Least Period Of Linear Recurring Sequences Over The Finite Field

Degree Project

2012-02-29

Author: Sajid Hanif Subject: Mathematics Level: Master

Course code: 5MA12E

Least Period of Linear Recurring Sequences

over a Finite Field

Abstract

This thesis deals with fundamental concepts of linear recurring se- quences over the finite fields. The theory of linear recurrence sequences (LRS) over finite field has great importance in cryptography, electric en- gineering and pseudo-random number generators.

Linear recurring sequences and polynomials over finite field𝔽^𝑞 are closely related. The least period of recurring sequences are discussed with the re- ducibility of corresponding characteristic polynomials. Few examples are constructed to find the least period of linear recurring sequences having reducible or irreducible characteristic polynomials.

Contents

1 Introduction 4

2 Preliminaries 4

3 Linear recurring sequences 5

3.1 Feedback Shift Register . . . 5

3.2 Periodicity of linear recurring sequences . . . 6

3.3 Impulse response sequences . . . 8

3.4 Characteristic polynomial . . . 9

4 Least period for irreducible characteristic polynomial 10 5 Least period for reducible characteristic polynomial 14 5.1 Algorithm for finding minimal polynomial m(x) . . . 17

6 Mathematica code 21

7 Conclusion 22

8 Bibiliography 23

1 Introduction

The applications such as spread-spectrum communications, security and en- cryption need the generation of random numbers. The most common way to implement a random number generator is a linear recurring sequence(LRS). In this project we will simulate the theory about linear recurring sequences given in chapter 6 of [1] and will focus that ”How can we find the least period of linear recurring sequences?”

In section 2 we will define field and finite field.

In section 3 we will study the generation of linear recurring sequences on switch- ing circuits called feedback shift registers and also will discuss its periodic prop- erties [1, p. 190]. And we will discuss the term impulse response sequence, and its periodic relation with periodicity of linear recurring sequences[1, p. 197].

In section 4 we will study how to find the least period of linear recurring se- quences having irreducible characteristic polynomial [1, p. 204] and will imple- ment on some linear recurring sequences.

In section 5 we will study how to deal with a linear recurring sequences hav- ing reducible characteristic polynomial. Here we will study minimal polynomial and will use it for finding least period of recurring sequences having reducible characteristic polynomials [1, p. 214] and will establish some examples.

2 Preliminaries

In this section we are going to define field and finite field.

Definition 2.1. A field (𝔽, +, ⋅) is a set 𝔽, together with two binary operations, denoted by + and⋅ such that:

1. 𝔽 is an abelian group with respect to both (+) and (⋅).

2. The distributive laws hold. That is, for all a,b,c∈ 𝔽, we have, 𝑎.(𝑏 + 𝑐) = 𝑎.𝑏 + 𝑎.𝑐 and (𝑏 + 𝑐).𝑎 = 𝑏.𝑎 + 𝑐.𝑎 Example 2.1. The set of all real numbersℝ is a field.

Definition 2.2. A field𝔽 is said to be a finite field if number of elements of field𝔽 are finite. A finite field can also be defined as:

For a prime p, let𝔽^𝑝 be the set {1, 2, ..., 𝑝 − 1} of integers and let

𝜙 :ℤ/(𝑝) → 𝔽^𝑝be the mapping defined by 𝜙([𝑎]) = 𝑎 for 𝑎 = 0, 1, 2, ..., 𝑝− 1.

Whereℤ/(𝑝) is residue class ring and [a] denotes the residue class of integer a, and 𝜙 is isomorphism. Then𝔽^𝑝, endowed with the field structure induced by 𝜙, is a finite field, called the Galois field of order p.

Example 2.2. The most simple and most important example is the finite field 𝔽². It has two elements 0 and 1 and operation table has the following form:

and

+ 0 1

0 0 1

1 1 0

. 0 1

0 0 0

1 0 1

3 Linear recurring sequences

In this section we will discuss about linear recurring sequences and we discuss how to generate linear recurring sequences on special switching circuit called feedback shift registers, how can we generate a periodic sequence by feedback shift registers? And we will discuss about impulse response sequence and prop- erties of corresponding characteristic polynomial of a linear recurring relation.

Definition 3.1. Let 𝑘 be a positive integer, and let 𝑎, 𝑎0, 𝑎1, . . . , 𝑎_𝑘−1be given elements of a finite field𝔽^𝑞. A sequence 𝑠0, 𝑠1, . . . of elements of𝔽^𝑞 satisfying the relation

𝑠𝑛+𝑘= 𝑎𝑘−1𝑠𝑛+𝑘−1+ 𝑎𝑘−2𝑠𝑛+𝑘−2+ . . . + 𝑎0𝑠𝑛+ 𝑎 for n = 0, 1, 2, . . . . (3.1) is called a linear recurring sequence in𝔽𝑞 with order 𝑘.

Definition 3.2. If 𝑎 = 0 in (3.1) then the linear recurrence relation is ho- mogeneous otherwise the linear recurrence relation is inhomogeneous. And the sequence 𝑠0, 𝑠1, . . . itself is called a homogeneous or in homogeneous, linear recurring sequence in𝔽𝑞 respectively.

3.1 Feedback Shift Register

A feedback shift register is a special kind of electronic switching circuit handling information in the form of elements of 𝔽𝑞. A feedback shift register consists of four types of devices. The first is an adder, which has two inputs and one output. The output being the sum in𝔽^𝑞 of two inputs. The second is a constant multiplier, which has one input and one output. The output being the product of input with a constant element in 𝔽^𝑞. The third is a constant adder, which is same as a constant multiplier, But adds a constant element of 𝔽^𝑞 to the input. The fourth type is a delay element, which has one input and one output and is regulated by an external synchronous clock so that its input at a particular time seems as its output after one unit time.

A feedback shift register is made by interconnecting a finite number of adders, constant multipliers, constant adders, and delay elements along a closed loop such that two outputs are never connected to each other. A feedback shift register that generates a linear recurring sequence satisfying (3.1) is shown in following example.

Example 3.1. Consider the following homogeneous linear recurrence relation.

𝑠𝑛+6= 𝑠𝑛+5+ 2𝑠𝑛+4+ 𝑠𝑛+1+ 3𝑠𝑛 for n = 0, 1, 2, . . . . (3.2) The circuit diagram of this linear recurrence relation is shown in following figure.

