2-Adic Complexity of a Sequence Obtained from a Periodic Binary Sequence by Either Inserting or Deleting k Symbols within One Period

(1)

2-Adic Complexity of a Sequence Obtained from a Periodic Binary

Sequence by Either Inserting or Deleting

k

Symbols within One Period

ZHAO Lu, WEN Qiao-yan

(State Key Laboratory of Networking and Switching Technology, Beijing University of Posts and Telecommunications, Beijing 100876, China)

Abstract. In this paper, we propose a method to get the lower bounds of the 2-adic complexity of

a sequence obtained from a periodic sequence over by either inserting or deleting

symbols within one period. The results show the variation of the distribution of the 2-adic

complexity becomes as increases. Particularly, we discuss the lower bounds when

respectively.

k

(2)

GF

k

=

1

Key words: Periodic Binary Sequence; 2-Adic Complexity; k-symbol deletion; k-symbol insertion

1 Introduction

Many modern stream ciphers are designed by combining the output of several LFSRs in

various nonlinear ways. Since 1955, large amount of efforts have been devoted into the study of

other (“nonlinear”) feedback architectures. In [1], Klapper and Goresky introduced a new feedback architecture for shift register generation of pseudorandom binary sequences called

feedback with carry shift register (FCSR) and discussed some basic properties of FCSR sequences,

such as the periods, the exponential representations and so on. Based on FCSR they presented the

concept of the 2-adic complexity (denoted

φ

( )

i

) as an important index of the security of pseudorandom sequences and summarized the rational approximation algorithm. For every

periodic sequence

S

, only knowledge of

⎡

_⎢

2 ( )

φ

s

⎤

_⎥

+

2

bits is sufficient to reproduce it using the

algorithm. In another word, the higher 2-adic complexity a sequence has, the more secure it is. In

recent years, more and more researchers show their interest in the 2-adic complexity of sequences.

It is well known that the linear complexity of a periodic sequence is unstable under small

perturbations, so -error linear complexity is proposed to study of the stability of stream ciphers.

Similar to -error linear complexity, Honggang proposed the concept of -error 2-adic

complexity and gave the expected value and variance of a periodic binary sequence’s 2-adic

complexity

k

[2]

. In[3], Chen Lanfang provided an analog of the extended Games-Chan algorithm

(2)

m n

p

. with period

2

Jiang Shaoquan and Dai Zongduo discussed linear complexity of a sequence by substituting,

inserting or deleting

k

symbols in a periodic sequence [4]

. Motivated by their work, we study the

lower bounds of the 2-adic complexity of a sequence obtained from a periodic sequence over

by either inserting or deleting symbols within one period. The method of this paper

is simple, can deal both the two different cases with the same manner and the results show the

variation of the distribution of the 2-adic complexity becomes as increases. Finally, we also

give out the lower bounds when .

k

(2)

GF

k

1 k

=

2 Definitions and Lemmas

m

_n₋₁

∑

1

n

a

₋

a

_n₋₂

a

_{n r}_{− +}₁

a

_{n r}₋

r

q

2

q

_r₋₁

1

q

An FCSR is determined by

r

coefficients

q q

₁

,

₂

,

q

_r, where

q

_i

∈

{0,1},

i

=

1, 2,

,

r

, and

an initial memory . The contents of the register at any given time consist of bits,

denoted and the memory is

1

r

m

₋

r

1 2 1

(

a

_n₋

,

a

_n₋

,

a

_{n r}_{− +}

,

a

_{n r}₋

)

m

_n₋₁. The operation of the shift register is defined as following:

1. Form the integer sum ₁

1

r

n k n k k

q a

m

σ

− n−

=

∑

+

;

2. Shift the contents one step to the right, while outputting the rightmost bit

a

_{n r}₋ ;

mod 2

n n

a

≡

σ

3. Put into the leftmost cell of the shift register;

1

n

m

₋

m

_n

=

(

σ

_n

−

a

_n

)

2

4. Replace the memory integer with .

The integer

q

= − +

1 q

₁

2 +

q

₂

2

+ +

q

_r

2

ris called the connection integer of the FCSR.

