# Linear Search based algorithm for Mth Lexicographic ordering of Mathematical Permutation and Combination

Following the discussion in "Generating
the mth Lexicographical Element of a Mathematical Combination"
, I
present here another algorithm for the Mth lexicographic ordering based on a
linear search procedure. The method is based on the fact that it is simple to
verify if Zero is the first element representing the Mth lexicographic ordering
of a permutation or combination. I’ll consider combinations first, followed by
permutation.

## Lexicographic Ordering of a Combination

Consider a combination of k elements from a set of n
elements.  Let

S(k,n) represent the set of all such combinations,

C(k,n) the number of elements in the set S(k,n)

Without loss of generality we shall deal with the universe
of Natural numbers including zero. Since two combinations are same if they
contain the same set of elements we reduce a combination to a canonical form
where it is represented as a sequence in which i-th element is greater than the
(i+1)-th element. In other words let the k-tuple < s1, s2,…,
sk >represent a combination. Then we can say that s1 >
s2  > … > sk.   Also the set
of all combinations are lexicographically
ordered
. In other words if P1 and P2 are two combinations from the same set
then

P1 ≠ P2 implies (P1> P2 or P2 >P1)

Thus the M-th lexicographical element has M-1 elements in
S(k,m) that are less than it and the remaining greater than it.

The Algorithm.

Note that ‘0’ will be the first element of the first
C(k-1,n-1) elements, because having chosen ‘0’ we are left with the problem of
choosing k-1 elements from n-1 elements. If ‘0’ is not chosen we have to choose
k elements from n-1 elements yielding the well known result

C(k,n) = C(k-1,n-1) + C(k,n-1)

Thus if M<=C(k-1,n-1) then we have effectively computed
the first element of the M-th sequence. If M>C(k-1,n-1) then we have reduced
the problem to finding the (M-C(k-1,n-1))-th element of the ordered set
S(k,n-1).

Thus the function to compute the Mth combination can be
described as

1.      MthCombination(M,k,n)

2.      If
(k=0)

3.      Return
<>

4.      If
(k=n)

5.      return
<0,1,…,k-1>

6.      if
(M < C(k-1,n-1))

7.      {

8.      Let
< s1, s2,…, sk-1 > =
MthCombination(M,k-1,n-1)

9.      return
< 0,s1 +1, s2 +1,…,sk-1 +1>

10.  }

11.  B else

12.  {

13.
Let  < s1, s2,…,
sk > = MthCombination(M-C(k-1,n-1),   k,  n-1)

14.  B B B B B B return <s1 +1,
s2 +1,…,sk +1>

15.  B }

16.  }

The first step is to return an empty sequence if k =0.

## Computing the Rank from a given Sequence.

Now consider the inverse problem of computing M given a
combination. We shall call this the Rank of the combination.  If the first
element is ‘0’then M<C(k-1,n-1) and hence the problem the is reduced to
finding the Rank of the sequence with the first element removed. If the first
element is not zero then the Rank is: C(n-1,k-1) + the rank of the combination
in the set S(k-1,n) with all elements in the sequence reduced by one.
More specifically the algorithm can be described as:

Rank(< s1 ,s2 ,…,sk >,
k, n)

1.      If
(k=0)

2.      Return
1

3.      If s1
=0 then

4.      Return
Rank(< s2-1, s3 -1, …,s-1> ,
k-1, n-1)

1.      Else

2.      Return
C(k-1,n-1) + Rank(< s1 -1,  s2  -1, …, sk
-1>, k,n-1)

The first step checks for an empty sequence in which case
the Rank is 1.

## Lexicographic Ordering of Permutations

A permutation P of N numbers can be denoted as an N-tuple (a
sequence of N items)  consisting of unique elements from the set P =
{0,1,…,N-1}. In this case the sequence itself is not ordered and
every element in the set P appears once and only once in the sequence. The set
of all permutations can also be linearly ordered and the M-th element defined
in like manner as for the combination

The Algorithm.

Note that there are at most factorial N (denoted as N!)
permuations of the set {0,1,…,N-1}; and ‘0’ is the first element in the first
(N-1)!  Permutations; ‘1’ is the first element in the next (N-1)!
elements and so on. Hence it is simple to identify the first element. Let the
first element be ‘j’. The problem is now reduced to finding the  (M –
(N-1)! * j)-th element of the set of permutations of (N-1) numbers.
Combining the two results consists of appropriately changing the second
sequence so that the number j is not repeated. This is easily done by
incrementing all si >= j by 1, where si is the i-th
element of the resulting sequence.

1.       PermutationM(
M, N)

1.      B {

2.      B B if
(N=0)

3.      B B B B return
<>

4.      else

5.      {

6.      B B B B m
=( N-1)!

7.      B B B B m0
= m

8.      B B B while
(M > m)

9.      B B {

10.  B B B B m = m+m0

11.  B B }

12.  B B Let  < s1,
s2,…, sN-1 > = PermutationM (M- (m-m0), n-1)

13.  B B Return <k, Ik(s1
), Ik(s2),…, Ik(sN-1 )>

14.  }

Here

Ik (i) = i if (i<k)

= i+1 if
(i=>k)

## Rank of a Permutation

Consider the inverse problem of computing M the rank of a
given permutation of N numbers.  The first element determines into which
block of (N-1)!  elements, the given permutation falls. The problem then
can be reduced to finding the rank of suffix of the permutation until we have
an empty sequence in which case the rank is 1. The algorithm is outlined in
more detail as follows:

1.       Rank(<s1,
s2,…, sN>,N )

1.        if (0 == n)

2.      return
1

3.      else

4.      k=s1

5.      return
(k-1) *(N-1)! + Rank(<Dk(s1 ), Dk(s2),…,
Dk(sN-1 )>,N-1);

where

Dk(i) B B = i if
i<k

= i-1 if
i>=k