Figure 1: Feed back shift register for (3.2).

Figure 2: The Building blocks of feedback shift registers. Adder, Constant multiplier by a, Constant adder for adding a, Delay element respectively.

3.2 Periodicity of linear recurring sequences

Definition 3.3. Let S be an arbitrary non empty set, and let 𝑠0, 𝑠1, . . . be a sequence of elements of S. If there exist integers 𝑟 > 0 and 𝑛0 ≥ 0 such that 𝑠𝑛+𝑟 = 𝑠𝑛 for all 𝑛 ≥ 𝑛⁰, then the sequence is called ultimately periodic and 𝑟 is called a period of the sequence.And the smallest number of all pos- sible periods of an ultimately periodic sequence is called least period of the sequence.

Lemma 3.1. Every period of an ultimately periodic sequence is divisible by the least period. [1, p. 193]

Lemma 3.2. The sequence 𝑠0, 𝑠1, . . . is periodic if and only if there exists an integer 𝑟 > 0 such that 𝑠𝑛+𝑟= 𝑠𝑛 for all 𝑛 = 0, 1, 2, . . .. [1, p. 193]

Theorem 3.3. Let𝔽^𝑞be any finite field and 𝑘 any positive integer. Then every 𝑘th-order linear recurring sequence in𝔽^𝑞 is ultimately periodic with least period 𝑟 satisfying 𝑟≤ 𝑞^𝑘, and 𝑟≤ 𝑞^𝑘− 1 if the sequence is homogeneous. [1, p. 193]

Theorem 3.4. If 𝑠0, 𝑠1, . . . is a linear recurring sequence in a finite field satis- fying the linearly recurrence relation (3.1) , and if the coefficient 𝑎0 in (3.1) is non zero , then the sequence 𝑠0, 𝑠1, . . . is periodic. [1, p. 194]

Definition 3.4. Let 𝑠0, 𝑠1, . . . be a 𝑘th-order linear recurring sequence in 𝔽^𝑞 satisfying (3.1). If 𝑛 is a non-negative integer, then after 𝑛 time units the delay element D𝑗 = 0, 1, . . . , 𝑘− 1 will contain 𝑠^𝑛+𝑗. It is therefore natural to call the row vector s𝑛 = (𝑠𝑛, 𝑠𝑛+1, . . . , 𝑠𝑛+𝑗−1) the nth state vector of the linear recurring sequence. The state vector s0 = (𝑠0, 𝑠1, . . . , 𝑠𝑘−1) is also referred as the initial state vector.

Example 3.2. Consider the linear recurrence relation 𝑠𝑛+4= 𝑠𝑛+1+ 𝑠𝑛, 𝑛 = 0, 1, . . . , in 𝔽² with initial state vector (0,0,1,1). Here we have initial values 𝑠0= 0, 𝑠1= 0, 𝑠2= 1, 𝑠3= 1.

Then,

𝑠4= 𝑠1+ 𝑠0= 0 + 0 = 0.

𝑠5= 𝑠2+ 𝑠1= 1 + 0 = 1.

𝑠6= 𝑠3+ 𝑠2= 1 + 1 = 0.

𝑠7= 𝑠4+ 𝑠3= 0 + 1 = 1.

𝑠8= 𝑠5+ 𝑠4= 1 + 0 = 1.

𝑠9= 𝑠6+ 𝑠5= 0 + 1 = 1.

𝑠10= 𝑠7+ 𝑠6= 1 + 0 = 1.

𝑠11= 𝑠8+ 𝑠7= 1 + 1 = 0.

𝑠12= 𝑠9+ 𝑠8= 1 + 1 = 0.

𝑠13= 𝑠10+ 𝑠9= 1 + 1 = 0.

𝑠14= 𝑠11+ 𝑠10= 0 + 1 = 1.

𝑠15= 𝑠12+ 𝑠11= 0 + 0 = 0.

𝑠16= 𝑠13+ 𝑠12= 0 + 0 = 0.

𝑠17= 𝑠14+ 𝑠13= 1 + 0 = 1.

𝑠18= 𝑠15+ 𝑠14= 0 + 1 = 1.

We obtain the string of binary digits 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, . . . of least period 15.

3.3 Impulse response sequences

We can isolate a homogeneous recurring sequence in𝔽^𝑞 satisfying a given 𝑘th order recurring relation such as (3.1), that yields the maximal value for the least period, in all possible homogeneous recurring sequences in𝔽^𝑞. This is the impulse response sequence 𝑑0, 𝑑1, . . . indomitable distinctively by its initial values 𝑑0= . . . = 𝑑_𝑘−2= 0, 𝑑_𝑘−1= 1(𝑑0= 1 if 𝑘 = 1) and the linear recurrence relation

𝑑𝑛+𝑘= 𝑎_𝑘−1𝑑_{𝑛+𝑘−1}+ 𝑎_𝑘−2𝑑_{𝑛+𝑘−2}+ . . . + 𝑎0𝑑𝑛 for n = 0, 1, 2, . . . . (3.3) Example 3.3. Consider the linear recurrence relation 𝑠𝑛+3= 𝑠𝑛+1+ 𝑠𝑛, 𝑛 = 0, 1, . . . , in 𝔽² with initial state vector (0,0,1).

Then,

𝑠3= 𝑠1+ 𝑠0= 0 + 0 = 0.

𝑠4= 𝑠2+ 𝑠1= 1 + 0 = 1.

𝑠5= 𝑠3+ 𝑠2= 0 + 1 = 1.

𝑠6= 𝑠4+ 𝑠3= 1 + 0 = 1.

𝑠7= 𝑠5+ 𝑠4= 1 + 1 = 0.

𝑠8= 𝑠6+ 𝑠5= 1 + 1 = 0.

𝑠9= 𝑠7+ 𝑠6= 0 + 1 = 1.

𝑠10= 𝑠8+ 𝑠7= 0 + 0 = 0.

The impulse response sequence 𝑑0, 𝑑1, . . . corresponding to above recurrence relation is given by the string of binary digits 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, . . . of least period 7.

Figure 3: Feedback shift register for Example 3.3.

Theorem 3.5. The least period of a homogeneous linear recurring sequence in 𝔽^𝑞 divides the least period of the corresponding impulse response sequence. [1, p. 197]

Example 3.4. Consider the linear recurrence relation 𝑠𝑛+5= 𝑠𝑛+1+ 𝑠𝑛, 𝑛 = 0, 1, . . . , in 𝔽².