(3)

Fig 1. Feedback with carry shift register

0

{ }

i i

S

=

s

∞₌

Any infinite binary sequence can be presented by a formal power series

called 2-adic number. Such power series forms the ring of 2-adic number, denoted

0

2

i

i i

s

α

∞

=

∑

Z

2.

Sis eventually periodic if and only if the 2-adic number

α

is rational, i.e. there exist integer

such that

,

p q

2

p

Z

q

α

= − ∈

. If S is strictly periodic with minimal period

T

, then

1

0

2

1

T i i i i i T i

s

p

s

q

α

−

∞

=

= −

−

∑

p

q

−

where

0 ≤ ≤

p

q

.

p

=

q

if and only if

S

is the all-1 sequence. If

gcd( , )

p q

=

1

, is called the reduced rational expression of

S

.

p

q

−

Definition 1. Let

S

be a periodic binary sequence with reduced rational expression , then

the 2-adic complexity

φ

( )

S

is the real number

log

₂

q

.

Obviously, if

S

is the all-0 sequence or the all-1 sequence, then

φ

( )

S

=

0

.

Lemma 1.[2] Let

p q

,

be two positive integers, where

0 < <

p

q

. Let be a nonzero integer and

h

mod

(

ph

)

_q

q

=

p q

'

, where the notation means the reduced residue

of modulo

q

, and . Then

mod

(

ph

)

_q

ph

0 <

p

'

<

q

'

gcd( ', ')

gcd( , )

q

p q

≤

p

q

The equality holds if and only if

gcd( ,

)

1 gcd( , )

q

h

p q

=

1

,

1

p

=

dp q

=

dq

gcd(

p q

₁

,

₁

) 1

=

Proof. Put

d

=

gcd( , )

p

q

. Then , and . We have

1 1

1 mod( ) 1 mod( ) mod

1 1

(

)

(

)

(

ph

)

q

dp h

dq

p h

q

=

dq

=

q

(4)

1

gcd( , )

q

p q

′

≤

=

′ ′

The equality holds if and only if

1

gcd( ,

)

gcd( ,

)

1 gcd( , )

q

h

p q

=

h q

=

■

Definition 2. Let be a periodic binary sequence,

then is the complementary sequence of , if

0 1 1 0

( , ,

_T

,

)

S

=

s s

s

₋

s

0 1 1 0

( , ,

,

_T

, , )

S

=

s s

s

₋

s

S

s

_i

+ =

s

_i

1, 0

≤ <

i

T

.

Lemma 2. Let be a periodic binary sequence, 2-adic complexity

of is

0 1 1 0

( , ,

_T

,

)

S

=

s s

s

₋

s

S

φ

( )

S

, if is the complementary sequence of , and 2-adic complexity of is ,

then .

S

φ

( )

S

( )

S

( )

S

φ

=

φ

p

q

−

p

q

−

Proof. Let the reduced rational expression of

S

and

S

be and respectively, that is,

1

0

2

1

T i i i

T

s

p

q

−

=

− = −

−

∑

1

0

2

1

T i i i

T

s

p

q

−

=

− = −

−

∑

and .

2

( )

S

log

q

φ

=

and

φ

( )

S

=

log

₂

q

. Thus

1 1

0 0

2 (2

1)

2

1

2

1

T T

i T

i i

T T

s

p

q

− −

= =

− −

− = −

= −

−

2

i

∑

Hence

1 1

0 0

2

1

2

1 gcd(2

1 2 , 2

1)

gcd(

2 , 2

1)

T T

T T T T T

i i

q

s

− −

= =

−

=

− −

∑

−

∑

−

=

S

Then we can get

φ

( )

S

=

φ

( )

