
            while current != j:
               current = p[current]
               cycle_length += 1
            if cycle_length == k:
               c += 1
               break
   return c

n = 3
k = 2
print(solve(n, k))

Input

3, 2

Output

0
raja
Updated on 25-Oct-2021 06:56:32

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