For least period of impulse response sequence we take (0, 0, 0, 0, 1) initial state vector and we will see that 21 is the least period of impulse response sequence.

Now for least period of recurring sequence we may take (1, 1, 0, 1, 1) initial state vector and we will get 3 as the least period of recurring sequence and if we take (1, 1, 1, 0, 1) initial state vector then we get 7 as the least period of recurring sequence. It is clear that 3 and 7 divides 21. We may take all other possible initial state vectors but every time we will see that least period of recurring sequence divides least period of corresponding impulse sequence.

By (Theorem 3.5) and above example we can make a conclusion about the relation between initial state vector and least period of recurring sequence that, If we take different initial state vectors for a linear recurring relation we will see that the least period of recurring sequence divides the least period of impulse response sequence. It means if least period of impulse response sequence is a prime number then every possible non zero initial state vector of that recurring relation gives the same least period equal to the least period of impulse response sequence.

3.4 Characteristic polynomial

Definition 3.5. Let 𝑠0, 𝑠1, . . . be a 𝑘th-order homogeneous linear recurring sequence in𝔽𝑞 satisfying the linear recurring relation

𝑠𝑛+𝑘= 𝑎𝑘−1𝑠𝑛+𝑘−1+ 𝑎𝑘−2𝑠𝑛+𝑘−2+ . . . + 𝑎0𝑠𝑛 for n = 0, 1, 2, . . . . (3.4) where 𝑎𝑗∈ 𝔽^𝑞 for 0≤ 𝑗 ≤ 𝑘 − 1. The polynomial

𝑓 (𝑥) = 𝑥^𝑘− 𝑎𝑘−1𝑥^𝑘−1− 𝑎𝑘−2𝑥^𝑘−2− . . . − 𝑎⁰∈ 𝔽^𝑞 (3.5) is called the characteristic polynomial of the linear recurring sequence. This polynomial depends on the given recurrence relation.

Example 3.5. The characteristic polynomial of linear recurrence relation 𝑠𝑛+6= 𝑠𝑛+5+ 𝑠𝑛+4+ 𝑠𝑛+3+ 𝑠𝑛+1+ 𝑠𝑛 for n = 0, 1, 2, . . .∈ 𝔽²

is 𝑓 (𝑥) = 𝑥⁶+ 𝑥⁵+ 𝑥⁴+ 𝑥³+ 𝑥 + 1.

Theorem 3.6. Let 𝑠0, 𝑠1, . . . be a 𝑘th-order homogeneous linear recurring se- quence in 𝔽^𝑞 that satisfies the linear recurrence relation (3.4) and is periodic with period 𝑟. Let 𝑓 (𝑥) be the characteristic polynomial of the sequence. Then the identity

𝑓 (𝑥)𝑠(𝑥) = (𝑥^𝑟− 1)ℎ(𝑥), (3.6)

holds with 𝑠(𝑥) = 𝑠0𝑥^𝑟−1+ 𝑠1𝑥^𝑟−2+ . . . + 𝑠𝑟−2+ 𝑠𝑟−1∈ 𝔽^𝑞 and

ℎ(𝑥) =

𝑘−1

∑

𝑗=0 𝑘−𝑗−1∑

𝑖=0

𝑎𝑖+𝑗+1𝑠𝑖𝑥^𝑗 ∈ 𝔽^𝑞. (3.7)

where we set 𝑎𝑘 =−1. [1, p. 201]

Definition 3.6. The smallest integer 𝑒 for which polynomial 𝑓 (𝑥) divides 𝑥^𝑒−1 is called order of polynomial 𝑓 (𝑥).

Theorem 3.7. Let 𝑠0, 𝑠1, . . . be a homogeneous linear recurring sequence in 𝔽𝑞 with characteristic polynomial 𝑓 (𝑥) ∈ 𝔽^𝑞[𝑥]. Then the least period of the sequence divides ord(𝑓 (𝑥)), and the least period of the corresponding impulse response sequence is equal to ord(𝑓 (𝑥)). If 𝑓 (𝑥)∕= 0, then both sequences are periodic. [1, p. 203]

4 Least period for irreducible characteristic poly- nomial

In this section we will discuss about the least period of recurring sequences having irreducible corresponding characteristic polynomials.

Definition 4.1. A polynomial 𝑓∈ 𝔽^𝑞[𝑥] is said to be irreducible polynomial in𝔽^𝑞[𝑥] if 𝑓 has a positive degree and 𝑓 = 𝑎𝑏 with 𝑎, 𝑏∈ 𝔽^𝑞[𝑥] implies that either 𝑏 or 𝑐 a constant polynomial, where 𝑞 is a field.

To check the reducibility of a polynomial over the finite field we can use Berlekamp’s algorithm or Zassenhaus algorithm. [3]

Theorem 4.1. Let 𝑠0, 𝑠1, . . . be a homogeneous linear recurring sequence in 𝔽^𝑞 with non zero initial state vector, and suppose the characteristic polynomial 𝑓 (𝑥)∈ 𝔽^𝑞 is irreducible over 𝔽^𝑞 and satisfies 𝑓 (0) ∕= 0. Then the sequence is periodic with least period equal to ord(𝑓 (𝑥)). [1, p. 203]

Proof. By (Theorem 3.7) the sequence 𝑠0, 𝑠1, . . . is periodic and its least period 𝑟 divides ord(𝑓 (𝑥)).

This implies that 𝑟≤ ord(𝑓(𝑥)).

Conversely from (3.6)

𝑓 (𝑥)𝑠(𝑥) = (𝑥^𝑟− 1)ℎ(𝑥),

𝑓 (𝑥) divides (𝑥^𝑟− 1)ℎ(𝑥). Since 𝑠(𝑥) is a non zero polynomial, therefore h(x) is also a non zero polynomial. As 𝑓 (𝑥) is an irreducible characteristic polynomial of the recurring sequence and deg(ℎ(𝑥)) < deg(𝑓 (𝑥)), therefore 𝑓 (𝑥) divides (𝑥^𝑟− 1). So, by definition of order of polynomial 𝑟 ≥ ord(𝑓(𝑥). Hence least period 𝑟 of recurring sequence is equal to ord(𝑓 (𝑥)).