■

3

k

-Symbol Insertion

Theorem 1. Let be a periodic binary sequence with periodT . Suppose

that the rational expression of is

0 1 1 0

( , ,

,

_T

,

)

S

=

s s

s

₋

s

p

q

α

= −

S

, where

0 < <

p

q

, . Let

be a sequence obtained from by -symbol

gcd( , )

p q

=

1 )

0 1 1 0

( , ,

,

_{T k}

,

)

(5)

1 2

0 ≤ < <

t

< <

t

k

T a

,

i

∈

{0,1},

i

=

1, 2,

,

k

insertion within one period, where , then we

have

2 2

log (2

T k+

− −

1)

φ

( ) (

S

−

t

_k

+ <

1)

φ

( )

S

≤

log (2

T k+

−

1)

1,0

max

i k

s i t k

m

i

= ≤ < −

=

Proof. Let ,

1 1,1

max

a i k

n

i

= ≤ ≤

=

(2)

'

2

1

T k T k

S

p

q

+

′

−

= −

−

S

′

cd( ', ')

p q

=

1

Let

'

be the rational expression of , where

g

,

0 <

p

'

<

q

'

, and

s

′

1 1

i

a

i

∑

a

i

1

0

(2)

2

T k T k i

i i

S

+ − +

=

′

=

∑

1 2

1 1 1

1 2 1

1 1

0 2 2

2

j k

i

j k k

t j t k

t t T k

t

i i j i k i k i

i i i i i

i i t i t j i t k i t k i

s

− −

− − −

− −

= = = − + = − + = − + =

=

∑

+

∑

+ +

∑

+ +

∑

+

∑

+

∑

If

n

< +

m k

, let

(2)

T k

(2) 2

k T

(2)

A

=

S

′

+

−

S

Then we have

2 1

1 1 1

2 1

1 1

( 2) 2 0 1

(2

2 )

2 (2

2 )

2 (2 2 )

2 (1 2 )

2

j k

i

k j

t j

t k t t k

t

k k i j k i k i k i

i i i

i t k i t j i t i i

s

− −

−

− − −

− −

= − − = − + = = =

=

−

∑

+ +

−

∑

+ + −

∑

+ −

∑

+

'

(2)

2 (2)

(2)

2 (2

1)

(2)

'

2

1

2

1 (2

1)

T k k T k T T k T k T k

p

S

A

p

qA

q

+

+ + +

′

+

− +

=

−

So

{ [2

(2

1)

(2)]}mod[ (2

1)]

[2

(2

1)

(2)]mod(2

1)

(2

1)

(2

1)

k T T k k T T k

T k T k

q

p

qA

q

p

qA

q

+ +

− +

−

− +

−

=

−

(2) (1 2 )

2

1

k T k

qA

p

+

+ −

=

−

2

1

2

1 '

gcd(2

1,

(2) (1 2 ) )

(2) (1 2 )

T k T k

T k k k

q

qA

p

qA

p

+ +

+

−

≥

−

+ −

From lemma 1 we have: (1)

(2) (1 2 )

k

qA

+ −

p

In the following, we discuss the value of .

For each

j

=

2,

,

k

, we have

2

1 1

2 2

(2

2 )

2 (2

2 )

2

j j

j j j j

j j

t j t j

t t j k t t j k j k i j k i

i

i t j i t j

s

− −

− + + + − + +

− −

= − + = − +

−

∑

≥

−

∑

=

−

+

,

(6)

1 1 1 1

1

1 1 2 1

1

2 2 2 2

[(2

2 )

2 ]

(2

2

2 )

2

j

j j j j j k

j

t j

k k k

t t j k t t j k t t t t j k i

i

j i t j j j

s

− −

−

− ₋

− + + + − + + + +

−

= = − + = =

−

≥

−

+

= −

−

∑

+

k

+

While

1 1

0 0

(1 2 )

2 (1 2 )

2

2 1 2

t t

t t k k i k i

i

i i

s

− −

+

= =

−

∑

≥ −

∑

=

−

So

1 1 1

(2)

