Massive Algorithms: Count Derangements (Permutation such that no element appears in its original position)

Count Derangements (Permutation such that no element appears in its original position) - GeeksforGeeks
A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of {0, 1, 2, 3} is {2, 3, 1, 0}.
Given a number n, find total number of Derangements of a set of n elements.

Input: n = 2
Output: 1
For two elements say {0, 1}, there is only one 
possible derangement {1, 0}

Input: n = 3
Output: 2
For three elements say {0, 1, 2}, there are two 
possible derangements {2, 0, 1} and {1, 2, 0}

Let countDer(n) be count of derangements for n elements. Below is recursive relation for it.

countDer(n) = (n-1)*[countDer(n-1) + countDer(n-2)]

How does above recursive relation work?
There are n – 1 ways for element 0 (this explains multiplication with n-1).

Let 0 be placed at index i. There are now two possibilities, depending on whether or not element i is placed at 0 in return.

i is placed at 0: This case is equivalent to solving the problem for n-2 elements as two elements have just swapped their positions.
i is not placed at 0: This case is equivalent to solving the problem for n-1 elements as now there are n-1 elements, n-1 positions and every element has n-2 choices

int countDer(int n)

{

    // Create an array to store counts for subproblems

    int der[n + 1];

    // Base cases

    der[0] = 1;

    der[1] = 0;

    der[2] = 1;

    // Fill der[0..n] in bottom up manner using above

    // recursive formula

    for (int i=3; i<=n; ++i)

        der[i] = (i-1)*(der[i-1] + der[i-2]);

    // Return result for n

    return der[n];

}

int countDer(int n)

{

   // Base cases

   if (n == 1)    return 0;

   if (n == 0)    return 1;

   if (n == 2)    return 1;

   // countDer(n) = (n-1)[countDer(n-1) + der(n-2)]

   return (n-1)*(countDer(n-1) + countDer(n-2));

}

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.

Extended:
http://rosettacode.org/wiki/Permutations/Derangements#Java

http://math.illinoisstate.edu/day/courses/old/305/contentderangements.html

Suppose we want to determine the number of derangements of the n integers 1,2,...,n for n bigger than 2. Let us focus on k and move it into the first position. We thus have started a derangement, for 1 is not in its natural position. Where could 1 be placed? There are two cases we could consider: either 1 is in position k or 1 is not in position k.

If 1 is in position k, here's what we know: The integers 1 and k have simply traded positions.

Prohibited Value	1	2	3	...	k-1	k	k+1	...	n
Derangement	k	?	?	...	?	1	?	...	?

Indicated by the question marks, there are (n-2) integers yet to derange. This can be done in D(n-2) ways.

If 1 is not in position k, we don't know as much. Note that we have shown 1 as a prohibited value twice. This is required for this case, because we cannot have 1 appear in the first position (its natural position) nor can 1 appear in position k.

Prohibited Value	1	2	3	...	k-1	1	k+1	...	n
Derangement	k	?	?	...	?	?	?	...	?

Indicated by the question marks, there are now (n-1) integers yet to derange. This can be done in D(n-1) ways.

Putting this together, we have D(n-1) + D(n-2) possible derangements when k is in the first position. How many different integers could we put in position 1 and carry out this process? All the integers except 1. that is, (n-1) different integers.

This yields the recursive formula D(n) = (n-1)[D(n-1) + D(n-2)]. As long as we know D(1) = 0 and D(2) = 1, we can generate subsequent values for D(n).

Derangements Number

intderangements_number( int n )

A derangement of n objects is a permutation with no fixed points. If we symbolize the permutation by $\sigma$ , then for a derangment, $\sigma(i)$ is never equal to