Theorem 4.2. Let 𝑓 (𝑥) ∈ 𝔽^𝑞[𝑥] be an irreducible polynomial over 𝔽^𝑞 with deg(𝑓 (𝑥)) = 𝑘. Then ord(𝑓 (𝑥)) divides 𝑞^𝑘− 1. [1, p. 204]

Example 4.1. Consider the linear recurrence relation 𝑠𝑛+3= 𝑠𝑛+1+ 𝑠𝑛, 𝑛 = 0, 1, 2, . . ., in𝔽²[𝑥]. The corresponding characteristic polynomial is 𝑓 (𝑥) = 𝑥³− 𝑥 − 1 = 𝑥³+ 𝑥 + 1∈ 𝔽².

The polynomial 𝑓 (𝑥) is irreducible over 𝔽². And order of 𝑓 (𝑥) is 7. And for

least period .

𝑠𝑛+3= 𝑠𝑛+1+ 𝑠𝑛. Take initial state vector (0, 1, 1). Then,

𝑠3= 𝑠1+ 𝑠0= 1.

𝑠4= 𝑠2+ 𝑠1= 0.

𝑠5= 𝑠3+ 𝑠2= 0.

𝑠6= 𝑠4+ 𝑠3= 1.

𝑠7= 𝑠5+ 𝑠4= 0.

𝑠8= 𝑠6+ 𝑠5= 1.

𝑠9= 𝑠7+ 𝑠6= 1.

So, the string of binary digits

0111001011 . . .

of least period 7. We can take (1, 1, 1), (1, 1, 0), (1, 0, 0) or (1, 0, 1) initial state vectors, but every time we will see 7 is the least period of above recurring relation.

Example 4.2. Consider an other linear recurrence relation 𝑠𝑛+5= 𝑠𝑛+4+ 𝑠𝑛+2+ 𝑠𝑛+1+ 𝑠𝑛 𝑛 = 0, 1, 2, . . . , in 𝔽².

The corresponding characteristic polynomial is 𝑓 (𝑥) = 𝑥⁵+ 𝑥⁴+ 𝑥²+ 𝑥 + 1 in 𝔽²[𝑥]. The polynomial 𝑓 (𝑥) is irreducible over𝔽² and 𝑓 (𝑥) divides 𝑥³¹− 1 so, by definition of order of polynomial ord(𝑓 (𝑥))=31.

Now we will find the least period of this recurring sequence, for this 𝑠𝑛+5= 𝑠𝑛+4+ 𝑠𝑛+2+ 𝑠𝑛+1+ 𝑠𝑛.

Take initial state vector (1, 1, 1, 1, 1). Then,

𝑠5= 𝑠4+ 𝑠2+ 𝑠1+ 𝑠0= 0.

𝑠6= 𝑠5+ 𝑠3+ 𝑠2+ 𝑠1= 1.

𝑠7= 𝑠6+ 𝑠4+ 𝑠3+ 𝑠2= 0.

𝑠8= 𝑠7+ 𝑠5+ 𝑠4+ 𝑠3= 0.

𝑠9= 𝑠8+ 𝑠6+ 𝑠5+ 𝑠4= 0.

𝑠10= 𝑠9+ 𝑠7+ 𝑠6+ 𝑠5= 1.

𝑠11= 𝑠10+ 𝑠8+ 𝑠7+ 𝑠6= 0.

𝑠12= 𝑠11+ 𝑠9+ 𝑠8+ 𝑠7= 0.

𝑠13= 𝑠12+ 𝑠10+ 𝑠9+ 𝑠8= 1.

𝑠14= 𝑠13+ 𝑠11+ 𝑠10+ 𝑠9= 0.

𝑠15= 𝑠14+ 𝑠12+ 𝑠11+ 𝑠10= 1.

𝑠16= 𝑠15+ 𝑠13+ 𝑠12+ 𝑠11= 0.

𝑠17= 𝑠16+ 𝑠14+ 𝑠13+ 𝑠12= 1.

𝑠18= 𝑠17+ 𝑠15+ 𝑠14+ 𝑠13= 1.

𝑠19= 𝑠18+ 𝑠16+ 𝑠15+ 𝑠14= 0.

𝑠20= 𝑠19+ 𝑠17+ 𝑠16+ 𝑠15= 0.

𝑠21= 𝑠20+ 𝑠18+ 𝑠17+ 𝑠16= 0.

𝑠22= 𝑠21+ 𝑠19+ 𝑠18+ 𝑠17= 0.

𝑠23= 𝑠22+ 𝑠20+ 𝑠19+ 𝑠18= 1.

𝑠24= 𝑠23+ 𝑠21+ 𝑠20+ 𝑠19= 1.

𝑠25= 𝑠24+ 𝑠22+ 𝑠21+ 𝑠20= 1.

𝑠26= 𝑠25+ 𝑠23+ 𝑠22+ 𝑠21= 0.

𝑠27= 𝑠26+ 𝑠24+ 𝑠23+ 𝑠22= 0.

𝑠28= 𝑠27+ 𝑠25+ 𝑠24+ 𝑠23= 1.

𝑠29= 𝑠28+ 𝑠26+ 𝑠25+ 𝑠24= 1.

𝑠30= 𝑠29+ 𝑠27+ 𝑠26+ 𝑠25= 0.

𝑠31= 𝑠30+ 𝑠28+ 𝑠27+ 𝑠26= 1.

𝑠32= 𝑠31+ 𝑠29+ 𝑠28+ 𝑠27= 1.

𝑠33= 𝑠32+ 𝑠30+ 𝑠29+ 𝑠28= 1.

𝑠34= 𝑠33+ 𝑠31+ 𝑠30+ 𝑠29= 1.

𝑠35= 𝑠34+ 𝑠32+ 𝑠31+ 𝑠30= 1.

So, the string of binary digits

111110100010010101100001110011011111 . . .

of least period 31. And we will get same least period if we take other possible initial state vector.

Example 4.3. Consider the following recurrence relation 𝑠𝑛+3= 𝑠𝑛+1+ 2𝑠𝑛 𝑛 = 0, 1, 2, . . . , in 𝔽³.

The corresponding characteristic polynomial is 𝑓 (𝑥) = 𝑥³+ 2𝑥 + 1 in𝔽³[𝑥]. The polynomial 𝑓 (𝑥) is irreducible over𝔽³and 𝑓 (𝑥) divides 𝑥²⁶− 1 so, by definition of order of polynomial ord(𝑓 (𝑥))=26. Now we will find the least period of this recurring sequence, for this

𝑠𝑛+3= 𝑠𝑛+1+ 2𝑠𝑛. Take initial state vector (0, 1, 2). Then,

𝑠3= 𝑠1+ 2𝑠0= 1.