2

j

(2

1)

2

i

2

j

(2

1)

k k k

t k t t k i

j i j

A

a

= = =

≥ −

∑

+

− +

∑

≥ −

∑

+

−

If

n

< +

m k

, then

A

(2)

<

0

So

1

1 1

(2) (1 2 )

(2) (2

1)

2

j

(2

1)(

)

2

j

2

k

k k

t t t

k k k

j j

qA

p

qA

p

q

p

q

+

= =

+ −

= −

+

−

≤

∑

−

<

∑

<

1

2

1

2

1 '

2 (2) (1 2 )

k

T k T k t k

q

qA

p

+ +

+

−

≥

>

+ −

From (1),

So

( )

log (2

₂ T k

1)

( ) (

1)

k

S

φ

_{′ >}

+

_{− −}

φ

₋

t

+

If

n

> +

m k

, let

S

and

S

′

be complementary sequence of

S

and

S

′

respectively.

By lemma 2, we have

φ

( )

S

=

φ

( )

S

and

φ

( )

S

′

=

φ

( )

S

′

.

p

q

′

−

p

q

−

S

′

Let and be the reduced rational expression of

S

and respectively, that is,

(2)

2

1

T T

p

S

q

− = −

−

(2)

2

1

T k T k

p

S

q

+

′

−

= −

′

−

gcd( ', ')

p q

=

1

,

0 <

p

'

<

q

'

and , where and

1

0

(2)

2

T k T k i

i i

S

+ − +

=

′

=

∑

s

′

1 1

i

a

where

1 2

1 1 1

1 2 1

1 1

0 2 2

2

j k

i

j k k

t j t k

t t T k

t

i i j i k i k i

i i i i i

i i t i t j i t k i t k i

s

− −

− − −

− −

= = = − + = − + = − + =

=

∑

+

∑

+ +

∑

+ +

∑

+

∑

+

∑

1

i i

a

+ =

a

Let

(2)

T k

(2) 2

k T

(2)

(7)

2 1

1 1 1

2 1

1 1

( 2) 2 0 1

(2

2 )

2 (2

2 )

2 (2 2 )

2 (1 2 )

2

j k

i

k j

t j

t k t t k

t

k k i j k i k i k i

i i i

i t k i t j i t i i

s

− −

−

− − −

− −

= − − = − + = = =

=

−

∑

+ +

−

∑

+ + −

∑

+ −

∑

i

+

∑

a

i

(2)

2 (2)

(2)

2 (2

1)

(2)

2

1

2

1 (2

1)

T k k T k T T k T k T k

p

S

A

p

qA

q

+

+ + +

′

₌

′

₌

+

₌

− +

′

−

Hence

{ [2

(2

1)

(2)]}mod[ (2

1)]

[2

(2

1)

(2)]mod(2

1)

(2

1)

(2

1)

k T T k k T T k

T k T k

q

p

qA

q

p

qA

q

+ +

− +

−

− +

−

=

−

(2) (1 2 )

2

1

k T k

qA

p

+

+ −

=

−

By lemma 1, we have

2

1

2

1 '

gcd(2

1,

(2) (1 2 ) )

(2) (1 2 )

T k T k

T k k k

q

qA

p

qA

p

+ +

+

−

≥

−

+ −

(

1 ′

)

(2) (1 2 )

k

qA

+ −

p

.

In the following, we discuss the value of

For each

j

=

2,

,

k

, we have

2

1 1

2 2

(2

2 )

2 (2

2 )

2

j j

j j j j

j j

t j t j

t t j k t t j k j k i j k i

i

i t j i t j

s

− −

− + + + − + +

− −

= − + = − +

−

∑

≥

−

∑

=

−

+

Then

1 1 1 1

1

1 1 2 1

1

2 2 2 2

[(2

2 )

2 ]

(2

2

2 )

