Cycles and fixed points

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In combinatorial mathematics, a cycle of length n of a permutation P over a set S is a subset { c₁, ..., c_n } of S on which the permutation P acts in the following way:

P(c_i) = c_{i + 1} for i = 1, ..., n − 1, and P(c_n) = c₁.

It is usual to write a cycle by its mapping:

( c₁ c₂ ... c_n ).

Every permutation may be written as a composition of disjoint cycles.

For example, let

be a permutation that maps 1 to 2 etc. Then P may be written as

P = ( 1 2 4 3 ) ( 5 ) ( 6 7 )

= ( 1 2 4 3 ) ( 6 7 ) ( 5 )

= ( 4 3 1 2 ) ( 7 6 ) ( 5 )

Note:

• There are different ways to write a permutation as a product of disjoint cycles, but the number of cycles and their contents is always unique.
• In this example, 5 is a fixed point of the permutation, i.e. 5 is mapped to itself. Every fixed point is a cycle with length 1.

Contents 1 Cycles 1.1 Properties 1.2 Reasons for properties 1.3 Some values 2 Fixed points 2.1 Properties 2.2 Reasons for properties 2.3 Some values 2.4 Alternate calculations 3 See also

[edit] Cycles

The unsigned Stirling number of the first kind, s(k, j) counts the number of permutations of k elements with exactly j disjoint cycles.

[edit] Properties

(1) For every k > 0 : s(k, k) = 1.

(2) For every k > 0 : s(k, 1) = (k − 1)!.

(3) For every k > j > 1, s(k, j) = s(k − 1,j − 1) + s(k − 1, j)·(k − 1)

[edit] Reasons for properties

(1) There is only one way to construct a permutation of k elements with k cycles: Every cycle must have length 1 so every element must be a fixed point.

(2.a) Every cycle of length k may be written as permutation of the number 1 to k; there are k! of these permutations.

(2.b) There are k different ways to write a given cycle of length k, e.g. ( 1 2 4 3 ) = ( 2 4 3 1 ) = ( 4 3 1 2 ) = ( 3 1 2 4 ).

(2.c) Finally: s(k, 1) = k!/k = (k − 1)!.

(3) There are two different ways to construct a permutation of k elements with j cycles:

(3.a) If we want element k to be a fixed point we may choose one of the s(k − 1, j − 1) permutations with k − 1 elements and j − 1 cycles and add element k as a new cycle of length 1.

(3.b) If we want element k not to be a fixed point we may choose one of the s(k − 1, j ) permutations with k − 1 elements and j cycles and insert element k in an existing cycle in front of one of the k − 1 elements.

[edit] Some values

k	j
	1	2	3	4	5	6	7	8	9	sum
1	1									1
2	1	1								2
3	2	3	1							6
4	6	11	6	1						24
5	24	50	35	10	1					120
6	120	274	225	85	15	1				720
7	720	1,764	1,624	735	175	21	1			5,040
8	5,040	13,068	13,132	6,769	1,960	322	28	1		40,320
9	40,320	109,584	118,124	67,284	22,449	4,536	546	36	1	362,880
	1	2	3	4	5	6	7	8	9	sum

[edit] Fixed points

The value f(k, j) counts the number of permutations of k elements with exactly j fixed points. For the main article on this topic, see rencontres numbers.

[edit] Properties

(1) For every j < 0 or j > k : f(k, j) = 0.

(2) f(0, 0) = 1.

(3) For every k > 1 and k ≥ j ≥ 0, f(k, j) = f(k − 1, j − 1) + f(k − 1, j)·(k − 1  − j) + f(k − 1, j + 1)·(j + 1)

[edit] Reasons for properties

(3) There are three different methods to construct a permutation of k elements with j fixed points:

(3.a) We may choose one of the f(k − 1, j − 1) permutations with k − 1 elements and j − 1 fixed points and add element k as a new fixed point.

(3.b) We may choose one of the f(k − 1, j) permutations with k − 1 elements and j fixed points and insert element k in an existing cycle of length > 1 in front of one of the (k − 1) − j elements.

(3.c) We may choose one of the f(k − 1, j + 1) permutations with k − 1 elements and j + 1 fixed points and join element k with one of the j + 1 fixed points to a cycle of length 2.

[edit] Some values

k	j
	0	1	2	3	4	5	6	7	8	9	sum
1	0	1									1
2	1	0	1								2
3	2	3	0	1							6
4	9	8	6	0	1						24
5	44	45	20	10	0	1					120
6	265	264	135	40	15	0	1				720
7	1,854	1,855	924	315	70	21	0	1			5,040
8	14,833	14,832	7,420	2,464	630	112	28	0	1		40,320
9	133,496	133,497	66,744	22,260	5,544	1,134	168	36	0	1	362,880
	0	1	2	3	4	5	6	7	8	9	sum

[edit] Alternate calculations

Example: f(5, 1) = 5×1×4! − 10×2×3! + 10×3×2! - 5×4×1! + 1×5×0!

= 120 - 120 + 60 - 20 + 5 = 45.

Example: f(5, 0) = 120 - ( 5×4! - 10×3! + 10×2! - 5×1! + 1×0! )

= 120 - ( 120 - 60 + 20 - 5 + 1 ) = 120 - 76 = 44.

For every k > 1:

f(k,0) = (k − 1)(f(k − 1,0) + f(k − 2,0))

Example: f(5, 0) = 4 × ( 9 + 2 ) = 4 × 11 = 44

For every k > 1:

Example: f(5, 0) = 120 × ( 1/2 - 1/6 + 1/24 - 1/120 )

= 120 × ( 60/120 - 20/120 + 5/120 - 1/120 ) = 120 × 44/120 = 44

where e is Euler's number ≈ 2.71828

[edit] See also

• Cyclic permutation
• Cycle notation