𝑠4= 𝑠2+ 2𝑠1= 1.

𝑠5= 𝑠3+ 2𝑠2= 2.

𝑠6= 𝑠4+ 2𝑠3= 0.

𝑠7= 𝑠5+ 2𝑠4= 1.

𝑠8= 𝑠6+ 2𝑠5= 1.

𝑠9= 𝑠7+ 2𝑠6= 1.

𝑠10= 𝑠8+ 2𝑠7= 0.

𝑠11= 𝑠9+ 2𝑠8= 0.

𝑠12= 𝑠10+ 2𝑠9= 2.

𝑠13= 𝑠11+ 2𝑠10= 0.

𝑠14= 𝑠12+ 2𝑠11= 2.

𝑠15= 𝑠13+ 2𝑠12= 1.

𝑠16= 𝑠14+ 2𝑠13= 2.

𝑠17= 𝑠15+ 2𝑠14= 2.

𝑠18= 𝑠16+ 2𝑠15= 1.

𝑠19= 𝑠17+ 2𝑠16= 0.

𝑠20= 𝑠18+ 2𝑠17= 2.

𝑠21= 𝑠19+ 2𝑠18= 2.

𝑠22= 𝑠20+ 2𝑠19= 2.

𝑠23= 𝑠21+ 2𝑠20= 0.

𝑠24= 𝑠22+ 2𝑠21= 0.

𝑠25= 𝑠23+ 2𝑠22= 1.

𝑠26= 𝑠24+ 2𝑠23= 0.

𝑠27= 𝑠25+ 2𝑠24= 1.

𝑠28= 𝑠26+ 2𝑠25= 2.

So, the string of binary digits

01211201110020212210222001012 . . . of least period 26.

5 Least period for reducible characteristic poly- nomial

Now in this section we will discuss about the least period of recurring sequences having reducible characteristic polynomials.

If 𝑓 (𝑥)∈ 𝔽^𝑞[𝑥] with deg(𝑓 (𝑥)) = 𝑘 is reducible, then ord(𝑓 (𝑥)) need not divide 𝑞^𝑘− 1. Consider 𝑓(𝑥) = 𝑥³+ 𝑥²+ 𝑥 + 1∈ 𝔽²[𝑥]. Then 𝑓 (𝑥) is reducible since

𝑥³+ 𝑥²+ 𝑥 + 1 = (𝑥²+ 1)(𝑥 + 1).

And ord(𝑓 (𝑥))=4, which is not a divisor of 2⁴− 1 = 15. So above example does not satisfy (Theorem 3.7).

To overcome this problem we will use minimal polynomial for finding least period of recurring sequences having reducible characteristic polynomial.

Given an arbitrary sequence 𝑠0, 𝑠1, . . . of elements of𝔽^𝑞,we associate with it its generating function, which is purely formal expression of the type

𝐺(𝑥) = 𝑠0+ 𝑠1𝑥 + 𝑠2𝑥²+ . . . + 𝑠𝑛𝑥^𝑛+ . . . =

∑∞ 𝑛=0

𝑠𝑛𝑥^𝑛 (5.1)

with an indeterminate 𝑥. The idea is that in 𝐺(𝑥) we have preserved all the terms of the sequence in the correct order, so that 𝐺(𝑥)should reflect the prop- erties of the sequence.

Definition 5.1. For a characteristic polynomial 𝑓 (𝑥) of a recurring relation, we can defined its reciprocal characteristic polynomial as

𝑓^∗(𝑥) = 1− 𝑎𝑘−1𝑥− 𝑎𝑘−2𝑥²− ⋅ ⋅ ⋅ − 𝑎⁰𝑥^𝑘 ∈ 𝔽^𝑞[𝑥]. (5.2) The characteristic polynomial 𝑓 (𝑥) and the reciprocal characteristic polynomial 𝑓^∗(𝑥) are related by

𝑓^∗(𝑥) = 𝑥^𝑘𝑓 (1

𝑥). (5.3)

Theorem 5.1. Let 𝑠0, 𝑠1, . . . be a 𝑘th-order homogeneous linear recurring se- quence in𝔽^𝑞 satisfying the linear recurrence relation (3.4), let 𝑓^∗(𝑥)∈ 𝔽^𝑞[𝑥] be its reciprocal characteristic polynomial, and let 𝐺(𝑥)∈ 𝔽^𝑞[𝑥] be its generating function in (5.1). Then the identity

𝐺(𝑥) = 𝑔(𝑥)

𝑓^∗(𝑥). (5.4)

holds with

𝑔(𝑥) =−

𝑘∑−1 𝑗=0

∑𝑗 𝑖=0

𝑎𝑖+𝑘−𝑗𝑠𝑖𝑥^𝑗∈ 𝔽^𝑞[𝑥], (5.5) where we set 𝑎𝑘 =−1.

Conversely, if 𝑔(𝑥) is any polynomial over𝔽^𝑞 with deg(𝑔(𝑥)) < 𝑘 and if 𝑓^∗(𝑥)∈ 𝔽^𝑞[𝑥] is given by

𝑓^∗(𝑥) = 1− 𝑎^𝑘−1𝑥− 𝑎^𝑘−2𝑥²− . . . − 𝑎⁰𝑥^𝑘 ∈ 𝔽^𝑞[𝑥], (5.6) then the formal power series 𝐺(𝑥) ∈ 𝔽^𝑞[𝑥] defined by (5.4) is the generating function of a 𝑘th-order homogeneous linear recurring sequence in𝔽^𝑞 satisfying the linear recurrence relation (3.4). [1, p. 211]

Definition 5.2. A polynomial in which the coefficient of the term of highest degree is +1 and the coefficients of the other terms are integers is called monic polynomial.