2

j

j j j j j k

j

t j

k k k

t t j k t t j k t t t t j k i

i

j i t j j j

s

− −

−

− ₋

− + + + − + + + +

−

= = − + = =

−

≥

−

+

= −

−

∑

+

k

+

While

1 1

0 0

(1 2 )

2 (1 2 )

2

2 1 2

t t

t t k k i k i

i

i i

s

− −

+

= =

−

∑

≥ −

∑

=

−

So

1 1 1

(2)

2

j

(2

1)

2

i

2

j

(2

1)

k k k

t k t t k i

j i j

A

a

= = =

≥ −

∑

+

− +

∑

≥ −

∑

+

−

If

n

> +

m k

, then

A

(2)

<

0

So

1

1 1

(2) (1 2 )

(2) (2

1)

2

j

(2

1)(

)

2

j

2

k

k k

t t t

k k k

j j

qA

p

qA

p

q

p

q

+

= =

+ −

= −

+

−

≤

∑

−

<

∑

<

1

2

1

2

1 '

2 (2) (1 2 )

k

T k T k t k

q

qA

p

+ +

+

−

≥

>

+ −

(8)

So

φ

( )

S

′ >

log (2

₂ T k+

− −

1)

φ

( ) (

S

−

t

_k

+

1)

, that is

( )

log (2

2

1)

( ) (

t

+

1)

.

T k

k

S

φ

_{′ >}

+

_{− −}

φ

₋

If

n

= +

m k

, we can get the conclusion with the same method above. ■

4

k

-Symbol Deletion

Theorem 2. Let be a periodic binary sequence with periodT . Suppose

0 1 1 0

( , ,

,

_T

,

)

S

=

s s

s

₋

s

p

q

α

= −

S

, where

0 < <

p

q

, .

Let be a sequence obtained from by -symbol

deletion within one period, where

gcd( , )

p q

=

1 )

1, 2,

,

0 1 1 0

( , ,

,

_{T k}

,

)

S

′

=

s s

′ ′

s

′

_{− −}

s

′

S

k

(

1, 2,

,

i

t

s i

=

k

1 2

0 ≤ < <

t

< <

t

k

T a

,

i

∈

{0,1},

i

=

k

, then we

have

2 2

log (2

T k−

− + −

1)

k

φ

( ) (

S

−

t

_k

+ <

1)

φ

( )

S

′

≤

log (2

T k−

−

1)

1,1

max

, , {1,2, , }

i k j

s i t i t j k

m

i

= ≤ < ≠ ∈

=

1,1

max

ti

s i k

n

i

= ≤ ≤

=

Proof. Let ，

(2)

'

2

1

T k T k

S

p

q

−

′

−

= −

−

S

′

Let

'

be the rational expression of , where , ,

and

gcd( ', ')

p q

=

1

0 <

p

'

<

q

'

1

k

i i

s

B

1

t

s

=

1

0

(2)

2

T k T k i

i i

S

s

− − −

=

′

=

∑

′

1

1 2

1 1

1 1 1

1 ( 1)

0 1 1 1

2

j k

t t

t t T

i i j i k i k

i i i i

i i t i t i t i t

s

+

−

− +

− − −

− − − − −

= = + = + = + = +

=

∑

+

∑

+ +

∑

+ +

∑

+

∑

Hence

φ

( )

S

′

=

log

₂

q

′

≤

log (2

₂ T k−

−

1)

If

n

< +

m k

, let

(2) 2

(2)

2 (2)

T k k T k

S

′

−

S

=

−

Then

1

1 2

1 1

1

0 1 1 1

2 [(2

1)

2 (2

1)

2 (2

1)

2 (2 1)

2 2 ]

j k

i i

j k

t t

t t k

t

k k i k i k j i i

i i i i

i i t i t i t i

s

+

−

− −

− − −

= = + = + = +

=

−

∑

+

−

∑

+ +

−

∑

+ + −

∑

−

∑

(2

1)

(2)

'

(2)

2 [

(2)

(2)]

(2

1)

(2)

'

2

1

2

1

2

2 (2

2 )

T

T k k T T

T k T k T k T k

p

B

p

S

B

q

p

qB

q

− −

−

+

′

+

−

=

−

(9)

So

{ [ (2

1)

(2)]}mod[ (2

2 )]

[ (2

1)

(2)]mod(2

2 )

(2

1)

(2)

(2

2 )

(2

2 )

2

T T k T T k k

T k T k T k

q p

qB

q

q p

qB

p

qB

q

− +

−

₌

− +

−

₌

− +

−

2

1

2

1 '

gcd(2

1, (2

1)

(2))

(2

1)

(2)

T k T k

T k k k

q

p

qB

p

qB

− −

−

≥

−

− +

From lemma 1 we have: (2)

For each

j

=

1,

,

k

−

1

, we have

j

1 1

(2

1)

2 (2

1)

2

j j

j j j

j j

t t

t k j t k j t t k j i k j i

i

i t i t

s

+ +

− −

+ − + − +

− −

= + = +

−

∑

≤

−

∑

=

−

+

Then

1

1 1 1 1

1

1 1 1

1 1 1 2

[(2

1)

2 ]

(2

2

2 )

2

j

j j j j k j

j

t

k k k

t k j t k j t t t t k t t k j i

i

j i t j j

s

+

+ +

−

− − −

+ − + − + + + +

−

= = + = =

−

≤

−

+

=

−

+

∑

While

1 1

0 0

(2

1)

2 (2

1)

2

1

t t

t k t k i k i

i

i i

s

− −

+

= =

−

∑

≤

−

∑

=

−

k

+

1 1

(2)

(1 2 )

(1

)2

j

(1 2 )

2

j

k k

t t

k k

t

j j

B

s

= =

≤ −

+

∑

−

≤ −

+

∑

So

If

n

< +

m k

, then

B

(2)>0

1

1 1

(2

1)

(2)

2

j

(

)(2

1)

2

j

2

k

k k

t t t

k k

j j

p

qB

q

p

q

+

= =

− +

≤

∑

− −

− <

∑

<

So

1

2

1

2

1

2

1 '

gcd(2

1, (2

1)

(2))

(2

1)

(2)

2

k

T k T k T k t

T k k k

q

p

qB

p

qB

q

− −

+ −

−

≥

−

− +

−

₋

>

From (2)

So

log (2

₂ T k−

− + −

1)

k

φ

( ) (

S

−

t

_k

+ <

1)

φ

( )

S

′

≤

log (2

₂ T k−

−

1)

If

n

> +

m k

, let

S

and

S

′

S

and

S

′

respectively.

By lemma 2, we have

φ

( )

S

=

φ

( )

S

and

φ

( )

S

′

=

φ

( )

S

′

.

p

q

′

−

p

q

−

S

′

Let and be the reduced rational expression of

S

and respectively, that is,

(2)

2

1

T T

p

S

q

′

− = −

−

(2)

2

1

T k T k

p

S

q

−

′

−

= −

′

−

cd( ', ')

p q

=

1

(10)

1

0

(2)

2

T k T k i

i i

S

s

− − − =

′

=

∑

′

1 1 2 1 1 1 1

1 1 1

1 ( 1)

0 1 1 1

2

j k

t t

t t T

i i j i k i k

i i i i

i i t i t i t i t

s

+ − − + − − − − − − − − = = + = + = + = +

=

∑

+

∑

+ +

∑

+ +

∑

+

∑

1 k i i

s

B

1 t

s

=

So

( )

log

₂

log (2

₂ T k

1)

S

q

φ

_′

₌

_′

_≤

−

₋

Let

(2)

2 (2)

T k k T k

S

′

−

S

=

−

So we have

1 1 2 1 1 1 1 1 1 1

0 1 1 1

2 [(2

1)

2 (2

1)

2 (2

1)

2 (2 1)