Theorem 5.2. Let 𝑠0, 𝑠1, . . . be a homogeneous linear recurring sequence in𝔽^𝑞. Then there exists a uniquely determined monic polynomial 𝑚(𝑥)∈ 𝔽^𝑞[𝑥] having the following property: a monic polynomial 𝑓 (𝑥)∈ 𝔽^𝑞[𝑥] of positive degree is a characteristic polynomial of 𝑠0, 𝑠1, . . . if and only if 𝑚(𝑥) divides 𝑓 (𝑥). [1, p. 214]

Proof. Let 𝑓0(𝑥) ∈ 𝔽^𝑞[𝑥] be the characteristic polynomial of a homogeneous linear recurrence relation satisfied by the sequence, and let ℎ0(𝑥) ∈ 𝔽^𝑞[𝑥] be the polynomial in (3.7) determined by 𝑓0(𝑥) and the sequence. If 𝑑(𝑥) = gcd(𝑓0(𝑥), ℎ0(𝑥)) is the monic polynomial in𝔽^𝑞[𝑥], then we can write

𝑓0(𝑥) = 𝑚(𝑥)𝑑(𝑥 (5.7)

and

ℎ0(𝑥) = 𝑏(𝑥)𝑑(𝑥) (5.8)

Where 𝑚(𝑥), 𝑏(𝑥)∈ 𝔽^𝑞[𝑥]. we will prove that 𝑚(𝑥) is the required polynomial.

Now let 𝑓 (𝑥) ∈ 𝔽^𝑞[𝑥] be an arbitrary characteristic polynomial of the given sequence, and let ℎ(𝑥)∈ 𝔽^𝑞[𝑥] be the polynomial in (3.7) determined by 𝑓 (𝑥) and the sequence. By applying (Theorem 5.1), we obtain that the generating function 𝐺(𝑥) of the sequence satisfies

𝐺(𝑥) = 𝑔0(𝑥)

𝑓^∗(𝑥) = 𝑔(𝑥)

𝑓^∗(𝑥) (5.9)

with 𝑔0(𝑥) and 𝑔(𝑥) determined by (5.5). Therefore 𝑔(𝑥)𝑓₀^∗(𝑥) = 𝑔0(𝑥)𝑓^∗(𝑥).

Now from (5.5)

𝑔(1 𝑥) =−

𝑘−1

∑

𝑗=0

∑𝑗 𝑖=0

𝑎𝑖+𝑘−𝑗𝑠𝑖𝑥^−𝑗. and

𝑥^𝑘−1𝑔(1

𝑥) =−𝑥^𝑘−1

𝑘−1

∑

𝑗=0

∑𝑗 𝑖=0

𝑎𝑖+𝑘−𝑗𝑠𝑖𝑥^−𝑗

=−

𝑘−1

∑

𝑗=0

∑𝑗 𝑖=0

𝑎𝑖+𝑘−𝑗𝑠𝑖𝑥^{𝑘−𝑗−1}

Put 𝑘− 𝑗 − 1 = 𝑙 then 𝑗 = 𝑘 − 𝑙 − 1 and since 𝑗 = 1, 2, ⋅ ⋅ ⋅ , 𝑘 − 1 , so 𝑙 = 𝑘− 1, 𝑘 − 2, ⋅ ⋅ ⋅ , 1. Then we have

𝑥^𝑘−1𝑔(1 𝑥) =−

𝑘−1∑

𝑙=0 𝑘−𝑙−1∑

𝑖=0

𝑎𝑖+𝑙+1𝑠𝑖𝑥^𝑙.

and using (3.7) we get

𝑥^𝑘−1𝑔(1

𝑥) =−ℎ(𝑥). (5.10)

Now using (5.10) and (5.3) we have

ℎ(𝑥)𝑓0(𝑥) =−𝑥^{deg(𝑓 (𝑥))−1}𝑔(1

𝑥)𝑥^{deg(𝑓0(𝑥))}𝑓₀^∗(1 𝑥)

=−𝑥^{deg(𝑓0(𝑥))−1}𝑔0(1

𝑥)𝑥^{deg(𝑓 (𝑥))}𝑓^∗(1 𝑥)

= ℎ0(𝑥)𝑓 (𝑥).

Therefore we have

ℎ(𝑥)𝑓0(𝑥) = ℎ0(𝑥)𝑓 (𝑥) (5.11) Since 𝑓0(𝑥) = 𝑚(𝑥)𝑑(𝑥) and ℎ0(𝑥) = 𝑏(𝑥)𝑑(𝑥).

Therefore (5.11) becomes

ℎ(𝑥)𝑚(𝑥) = 𝑏(𝑥)𝑓 (𝑥). (5.12)

and since 𝑚(𝑥) and 𝑏(𝑥) are relatively prime, this means that 𝑚(𝑥) divides 𝑓 (𝑥).

Conversely, suppose that 𝑓 (𝑥) is a monic polynomial with positive degree in 𝔽^𝑞[𝑥]. And 𝑓 (𝑥) is divisible by 𝑚(𝑥), so we have 𝑓 (𝑥) = 𝑚(𝑥)𝑐(𝑥). Where 𝑐(𝑥)∈ 𝔽^𝑞[𝑥]. Now in reciprocal polynomial form we have 𝑓^∗(𝑥) = 𝑚^∗(𝑥)𝑐^∗(𝑥).

Also from (5.12) we have ℎ0(𝑥)𝑚(𝑥) = 𝑏(𝑥)𝑓0(𝑥) and using relation (5.10) and (5.3), we get

𝑔𝑜(𝑥)𝑚^∗(𝑥) =−𝑥^{deg(𝑓0(𝑥))−1}ℎ0(1

𝑥)𝑥^{deg(𝑚(𝑥))}𝑚(1 𝑥)

=−𝑥deg(𝑚(𝑥))−1𝑏(1

𝑥)𝑥^{deg(𝑓0(𝑥))}𝑓0(1 𝑥).

Now in above relation first two factors including with negative sign becomes an other polynomial, let say 𝑎(𝑥),here 𝑎(𝑥)∈ 𝔽^𝑞[𝑥] and using (5.3) we arrive at

𝑔𝑜(𝑥)𝑚^∗(𝑥) = 𝑎(𝑥)𝑓₀^∗(𝑥). (5.13) From (5.13) and generating function described in (Theorem 5.1) and, we get

𝐺(𝑥) = 𝑔0(𝑥)

𝑓₀^∗(𝑥) = 𝑎(𝑥) 𝑚^∗(𝑥). 𝐺(𝑥) = 𝑎(𝑥)𝑐^∗(𝑥)

𝑚^∗(𝑥)𝑐^∗(𝑥) =𝑎(𝑥)𝑐^∗(𝑥) 𝑓^∗(𝑥) . Because 𝑓^∗(𝑥) = 𝑚^∗(𝑥)𝑐^∗(𝑥).

Since

deg(𝑎(𝑥)𝑐^∗(𝑥)) = deg(𝑎(𝑥) + deg(𝑐^∗(𝑥)) < deg(𝑚(𝑥)) + deg(𝑐(𝑥)) = deg(𝑓 (𝑥)).

Because 𝑓 (𝑥) = 𝑚(𝑥)𝑐(𝑥). Now from the converse statement of (Theorem 5.1) 𝑓 (𝑥) is a characteristic polynomial of the sequence. and it is clear that 𝑚(𝑥) is unique with desired properties.

Definition 5.3. The uniquely determined polynomial 𝑚(𝑥) over𝔽^𝑞 associated with the sequence 𝑠0, 𝑠1, . . . according to (Theorem 5.2) is called the minimal polynomial of the sequence.

Theorem 5.3. Let 𝑠0, 𝑠1, . . . be a homogeneous linear recurrence sequence in𝔽^𝑞 with minimal polynomial 𝑚(𝑥)∈ 𝔽^𝑞[𝑥]. Then the least period of the sequence is equal to ord(𝑚(𝑥)). [1, p. 216]

5.1 Algorithm for finding minimal polynomial m(x)

To find the minimal polynomial of a homogeneous recurrence relation we will do following steps as we have done in (Theorem 5.2).

1. First we will find the characteristic polynomial 𝑓 (𝑥) according to recur- rence relation.

2. Then we will expand the (3.7) to find the polynomial ℎ(𝑥). Here (𝑠0, 𝑠1, . . . , 𝑠𝑛) is the initial state vector and 𝑎0, 𝑎1, . . . , 𝑎𝑛 are the coefficient of polyno- mial 𝑓 (𝑥).

3. After finding ℎ(𝑥), we will find 𝑑(𝑥) = gcd(𝑓 (𝑥), ℎ(𝑥)).

4. And then in the last we will get minimal polynomial 𝑚(𝑥) by dividing 𝑓 (𝑥) by 𝑑(𝑥).

5. In the end we will find the least period of 𝑚(𝑥) and using (Theorem 5.3) we will get the least period of given linear recurring sequence.

Example 5.1. Consider the recurrence relation

𝑠𝑛+4= 𝑠𝑛+3+ 𝑠𝑛+1+ 𝑠𝑛 𝑛 = 0, 1, 2, . . . , in 𝔽²

The corresponding characteristic polynomial is 𝑓 (𝑥) = 𝑥⁴+ 𝑥³+ 𝑥 + 1 in𝔽²[𝑥].

The polynomial 𝑓 (𝑥) is reducible in𝔽²[𝑥]. Since 𝑓 (𝑥) = (𝑥 + 1)(1 + 𝑥)(1 + 𝑥 + 𝑥²) for this polynomial first we will find minimal polynomial 𝑚(𝑥) of 𝑓 (𝑥) by following method.

For finding minimal polynomial we have to find the polynomial ℎ(𝑥). And we can find the coefficient of ℎ(𝑥) by using (3.7).

ℎ(𝑥) =

𝑘−1

∑

𝑗=0 𝑘−𝑗−1∑

𝑖=0

𝑎𝑖+𝑗+1𝑠𝑖𝑥^𝑗 ∈ 𝔽^𝑞.

If we take (𝑠0, 𝑠1, 𝑠2, 𝑠3) = (1, 1, 0, 1) as initial state vector. and we have coeffi- cient of polynomial 𝑓 (𝑥) as 𝑎3= 1, 𝑎2= 0, 𝑎1= 1, 𝑎0= 1 and deg 𝑓 (𝑥) = 𝑘 = 4.

So, 𝑎𝑘 = 𝑎4=−1 as described in (Theorem 3.6). Then, For 𝑗 = 0 above relation gives

4−1

∑

𝑖=0

𝑎𝑖+1𝑠𝑖.

= (1, 0, 1,−1)(1, 1, 0, 1) = 0.

For 𝑗 = 1

𝑥

3−1∑

𝑖=0

𝑎𝑖+2𝑠𝑖.

= 𝑥(0, 1,−1)(1, 1, 0) = 𝑥.

For 𝑗 = 2

𝑥²

2−1

∑

𝑖=0

𝑎𝑖+3𝑠𝑖.

= 𝑥²(1,−1)(1, 1) = 0.

For 𝑗 = 3

= 𝑥³(1)(−1) = −𝑥³.

Therefore by above calculation we get ℎ(𝑥) =−𝑥³+ 𝑥 = 𝑥³+ 𝑥 ∈ 𝔽²[𝑥].

Now,

𝑑(𝑥) = gcd(𝑓 (𝑥), ℎ(𝑥)) = 𝑥²+ 1.

and

𝑚(𝑥) = ^{𝑓 (𝑥)}_𝑑(𝑥) = 𝑥²+ 𝑥 + 1.

and 𝑚(𝑥) = 𝑥²+ 𝑥 + 1 satisfy the linear recurrence relation 𝑠𝑛+2= 𝑠𝑛+1+ 𝑠𝑛.

since 𝑚(𝑥) is irreducible and 𝑚(0) ∕= 0 and ord(𝑚(𝑥)) = 3. Therefore by (Theorem 4.1) 3 is the least period of 𝑠𝑛+2 = 𝑠𝑛+1+ 𝑠𝑛. And least period of 𝑠𝑛+4= 𝑠𝑛+3+ 𝑠𝑛+1+ 𝑠𝑛 will be 3 by (Theorem 5.3).

Example 5.2. Consider an other recurrence relation

𝑠𝑛+6= 𝑠𝑛+4+ 𝑠𝑛+3+ 𝑠𝑛 𝑛 = 0, 1, 2, . . . , in 𝔽2

The corresponding characteristic polynomial is 𝑓 (𝑥) = 𝑥⁶+ 𝑥⁴+ 𝑥³+ 1 in𝔽²[𝑥].

Since 𝑓 (𝑥) = (𝑥 + 1)(𝑥⁵+ 𝑥⁴+ 𝑥³+ 1), therefore 𝑓 (𝑥) is reducible in𝔽²[𝑥]. So firstly we have to find minimal polynomial 𝑚(𝑥) as we have done in previous example. For finding minimal polynomial we have to find the polynomial ℎ(𝑥).