2 2 ]

j k

i i

j k

t t

t t k

t

k k i k i k j i i

i i i i

i i t i t i t i

s

+ − − − − − − − − = = + = + = +

=

−

∑

+

−

∑

+ +

−

∑

+ + −

∑

−

∑

(2

1)

(2)

'

(2)

2 [

(2)

(2)]

(2

1)

(2)

'

2

1

2

1

2

2 (2

2 )

T

T k k T T

T k T k T k T k

p

B

p

S

B

q

p

qB

q

− − − −

−

₊

′

+

−

=

−

+

Therefore,

{ [ (2 1) (2)]}mod[ (2 2 )] [ (2 1) (2)]mod(2 2 ) (2 1) (2)

(2 2 ) (2 2 ) 2 2

T T k T T k k

T k T k T k

q p qB q q p qB p qB

q q

− + − − + − − +

= =

− − −

2

2 '

gcd(2

2 , (2

1)

(2))

(2

1)

(2)

T k T k

T k k k

q

p

qB

p

qB

−

≥

−

− +

(

2 ′

)

By lemma 1, we have

For each

j

=

1,

,

k

−

1

, we have

j 1 1 1 1 1 1 1 1 1 1

(2

1)

2 (2

1)

2

j j

j j j

j j

t t

t k j t k j t t k j i k j i

i

i t i t

s

+ + + + − − + − + − + − − = + = +

−

∑

≤

−

∑

=

−

+

+ Then 1

1 1 1 1

1

1 1 1

1 1 1 2

[(2

1)

2 ]

(2

2

2 )

2

j

j j j j k j

j

t

k k k

t k j t k j t t t t k t t k j i

i

j i t j j

s

+ + + − − − − + − + − + + + + − = = + = =

−

≤

−

+

=

−

+

∑

While 1 1 1 1 1 1 0 0

(2

1)

2 (2

1)

2

1

t t

t k t k i k i

i i i

s

− − + = =

−

∑

≤

−

∑

=

−

k

+

1 1

(2)

(1 2 )

(1

)2

j

(1 2 )

2

j

j k k t t k k t j j

B

s

= =

≤ −

+

∑

−

≤ −

+

∑

So

(11)

1

1 1

(2

1)

(2)

2

j

(

)(2

1)

2

j

2

k

k k

t t t

k k

j j

p

qB

q

p

q

+

= =

− +

≤

∑

− −

− <

∑

<

So

1

2

2 '

gcd(2

2 , (2

1)

(2))

(2

1)

(2)

2

k

T k T k T

t

T k k k

q

p

qB

p

qB

q

+

−

≥

−

− +

k

−

>

From (2)

So

log (2

₂ T k

1)

( ) (

1)

( )

log (2

₂ T k

1)

k

φ

S

t

φ

S

−

_{− + −}

₋

_{+ <}

_′

_≤

−

₋

That is

log (2

₂ T k

1)

k

φ

( ) (

S

t

k

1)

φ

( )

S

log (2

₂ T k

1)

.

−

_{− + −}

₋

_{+ <}

_′

_≤

−

₋

If

n

= +

m k

, we can get the conclusion with the same method above. ■

5 One Symbol Insertion or Deletion

In this section, we look into a special case. We get the lower bounds of the 2-adic complexity for a

sequence obtained from a periodic binary sequence by either inserting or deleting 1 bit.

0 1 1 0

( , ,

,

_T

,

)

S

=

s s

s

₋

s

p

q

α

= −

S

, where

0 < <

p

q

, .

Let be a sequence obtained from by 1 symbol insertion within one

period, then we have

gcd( , )

p q

=

1

0 1 0

( , ,

,

_T

,

)

S

′

=

s s

′ ′

s s

′ ′

S

a

1 1

2 2

log (2

T+

− −

1)

φ

( )

S

<

φ

( )

S

′

≤

log (2

T+

−

1)

Proof. It is well known that 2-adic complexity of a sequence obtained from a sequence by shifting

is not changed, without losing generality we insert 1 symbol behind the last bit within one period.