And we can find the coefficient of ℎ(𝑥) by using (3.7).

ℎ(𝑥) =

𝑘−1

∑

𝑗=0 𝑘−𝑗−1∑

𝑖=0

𝑎𝑖+𝑗+1𝑠𝑖𝑥^𝑗 ∈ 𝔽^𝑞.

If we take (𝑠0, 𝑠1, 𝑠2, 𝑠3, 𝑠4, 𝑠5) = (1, 1, 0, 0, 1, 1) as initial state vector. and we have coefficient of polynomial 𝑓 (𝑥) as 𝑎5 = 0, 𝑎4 = 1, 𝑎3 = 1, 𝑎2 = 0, 𝑎1 = 0, 𝑎0 = 1 and deg 𝑓 (𝑥) = 𝑘 = 6. So, 𝑎𝑘 = 𝑎6 = −1 as described in (Theo- rem 3.6). Then,

For 𝑗 = 0 above relation gives

∑5 𝑖=0

𝑎𝑖+1𝑠𝑖

= (0, 0, 1, 1, 0,−1)(1, 1, 0, 0, 1, 1) = 1.

For 𝑗 = 1

∑4 𝑖=0

𝑎𝑖+1𝑠𝑖𝑥

= (0, 1, 1, 0,−1)(1, 1, 0, 0, 1)𝑥 = 𝑥.

For 𝑗 = 2

∑3 𝑖=0

𝑎𝑖+1𝑠𝑖𝑥²

= (1, 1, 0,−1)(1, 1, 0, 0)𝑥²= 0.

For 𝑗 = 3

∑2 𝑖=0

𝑎𝑖+1𝑠𝑖𝑥³

= (1, 0,−1)(1, 1, 0)𝑥³= 𝑥³. For 𝑗 = 4

∑1 𝑖=0

𝑎𝑖+1𝑠𝑖𝑥⁴

= (0,−1)(1, 1)𝑥⁴= 𝑥⁴. For 𝑗 = 5

(−1)(1)𝑥⁵= 𝑥⁵.

Therefore by above calculation we get ℎ(𝑥) = 𝑥⁵+ 𝑥⁴+ 𝑥³+ 1 ∈ 𝔽²[𝑥].

Now,

𝑑(𝑥) = gcd(𝑓 (𝑥), ℎ(𝑥)) = 𝑥 + 1.

and

𝑚(𝑥) = ^{𝑓 (𝑥)}_𝑑(𝑥) = 𝑥⁵+ 𝑥⁴+ 𝑥²+ 𝑥 + 1.

and 𝑚(𝑥) = 𝑥⁵+ 𝑥⁴+ 𝑥²+ 𝑥 + 1 satisfy the linear recurrence relation 𝑠𝑛+5= 𝑠𝑛+4+ 𝑠𝑛+2+ 𝑠𝑛+1+ 𝑠𝑛.

since 𝑚(𝑥) is irreducible and 𝑚(0)∕= 0 and ord(𝑚(𝑥)) = 31.Therefore by (The- orem 4.1) 31 is the least period of 𝑠𝑛+5= 𝑠𝑛+4+ 𝑠𝑛+2+ 𝑠𝑛+1+ 𝑠𝑛. And least period of 𝑠𝑛+6= 𝑠𝑛+4+ 𝑠𝑛+3+ 𝑠𝑛 will be 31 by (Theorem 5.3).

6 Mathematica code

(* The least period of impulse r e s p o n s e s e q u e n c e . *) (* P o l y n o m i l *)

f [ x ] = x ˆ6 + x ˆ4 + x ˆ2 + x + 1;

(* Field *) F = 2;

(* Degree of p o l y n o m i a l *) d = E x p o n e n t [ f [ x ] , x ] 6

(* R e c u r r e n c e r e l a t i o n a c c o r d i n g to p o l y n o m i a l *) a [ n_ ] := a [ n ] = Mod [ a [ n + 1] + a [ n + 2] + a [ n + 4]

+ a [ n + 9] , F ];

(* Initial state vector *)

a [0] = 0; a [1] = 0; a [2] = 0; a [3] = 0; a [4] = 0;

a [5] = 1;

a [2] = 0 0

(* Least period *) lista = {};

Do [ A p p e n d T o [ lista , a [ k ]] , {k , 0 , 6}]

i = 1;

l a s t p a r t = Take [ lista , -6];

While [ l a s t p a r t != {0 , 0 , 0 , 0 , 0 , 1} , A p p e n d T o [ lista , a [ i + 5]]; i ++;

l a s t p a r t = Take [ lista , -6]];

i 22

(* Impulse r e s p o n s e s e q u e n c e *) lista

{0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 0 , 1 , 1 , 0 , 0 , 1 , 0 , 1 , 1 , 0 ,0 , 0 , 0 , 0 , 1}

l a s t p a r t = Take [ lista , -6]

{0 , 0 , 0 , 0 , 0 , 1}

7 Conclusion

By this project, now we have found two different corresponding ways for find- ing least period of linear recurring sequences having reducible or irreducible characteristic polynomials. If we have to find the least period of a linear re- curring sequence having irreducible characteristic polynomial then instantly we will find order of that corresponding polynomial, which is the least period of that sequence by (Theorem 4.1). And if we have to find the least period of a lin- ear recurring sequence having reducible characteristic polynomial then we will find minimal polynomial of that characteristic polynomial and use the further process as we have done in examples of section 5.

This work will helpful in cryptography, radar and communication system, se- curity systems and in many branches of electrical engineering. In all these applications the least period of recurring sequence has great importance. For example if we study the radar and communication system we have to use lin- ear recurring sequence with least period 𝑟 in order to measure the distance of an object or aircraft from the radar station. In this system the result will be accurate if we will use linear recurring sequences having long periods[2, p. 365].

8 Bibiliography References

[1] RUDOLF LIDL and HARALD NIEDERREITER. Introduction to finite fields and their applications. Cambridge university press,2000.

[2] RUDULF LIDL and GUNTER PILZ. Applied abstract algebra. Springer, 1998.

[3] Sajid Hanif and Muhammad Imran. Factorization Algorithms for Polynomi- als over Finite Fields. Linnæus university ,2010. http://lnu.divaportal.

org/smash/record.jsf?pid=diva2:414578.

SE-351 95 Växjö / SE-391 82 Kalmar Tel +46-772-28 80 00

dfm@lnu.se Lnu.se/dfm