1

(2)

2

1

T T

S

p

q

+

′

−

= −

′

−

S

′

If

a

=

0

, let is the rational expression of , where

1 1

0 1 1

0 0

(2)

2

2 (2)

T T

T T T i T i T

T i i

i i

S

s

a

s

a

s

S

− −

+ −

−

= =

′

= +

+ +

+

=

∑

+

=

∑

=

1 1

2

1 (2)

2

1

2

1

T T

p

S

q

+ +

−

′

=

′

−

So

Since

gcd(2

T

−

1, 2

T+1

− =

1) 1

, so

1 1

1

2

1

2

1

2

1 gcd(2

1, )

T T T

T

q

1

p

q

+ +

+

−

′ =

≥

>

(12)

Therefore

( )

log

₂

log (2

₂ T 1

1)

( )

S

q

φ

_′

₌

_′

_>

+

_{− −}

φ

S

If

a

=

1

, let

S

and

S

′

S

and

S

′

respectively.

By lemma 2, we have

φ

( )

S

=

φ

( )

S

and

φ

( )

S

′

=

φ

( )

S

′

.

Because , we have

( )

log (2

₂ T 1

1)

( )

S

φ

_{′ >}

+

_{− −}

φ

0 a

=

That is

φ

( )

S

′ >

log (2

₂ T+1

− −

1)

φ

( )

S

■

0 1 1 0

( , ,

,

T

,

)

S

=

s s

s

₋

s

p

q

α

= −

S

, where

0 < <

p

q

, .

Let be a sequence obtained from by 1 symbol deletion within one

period, then we have

gcd( , )

p q

=

1

0 1 2 0

( , ,

,

T

,

)

S

′

=

s s

′ ′

s

′

₋

s

′

S

a

1 1

2 2

log (2

T

1)

( )

log (2

T

1)

S

φ

−

_{− −}

_<

_′

_≤

−

₋

Proof. The method is similar to theorem 3.

6 Conclusion

In communications, noise can easily cause modification to the original key stream, even

compromise the security. In this paper, we discuss the lower bounds of the 2-adic complexity of a

sequence obtained from a periodic sequence over by either inserting or deleting

symbols within one period. Interestingly, two types of changes can be treated with the same

method. The results show the stability of the stream ciphers which generated by FCSR.

Particularly, we give out the lower bounds of 2-adic complexity when

k

(2)

GF

1 k

=

respectively.

References

1. A. klapper, M. Goresky. Feedback shift registers, 2-adic span, and combiners with memory, J.

Cryptogoly, Vol.10, pp.111-147, 1997.

2. Honggang Hu, Dengguo Feng. On the 2-Adic Complexity and the k-Error 2-Adic Complexity

of Periodic Binary Sequences. IEEE Trans. Inform. Theory, Vol.54, No.2, pp.874-883, 2008.

3. Lanfang Chen, Wenfeng Qi. 2-Adic Complexity of Binary Sequences with Period2m n.

(13)

Journal on Communications, Vol.26, No.6, 2005.

4. Shaoquan Jiang, Zongduo Dai et al. Linear Complexity of a Sequence Obtained from a

Periodic Sequence by Either Substituting, Inserting, or Deleting k Symbols Wihtin One

Period. IEEE Trans. Inform. Theory, Vol.46, No.3, pp.1174-1177, 2000.

5. Lei Wang, Mian Cai et al. On Stability of 2-Adic Complexity of Periodic Sequence. Journal

of Xidian University, Vol.27, No.3, pp.348-350, 2000.

6. Lihua Dong, Yupu Hu et al. k-Error 2-Adic Complexity Algorithm for the Binary Sequences

with Period2n. Chinese Journal of Computers, Vol.29, No.9, pp.1590-1595, 2006.

7. Yaodong Zhao, Wenfeng Qi. The Linear Complexity of the Period Sequences Obtained from

the Binary Sequences by One-Symbol Delete or Insert. Journal of Information Engineering